| Author | Message | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 566 Registered: 2-2003 |
You know, Mitch, you're just plain wrong. You haven't shown anything, really, you haven't even explained your own post showing acceleration maximum far past torque peak. Despite excellent advice from Peter Sedlak, you continue with personal perjoratives. Similarly, the whole "in Mitches [sic] favor and in Efwun's favor thing is exceedingly childish, why stoop to that? I've derived equations that show why Power is important, and those equations also explain the variations even your own posts show. You've merely attacked those equations and put forth the idea that you can somehow slice D down to the point where no Work occurs. I've owned up to the errors I've made, and evolved from there. You are simply ignoring things you've posted that make no sense, and also that actually prove you wrong. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 551 Registered: 4-2002 |
"N.B. to Alsup: My equations were derived through honest thought without regard to proving myself right, and in fact, they�ve proven to me that my original construct is wrong." Ah, at least some progress has occurred. Proven can only be used when there is no means to show that something is incorrect any longer. You, sir idiot, have not proven anything. However, you have shown that you are a jackass, and an imbicile. Whether they have proven anything to you is irrelevent, the issue is whether they lead to a different conclusion. They do not. Whether you want to believe it or not. "However, they�ve also shown that your original construct is wrong." Only in your mind. I still stand my F=M*a derivation, and I have shown that the E=1/2*m*v**2 derivation also leads to the same conclusion. At this point, you have to find the errors in these derviations. I do not need to derive a W=F*D derivation, however, I have demonstraited the ablity to do the math (that you have not). I have derived two different ways to achieve the same conclusion, you postulate a third (W=F*D). Physics has the principle that if there are more than one ways to solve a problem, that all solutions should lead to the same conclusion. All you posit is that this new equation will put my previous work assunder. Yet you are unlearned enough to not to be able to dervie such a solution. (I won't hold my breath). "You, on the other hand, have thrown up a haze of manipulations in an attempt to beat down my arguments and prove yourself right at all costs." For the enlightenment of those other readers of this column; the haze of manipulations is called calculus it was invented to by Sir Newton in the 1660's to deal with exactly these kinds of problems (however, Libenezt also deservs credit for some of the foundational work on the math side, while the mechanics side was all Newtoon). When an argument is wrong, it deserves to be beaten down. If an argument holds water, it can stand on its own. The only cost I care about is the education of EFWUN with respect to Physics and Math. And at the current rate of progress, we will be here until 2047 getting EFWUN to understand how calculus works. With hope, a few years later he will understand Newtonian mechanics. It is not I who an right, it is the math and physics that are right. And you never answered the question as to why all the listed web sites are in agreement with my position. Why do they agree with me? Did I, somehow, create all those web sites? Is there a big conspiracy to posit an untenable position, worldwide? You seem to think that if there is onle tiny flaw in my argument that the whole house of cards falls apart. {You will, however, find that fragility in your argument.} So lets say there is a tiny flaw in my argument. Can you find it? Can your so called professors find it? If not why not? Can they also find the flaw that causes all the posted web sites from achieving the same position? "You�ve made this very unpleasant, for no particular reason other than your own peculiar personality." If, at the beginning of this thread, have been educated sufficiently to understand math and physics, then you would have been able to modify your position (after a couple of postings defending you position), and changed you mind to be in synch with the rest of people who actually understand this kind of stuff. If you would have gotten on board at this point, this thread would have been over 2 weeks ago. Since this is one of my areas of expertise, I took you to task--you failed to come up with any defensible position. Lets see a quote from my previous posting should suffice here: "Let see: In Mitches favor: graphs showing that F=ma leads to max acceleration at peak TQ graphs showing that E=1/2mv**2 leads to max acceleration at peak TQ equations based on F=Ma leads to max acceleration occuring at peak TQ equations showing that E=1/2m*v**2 leads to max acceleration occuring at peak TQ Numerous web sites substantiating my position solving the calculus that shows that F=Ma and E=1/2mv**2 for a loop. Then in EFWUNs favor I don't think I'm wrong I still don't think I'm wrong I don't thingk that any data will ever show me to be wrong." | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 563 Registered: 2-2003 |
You know what? When you're right, Peter, you're right. I'm not going to kiss Alsup and make up, but I am going back to the high road, and I apologize for my comments. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 562 Registered: 2-2003 |
Peter, I've got to agree with you, and apologize for my over-the-top comments. They resulted from frustration with what I perceive to be intellectual dishonesty, and | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 226 Registered: 1-2003 |
EFWUN & Mitch You guys need to kiss and make up. I thought this forum is for constructive comments, not a pissing contest and name calling. I would think that newcomers to FChat would be attracted to a 300+ post on a single topic enough to check it out, only to find less than appropriate venacular of members that should show a bit more class...This is not www.ChevyTruckChat.com! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 560 Registered: 2-2003 |
Victor, I understand what you're saying. Nevertheless, I don't think you can tell me with a straight face that F*D can be a number other than zero without Work being performed. This is what Alsup is saying. Further, if you evaluate his equations, you'll see that he takes quite a few liberties with other equations. KE=1/2MVSquared doesn't "imply" anything about "monotonically" decreasing Power. You may be able to slice "F", but you cannot slice "D" into any number so small that F*D doesn't equal zero, yet F*D doesn't yield W. | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 13 Registered: 1-2003 |
EFWUN Actually calculus is what you use to use F=ma when F is not a costant. That is what inegral calculus is used for. Calculus lets you perform these calculations when things are changing. You use algebra when are in steady state. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 557 Registered: 2-2003 |
Victor, to reiterate, I'm not saying that differential calculus cannot define X as t goes to zero, I'm saying that F=MA cannot obtain where F is not constant; you need to examine delta KE to reach an understanding of what's going on. I'm also saying that as t goes to zero, you cannot simply say, as alsup did, that these slices are so minute that no Work is accomplished, rather, unless t actually reaches zero, which it doesn't, you have accomplished an infinitesimal amount of Work because F*D cannot be zero! I am not dismissing calculus, I am dismissing the ridiculous notion that D can be so tiny that F*D results in no W. Don't be so quick to pile on, I thought you were a deeper thinker than that. TimN, I'm not "spewing" the same nonsense as Alsup, I'm saying that his equations are a fraud, (just as he ignored several of my defined terms previously to reach his conclusions) and that, while I've publicly admitted when I've posted "silly" responses, his misunderstandings run so deep that he has told us, unrepentantly, that no wear occurs in a 355 at 7,000rpm, and that manufacturers set redlines where stuff begins to break for "know-nothing" customers, and that Distance can be so small that F*D, while not equal to zero, is nevertheless no W!! Finally, he posted a graph showing a car accelerating at maximum well past torque peak, and first disavowed it because "it was Motor Trend's graph", then simply lied about where the Turbo's torque peak occurs. You've developed quite a bit of animosity toward me; don't let that cloud your thinking. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 556 Registered: 2-2003 |
I'm simply stating that alsup's statement that these slices represent such minute amounts of movement that F*D doesn't result in Work is ridiculous. Before you jump on, remember I said I've long forgotten calculus, and I was happy to forget it. Are you really arguing that delta KE isn't equal to P?? Can you explain why alsup's graph shows a vehicle accelerating maximally well past torque peak? Examined with logic, instead of heated passion, the following obtains: F=MA works only in the construct of the application of a force as in an inelastic collision, because if that force is more than the instantaneous overcoming of inertia, there would be an increasing rate of acceleration, and distance and time become factors. Remember, F*D =Work. Therefore, one physical construct that works for acceleration over distance and time is the change in kinetic energy of the vehicle, or W=delta KE. Then, it follows without arcane manipulation that Delta KE/time=W/Time or POWER. Let me finish by stating that my final understanding of this conundrum is that a motor that can hold within percentage points of its torque peak as revolutions ascend will accelerate hardest at the point where ascending revolutions optimally intersect with descending torque. Those vehicles accelerate hardest well above torque peak, perhaps even at �power� peak. Vehicles with steeply descending torque curves reach that optimal combination of torque and revolutions much earlier, and therefore accelerate hardest closer to, or at, torque peak. Remember, torque is the measured resistance on a dyno, while power is that output calculated as the ability to accomplish a set amount of work in a discrete amount of time. Therefore, P=W/T= Delta KE works for all situations, and explains why some cars accelerate hardest near torque peak, and why some accelerate hardest at power peak and the vast majority are somewhere in between. To reiterate, �it depends.� N.B. to Alsup: My equations were derived through honest thought without regard to proving myself right, and in fact, they�ve proven to me that my original construct is wrong. However, they�ve also shown that your original construct is wrong. You, on the other hand, have thrown up a haze of manipulations in an attempt to beat down my arguments and prove yourself right at all costs. You�ve made this very unpleasant, for no particular reason other than your own peculiar personality. | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 11 Registered: 1-2003 |
EFWUN Diferrential calculus absolutely lets you think about slices of time or anything else you can define with a number that become infinately small . This is why engineers and physical scientists use it to define manipulate and analyse all sorts of things. As a high school debater I learned if you couldn't attack the argument you should attack the person. Before you call any more people names you should learn more about math than high school algebra | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2884 Registered: 6-2001 |
EFWUN, you are spewing nonsense too. Where did these physicists get their degrees from?? EFWUN's upstairs physics college??? | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 10 Registered: 1-2003 |
Mitch is correct in his use of calculus to prove that you can go from F=ma to E(k)=1/2mv^2 as force is used to accelarate a mass to move it over a distance over a certain period of time to produce work and and increase the kinetic enrgy of an object. Some guy named Newton figuered it out several hundred years ago. He also invented the math so he could describe what was happening. The math is called calculus of which there are two forms. One is called integral calculus which figueres out the area under the curve. This is important because it lets you calculate things like the instanraneous velocity of a car or plane or rocket when all you know is the accelaration which can vary as a funtion of time or speed or what have you. To say it is only usefull to figure out the volume of a swimming pool implies a certain amount of ignorance about the subject. Differential calculus goes the other way. It will tell you the slope or rate of change of a curve. A real world application of this is being able to calculate the instantaneous acceleration of af a car when all you have is a plot of its velocity as a function of time. EFWUN before you insult anyone else regaurding their understanding of math, I suggest you go to your local college bookstore and buy a text book of calculus and read the section on derivatives. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 555 Registered: 2-2003 |
Alsup, your fraudulent arguments are descending into gibberish. The torque peak of the Turbo S is at 1,950rpm, source, R & T. The graph you posted shows peak acceleration at 3,000rpm. Your suggestion that there can possibly be a fraction of movement such that no work is done when torque creates that movement is unbelievably dumb, even coming from you. This alone is an indication that you don't understand the concepts you're reciting. Your intellectual dishonesty and inability to even explain your own posts disables your arguments, and your personal attacks point to your arguments' lack of real substantive force. My equations (as corrected by real physics doctorates) explain the variations observed and the pronouncement from R & T, to wit: "it depends." Your equations explain nothing (not even your own graph) and your intellectual dishonesty in fraudulently deriving them points to your "win at all costs" mentality (ridiculous, really, because this is a discussion!) This is in contrast to my acceptance that not all vehicles accelerate maximally at power peak, or at torque peak, but rather on a case by case basis, "it depends." I had the honesty to admit that my original statement about 180degree cranks was "silly." I admitted when I rushed to post something without thinking it through that the result made me look like I was suffering Alzheimer's. I've spent several weeks evolving these equations, and I'm confident of the answer, "IT DEPENDS." What about you? Shall we look at some of your idiocies? No wear at 7,000rpm? (conspiracy!) Redlines set by manufacturers to deceive "know nothing" customers? (more conspiracy!) Your post of that graph disproving your theory "that was Motor Trend's graph, not mine." (Motor Trend is in on it too!) Slices of time so small that the movement is so minute that no work occurs? (Let's dig up Newton and villify HIM, because he must be in on the conspiracy too!Pure Horseshit. Can you tell us who killed Kennedy, or where they buried Jimmy Hoffa? Why not? After all, you're making this stuff up as you go. I'm genuinely sorry that I've descended to your puerile, name-calling level. You've made an intellectual discussion truly unpleasant, and that's a shame. Nevertheless, the equations I've evolved here really do explain the variations that are observed. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 549 Registered: 4-2002 |
"As I remember, you can use calculus to find the area under curves by creating smaller and smaller slices," This is the numerical method of performing this task. However, the progression from deltas to differentials requires limit theory (as t->0 and such). "and the volume of swimming pools with compound curves," You are confusing integral and differential calculus. "I don't remember using it to slice "time" into units that would turn torque times distance back into pure instantaneous force without any component of time." dt is the instantaneous unit of time, therfore it is not absent. But you need to remember, that in T*D, D is created by velocity, which is created by acceleration which is F=M*a. Therfore when accelerating anything, D is not separate from a (or v). And thats what calculus does, puts enogh equations on the table so you can solve n equations with n unknonws. At least if you haven't fogotten how..................... | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 548 Registered: 4-2002 |
"In the meantime, one glaring fault appears obvious, you state: "Yet E=1/2*m*v**2 implies that at constant power, the rate of acceleration is proportional to 1/(sqrt(2e/m)) which is NOT constant (but is monotonically decreasing!) someting that is monotonically decreasing is not in constant acceleration! therfore your equations are wrong." KE=1/2MVsquared implies NOTHING about acceleration whatsoever!!" E=1/2m*v**2 only indicates the amount of energy associated with a moving mass. Then I go on to say "at constant power the rate of acceleration" so it is only after I start adding (or subtracting) power that accleration (or decleration) occurs. "KE=1/2MVSquared implies that Kinetic Energy equals 1/2 the Mass times the square of velocity, implying NOTHING about acceleration." Right, just as I said. "Can you say FRAUD? (or spell it?) " Yes, fraud. However the fraud is on your part. "The Beetle's torque peak was at 1,950rpm. Yet, acceleration was significantly harder at 3,000rpm. Explanation? Motor Trend's conspiracy? " Looks to me that peak TQ is at 3000 RPMs. perhaps peak TQ of the nonturbo engine peaks at 1800. "Oh, Mitch, are you really going to say that there is a slice of time and distance so small that torque can convert mass at rest into movement without accomplishing Work? " No, I am saying that TQ (a force applied over time) can convert a mass at rest to a moving mass. Work is performed, but I specifically never said anything about work. But if you really want me to go there, I can develop a solution to this problem based on work that also shows that peak acceleration occurs at peak TQ. "Do you really want to say something that ridiculous?" I didn't say that, I made no mention of work. "That statement alone invalidates your argument," But what invalidates all the other web sites agruments that back me up? "because a person who could say that obviously has no understanding of these concepts." Speaking for your self again? "while your's don't even explain your own graphs." Free invitation to anyone other than EFWUN to indicate any errors made in my previous graphs, any inaccuracies (I know of 3), and any missing information pertinate to this discussion (I know of 1). Inaccuracies: First graph: Thrust axis is imporperly scaled. This is corrected in the second set of graphs. Second: I used the Z06 transmission ratios rather than the regular corvette transmission ratios. Third: I used a biased computation when computing acceleration from HP. Instead of (MPH-0.5) to (MPH+0.5) as 1 MPH steps, I simply used (MPR) and (MPH+1). You can see that at very low velocity this tilts the acceleration axis upwards. Missing information: I did not account for the losses in the driveline, applying 100% HP/TQ at the rear wheels where 85% is the generally recognized number. "You still maintain that "peak A must be ALWAYS at peak torque" despite your OWN graph showing that it just plain ain't so!!" A) you still think Motor Trend's graph is mine, it is not. B) I aluded to the lack of air resistance, and that air resistance will make the real point of peak acceleration even lower than at peak TQ. C) the graph you speak of, actually backs my position, not yours--or are you trying to say this engine has peak power at 3000 RPMs? "Who is the stubborn, unpleasant prick here?" You! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 554 Registered: 2-2003 |
Mitch, First, no, I'm not an unpleasant prick, I've tried to be pleasant to everyone on this forum; you condescend to people on a regular basis. Second, you can't even interpret your own graphs. The Beetle's torque peak was at 1,950rpm. Yet, acceleration was significantly harder at 3,000rpm. Explanation? Motor Trend's conspiracy? Third, you state: "But is has created the first minutia of acceleration, which in later time steps will create the first minutia of velocity." Oh, Mitch, are you really going to say that there is a slice of time and distance so small that torque can convert mass at rest into movement without accomplishing Work? Do you really want to say something that ridiculous? That statement alone invalidates your argument, because a person who could say that obviously has no understanding of these concepts. Fourth, you've filled your previous posts with somewhat snide invective, and unfortunately you've finally drawn me down to your level. However, if rational evaluation is done, my equations explain the variations possible, while your's don't even explain your own graphs. Fifth, I've changed my posture from the absolute "it must be peak hp" after discussion with real physics professors, who second R & T and told me that "it depends" and corrected my equations to those you see here. You still maintain that "peak A must be ALWAYS at peak torque" despite your OWN graph showing that it just plain ain't so!! Who is the stubborn, unpleasant prick here? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 553 Registered: 2-2003 |
And a sense of humor conspicuously lacking elsewhere!! Thanks, Sean! Mitch, I'm pretty sure that your "calculus" is merely intended to intimidate rather than actually prove anything and is similarly lacking in any veracity. I will have a definitive answer by Friday. In the meantime, one glaring fault appears obvious, you state: "Yet E=1/2*m*v**2 implies that at constant power, the rate of acceleration is proportional to 1/(sqrt(2e/m)) which is NOT constant (but is monotonically decreasing!) someting that is monotonically decreasing is not in constant acceleration! therfore your equations are wrong." KE=1/2MVsquared implies NOTHING about acceleration whatsoever!! KE=1/2MVSquared implies that Kinetic Energy equals 1/2 the Mass times the square of velocity, implying NOTHING about acceleration. Can you say FRAUD? (or spell it?) As I remember, you can use calculus to find the area under curves by creating smaller and smaller slices, and the volume of swimming pools with compound curves, I don't remember using it to slice "time" into units that would turn torque times distance back into pure instantaneous force without any component of time. Perhaps you should look up the definition of an inelastic collision? Your purported analysis cannot even explain your own graph, where a Beetle accelerated harder after peak torque was reached. Why? If you're right (joke, joke!), then the Beetle should have accelerated hardest at max torque, but it didn't. Is that another of your conspiracy theories? My equations seem to indicate that in certain cars, peak A will occur at or near torque peak, while in other cars, peak A will occur at or near power peak. Seems to me that empirically, my derivations are correct, and your's are, well, INCORRECT. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 546 Registered: 4-2002 |
"You know, Mitch, you're really an unpleasant prick." Is the pot calling the kettle black? And I will continue to pick on you until you realize the the physics of this situation indicates I am right. "You posted a graph that showed that the VW turbo S actually accelerated much harder above peak torque, and yet you've never returned to that faux pas, and you continue to argue in absolutes." No, First, it was Motor Trend's graph Second, it shows maximum acceleration around 3000 RPMs Third, if you look in the magazine, you will find peak HP around 5000 RPMs. Fourth, you are more likely lokking at the Pontiac curve with variable valve timing, Fifth, the Pontiac has max TQ at the shown peak. "and yet you've never returned to that faux pas, and you continue to argue in absolutes. " Now that I have returned are you still unsatisfied? "You call me out, yet you haven't even gone out to your 355 and posted your own experimental data." I don not have timing equiptment capable of measuring the difference we are talking about. So any measurement I gave would be suspicious. "You've been wrong before, (as have I), and it wouldn't surprise me if after evaluation, you're wrong again!" I have been wrong before, I will be wrong again. However on this issue, I am not. Let see: In Mitches favor: graphs showing that F=ma leads to max acceleration at peak TQ graphs showing that E=1/2mv**2 leads to max acceleration at peak TQ equations based on F=Ma leads to max acceleration occuring at peak TQ equations showing that E=1/2m*v**2 leads to max acceleration occuring at peak TQ Numerous web sites substantiating my position solving the calculus that shows that F=Ma and E=1/2mv**2 for a loop. Then in EFWUNs favor I don't think I'm wrong I still don't think I'm wrong I don't thingk that any data will ever show me to be wrong. "Do you REALLY think I give a crap what you think?" If the problem is in physics, ther is no room for "I think it has to happen this way". Physics is about absolutes (at least in the time scales and velocitys we are talking about). And by the way, It's not what I think, its what the equations say. "You're a chauvinistic prick who proudly posted about how you were speeding through a neighborhood and when your wife objected, you stamped on the brakes and gave her a sore neck for three days." Attempt at diversion: the mark of a poor argument Notice my fault speding at 40 in a 35. "I don't have time right now to read through your equations, but because you're the opinionated jackass" As opposed to your argument--I can't be wrong, no mater what the math says. You have not supplied a single shred of evidence supporting your position. I have supplied massive amounts. "that told us wear was essentially nil at 7,000rpm, I'm sure they're as flawed as the rest of your argument." So, if you professor comes back and indicates I'm right on acceleration, will you belive that I'm also right on wear? "Also, I said I'd hate to have to revive my calculus which is happily forgotten over these last 20 years." If seems you never learned it. Or just learned it good enough to get out. I happen to use it on a regular basis. "Instantaneous force creating an average acceleration" I say it again, instantaneous <snip> average. An instantaneous force creates and instantaneous acceleration. An average force (continuously differentiable over the interval, you understand) creates and average acceleration. As for me responding to my errors see my post number 536--last line. "You cannot slice time down to zero, where torque has accomplished no movement" But is has created the first minutia of acceleration, which in later time steps will create the first minutia of velocity. This is what differential calculus is all about--and its the easy side--that why the teach it first--seems to be the one you forgot first. "I look forward to hearing their laughter." Please come back and appologize when you understand you are wrong. | ||
| Sean F (Agracer)
Junior Member Username: Agracer Post Number: 96 Registered: 2-2003 |
"F is in Newtons, Torque is in foot-pounds (Yes, Sean!!)" Always glad to help! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 550 Registered: 2-2003 |
That's the flaw (at least on first view) in your algebra and "calculus." You cannot slice time down to zero, where torque has accomplished no movement, and therefore no work. I will research this myself, and send your 'equations' to Penn, and I look forward to hearing their laughter. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 549 Registered: 2-2003 |
Not "instantaneous average"; you can't even read!! Instantaneous force creating an average acceleration, as in an inelastic collision; freshman physics. Who needs their diploma revoked?? You can integrate or differentiate all you want, but you cannot argue (well maybe YOU can) that in the moment that torque creates movement, you have F*D, which (despite your haze of 'calculus') is Work, plain and simple. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 548 Registered: 2-2003 |
You know, Mitch, you're really an unpleasant prick. You posted a graph that showed that the VW turbo S actually accelerated much harder above peak torque, and yet you've never returned to that faux pas, and you continue to argue in absolutes. You call me out, yet you haven't even gone out to your 355 and posted your own experimental data. I will submit your equations to Penn, if you like. You've been wrong before, (as have I), and it wouldn't surprise me if after evaluation, you're wrong again! How can one "loose" (assuming you meant "lose") in public in a "discussion" forum? I am a happily married man, with a successful career where I behave with absolute ethics, and the love of family and friends. Do you REALLY think I give a crap what you think? You're a chauvinistic prick who proudly posted about how you were speeding through a neighborhood and when your wife objected, you stamped on the brakes and gave her a sore neck for three days. Then, when I called you on speeding through areas where kids might run out, you said you were only doing 30 or so. Wonderful thought processes, and ample reason for me to concern myself with your opinions. I don't have time right now to read through your equations, but because you're the opinionated jackass that told us wear was essentially nil at 7,000rpm, I'm sure they're as flawed as the rest of your argument. Also, I said I'd hate to have to revive my calculus which is happily forgotten over these last 20 years. When I was wrong, I posted that I had made a silly mistake. Where is your post owning up to your own errors? Why the personal attacks? Tying your adequacy as a man to this forum? | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 545 Registered: 4-2002 |
"F=MA yields acceleration, but without regard to any time frame. You cannot slice T small enough to create a series of F=MA that result in a curve," Thus proving you don't understand calculus! Should we contact your alma Mater and have them recind your diploma? "rather, F=MA exists as the instantaneous application of a force that creates an average of acceleration," An instantaneous average--hmm--last time I heard an agrument as malformed as this I was arguing with a creationist. Instantaneous Force creates and Instantaneous acceleration (you know F=m*a): when F is instantaneous so is acceleration, when F is average so is acceleration. The reason calculus works with deriviatives (dt approximately zero) is that then one can consider each instant in time uniquely, and then integrate (weighted sum addition) to find what happens on human time scales. "validate my theoretical postulate" Your postulate has been disproven, not just discredited, but disproven. It cannot be resurected, it doesn't fly! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 544 Registered: 4-2002 |
"I am not arguing the maximum torque isn't applied to the contact patch at peak torque," You accept, here, that peak acceleration occurs around peak TQ in an instantaneous sense....... "I'm arguing that over any discrete interval of time, acceleration is a function of F*D/T, or Power." ....Then go one to argue, here, that one cannot integrate instantaneous events to achieve an agregation of events. This is just plain bogus: calculus (another thing I seem to understand that you don't) has been santified for doing this kind of thing since the middle 1600's. But either through F=M*a or through E=1/2*m*v**2 we arrive at the same answer. Know why? E = 1/2*m*v**2 P = dE/dt = m*v*dv/dt F = dP/dt = m*dv/dt a = dv/dt F = m*a So this set of equations forms a complete loop, you can differentiate from E or you can integrate from F. You HAVE to get the same answer or they system of equations you set up is inherently faulty--as is your logic--let alone your stubbornness. "Power = 1/2MV Final Squared minus 1/2 MV Original Squared) all divided by Time. Therefore, we can see that: Power = Delta KE/T, or acceleration. " Here is the problem with your math--and the doctorial helper should be flunked--- Under the last equation, a constant supply of power should result in a constant rate of acceleration. Yet E=1/2*m*v**2 implies that at constant power, the rate of acceleration is proportional to 1/(sqrt(2e/m)) which is NOT constant (but is monotonically decreasing!) someting that is monotonically decreasing is not in constant acceleration! therfore your equations are wrong. The rate at which power has to be increased to achieve constant acceleration (E=1/2*m*v**2) is in fact P = dE/dt = m*v*a; thus power has to increase linearly to achieve constant acceleration! "Mitch, you've done some elegant derivations," Nothing that your physics professor should not have done! BTW did you ask him about my equations? Do you wnat to respond to my challenge and have him contact me? why not? afraid to loose in public? "but what you've left out is the time frame, e.g., DV/DT as t goes to what value?" First its dv/dt; Second its a deriviative and t does not have to go to some value; Third you are thinking about integration when you want t to go to some value. Showing that you no longer understand calculus. "I've driven my 355s probably 10K miles, and I never felt them to accelerate harder at 5,500rpm than at, say 7,500rpm" This calls in to question whether any seat-of-the-pants measurements you have ever done are trustworthy. I find my blood pressure accelrates faster at 8000 than at 5000, but the my F355 does otherwise. "Yes, if you make an instantaneous evaluation, (F=MA) the maximum force is applied to the contact patch at max torque, HOWEVER, that doesn't translate to an ongoing delta KE, or ongoing delta V, because to create that delta KE, Power is your essential inquiry, and power peaks at, well, power peak!! " As I have shown, simulated, and solved in calculus form, you are wrong. The amount of delta KE that is required to accelerate from 32 MPH to 33 MPH is less than the amount of delta KE required to accelerate from 44 MPH to 45 MPH. After putting this modification into YOUR original equations, one (who can use algebra) will find that max acceleration occurs at max TQ in any given gear. Ad in wind resistance and peak acceleration occurs BELOW max TQ in any given gear. Are you trying to argue that you no longer understand algebra? You've already proven you don't understand physics, nor calculus. Want to make it a clean sweep? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 547 Registered: 2-2003 |
No, Brian, what I maintain is that the closest we'll come to an answer to this question is, "it depends." I absolutely disagree that an Atlantic accelerates hardest at peak torque, while I agree that some MomBomb minivan might well accelerate best at peak torque. I thought you understood the equations better; the concept you're calling "dynamic torque" is really work, and the concept I'm explaining is that work/time equals power, the ability to create a change in KE more quickly. F=MA yields acceleration, but without regard to any time frame. You cannot slice T small enough to create a series of F=MA that result in a curve, rather, F=MA exists as the instantaneous application of a force that creates an average of acceleration, not a dynamic graph of changing KE. Once that F becomes, as you put it, "dynamic", it is F*D, or Work. Then, the analysis is W/T, or Power. P = W/t = F*D/t = 1/2mvfinal squared minus 1/2mvinitial squared = Delta kinetic energy/time is the equation best suited to evaluating vehicle acceleration IN THIS CONSTRUCT. It is also the only equation capable of conforming with the experimental data available, simply put "it depends." | ||
| Brian Kennedy (Kennedy)
Member Username: Kennedy Post Number: 264 Registered: 3-2002 |
EFWUN writes: "Similarly, a simple stop watch evaluation of the 550 yielded virtually twice as quick a transition from 6k-7.5k than from 4k-5.5K." Given thumbs and stopwatches have precision measured in tenths of seconds and your total time in that range of RPMs is maybe half a second... not really too useful. "I'll wait until I can get my Funk and Wagnalls independent DESignated tester to confirm those values." I'd recommend getting either a Gcube or a G-Tech device. Each device is essentially an accelerometer... measures acceleration directly. Interestingly, they have features to then compute HP and Torque from the acceleration based on other car data provided. On that note, the physics of the acceleration of cars has been wholly solved by others... and in one case it is available low cost to all of you. If you are interested in playing with car parameters (like torque and horsepower and gearing) and seeing the effect on the car's acceleration, you can get the shareware simulator called CarTest by Patrick Glenn. People on many car lists all over the net have validated and confirmed its accuracy via independent tests of reality. You can try it out for free; but support the author by registering it for $20. If I knew how to take a screen snapshot of a DOS program running under Windows 2000, I could show you some great acceleration curves that clearly show peak acceleration occurring at peak torque and decling from there to peak HP and beyond. "My posit is that F=MA is best for determination of average acceleration as the function of an inelastic collision between F and M, while ongoing acceleration, delta KE/T, is a function of F*D/T, or power." And the rest of us would argue: used properly, both equations can be used for both situations and will yield identical results. "I think movement toward settlement would come from the realization that cars with optimized breathing invoke these equations to varying degrees because F stays within mere percentage points of F max as D/T (or RPM) increase, while cars that have steep torque curves max out at or near torque peak, because as D/T maxes out, F is falling dramatically." I think you just agreed with us. | ||
| DES (Sickspeed)
Advanced Member Username: Sickspeed Post Number: 3523 Registered: 8-2002 |
"I'll wait until I can get my Funk and Wagnalls independent DESignated tester to confirm those values." Obviously, there are few whose luck surpasses my own.  | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 541 Registered: 2-2003 |
Brian, no insult taken. However, plugged into a race car (where I've had fairly extensive experience) with earplugs in and the "feel" of speed essentially an ignored quantity, my "butt dyno" nevertheless felt that Atlantic cars were accelerating harder at 9,000rpm than at 7,500rpm (peak torque). E.g., stutter out of the hairpin at Long Beach with a 15/42 first, and wait while the needle slowly climbs past 7,500rpm, and the growing waaaaaaaaaAAAAAAAAAHHHHHHHHHHHH as it finally gets to 8,500 and things happen! Also, remember, I'm not saying peak A occurs at redline, because it was quite obvious to me that the only reason I hung onto a gear past 9,200 or so was to get a leg up on the next gear, where optimal rev-drops put you at about 8,500, or 800rpm or so past peak torque. Similarly, a simple stop watch evaluation of the 550 yielded virtually twice as quick a transition from 6k-7.5k than from 4k-5.5K. Given the contentious nature of this debate (not sure why some people are making this so unpleasantly personal?), I'll wait until I can get my Funk and Wagnalls independent DESignated tester to confirm those values. Further, I have stated that the answer from Road & Track "it depends" is probably the only correct answer to this conundrum, rather than my earlier "It must be at peak HP" posit, or those of several posters who've categorically stated "it must be at peak torque." A car with excellent breathing e.g., Atlantic, or S2000, or 355, or 550 may well hold enough torque as rpm climb to validate my theoretical postulate, while a car with optimized torque and little breathing (rev) capacity may have such a sharp drop-off in torque as rpm climb that "power" delivery is truncated. This accords with Brian's statement, with which I agree, that acceleration depends on the area under the torque curve. Taken to its logical conclusion, that dynamic "area" is really power delivery, or the ability to do work (torque times distance) as a function of time, W/T=Power. Remember, Brian, "dynamic torque" or torque times distance, is work. My posit is that F=MA is best for determination of average acceleration as the function of an inelastic collision between F and M, while ongoing acceleration, delta KE/T, is a function of F*D/T, or power. I think movement toward settlement would come from the realization that cars with optimized breathing invoke these equations to varying degrees because F stays within mere percentage points of F max as D/T (or RPM) increase, while cars that have steep torque curves max out at or near torque peak, because as D/T maxes out, F is falling dramatically. Thanks to those who have taken the time to read my evolving posts without resort to doctrinal responses or personal attacks. | ||
| Brian Kennedy (Kennedy)
Member Username: Kennedy Post Number: 263 Registered: 3-2002 |
Oh, and on the butt-dyno comments... My 360, 911, M3, 280ZX, and so on, all showed drop in acceleration as they go from peak torque to the rev limiter, even though HP in all was increasing. My 360 and M3 in particular show this dramatically well. HOWEVER, it is a well-known fact that *sound* plays a huge role in most people's butt-dynos. Put a new louder exhaust that actually reduces torque, power, and acceleration... most people will judge the car to be accelerating faster. Now combine that with the fact that your 355 and my 360 both scream and wail from 6K to 8K... as a result, there is no doubt that most of my friends are more wowed in that portion of the rpm band than they are by the briefly higher G's experienced at peak torque. Don't mean to insult your butt-dyno, EFWUN... so I'll just insult all butt-dynos... the precision of a butt-dyno is not adequate to measure the acceleration differences between peak torque and peak horsepower in most cars (since torque is fairly flat in between), and the butt-dyno distortion due to sound & velocity effects is much greater than the acceleration delta... rendering butt-dynos useless in this discussion. | ||
| Brian Kennedy (Kennedy)
Member Username: Kennedy Post Number: 262 Registered: 3-2002 |
Oooo... making progress, I think... EFWUN says: "I am not disputing with you that in an instantaneous snapshot, torque peak creates max force on the contact patch. " And thus, do you also not dispute that peak *instantaneous* acceleration will also occur at peak *instantaneous* torque? If so, then we can put that whole line of discussion behind us. Your next point could have some validity in certain situations. A common analogous situation is in electrical circuitry... where you can have instantaneous "force" (aka Voltage) that doesn't equate to the expected instantaneous "flow" (aka Current, Amperage) due to the lack of adequate Power or Energy to really drive it. In layman's terms, a battery may measure 12V but actually be unable to drive the device you are wanting to run (the starter motor). HOWEVER, in the case of typical automotive engines, this is a non-issue. The measured instantaneous torque is the same as the torque over time. If you hold an engine at the same RPM, the torque does not fall off over time as more and more work is done. It stays the same (until it runs out of fuel). This is reflected in the fact that automotive HP is never (practically) really measured... rather, it is just COMPUTED FROM TORQUE... that works in automotive applications because there is no separate power issue vs. torque. Thus, a physicist just stating general physics principles could show that peak acceleration would NOT NECESSARILY be achieved at peak *instantaneous* torque, and rather you'd need to look at peak *dynamic* torque (which is power limited; your system may not have adequate power to sustain the torque value over time). HOWEVER, that would be irrelevant here as nobody measures the peak *instantaneous* torque... when you put an engine on a dyno and measure torque/HP, you are measuring dynamic torque! Peak torque is just the highest point in the dynamic torque curve! And then the HP curve is computed (simple math) from the torque curve. Thus, "integrating dv/dt over time" will just be the area under the torque curve. And that will be greatest under the highest part of that torque curve... and that is at PEAK TORQUE. EFWUN, any chance that explanation resolves the issues you've been hung on? | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 175 Registered: 10-2001 |
I must say the EFWUN has clarified his position quite well now and is actually making sense to a layman like myself. Looking forward to how this all plays out. I guess what we are looking for in the end is a practical application for the results. And an explanation of why some cars, apparently EFWUN's among them, seem to (or do?) accelerate faster way past torque peak, while other cars seem to (or do?) accelerate the most near torque peak. The torque curve being so flat on the Ferrari would be why acceleration could be about the same at 7500 and 5000rpm, but if it's actually greater at higher rpms than at its torque peak, we need some other explanation. EFWUN is the only one trying to offer that. Anyone else have any ideas on that? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 538 Registered: 2-2003 |
Mitch, you've done some elegant derivations, but what you've left out is the time frame, e.g., DV/DT as t goes to what value? I don't want to now start to remember Differential Calculus also (argh), I'd rather trust you to look at the stuff I posted over your post, and consider that it is the very essence of "continuing" acceleration as a function of time that differentiates our equations. I am not disputing with you that in an instantaneous snapshot, torque peak creates max force on the contact patch. My point is that when acceleration is evaluated over any discrete (used for its meaning in math) interval of time, F*D/T, or W/T or Power, is the essential inquiry. I've driven my 355s probably 10K miles, and I never felt them to accelerate harder at 5,500rpm than at, say 7,500rpm. Yes, if you make an instantaneous evaluation, (F=MA) the maximum force is applied to the contact patch at max torque, HOWEVER, that doesn't translate to an ongoing delta KE, or ongoing delta V, because to create that delta KE, Power is your essential inquiry, and power peaks at, well, power peak!! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 537 Registered: 2-2003 |
F=MA is the perfect means of determining the average acceleration of a mass after an inelastic collision with a force in Newtons. It is an instantaneous determination, not an explanation of the ongoing input of force to create continuing acceleration. By analogy, it is a snapshot, rather than a video. Heavy hitters have weighed in, e.g., the Physics dept. at the University of Pennsylvania; their pronouncements have been laughed off by several posters. Understand that this is a complex question, not a simple determination of F=MA, and therefore case closed. Mike, No intent to offend, sorry for any offense you took. There are a mix of posters here; I find myself offended by posters' unkind remarks made as a substitute for intellectual discussion. People like PSK (Pete) state that they have "proven" peak A occurs at Peak Torque, but in reality, all they've done is continue to repeat doctrinaire statements and decry my "stubborness" or "contrary-ness." I didn't mean to lump you with the crowd. Mitch Alsup's graph clearly shows that acceleration in the VW Turbo S peaks AFTER torque peaks, clearly demonstrating that "as power ramps up", acceleration G increases. However, that post seems to have been disregarded in the rush to denigrate my equations (approved by Dr. Kalafut and his doctoral candidate) Once again, remember: F times D = W, or delta KE. (1/2MV Final Squared minus 1/2MV Original Squared). Power equals W/T or: Power = 1/2MV Final Squared minus 1/2 MV Original Squared) all divided by Time. Therefore, we can see that: Power = Delta KE/T, or acceleration. Remember, we're talking about ongoing change in KE, not an instantaneous snapshot of average acceleration as in F=MA. I am not arguing the maximum torque isn't applied to the contact patch at peak torque, I'm arguing that over any discrete interval of time, acceleration is a function of F*D/T, or Power. The longer the interval of time, the more important Power is, and the less important the instantaneous snapshot of F=MA becomes. This is why the old saw holds true that torque is responsible for low et, while power is responsible for high MPH. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 542 Registered: 4-2002 |
The question was: at what point is maximun acceleration achieved? F=M*a F is proportional to TQ M is constant Therefore a is a direct function of TQ E=1/2*m*v**2 More HP is delived into E at max HP than at max TQ However: dE/dt = m*v * dv/dt = HP Therefore: a = dv/dt = HP/(m*v) v is proportional to RPM (in a single gear) Therefore: a is proportional to HP/RPM HP = TQ*RPM*5252 or TQ = HP/RPM*5252 TQ is proportional to HP/RPM Therefore: max acceleation occurs at max TQ. Both sets of equations achieve the same result. Answer: max acceleration occurs at peak TQ in any gear. Both starting points achieve the same result. Extra credit: When v is NOT proportioal to RPM (like in a CVT), then max possible acceleration occurs continuously based on E=1/2*m*v**2 formulae. E=1/2*m*v**2 v = sqrt(2*E/m) a = dv/dt = sqrt(m/2E)dE/dt At constant dE/dt, E increases linearly a = sqrt(m/2E), a decreases as 1/sqrt(E) | ||
| Mario B (Lawwdog)
New member Username: Lawwdog Post Number: 14 Registered: 12-2002 |
See I said this was looking like a geometry book and now all of the authors are fighting for book rights. Who's the teacher...anyone....anyone...? Bueller...Bueller....Bueller | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 214 Registered: 1-2003 |
Dang! I lost the office pool on this post. I guessed 255 posts by 4/31. There are four people ahead of me with 289, 290, 305 & 325! | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2868 Registered: 6-2001 |
EFWUN, dont be ridiculous. With your F1 example, you say that POWER makes the car spin its tires. It is TORQUE that smokes tires. Go take an S2000 with 240hp and no torque and do a redline clutch drop. The tires will barely spin. Then do the same in a C4 vette with 245hp, but alot of torque (and wider tires) and watch the tires light up. The torque is what breaks the tires loose. its the torque which also makes the car accelerate. Mike, apparently the entire physics department at MIT does not have as good a grasp on this physcis than does EWFUN. And i thought I was stubborn... | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 514 Registered: 6-2002 |
Wm Hart - You present a simple request that I second. However, through all of the mud slinging, I can't remember what the question is that everyone is debating. Forgive my attempt, but I believe the question is: "Is acceleration greater at peak torque or peak horsepower?" Assumptions include: Single gear discussion. Acceleration, NOT velocity. A simple answer is all that is required, and that answer is - ? My understanding of physics suggests that Newton answered this question 250 years: Peak acceleration will occur at peak force (constant mass). Peak torque equals maximum force at the tire/asphalt interface, which means maximum acceleration. We may be dealing with semantics here, as others have suggested a linear mathematical relationship between horsepower and torque. Therefore, at any particular RPM, they differ merely by a constant. This concept, however, is not supported by emperical data (torque peaks and horsepower peaks do not occur at the same RPM). Thank you. Jim S. | ||
| Mike B (Srt_mike)
Junior Member Username: Srt_mike Post Number: 116 Registered: 12-2002 |
Bill, That is a great idea! I have a friend who is a professor at MIT and I am sure he could get a "heavy hitter" to weigh in on the matter. I don't think it looks like any consensus is going to be reached here, but maybe if both sides can agree to stand by the conclusion presented by the heavy hitter, it would be good enough? It would be for me! | ||
| wm hart (Whart)
Member Username: Whart Post Number: 946 Registered: 12-2001 |
I haven't checked this thread for a while, and i am fascinated by the efforts of many of you to persuade thru logic and formulae that exceed my grasp of physics, math, etc. Therefore, i have two modest, simple requests to make of you guys, collectively, so that we lesser mortals can benefit from your hard work. First, after you have reached some basic conclusions, could you present those, in the form of an executive summary, without heavy math, to those of us who desire an understanding of the results, but not of the process? Second, if you guys cannot agree, can we submit a "certified question" to the heaviest geeks we can find? I can probably muster up some folks at MIT, i would assume that JPL or whatever the hell its called these days has some good folks, hell, Stanford, wherever. You can even reach a consensus on who to submit the problem to,if you like. I'm not trying to broker a compromise, cause i'm sure you're all having fun, but i'd be interested in knowing if there is an "answer." Your mathematically challenged colleague..., bill | ||
| Mike B (Srt_mike)
Junior Member Username: Srt_mike Post Number: 115 Registered: 12-2002 |
EFWUN, I take mild offense to your insinuation that I "don't understand physics". It is you who does not understand. You are the most stubborn person I've ever seen! I think Stephen Hawking could be here explaining this to you and you would still claim he is wrong because you can't seem to take the hit to your ego that you just don't know what you are talking about. As for your F1 example, you just gave an example of WHY we want high-revving engines if the hardest acceleration comes at the torque peak - because then we can use the gears. That F1 car may have a 30:1 first gear ratio (just using random numbers) while a Modena may have more like a 6:1 first year ratio. So while they may both make 250lb-ft, the F1 car would be putting 7,500lb-ft to the wheels whereas the Modena would be 1,500lb-ft to the wheels. Since the F1 car is also alot lighter, it's a LOT faster - and hence is able to smoke the tires at 100mph. EFWUN, you are going out of your way to "prove" that it's not just F=ma. Ok, so why don't you mathematically prove F=ma isn't where it's at? You said something about "instantaneous acceleration" and then about the rate of energy being added. Prove F=ma wrong. You can't, you just come up with some silly comment that anyone who doesn't agree with you obviously does not understand physics. I'd thoroughly enjoy seeing you eat crow. But you won't because you will cling to your belief until the bitter end, regardless of how much evidence contradicts you. That is not something to be proud of. | ||
| Mario B (Lawwdog)
New member Username: Lawwdog Post Number: 12 Registered: 12-2002 |
Good Lord...this thread looks like a geometry text book! : ) | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 174 Registered: 10-2001 |
EFWUN wrote: " F/1 cars smoke the tires from 100mph rolls; do you really think they do that with the same "output" as a Modena? ... F/1 cars run from 0-180mph in less than 9 sec, not just because of lightness and gearing advantage, but also because of extremely high "power!" Are you sure? Gear advantage, i.e. the ability to travel 100mph in the same gear as a Modena goes 30mph and thereby be putting the same torque to the rear wheels (if f1 torque curve is decently flat enough) and the fact that the f1 car is much lighter seem to be perfectly reasonable explanations for the phenomenon you describe. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 541 Registered: 4-2002 |
"Further, if you're going to be intellectually honest, you'll admit that your equation actually subsumes the fact that there is more "power" available to create delta KE at power peak than at torque peak." If you're going to be intellectually honest, you'll admit that there is more acceleration at a point other than peak HP. The reason is that, while there is more power at peak HP, the energy needed to acclerate has also gone up, and gone up faster than the HP has gone up. Therefore acceleration is faster somewhere other than at peak HP. But if you really wanted to be intellectually honest, you would admit that you are just plain stupid when it comes to physics. | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 9 Registered: 1-2003 |
EFWUN In the USA torque is measured in lbft. In the rest of the world torque is measured in newtonmeters. Just check out any brochure for a European car. They are EQUIVALENT ways of measuring the same thing!!!! I think we can all agree that feet and meters are measures of length. The pounds in torque and the newtons in torque must both measure the same thing. FORCE!!!! Newtons have units that include time so pounds force, which is what the pounds in torque are, must also include time. This is a concept that you seem not to be able to grasp. When dealing with pounds of weight or pounds of force you must include a factor for "g" to convert from weight to mass. You tell people they shouldn't talk about physics if they don't understand basic priciples. What is it they say about people in glass houses? | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 540 Registered: 4-2002 |
In the graph you will see that delta E is the amount of energy that has to be added to the car to achieve 1 more MPH of velocity. "Your equation is too simplistic," It is YOUR equation "E=1/2*m*v**2" that I used. All I did was to frame this equation in a realistic setting and graph the result. "F/1 cars smoke the tires from 100mph rolls; do you really think they do that with the same "output" as a Modena? It is that torque, times distance in radians, times 18,000rpm that creates a tremendous RATE of delivery of power" No, it is TQ*gears*diff*rolling_radius that accelerates the car. "F/1 cars run from 0-180mph in less than 9 sec, not just because of lightness and gearing advantage, but also because of extremely high "power!" And it is the TQ at high RPMs that creates this power. The high RPMs allow you to gear the car higher and make use of the TQ. "Just the fact that Mitch's graph shows a car accelerating much harder past its torque peak" You have not looked at the graph, it indicates no such thing. Look for the line attached to "2". "should explain why I'm increasingly confident " Since you have not bothered to look at anything in disagreement with your point of view, this statement does not surprise me. At least I went entirely through your point of view and achieved the same conclusion as the original F=Ma based point of view. At this point: A) you don't understand arithmetic (let alone calculus) B) you don't understand newtonian physics C) you don't accept points of view contrary to your own D) you are just plain stuborn. But I digress: EFWUN; you said you took this problem to a local physics professor. I'll make you a deal. Send my e-mail address to him, let us discuss this issue off line. Then I will let HIM deliver the verdict and stand by whatever we work out; after we have had adequate time to fully discuss this matter. DEAL? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 532 Registered: 2-2003 |
Ah, what did you guys do over your Passover/Easter break? Mitch, pretty good, but remember that delta KE is Work, while "power" is our derivation of how fast "Work" can be delivered (how fast we can accomplish delta KE) Further, if you're going to be intellectually honest, you'll admit that your equation actually subsumes the fact that there is more "power" available to create delta KE at power peak than at torque peak. Your equation is too simplistic, but never mind, because I think the real answer to the underlying conundrum is "it depends." Please read further. Pete: I'm not waiting for DES to time the 550 so that I can prove my 550 doesn't obey the laws of physics, but that sure was funny; not sure how your emails can be so pleasant, while your posts are so snide? I'm waiting for DES to validate the timing I've already done, which indicates that in this particular example, max G occurs well past torque peak. Further, Pete, you indicate the absurdity of your absolutist position with your F/1 analogy. F/1 cars smoke the tires from 100mph rolls; do you really think they do that with the same "output" as a Modena? It is that torque, times distance in radians, times 18,000rpm that creates a tremendous RATE of delivery of power, and the ability to impart kinetic energy to the car in an extremely short period of time. F/1 cars run from 0-180mph in less than 9 sec, not just because of lightness and gearing advantage, but also because of extremely high "power!" Just the fact that Mitch's graph shows a car accelerating much harder past its torque peak should explain why I'm increasingly confident in R & T's answer, "IT DEPENDS!", vindicating neither my absolutist position nor the absolute position you share with Mitch. In my ML500, my wife and I noticed that acceleration was harder below power peak; while a ride in our 550 this morning, it was obvious that the car literally yanked us forward after 6,000rpm. My theoretical physical derivations are correct, despite attempts to refute them, but I think we all need to realize that in the real, non-theoretical world, the widely varying torque curves for myriad vehicles produce different acceleration curves. For those who say "F=MA" pure and simple, leave the physics to those of us who have the background. F is in Newtons, Torque is in foot-pounds (Yes, Sean!!) without any unit of time. If you don't understand physics, why make comments? I think the final answer is that for different motors and vehicles, max acceleration is at peak power or at peak torque, or in an infinite variety of places in between those extremes on the rev counter, DEPENDING ON THE VEHICLE. As R & T said, "it depends!" | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 538 Registered: 4-2002 |
To the casual observer, one might wonder why this topick continues to be beaten to death. To one with a casual understanding of physics, you should understand that if there are two differnet ways of computing some phenomenom, that both ways better come up with the same results. Earlier, I established a means to compute acceleration based on F = M*a. EFWUN contends that we can compute acceleration from energy using E=1/2*m*v**2. Acceleration is then computed by the change in velocity, which is related to the change in energy. It is important that both methods achieve the same result, otherwise physics itself becomes untrustworthy. Lets see how much energy it takes to increase the speed of the car by 1 mile per hour at our two chosen velocities. E(32) = 1/2*m * (32*32) E(33) = 1/2*m * (33*33) E(44) = 1/2*m * (44*44) E(45) = 1/2*m * (45*45) E(34) - E(33) = 1/2*m * ( (33*33) - (32*32) ) = 1/2*m * ( 65 ) E(45) - E(44) = 1/2*m * ( (45*45) - (44*44) ) = 1/2*m * ( 89 ) So it takes 65 units of energy to accelerate from 32 to 33 MPH So it takes 89 units of energy to accelerate from 44 to 45 MPH At 32 MPH there are 268 HP available to deliver energy into the car At 44 MPH there are 380 HP available to deliver energy into the car 65 / 268 HP = 0.2425 units of time 89 / 380 HP = 0.2342 units of time Since it takes less time at 32 MPH to accelerate to 33 MPH than it takes from 44MPH to 45 MPH, acceleration was greater at 32 MPH than at 44 MPH. This shows that the acceleration is higher someplaces than at peak HP, So EFWUN's argument is refuted. However, it does not show that that the maximum point of acceleration occurs at peak TQ, my argument.  So for each data point, I plot the HP and dE and rapidity of change 1 This point is the point of Max HP 2 This point is the point of max Acceleration The shape of acceleration 'looks' just like the shape of the TQ curve. Yet the TQ curve was not used anywhere in this computation. One could in principle figure out all the multiplicative constants to change this reletive acceleration curve into either thrust or TQ, I leave this as an excersize in futility. Therefore, both the forced based (F=M*a) and energy based (E=1/2*m*v**2) methods indicate that maximum acceleration occurs at the point of peak TQ. And physics can rest easily through the night. | ||
| Frederick Thomas (Fred)
Member Username: Fred Post Number: 692 Registered: 2-2001 |
Wow, I am impressed! I saw the post count on this thread and thought surely it must have moved on to some other tangent(sp) but its all still on the same subject only some of it is written in chinese | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 537 Registered: 4-2002 |
EFWUN wants everyone to believe that the point where energy is being added at its maximum rate causes velocity to increase at its maximum rate. This is not true: consider the F355 engine (just because we have data already established. The velocity at max TQ in 1st gear is 32 M/hr and the corresponding HP is 268 HP The velocity at max HP in 1st gear is 44 M/hr and the corresponding HP is 380 HP E = 1/2*m * (32*32) at peak TQ E = 1/2*m * (1024) while E = 1/2*m * (44*44) at peak HP E = 1/2*m * (1936) So it takes 1936 units of energy to increase velocity by 1 unit at 44 M/hr while it only takes 1024 units of energy to increase velocity by 1 unit at 32 M/hr So if there is more than 1024/1936 units of energy at 32 than at 44 then the acceleration at 32 M/hr is faster than the acceleration at 44 M/hr There is 268 HP at 32 and 380 at 44. 268/380 = 0.7053 While 1024/1936 = 0.5289 Since 0.7053 is greater than 0.5289 the acceleration is greater at 32 M/hr than at 44 M/hr. Even using EFWUN's own equations! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 536 Registered: 4-2002 |
"= 245 ft lbs This is only approximately the same torque produced as a 360 Modena engine!!! " F360 = 280 lb*ft F355 = 256 lb*ft What did you expect out of a natrually asperated 3.0 litre engine? With 14:1-15:1 compression? I acknowledge the error in my previous posting: I converted rpm into ft/sec whereas it should only be 1/sec (or count/sec : wehre count is dimensionless). Whoops. | ||
| Sunny Garofalo (Jaguarxj6)
Member Username: Jaguarxj6 Post Number: 325 Registered: 2-2003 |
Pete, just making sure that #1 and #2 as you state aren't getting lost in the fray ;) Sunny | ||
| PSk (Psk)
Member Username: Psk Post Number: 382 Registered: 11-2002 |
Hmmmm, I was thinking about Torque again as I was driving my family to the shops and F1 engines, and I thought I would calculate the engine torque of a F1 engine at peak rpm and power. Lets say we have an 850hp Ferrari F1 engine that spins to 18000 rpm. For the sake of making life easy lets assume that it makes peak power at this rpm at not some value slightly under that rpm. Thus using Tq = Hp/(rpm/5252) we have: = 850/(18000/5252) = 245 ft lbs This is only approximately the same torque produced as a 360 Modena engine!!! Thus my theory is that these extremely high revving engines are not as scarey as one thinks, or put another way the higher you make an engine rev and push the power up the rpm band the less intimidating the torque curve will be. Now I am not saying that F1 drivers are not brave chaps, I am just saying that they torque curve 100% suits racing, as high torque at low rpm just causes wheel spin ... and that the Can Am cars of the 60s-70s with their huge v8 motors producing 700hp at 6000rpm would be more intimidating than a F1 motor ... Now that should cause a few rebutals Pete | ||
| PSk (Psk)
Member Username: Psk Post Number: 381 Registered: 11-2002 |
Sunny, Please read about 100 posts down when we proved that peak acceleration does occur at peak torque, and the only reason you rev past this point (as does an automatic transmission) is to use the mechanical advantage of gear ratio torque multiplication. But while revving past peak torque, acceleration (ie. m/s^2) does actually decline as Mitches graphs show, but the overall acceleration including gear changes is optimal. Note: there are two different things here. 1. Maximum acceleration without gear changes occurs at peak torque PROVED 100% correct and will never be disproven!. 2. Gear change points to optimise acceleration are actually above peak torque and for lower gears could even be above peak power. Note: has nothing to do with peak power or power at all. This is because the lower gear multiplies the engines torque more and thus even while revving out more you still are producing more torque at the rear wheels than the next higher gear will ... thus the most torque at the rear wheels will provide peak car acceleration. Thus Sunny, please read the facts before taking us (er, EFWUN ) backwards. The research is there for all to see, and yet you come in with that comment. BTW Sunny automatic transmissions are hardly what I would consider clever enough to be used as reference for this theoretical conversation. Okay some of the new electronic ones may actually be programmed to change at approximately the right time but most (I think) just make sure the motor does not spread it's insides all over the road Again maximum force at the rear wheels will accelerate the car hardest AND thus fastest at any point. One only has to get in a car that has a very high revability but low peak torque to FEEL this. I used to own a Simca 1000 when I was at Engineering school and this thing could rev to around 7000 rpm but really stopped doing anything at all past 6000 rpm. I do not know where peak torque was but probably around 5000 rpm or something and you could feel the car slow in acceleration as I continuously abused it right up to 7000 thinking I was Alain Prost ripping down to La Source (sp?). Thus the only reply we are waiting for is EFWUN's proof that his cars do not obey the laws of physics and everybody else is wrong, including all those knowledgeable sites that Mitch has posted links to. Pete | ||
| Sunny Garofalo (Jaguarxj6)
Member Username: Jaguarxj6 Post Number: 324 Registered: 2-2003 |
Thread is way out of hand. Your over complicating this by trying to prove max acceleration, **the rate of change of velocity with respect to time**, is at TQ peak when it occurs above it. GEEEEEZZZZZZZZUUUSSS. Don't substitute quickest for hardest. Auto tranny's are perfect for this test. Put the pedal down to the floor and observe what the shift points are. They're above TQ peak, aren't they? Not a post count record but sure as hell fun! Sunny | ||
| Brian Kennedy (Kennedy)
Member Username: Kennedy Post Number: 259 Registered: 3-2002 |
"Okay, the following physical equations prove my point." [chuckle] EFWUN is just messing with you guys... didn't you guys have Physics professors in college who were constantly presenting long formulaic proofs of all kinds of ridiculous things? Geez, he used a dozen formulas combined in obtuse (but clever) ways, when simple "a = F/m" would do just fine. But a = F/m alone would not allow him to prove his contrary position. Towards the end of my Freshman physics class we were asked on a test to prove that a horse would run faster on two legs. Physics buffs seem to get a sick pleasure out of obtuse proofs of nonsense. And then there were my Philosophy classes... proofs no longer being restricted to mathematical formulae... and they seemed to enjoy proofs that just circle back on themselves in reverse (paradoxes). Ahhh, those were the days... a fair amount of work, but no real responsibility... | ||
| Steve Magnusson (91tr)
Intermediate Member Username: 91tr Post Number: 1676 Registered: 1-2001 |
the units for rpm are radians/sec (not ft/sec)... | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 533 Registered: 4-2002 |
I see where EFWUN is problem is by his own derivation, but don't have a numeric solution in his set of equations yet. The problem: dimensionality is wrong! E = 1/2*m*v**2 m in units of lb/G v in units of ft/sec G in units of ft/sec**2 Thereby: M in units of 1b*sec**2/ft E in units of lb*ft So energy represents the work done in moving x lbs y ft. Notice that Kinetic energy is in the same units as potential energy (timeless). While: HP in units of lb*ft*rpm rpm in units of ft/sec Thereby: HP in units of lb*ft**2/sec and HP*sec in units of lb*ft**2 Since HP times time is not in the same units as energy, it is not energy, but energy times time. This is the mistake EFWUN has made--confusing a given amount of HP over a certain amount of time with energy--it is not. If you use: HP*sec/ft instead of HP*sec you will end up with energy. Never even had to resort to TQ to prove that EFWUN is (and continues to be wrong) on this. Back later with coefficients to make this all work. | ||
| Mike B (Srt_mike)
Junior Member Username: Srt_mike Post Number: 108 Registered: 12-2002 |
The scenario: A car with only one gear, redline at 8,000RPM, torque peak at 3,000RPM, and power peak at 7,000RPM. Psk and Mitch are saying that the car will accelerate the hardest (i.e. produce the most G's on the driver) at 3,000RPM. EFWUN is saying that the car will accelerate the hardest (produce the most G's on the driver) at 7,000RPM. Is that correct? EFWUN, you are completely, 100% wrong. Maximum acceleration occurs at the torque peak. I don't know what all the ballyhoo is about physics professors or what not, but any physics professor would agree that the car described above will produce the most "G's" at it's torque peak. EFWUN, I know what you're saying about the time component and how you may produce less torque but it's at a higher RPM, but this does not relate to how quick you are accelerating. It's REALLY simple. F=ma. a=F/m. F is measure in lb-ft. m does not change. If F is higher, a is higher. No ifs, ands or buts. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 531 Registered: 2-2003 |
Finally, Mitch, note that the Volkswagen New Beetle Turbo S torque peaks beginning at 1,950rpm where it is at its maximum of 173ft-lbs. (source, R & T) Nevertheless, your OWN CHART GRAPHICALLY DEPICTS ACCELERATION STEEPLY INCREASING AFTER PEAK TORQUE, as, in your OWN chart's wording "Power ramps up!" More force per unit of time, or increasing POWER (Power ramping up as your chart states) equates to harder acceleration. LOOK AT YOUR OWN CHART. Your examples are poorly chosen in general, and you should have figured out that your own chart disproved your thesis!! A Turbo charged car that despite forced induction does NOT reach peak acceleration at peak torque, but rather significantly later "as power ramps up"; a Supercharged car (again, forced induction, not the subject of this discussion) that has enormous parasitic losses as rpms climb, and an ECU enhanced car with variable cam timing which nevertheless shows a tremendous spike of acceleration late in the "power" curve. My theoretical derivations apply to motors without computer or forced induction enhancements. (Even though your own posting shows the Turbo S proves my theory and disproves yours!) These non-enhanced cars like our wonderful 355s accelerate hardest past peak torque as POWER ramps up!! Review the equations, and look at YOUR OWN CHART!! As I said, my own timing yielded virtually twice as quick a time from 6-7500 as from 4-5500 in the 550, but I will get independent verification. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 530 Registered: 2-2003 |
Mitch, I'm not really interested in opinions on the net, I'm more impressed with the opinions of two Ph.Ds in Physics. Those equations are correct. Please note that neither of those three graphs states that max acceleration is occuring at max torque. I'm told that in real life, you're a nice fellow, please prove that out by being academically honest. I'll (or DES will) post results of experimental trials next week. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 529 Registered: 2-2003 |
Mitch, I accept that the Dinan graph represents a real car, faced with high aero drag increasing as a square of increase in velocity. If you calculate the differential delta, I think you'll see that the theoretical model is proven, e.g., that cars accelerate at maximal G at power peak, without reference to wind resistance and other outside factors. It isn't the same to say that cars accelerate hardest at peak torque, and then represent that thesis with a graph that clearly demonstrates that despite vastly increased wind resistance, the car is accelerating a nearly identical delta V at power peak. Also, the equations are fine, and prove the point (let's not denigrate Dr. Kalafut, who IS a mathematician), rather, your understanding of the physical math is inadequate; the quadratic is there, you just didn't recognize it. Finally, as I've said for quite some time now, there are motors like those in a minivan that can't breathe worth crap that max out near torque peak, because torque falls off too quickly to gain the advantage of higher revving, and there are cars with such broad torque peaks that revs are virtually unimportant. As R & T said, "it depends." It is the area under the power curve (torque as a function of rpm and time) that is all important. Nevertheless, as your Dinan example proves quite clearly, despite wind resistance increasing as the square of speed, the acceleration is nearly identical, proving that there is considerably more Power pushing that car forward against resistance at power peak than at torque peak. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 528 Registered: 4-2002 |
Found this graph in Motor Trend http://www.motortrend.com/roadtests/coupe/112_0212_fun/index8.html  Notice the downward slope of the curve. Notice that MT agrees that max acceleration occurs at mac TQ. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 525 Registered: 4-2002 |
"the following physical equations prove my point." Proof is a much too strong a word--demonstraight at best your memtal model of what you thingk is happending. However, proof is left to mathematicians. "Combining these physical definitions, we get Work = M(V-V0)/t times (V + V0)/2t which equals Mass times (V squared - V0 squared)/2 or simplified : " Arithmetic error here: Mass * (V - V0)**2 / 2 (V - V0)**2 -> (V**2 - 2*V*V0 + V0**2) not (V**2 - V0**2) "However, in pure theoretical physics, the derivation is clear, Torque creates change, Power creates the RATE of that change." What ever happened to F=M*a? It is the Force that causes acceleration; acceleration is the rate of change in velocity. But the real question is: <drum> Do you (or do you not) accept that the Dinan acceleration graph represents a real car? under usefully realistic acceleration senario? Do you (or do you not) accept that the slope of acceleration at peak TQ RPMs is actually greater than the slope at peak HP RPMs? If so, then does this not render your argument moot? Can you find anything on the net backing your position? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 528 Registered: 2-2003 |
Han's post is particularly apposite to this discussion, by the way. The reason the motor in Han's theoretical CVT rushes past torque peak to power peak, and parks there, is because the output of the motor, (measured in torque as a function of distance and time), is greatest at power peak. Dynos measure torque, but output is that measured value calculated as a function of distance and time, and output is greatest (redundant) at power peak. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 527 Registered: 2-2003 |
Okay, the following physical equations prove my point. I won�t periodically condescend to everyone by asking if everyone is �still with me�, rather, I�ll assume you�re all uniformly intelligent. We�ve agreed that torque is a force that when multiplied by distance equals Work. (W= F x D) Acceleration is defined in Newtonian physics as A = Velocity(V) - Velocity Original (V0) all divided by time A = (V-V0)/t. Distance traveled at Average Velocity, D = t(V + V0)/2 Remember Work equals F times D, and F equals Mass times Acceleration. Combining these physical definitions, we get Work = M(V-V0)/t times (V + V0)/2t which equals Mass times (V squared - V0 squared)/2 or simplified : � (M times V squared) minus � (M times V0 squared) = Work � MV squared, remember, is Kinetic Energy, and the difference between final Kinetic energy: � MV(final) squared, and initial Kinetic Energy, or � MV (Original) is the Work accomplished. Torque times Distance equals Work, (the ability to change KE initial into KE final), While Power is the amount of Work possible in any discrete time interval. (Definitions to follow after Easter Sunday). Or see my post deriving the equation for HP (a unit of power). Therefore, increasing W equals the ability to create a greater delta KE, while increasing POWER equals the ability to create that delta KE more quickly. Power is simply the derivation of Torque as a function of distance and time. Again, in real world situations, whether the car accelerates more quickly at peak torque, or at the measured derivation of that torque as a function of time, is clouded by exponentially increasing aero resistance, and the particular characteristics of the individual vehicle's power curve. However, in pure theoretical physics, the derivation is clear, Torque creates change, Power creates the RATE of that change. Experimental data will follow as soon as I can hook up with DES after Easter. Good Holiday to all. | ||
| Rob Schermerhorn (Rexrcr)
Member Username: Rexrcr Post Number: 522 Registered: 11-2002 |
Mitch, I was working on presenting the exact same data, but with 360 Challenge data. Well done, perfect, strong work! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 521 Registered: 4-2002 |
I found this graph on the Dinan BMW site.  Notice the red lines and the blue lines. The red lines are tangent to the RPM curve at maximum torque. The blue lines are tangent to the RPM curve at maximum HP. Notice that the slope of the change in RPM/time (aka acceleration) is greater at maximum TQ than at maximum HP. A greater slope means greater acceleration. Thus ending the debate. | ||
| Brian Kennedy (Kennedy)
Member Username: Kennedy Post Number: 253 Registered: 3-2002 |
> Brian, the only dispute is my contention (supported by physics) that where > any distance is involved, horsepower creates a greater rate of delta V than torque. This statement, strictly speaking, is non-sense... the torque curve and the horsepower curve are measures of the same property of the car... simple mathematical formula converts between the two. They both relate to exactly the same acceleration. I assume you meant to say something like: "Peak horsepower is a better indicator of max acceleration than peak torque." (which I would agree with, though both are relatively poor indicators compared to either *curve*). Or perhaps you meant to say something like: "When accelerating a long enough distance that you are moving through the gears, you'll want to shift to keep the engine RPMs up around peak horsepower, not around peak torque, in order to maximize acceleration." (which I would also agree with) However, the point of maximum instantaneous acceleration will undoubtedly be at the point of peak torque while in first gear. But of course, since you are accelerating, you won't spend much time there at all! | ||
| DES (Sickspeed)
Advanced Member Username: Sickspeed Post Number: 3402 Registered: 8-2002 |
Psk, i'm trying to stay out of this entire debate, 'cause, well... i don't have two cents worth to put into it... However, i'll second your notion that EFWUN should do it in my car (as well as his Ferrari ) so we can show results from a 1.9 liter engined car and a dream car... i'd be honored to have him drive my car, ensuring that i'll most definitely learn something that day... EFWUN...? Wanna make that day twice as honorable for me...? i'm so lucky, it's not even funny... | ||
| PSk (Psk)
Member Username: Psk Post Number: 378 Registered: 11-2002 |
Well put Hans. DJParks, Hmmm becareful, EFWUN thinks Torque does not actually do anything and is just a theoretical joke that the rest of the world use. EFWUN, Please read Brians comment regarding how Power and Torque are just a mathematical representation of each other (which I stated 1000 posts ago ). Again a dyno measures Torque and we plug that in a formula to come up with a different way of looking at the SAME value. Yawn, where is that brick wall. BTW: EFWUN please do your test in DES's car not your Ferrari. No disrespect to your car DES (if I could take EFWUN for a ride in my people mover he will see exactly what we are talking about, as its torque curve dives so fast it is though a soft rev limiter has been hit), but I assume that your car has a normal engine designed to last a reliability long time and thus engine breathing has been compromised for manufacturing costs, etc. and thus the torque curve might full off much more than EFWUN's 550 and thus he will feel the push reducing. Also your heart will not be getting so excited blurring the results ... Pete | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1110 Registered: 4-2002 |
Believe it or not, but I actually think I'm beginning to understand all of this! What is probably the most helpful (at least to me) is to consider the example of a CVT, where you have the 'go pedal' buried, and the tranny is properly calibrated for max performance. The tach will park itself on the HP peak, rather than the torque peak. Why? For a given road speed, the torque multiplication in the tranny will be greater with the engine at the HP peak. The engine may have more 'grunt' at the torque peak, but if the tranny was using a ratio to put the engine there, it would have less torque multiplication. Here's what hit home for me: The HP peak is the optimum point for the CVT to shoot for to provide the maximum "(engine torque)*(gear ratio)". In other words, you're not trying to achieve max engine torque, you're trying to achieve max wheel torque. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 516 Registered: 4-2002 |
"I understand the basic theory presented here and also understand the effects external factors would have on the basic theories which would in turn produce an endless variety of results which would lead me to believe that alot of this 'just depends'." There is no room in physics for 'I think it goes like this' In physics, it follows the rules (Newtons laws mostly) or it doens't (ahem) fly. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 515 Registered: 4-2002 |
"Brian, the only dispute is my contention (supported by physics) that where any distance is involved, horsepower creates a greater rate of delta V than torque." This is NOT supported by physics. I will admit that you have a lot more experience in racing cars than I do, have raced more than me, and may have even won your fair share of races. Your mental model of how a car works is sufficiently accurate that (i purport) that you can set up a car to get the most out of your driving style on any traack you visit. However, your mental model is incorrect with respect to the physics of (straight line) acceleration. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 514 Registered: 4-2002 |
Just for fun, I googled 'maximum acceleration rpm' http://www.4wdonline.com/A/Power.Torque.html Substantiates my position; http://www.v8914.com/Horsepower-v-torque.htm Substantiates my position, but also includes an excersize where the transmission is a CVT, and concludes that max acceleration doe indeed occur at max HP. However, read the caveats and my position is still substantiated for cars with a fixed number of gears. Top fuel cars (7000lb-ft of TQ and 6000 HP both at unspecified RPMs) are launched on the clutch, in fact until about 250 MPH, the clutch is still modulating the power deliverd to the rear wheels to keep from smoking them. The typical programming is to keep the engine at peak power (HP) and burn the clutch transmitting only that power that the rear wheels can absorb without spinning. As such, they are operating like a CVT and unlike a geared car. Anyone ever notice how they tend to really accelrate at 3/4 track? This is where the clutch finally locks down and the engine is no longer limted in HP delivered to the wheels. http://216.239.57.100/search?q=cache:FFjuSJqO5OsC:www.allpar.com/eek/hp-vs-torque.html+maximum+acceleration+rpm&hl=en&ie=UTF-8 http://216.239.57.100/search?q=cache:VkXR7KyRI6YC:www.renthal.com/website/resources/guide.htm+maximum+acceleration+rpm&hl=en&ie=UTF-8 And one to satisfy even EFWUN: http://216.239.57.100/search?q=cache:G8dfvr4Mf_AC:www.hondaclub.com/Articles/ArticlesView.cfm%3FarticleID%3D57+maximum+acceleration+rpm&hl=en&ie=UTF-8 which states: Maximum acceleration at any speed occurs at the HP peak. Maximum acceleration in any gear occurs at the torque peak. I think this is what you are trying to get to. Right? So look at the first sentence like this: "maximum acceleration occurs in the gear where more HP is available at any speed." So given a choice between 2nd gear at 8250 RPMs and 3rd geat at 6700 RPMs, there is more HP available in 2nd gear, and that gear delivers more acceleration than 3rd gear. However, in 2nd gear, maximum acceleration occus at peak TQ, and in 3rd gear maximum acceleration occurs at peak TQ. Moral: HP tells you when to shift, TQ accelerates the car. http://216.239.57.100/search?q=cache:24gZFQ9K-FAC:www.goldenbig.co.kr/Part_2_files/p2-04-04.htm+maximum+acceleration+rpm&hl=en&ie=UTF-8 Which goes on to say: "Torque and the weight of cars have more influence on the PW than on the horsepower. In consequence, the performance of acceleration is determined by torque and engine response. " And for maximum acceleration with a slushbox: http://216.239.57.100/search?q=cache:uijuGDJ4i2kC:www.protorque.com/protorque_new/techi/ti_faq.htm+maximum+acceleration+rpm&hl=en&ie=UTF-8 which states: "Theoretically, for maximum acceleration the stall speed of the torque converter should match the peak torque rpm of the engine. A good explanation for the way it works is this: when you go outside jogging you start to breathe in and out faster and harder. Well the same thing goes for a performance engine. The engine is breathing in and out harder and faster, at a higher rpm. If a high performance engine makes power at a higher rpm, then a higher stall speed torque converter is what you need to put more power to the ground quicker." This next one creates charts just like I did. http://216.239.57.100/search?q=cache:h1ase-DqBZ0C:kennedyp.iccom.com/pat/contour-gearing.htm+maximum+acceleration+rpm&hl=en&ie=UTF-8 Substantiates my position. SO don't respond that we don't know what we are talking about--everything I find on the net substantiates my position. Why not spend as much time searching for things that support you position as you spend writing rebuttals. Who knows you might learn something. | ||
| DJParks (Djparks)
Junior Member Username: Djparks Post Number: 148 Registered: 2-2003 |
No offense taken, meaning in comparison to to the level of technicality of the rest of the thread. I understand the basic theory presented here and also understand the effects external factors would have on the basic theories which would in turn produce an endless variety of results which would lead me to believe that alot of this 'just depends'. Just my two cents. The information is worth it's weight in gold. DJ | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 526 Registered: 2-2003 |
Brian, the only dispute is my contention (supported by physics) that where any distance is involved, horsepower creates a greater rate of delta V than torque. DES will post the experimental results of the acceleration timing with regard to one specific car. I will post the equation derived by Dr. Kalafut's doc candidate proving my theory. In the meantime, I'm going home to drive my blu car and enjoy. A very happy Passover/Easter holiday to all. | ||
| Brian Kennedy (Kennedy)
Member Username: Kennedy Post Number: 252 Registered: 3-2002 |
There's a whole lot of disputes on general physics principles in this thread... various people slamming others for imprecisions in their physics dissertations... but its not clear there are any real car questions left unanswered... just checking... the physics of a car are much simpler than a lot of the general physics being explored in this thread... 1) Peak acceleration will coincide with peak real wheel torque (assuming no traction loss), and that will coincide with peak engine torque while the car is in first gear. 2) For most any normal car, acceleration at peak horsepower in first gear will exceed acceleration at peak torque in second gear (or any other point in second gear). The same is generally true for all successive gears... and this is why you shift at or near redline, in most any reasonably designed car, for max acceleration. 3) The torque curve and the horsepower curve of a car have a direct mathematical relationship... they are two ways of measuring the same attributes of a car... you can use either curve for any computation... you are essentially just changing units. 4) Most people (the target of automotive marketing) don't think in curves, they prefer single numbers. Thus, we've developed the habit of describing the torque/hp curve (they are the same curve, essentially) via two points on the curve: the peak HP and its RPM and the peak Torque and its RPM. Simulation software is often designed to interpolate the torque/HP *curve* from those two points... making numerous assumptions about the shape of most such curves. Engineers, however, use the curve, not the two points. You design an engine for a good curve (either curve, same result/goal)... you only worry about the two points for marketing reasons. 5) Driving near redline puts far more wear on the engine (the original question, right?). By spinning 3-5x faster, it is not only getting 3-5x more wear, but it is doing so near a variety of mechanical limits of the engine, such as the forces on the valves and linkages and the amount of heat needing to be dissipated. Driving around town at redline would definitely degrade engine life. 6) If you own a Ferrari, you own a piece of automotive art that was meant to be driven under high G's... high G's are great fun... that fun is best produced up near redline. So, although it degrades engine life, and automotive part life in general, I highly recommend driving your Ferrari at High G's (preferably on a closed track)... I'd rather use up a Ferrari after 60K miles of fun than waste a Ferrari by never using it as Enzo intended it!! (Historical Ferraris are excluded from that statement.) Redline is good! Now, are there any other *car* or *Ferrari* questions that are still being disputed?? If not, I'll leave the physics disputes for you guys to hash out. ;) Ciao, Brian | ||
| DES (Sickspeed)
Advanced Member Username: Sickspeed Post Number: 3380 Registered: 8-2002 |
i'm so lucky... | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 523 Registered: 2-2003 |
DJ, no intent to call it simplistic! No offense either, I'm battling pretty hard here! Rob: You're absolutely right, a top fueler, with virtually unlimited torque achieves maximum G at launch, with the aid of carefully calculated clutch slip and with tires capable of translating force sufficient to achieve 5-6G of load. They also transition VERY quickly to speeds where aero drag becomes the limiting factor. I have been hung out on a straight, having fluffed a shift or something, and accelerated up through the revs in very peaky cars, and my experience was that those cars came on like banshees WELL past torque peak. As I tried (perhaps inelegantly) to say earlier, the shorter the distance, the more "torque" is the arbiter, e.g., of elapsed time, while the longer the distance, the more "power" is the arbiter, e.g., the ability to impart more of that "torque" per any discrete unit of time and create a higher rate of delta V. As I said, I'll have DES time the 550 over various rev ranges, and we'll do multiple trials, until the results coincide within a statistical probability of error. This should provide empirical evidence of my theory. If it doesn't, I will surely admit it. If it does, I hope the gentlemen on here will similarly admit that, or at the very least, admit that, as R & T said, "it depends." | ||
| DJParks (Djparks)
Junior Member Username: Djparks Post Number: 147 Registered: 2-2003 |
WOW! What a thread. Where else can a subject be so thoroughly investigated and discussed in such a short amount of time by so many. Great site. EFWUN, thankyou for the clarification of my simplistic view of torque vs power, a poor choice of words on my part. DJ | ||
| Rob Schermerhorn (Rexrcr)
Member Username: Rexrcr Post Number: 520 Registered: 11-2002 |
Ps, please don't post 'Rob prove's me right/ you wrong', I'm not interested in participating in that type of discussion. It's all good. | ||
| Rob Schermerhorn (Rexrcr)
Member Username: Rexrcr Post Number: 519 Registered: 11-2002 |
Though this thread is most interesting, for me it's interesting for other than scientific reasons Here's the deal: If the question is 'When does a car achieve maximum acceleration?' The answer is at or near the torque peak in low gear. Here are the assumptions: Proper gearing (which most manufactures do, most racers, too.) Not traction limited (eliminates friction discussions and debates over tire characteristics, etc.) Torque accelerates a car, horsepower HELPS determine top speed (other design variables too.) My computer is filled with data that substantiates my experiences. If one wants to use different assumptions or different limiting parameters, this may change the result. If we're talking about real cars that normal people drive every day, this is the answer. Even if you look at the ultimate in automotive acceleration equipment, a Top Fuel dragster, they pull maximum positive delta vee right off the line (torque), not at the end of the run. Ps, thanks for the engineering entertainment, this thread is why I keep coming back every few hours to see what's up Best regards, Rob | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2838 Registered: 6-2001 |
it wasnt an insult, pal. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 522 Registered: 2-2003 |
Gee Tim, ouch! | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2837 Registered: 6-2001 |
EFWUN and iraqi info minister.....same person? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 521 Registered: 2-2003 |
Pete; you seem unwilling to explore the fact that torque is a static force. Once rotation begins, you have that force as a function of angular rotation, (Omega, I think) OKAY: Last night, admittedly timing on my own, 4,000rpm to 5,500rpm in 2nd, a little over a second, 6,000rpm to 7,500rpm in 2nd, a little more than 0.5sec. Recognizing the need for independent verification, I will ask a DESignated timer to join me for a bunch of experimental runs, and then ask him to post the results. I'm sorry you and Mitch agree that you're right, because while the physical concepts are difficult, they are accessible to someone of your intelligence, if you'd only be willing to explore them. Finally, Dr. Kalafut is not my "friend" but rather an adjunct professor to Tom Lubensky, the chairman of "Condensed Matter" Physics at an Ivy League University, Penn, my alma mater. Many years ago, Dr. Lubensky offered to sponsor me for a Fulbright fellowship, but following my failure to pursue that "honor" I doubt he remembers me today. He did, however, reply through Dr. Kalafut, stating that my theory was correct, and supported that statement with some incomprehensibly dense equations. I asked for clarification, and Dr. Kalafut's teaching fellow/doctoral candidate agreed to write back when time permits. Finally, Mitch, what the esteemed Mr. Smith meant (I think) is that many people fail to take advantage of a broad torque curve, gearing the vehicle too short to take advantage of the broad "power" available. The early 930 is a perfect example of relying on the breadth of the power band (yes, the area under the torque curve); that car (as I remember) used only 4 speeds, because Porsche felt it didn't need any more! Nevertheless, the breadth of the measured torque curve results in an equally broad "power band." My numbers above show that the motor will put more G on the car at peak "power" than at peak "torque", but I understand that this vehement an argument requires independent proof. Why not wait for DES to post the experimentally derived data, rather than continue to argue theory?? | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 513 Registered: 4-2002 |
"Tune to Win" Carroll Smith chapter 12 section 'gearing' last paragraph: Quote: "The two most common mistakes that racers make with respect to gearing is running too short a low gear and twisting the engine too tight. In both cases the driver is probably confusing noise with power and wheelspin with forward bite." unQuote Q.E.D. | ||
| PSk (Psk)
Member Username: Psk Post Number: 377 Registered: 11-2002 |
Mitch, I love your last few posts. Yep as you, and I have stated the case is closed. I also owe you an appology. I thought you had been arguing that maximum acceleration was obtained at peak power rpm as EFWUN is. The only area you and I disgreed in was that my EARLIER posts were ignoring gearboxes and thus maximum acceleration occurs at peak torque. Once gearboxes are included you are 100% right ... but I am a really theoretical guy and was discussing eggs against eggs with Rich, thus a car with the most torque and identical everything else (including gear and diff ratios) will accelerate faster ... being careful with what I say here: produce the highest acceleration figure (m/s^2) and thus DID accelerate faster at some point of time. EFWUN, I appologise for being condensending in my earlier post. You have more mail ... but I am struggling to understand what and where you are trying to go or prove. Many, many times I have proved acceleration via Torque, and many others (no doubt with higher qualifications) have backed this up or taken it further ... and yet you still continue with your power = max acceleration mistake. Even Road & Track have confirmed what we are saying ... I personally think you are confusing the torquing of a bolt with the rotational force concept of torque. The only reason we talk about torquing a bolt is because it takes this much rotation force to do the bolt up ... it is not static or instantious but a REAL continuous rotational force. I personally am no longer interested, but will be interested in any formula your Professor friend provides. But remember we are not interested in work done over time but acceleration figures. Pete | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 512 Registered: 4-2002 |
"Race Car Engineering" by Paul VanValkenburgh, Chapter 9: Gearing and Differentials Page 192: Quote: "By this stage the racer should know nearly all that is necessary about a given car to make some good mathematical pedictions. The necessary data includes; an engine Torque Curve, all numerical gear ratios available, tire rolling radius, and vehicular air drag. It is also a useful idea to have a rough idea of the rear tire <traction> coefficient and the center of gravity height" UnQuote. Se, HP curve not required. He then oges on to build the SAME graphs I did below. Now for the clincher: Quote: "The use of these curves should be apparent. Any arbitrary gear is selected and run up to its maximum speed in rpm (or the point at which thrust intersects the next ratio curve), and it then shifted down to another higher ratio." | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 511 Registered: 4-2002 |
EFWUN: "Race Car Engineering" by Paul VanValkenburgh, chapter 2 page 14 'Ellipse representing peak tire traction in any direction'. "Race Car Vehicular Dynamics" Millikena nd Milliken. Chapter 2 figure 2.16 'typical traction slip ratio curve' accleration. Chapter 2 figure 2.17 'Typical Braking--slip ratio curve' braking. Chapter 2 figure 2.22 'Resultant force vs resultant slip' sideways. Section 2.7 Friction circly and Ellipse Chapter 2 figure 2.31 'Friction Circle Diagram' Still with me? Ok, on 2.31 the cornering axis and the braking axis and the accelerating axis are denoted in pounds (lbs). Notice that the cornering axis has the same amout of total force that the accelerrting axis has (as with the braking axis). Still with me? This shows that the a set of tires that can allow a car to corner at 1Gs will allow that car to accelerate (*1) or decelerate (*2) at 1Gs. Agreed? *1) If more thrust is delivered to the tire it will not be able to sustain 1Gs *2) Heavier braking ends up locking a wheel up--this does nothing good to acceleration in a negitive direction. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 507 Registered: 6-2002 |
EFWUN - Clarify something for me. You refer to torque as a "static" force, not capable of work. My understanding is that torque is to rotational dynamics as force is to linear motion. They are analogous. If one applies a constant 10 pound-feet to a shaft (using steam, or electicity, or falling water, or donkey - the donkey pulling a calibrated spring which reveals 10-pounds of force, and it is tethered to a wheel 1-foot in radius), and it raises a 10-pound load of bricks 10 feet off the floor, then 100 pound-feet of work has been done. If one slides a box (weight irrelevant) 10 feet along a floor, and the surface (and weight) create a frictional force of 10 pounds, then 100 pound-feet of work has been performed. Why are these different? Jim S. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 510 Registered: 4-2002 |
"I'm loathe to say you don't "understand" so I'll say that you're just not willing to explore the concept that torque is the instantaneous force necessary to overcome resistance, Ok, so we have TQ is a FORCE. Its force independent of whether the force works against a resistance or it works against an unmovable object or it works against nothing whatsoever. It is a force. (PERIOD) Well, Ok, lets explore: F = M*A M is the mass (weight) of the vehicle; so Force controls the acceleration--right? Hint #1 weight is constant. So F(time) = TQ(time) * gear(time) * diff / radius Since TQ, and gear change over time--still with me here? "while power is that force as a function of distance and time. W=FD; the concept is difficult, but accessible if you stop arguing and explore. " Now W = F*D Following along so far? D(time) = Integral( Integral( A(time) ) ) So: TQ(time) determines acceleration(time) acceleration(time) determines velocity(time) velocity(time) determines distance(time) distance(end - start) determines work. So it all comes down to TQ(time). Therefore, you actual argument supports my position! | ||
| arthur chambers (Art355)
Intermediate Member Username: Art355 Post Number: 1282 Registered: 6-2001 |
EfWUN: You're damn right your a dotting old fuddy duddy, just like me. We raced 20, 30, 40 years ago. Some of these kids weren't born then. Regards, Art | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 509 Registered: 4-2002 |
Once more with units: F = M * a M = 3200 lb/G :: the car weighs 3200lbs in a 1 G environment F = TQ (in lb-ft) * gear (dimensionless) * diff (dimensionless) / roling_radius (ft) Therefore: F = TQ * constant (in lb) In particular for 1st gear: F = 3400 lb (of force in the forward direction at peak TQ) A = F/M A = (3400 lb) --------- (3200 lb/G) A = 1.07 G To argue further, you need to provide equations with dimensions. Otherwise, case closed. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 508 Registered: 4-2002 |
"Mitch, your misunderstanding of the concept of tire friction explains your misunderstanding of the difference between torque and power." No the misunderstanding is yours. "A tire doesn't have friction in "G." G is meters/sec/sec, a rate of change (acceleration). Tires have friction as force Normal (N) straight down through the contact patch (times) the coefficient of friction." A single tire does not--however 4 tires in a car with descent suspension does! "Breaking a tire loose doesn't mean you're "accelerating" at 1G, but rather imparting sufficient torque in foot pounds to overcome (coefficient of friction )x (force N), a static number." Braking loose a tire means you have exceeded the force it can transmit to the surface of the road. I will show below why you are wrong. But first to the problem at hand----- I needeth not your misrepresentation of physics. F=ma :: Newtons second law Following along so far? any question on equation 1? F = thrust = engine Tq * ( tranny_ratio * differential_ratio / Rolling_radius ) Therefore, in any particular gear; F is proportional to TQ. If not, then you need to supply an equation that represents what you think is supposed to happen--be sure that the units match. M = weight_of_car Any arguments here? A = M/F :: simple physics--arithmetic actually As you see, torque (with a bunch a multiplication constants) causes acceleration. Not power. Power tells you WHEN TO SHIFT, not how fast you are accelerating. POWER tells you WHEN there is more thrust in the next gear than in the curreng gear. POWER DOES NOT ACCELERATE THE CAR. {{And at this point we are still ignoring the degredation caused by having to accelerate the rotating and reciprocating engine and driveline components as they are a small fraction of the weight of the car (not so when real racing vehicles are used -- like 800 HP with 1200 lbs -- here the roating inertia of the engine and driveline limit the acceleration of the car in 1st and second gears: See "Engineer to win" Carroll Smith).}} Now back to the new problem generated----- Have you ever 'looked' at an acceleration graph? Have you ever 'looked' at a deceleration graph? Have you ever read a book on the physics of tires? Have you ever read a book on suspension design? If you have done any of the above, you will see that durring braking the car is traction limited by all 4 tires (assuming the brakes don't cook and assuming that the brake bias is set correctly). If you look at the deceleration graph you will see that it is a straight line with a slope of ~32 ft/s/s ~= 1Gs. I am purposely ignoring the difference of 5% level stuff here. Routine measurements by car magazines show that many high performance sports cars are capable of cornering at 1Gs of sideways acceleration. Our Ferraris with modern rubber and correct alignments are in the right ballpark. I have thus, established that 4 tires on a car with reasonable suspension can indeed get 1 Gs of force into the car. I postulate that 1Gs of force is also possible in the forward direction. A notion supported by Milliken and Milliken (I think in chapter 14). They compute slightly different shapres for frontward, rearward and sideways traction, but unless you are woried about 1%-2% level stuff, rounding off to 1Gs suffice for this level of misunderstanding--yours. "I've asked several times for someone to do an empirical analysis!!" If you look down about 50 posts, you will see that I did address this very issue--however, the result was not to your liking, so you dismiss the data point. Not very scientific was that? One of the problems with your through train is that racing engines are very much liike Ferrari engines--as shown in my graphs. They hold onto their TQ as RPMs rise, degrading only slowly. This makes these cars FEEL like they are continuing to accelerate at the same rate as peak TQ, and indeed, are only a few Ft-lb short at peak power. However they are still only accelerating at 97-98% as fast as they were at peak TQ. So your argument comes down to believing that 97% is bigger than 100% -- right? If you look at the american V8, you see that TQ falls rapidly with RPMs. Not even in your wildest notion are you arguing that this shape TQ curve accelrates faster at peak power than at peak TQ (in any gear you choose). | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 168 Registered: 10-2001 |
"Casting aspersions on my (admittedly LONG ago) racing career isn't necessary, is it?" Sorry Efwun - didn't mean to offend. Just thought a test done today would be more accurate than a memory. And compared to Lawrence, my comments are pretty tame, ha! | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 578 Registered: 4-2002 |
Yes, I have what I say I have as far as degrees go. And I proved that the acceleration is the greatest at peak torque several days ago. I didn't offer it without proof. You forgot about it or did not comprehend it. In my last post I'm saying that it takes more energy to change the speed from 102 from 100 than it does from 80 to 82. The peak horsepower is generated at higher rpm. Therefore the vehicle is going faster. Therefore each difference in speed takes more power at the higher speed. Surely even you can understand this. I just read some old posts. You're the one who thinks Ferraris have big Weed Eater engines in them (2 cycle). If I ever made a mistake like that in public, I would never post again. You have a real ability to demonstrate your lack of knowledge. Had I known it was you I would not have wasted my time. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 576 Registered: 4-2002 |
Yes, I have what I say I have as far as degrees go. And I proved that the acceleration is the greatest at peak torque several days ago. I didn't offer it without proof. You forgot about it or did not comprehend it. In my last post I'm saying that it takes more energy to change the speed from 102 from 100 than it does from 80 to 82. The peak horsepower is generated at higher rpm. Therefore the vehicle is going faster. Therefore each difference in speed takes more power at the higher speed. Surely even you can understand this. I just read some old posts. You're the one who thinks Ferraris have big Weed Eater engines in them (2 cycle). If I ever made a mistake like that in public, I would never post again. You have a real ability to demonstrate your lack of knowledge. Had I known it was you I would not have wasted my time. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 520 Registered: 2-2003 |
Additionally, Rich, the reason F/Atlantic and F/2 cars are relevant (you're right, they were a long time ago) is because they were the quintessential "Peaky" motors, little 1.6 and 2.0 liters, belting out Herculean power, with VERY little torque. Therefore, the surge after their torque peaks was Dramatic, whereas the more modern stuff is better engineered, to have broader torque and therefore, more available units of power over a larger rev range! Finally, most of us don't have a straight where we can max out 3rd and change to 4th (I tried it this morning and I was WAY to busy to evaluate), and in racing, you're often sitting on a straight at 130 or so, waiting for the damn thing to GO!! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 519 Registered: 2-2003 |
Rich, I've asked several times for someone to do an empirical analysis!! Casting aspersions on my (admittedly LONG ago) racing career isn't necessary, is it? I'm not some doddering effing old fogie, it was simply that I raced between the ages of 19 and 24. You're right, I think the variable intake lengths on a Modena and the concomitant broad torque curve might well mask some of the difference, as well as wind resistance increasing as the square of any speed where time would be measurable. I will be driving a 355 this weekend, and I'll attempt to measure. Nevertheless, the math shows that work creates the change in kinetic energy, and torque is incapable of work, because it is a static force. Therefore, the fact that there are nearly double the units of work available at 8,200rpm than at 4,750rpm means the car is accelerating harder (more G) at 8,200rpm. And YES, I plan to drive my 550 home today, and enjoy it as a break from arguing physics!! | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 166 Registered: 10-2001 |
I still think that most members of this site are more interested in driving cars than doing math. Therefore, will someone please go get in your car and measure the time it takes to go 1000rpm up from your torque peak and then do it again for the last 1000 rpm to power peak? Do not change gears. Use 1st or 2nd to lessen aerodynamic penalty for the higher speed the car must travel near redline. Use a flat road. Etc. Then we will know a "real world" case not a theoretical one. (memories of formula atlantic cars or whatever they were don't count - let's measure a Ferrari now). My suspicion with a current Ferrari is that since the torque curve is so flat, the times will be nearly identical (because near redline the torque is almost the same as at the peak). Perhaps one of the guys with the older cars would be a better test driver and produce the results that acceleration time (as defined as the amount of time it's going to take the car to raise its rpm by 1000 rpms) is shorter near torque peak than at hp peak. | ||
| Sean F (Agracer)
Junior Member Username: Agracer Post Number: 90 Registered: 2-2003 |
Efwun is right, torque does not do work ...but it is still lbf ... HA ha, just kidding ya! | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 517 Registered: 2-2003 |
What is confusing you LC (and I note you changed your profile to indicate PH.D in Mechanical Engineering), is that Torque does no work at all. Should I post my National Merit Scholarship, or my Westinghouse runner-up? Rather, as you actually state, Work creates the rate of change in kinetic energy. Available power at peak torque, say 250units at 4,750 is significantly less than the 400units of power available at 8,200rpm. Having admitted that work creates the change in kinetic energy, your posit that acceleration is maximal at torque peak is bankrupt. Go open a moldy text on angular momentum, and tell me if "torque" can possibly do "work." If you're scientifically honest, you'll admit it can't. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 516 Registered: 2-2003 |
LC, that is the most transparent crap I've seen in a while. You state what in law is called an ipsa dixit, the thing is so because you say so. After positing the correct equations for acceleration, you then simply say that A is greatest at peak torque. Wow, elegant derivation. Please stop talking down to me! When Dr. Kalafut's teaching assistant/doctoral candidate unravels his too complicated equations, I will demonstrate to you that Torque itself creates NO ACCELERATION AT ALL, it is a STATIC FORCE. In the instant that you have rotation, torque becomes work, of F x D. Torque is the ability to overcome static resistance, power is the ability to create a rate of change. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 575 Registered: 4-2002 |
In my little demonstration of the classical equations from both torque and Newton's second law and then from a power equated to rate of change of kinetic energy, I clearly showed what nearly everyone knows. Acceleration is maximum at peak engine torque, providing you are in the same gear. Let's assume the acceleration remains constant. Standard kinematics says the new velocity will be V = V1 + A * DT V is new velocity based on an acceleration A for a time duration of DT V1 is old velocity DT is time interval With DT, V1, and A positive, the equation with the larger A will produce a larger V for the given time DT. Peak torque produces the greatest acceleration. Therefore you gain the most speed per unit time. What is confusing you, EFUN, is that while more power enables you to do more work (more change in kinetic energy) in a give time, you have to do it at a higher speed if you are in the same gear. Take any two speed intervals, say 82 and 80 and also 102 and 100. Look at the difference in kinetic energy that is proportional to speed squared. Kinetic energy is proportional to speed squared. Whether they are feet per second or mph, the point is still made. 82 * 82 - 80 * 80 = 324 102 * 102 - 100 * 100 = 404 So you can observe that to change the speed two miles per hour at a higher speed requires more energy. To do it in the same amount of time requires more power. That is why the car does not accelerate as fast as one goes through the peak power point of an engine even though the rate of doing work (changing kinetic energy) is greater. You have more change to overcome at higher speeds. | ||
| Rob Schermerhorn (Rexrcr)
Member Username: Rexrcr Post Number: 509 Registered: 11-2002 |
And the Honorary Doctorate in Physics from The Masseteusets Institute of Technology goes to... | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 515 Registered: 2-2003 |
Close, DJ, but torque is not work, because work has a necessary component of distance, be it angular rotation or linear. W=FD. Torque is F, D is angular rotation (or distance) and W is work. Horsepower is how much "power" is produced as a function of time. Please see derivation below. Torque is the ability of a motor to strain against (or overcome) static resistance, while horsepower is the same motor's ability to create a rate of change as a function of time. | ||
| DJParks (Djparks)
Junior Member Username: Djparks Post Number: 140 Registered: 2-2003 |
In the simplest terms it was explained to me thus, 'Torque is the amount of measurable work an engine can do and Horsepower is how fast it can get it done.' Probably not the most accurate explanation but gave me a foot hold on the two concepts when I was 10 years old. DJ | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 513 Registered: 2-2003 |
Mitch, your misunderstanding of the concept of tire friction explains your misunderstanding of the difference between torque and power. A tire doesn't have friction in "G." G is meters/sec/sec, a rate of change (acceleration). Tires have friction as force Normal (N) straight down through the contact patch (times) the coefficient of friction. Breaking a tire loose doesn't mean you're "accelerating" at 1G, but rather imparting sufficient torque in foot pounds to overcome (coefficient of friction )x (force N), a static number. Torque is the ability to strain against static resistance, e.g., we "torque" wheels, or impart say, 84pound-feet of torque to tighten a nut. Similarly, torque is the instantaneous force necesary to overcome static resistance, in your example, (N)x (cf of Friction). Once your tires begin to rotate, or your car begins to acceleration, you're producing work, the product of force times distance. Acceleration is the rate of that change as a product of time. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 512 Registered: 2-2003 |
Mitch, I'm loathe to say you don't "understand" so I'll say that you're just not willing to explore the concept that torque is the instantaneous force necessary to overcome resistance, while power is that force as a function of distance and time. W=FD; the concept is difficult, but accessible if you stop arguing and explore. Torque is expressed as pound-feet because the concept is, for example 275lbs of force on a moment arm of 1-foot, equals 275lb-ft of torque. Once that torque creates rotational speed (omega, I think?) you have work, or (force times distance in radians/sec). Let's wait for Dr. Kalafut as to the derivation of these equations; I can't find my physics texts, and if I suffer early onset Alzheimers, and get something wrong, you'll pounce on that one error without realizing that the concept is sound. Torque creates the change, e.g., overcomes resistance, while Power creates the rate of change, e.g., increasing speed as a function of time. This is the definitive answer to why more torque equals lower elapsed time and more horsepower equals higher mph at distance. The longer the distance, the less influence torque has, and the more power is important. Distance is a variable, as are gearing and different torque and horsepower curves. As the editor of R & T said, "it depends." | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 507 Registered: 4-2002 |
"My argument is that the car will put increasing G on you until you reach power peak, then tail off toward redline." Your argument is wrong. "To include an analogy from the intitial reply from R & T, the car will break the tires loose more easily at peak torque." If the engine can break the tires from their grip on the pavement better at peak TQ but can't at peak power; does this not mean that more accelerative force is available at peak TQ where the tires break free? Otherwise, yu would break the tires free at peak HP also. For example: Lets take a tire that allows you to corner at 1Gs. Various studies have shown that this tire can also brake at 1Gs. It makes sense that this tire can also accelerate that car at 1Gs. So if you have engine TQ to break this tire from its grip on the road at peak TQ, you are accelerating at 1Gs when the tire breaks free. If you don't have the TQ at peak HP RPMs to break the tire free from its grip on the road at peak HP then you CAN NOT be accelerting at 1Gs (must be lower). Therefore, peak acceleration occurs at peak TQ. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 506 Registered: 4-2002 |
"have you ever tried to move your car off the line in 4th (1:1) gear. You just do not have enough torque!" Actually, I have, and with correct clutch slippage, it is not hard at all, not fast, but not hard. {Hint: slipping the clutch also causes TQ multiplication}. I don't recommend it. In traffic, I often slip away from a stop in 2nd. "What I will give you, is a big pat on the back for making the mindset jump that EFWUN has yet to do or understand ... but it may happen " If you reRead my posts, you will see that my mindset has not wavered whatsoever. "Reciprocating motion ... hmmmm, we are splitting hairs here again " No, actually, we are not splitting hairs here. If our pistons were connected to the crankshaft with a scotch yoke (look it up), then we have a situation where linear motion is being converted into rotational motion. However, we have finite length connecting rods. Finite length connecting rods change the geometry and the efficiency of power conversion and create secondary and tertiary forces that the rest of the engine has to withstand. "I was only talking about where the torque comes from " TQ comes from igniting a mixture in an enclosed container that is then allowed to expand while work is extracted. SO TQ comes from the mixture. The piston is only the delivery means to the con rod, and the con rod is the delivery means to the crankshaft. "going up does not produce any torque ... infact it is a loss due to work required to be done" This is basically correct, however it is only correct for the power stroke and misrepresents the 4 cycle process. At low throttle openings the vacuum in the inlet can add power to the piston as it rises. I have an engine simulator written in eXcel that can simulate this pressure curve. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 511 Registered: 2-2003 |
Pete: Gee, your email was quite a bit more polite than your post! Just because I disagree with you doesn't mean I don't "understand" you, that's really rather arrogant. I decline to be similarly condescending. Pete, I completely understand that you're saying that torque at the rear wheels is maximal at peak torque (sort of redundant, but nevertheless quite true). What you are ignoring is that at peak horsepower, the force delivered by the rear wheels is greater over any discrete period of time. Again, as an over simplification of the initial (way too complicated) response from my physics prof, torque creates change, (overcoming static resistance) while power creates the rate of change as a function of time. To paraphrase the initial reply from R & T, elapsed time is a function of torque, while mph is a function of horsepower. (we've all heard this in one form or another). At peak torque, your car may well cover a short distance more quickly, but over increasing distances, where the accumlation of mph is important, your car at peak horsepower will be create a greater rate of change and be faster. Similarly, a car with boatloads of torque may run 12.0 @ 90mph in the quarter, while the peaky, high-revving horsepower car will run 13.0 at 120mph (simplifications). Again, empirically, take your car out, and in third (to give time for analysis) accelerate from 500rpm below peak torque, to redline. My argument is that the car will put increasing G on you until you reach power peak, then tail off toward redline. To include an analogy from the intitial reply from R & T, the car will break the tires loose more easily at peak torque. However, that doesn't translate into creating a RATE of change, only the change itself. Mitch, the above empirical demonstration should be simple enough even for you to try. I apologize for the condescension! (my dispute with you is that you can't convert "torque" into "thrust", because in the instant that any rotation occurs, you have power or work "W=FD" rather than "torque.") The final understanding is, then, that if you're talking about, e.g., 0-10ft, torque is the only arbiter, while as your distance extends, and the rate of increasing MPH as a function of time becomes more important, power is the ultimate arbiter. This comports with the second point made by R & T, "it depends!" | ||
| PSk (Psk)
Member Username: Psk Post Number: 375 Registered: 11-2002 |
Mitch, I give up with you, it is almost impossible to write something with out you miss-understanding what I was saying ... maybe I am writing it a confusing natural? fasteR versus fastest???, I thought if it was faster then it was also the fastest ... but I am not anal retentive While I know that you were trying to be funny, the first and most important reason for a gearbox is to multiply torque not to keep revs within a sensible rev range. If we did not have a gearbox then your car would not move ... have you ever tried to move your car off the line in 4th (1:1) gear. You just do not have enough torque!, again EFWUN torque is everything ... not power My comment regarding torque is still building was referring to 'compared to the next highest gear', but yes your point is correct ... and I argued with you and EFWUN about this about 100 posts ago ... ie. torque and thus push dropping off as we approach peak Hp and RPM. Reciprocating motion ... hmmmm, we are splitting hairs here again but I was only talking about where the torque comes from ... the piston going up does not produce any torque ... infact it is a loss due to work required to be done. What I will give you, is a big pat on the back for making the mindset jump that EFWUN has yet to do or understand ... but it may happen See ya, I'll let EFWUN and you argue that the world is flat, but I have lost interest, but gained some interesting insites via the process ... isn't engineering great Pete | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 171 Registered: 1-2003 |
My office has a $100 pool as to how many posts this thread will get by 4/31 | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 504 Registered: 4-2002 |
"WITHOUT gears (or while revving out a gear) a car will accelerate fastest at peak torque." Correct. "WITH gears a car will accelerate fastest at somewhere around peak rpm." Incorrect--but close: With gears a car will accelerate fasteR when shifting near peak RPMs. Or A car will accelerate faser in 2nd at max RPMs than in 3rd (at the same road speed that corresponds to max RPMs in 2nd gear.) "maximising torque at the rear wheels is why we have a gearbox" {{futile attempt at injecting humor into discussion}} I thought we had a gearbox to keep the engine in a reasonable RPM band. {{end futile attempt}} "Thus revving a car to peak rpm has little to do with peak power" Other than peak power occurs near peak RPMs. "more to do with the fact that your torque is still building and due to the mechanical advantage of the lower gear you are ahead of the next gear." The TQ is (in fact) diminishing (as RPM increase above peak TQ RPMs), but diminishing slower than the RPMs are building--that why power goes up. Its the product of the diminishing TQ and the increasing RPMs that makes the power (go up). "Torque is the only thing an engine produces," Incorrect--heat, exhaust,... but basically acceptable. "as this is what the crankshaft is for, turning a linear force into a rotating force." By the way--a crankshaft turns a reciprocating motion into a rotating motion. Torque does the work, HP gets the credit. Carrol Shelby (circa 1962-3) | ||
| PSk (Psk)
Member Username: Psk Post Number: 374 Registered: 11-2002 |
EFWUN, I was not going to post this ... instead send you a mail (which I have) but this needs to be sorted out for everyone. WITHOUT gears (or while revving out a gear) a car will accelerate fastest at peak torque. WITH gears a car will accelerate fastest at somewhere around peak rpm. (Refer ps2 below) I have proved this with my 360 Modena analysis that shows that at peak rpm the torque AT THE REAR wheels is higher in the lower gear (than the higher gear at peak torque) thus you will be accelerating faster. Do not confuse this EFWUN, I am not saying that an engine produces more torque at peak rpm or power just that due to the mechanical advantage of the lower gear the rear wheel torque is higher at peak rpm than the next gear. The variables in this are where your torque drops off versus peak rpm. So please let this rest. WE ALL KNOW THAT A CAR WILL ACCELERATE FASTEST IF YOU CHANGE AROUND PEAK RPM WHEN GEARS ARE INVOLVED!!!!! (refer ps2 below) Torque is for acceleration, and maximising torque at the rear wheels is why we have a gearbox. Thus revving a car to peak rpm has little to do with peak power more to do with the fact that your torque is still building and due to the mechanical advantage of the lower gear you are ahead of the next gear. Please forget about your reference to Power, power is just another mathematical way of representing torque versus rpm. Torque is the only thing an engine produces, as this is what the crankshaft is for, turning a linear force into a rotating force. Thus EFWUN we ALL agree with you that revving to maximum rpm in a Ferrari will obtain overall maximum acceleration. Refer to my 360 Modena post where I worked out that I was wrong!!! (regarding that overall acceleration would be maximum if we changed gear so that the engine ended up at peak torque. I had forgotten about the mechanical advantage of the gear box), yep I was , and the acceleration will only be maximum while passing peak torque. Thus end of this debate Pete ps: we have 2 completely different situations, gears are involved and not ... but it is still all about maximum torque at the rear wheels and mechanical advantage. ps1: Gosh I hope I have made this clear ... just calculate rear wheel torque EFWUN and you will get it. ps2: Each cars (and for changes in gear ratios) gear change points need to be calculated independently. So my comment that any car will accelerate fastest at around peak rpm above, is not really valid. It 100% depends on the engines torque curve and the cars gear ratios. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 501 Registered: 4-2002 |
EFWUN: "I'm assuming you'll agree that the fastest way to accelerate is to shift at the redline " This has always been my position. "If the fastest way to accelerate is to shift at the redline, and that drops you into the 6s and 7s (depending on gear), then why isn't the car engineered to stay within range of the torque peak?" Because with the use of all 6 gears, you can get more acceleration by shifting to slightly above peak TQ and use the additional TQ multiplication of the slightly higher gear ratio. Notice that the essentially flat TQ curve of the Ferrari makes shifting at or slightly above peak TQ not an issue. "In higher gears, you never wanted to go from 9,800rpm (3-400 over peak power) to 7,500rpm (approximate peak torque). You'd go for a drop of 800-1,200rpm, max." This is completely consistent with my points. "I suggest that your 355 may have similarly close gearing, indicating that the factory, in engineering the car to accelerate as quickly as possible, worried about peak power (and area under the power curve) more than peak torque." I agree. "My daughter's Z-06 (!!) accelerates most quickly when shifted at the redline" As will any car without a seriously stupid TQ curve or an overly agressive red line. However, notice on the F355 HP graph, in 4th->5th and 5th->6th gear changes should be done just a smidgeoun below red line, while those of 1st->2nd, and 2nd-3rd should be done just above redline (maintanence notwithstanding) "A Modena has a relatively broad torque curve as the result of the ECUs changing effective intake length." I agree, I could have plotted the Modena charts but the biggest part of the Modena TQ curve (after normalizing out the displacement) occurs below the TQ peak in the F355. | ||
| TomD (Tifosi)
Advanced Member Username: Tifosi Post Number: 3252 Registered: 9-2001 |
yeah it was over 500 - it got bifucated http://www.ferrarichat.com/discus/messages/21/4043.html | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 506 Registered: 2-2003 |
Ah, too bad! | ||
| TomD (Tifosi)
Advanced Member Username: Tifosi Post Number: 3251 Registered: 9-2001 |
there is a long way to go before the record which I beleive is over 500 post maybe over 700 | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 505 Registered: 2-2003 |
Is there a record, or does the DESinator hold that by default?? | ||
| Rob Schermerhorn (Rexrcr)
Member Username: Rexrcr Post Number: 500 Registered: 11-2002 |
Come-onnnn 200, I know we can do it!!! I just did this to get me to 500 posts, if it helps get this thread to 200, all the better. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 504 Registered: 2-2003 |
Mitch, with all due deference to your "charts", you've got it backwards. A Modena has a relatively broad torque curve as the result of the ECUs changing effective intake length. Your 355 does not have such a broad curve, and is more like an Atlantic. It is the broadness of the area under the hp curve that makes an American V-8 (or a Can Am car v. an Atlantic) relatively indifferent to rpm as an accelerative necessity. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 503 Registered: 2-2003 |
Mitch, I was going to wait until I heard from R & T or Dr. Kalafut. Nevertheless, if you look at the rev drops in your 355, e.g., shift from 2nd to 3rd, say, at 8,500rpm, the revs in the following gear never get anywhere near as low as the torque peak at 5 or so for the 355. I don't remember the actual rev-drops from my 355s. I had the wonderful fun of driving my 550 to the office today on a beautiful Spring morning, and I maxed out 2nd (7,600rpm) and checked the revs in 3rd, 6,200rpm. Way above the torque peak. I'm assuming you'll agree that the fastest way to accelerate is to shift at the redline (I invite you to run several laps at your next outing short-shifting, and compare). If the fastest way to accelerate is to shift at the redline, and that drops you into the 6s and 7s (depending on gear), then why isn't the car engineered to stay within range of the torque peak? The task of gearing an Atlantic or F/2 car was to minimize the rev-drops, consistent with having a gear for the slowest corner, and any "in-between" corners. In higher gears, you never wanted to go from 9,800rpm (3-400 over peak power) to 7,500rpm (approximate peak torque). You'd go for a drop of 800-1,200rpm, max. I suggest that your 355 may have similarly close gearing, indicating that the factory, in engineering the car to accelerate as quickly as possible, worried about peak power (and area under the power curve) more than peak torque. Finally, Ferrari engines are not that different from any others. My daughter's Z-06 (!!) accelerates most quickly when shifted at the redline, as does our Turbo Beetle, which has a nearly flat torque curve beginning at 1,700rpm. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 499 Registered: 4-2002 |
EFWUN: You are confusing two different issues; A) The way Ferrari engines hold onto the TQ curve b) The way other engines loose the TQ curve If you go to the thrust graph presented below, you will see that the American V8 heads straight down hill after its peak TQ. The Ferrari holds onto (most of) its acceleration to the end of that gear. In comparison, the Ferrari holds onto that rush of power, but the acceleration IS less than at peak TQ, it is still bigger than the american V8 style of engine delivery, but still less than at peak TQ nonetheless! "they didn't have that dizzying rush after torque peak that the little fellas did, but seemed to accelerate pretty much the same all the way up" If you look at the thrust graph, you will see why; indeed, the american V8 does accelerate at about maximum power (350 HP) for essentially the whole set of gears. While the Ferrari looks like stepping stones, the american V8 looks like a continuous line. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 501 Registered: 2-2003 |
Rich, I'm not talking about redline, I'm talking about power peak. I agree that power falls off pretty quickly as you pass, e.g., from 8,200 to 8,500. I'm suggesting that acceleration from peak torque to some arbitrary number soon thereafter, e.g., 4,750 to say, 5,500 will take longer (ignoring wind resistance) than from 7,500 to 8,250. I haven't watched a lot of racing, I've driven several seasons in F Atlantic and F-2. These motors don't do anything until, yes, torque peak, but from there on up to power peak, they acceleration in a growing rush of power. Yes, they fall off damn dramatically after 9,500 or wherever the builder has put the power peak. And yes, the F-5000 based Can Am cars were far more indifferent to gearing than F/2 or F/Atl, because they had massive torque and huge areas under their horsepower curves as a result. Similarly, because they were grunty and a little short of breath over 8,000, they didn't have that dizzying rush after torque peak that the little fellas did, but seemed to accelerate pretty much the same all the way up, falling off dramatically after power peak. Everyone gears (generally) so that you exit the corner heading onto your most important straight at around torque peak, and finish at redline in top (whatever that is). At Mosport, for instance, in a weak sister like an Atlantic, you have plenty of time to watch the revs rise in (back then) fifth. You don't stack, say a 21/27 fifth on to just reach peak torque, rather, you go with a 20/28 and go for redline, or at least power peak. Please do me the favor of waiting for the heavy-hitters I've enlisted to write back; if I'm wrong, I will certainly fess up. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 500 Registered: 2-2003 |
Gentlemen! I've emailed a precis of these arguments to the engineering editor of R & T, (an occasional penpal) and the chair of the department of physics at my alma mater (Penn). If I'm wrong, I will post here, and if I'm correct, I will similarly post here. Talk soon. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 491 Registered: 6-2002 |
Mitch - excellent analysis and demonstration. Jim S. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 497 Registered: 4-2002 |
Now some more numverical integration to convert speed to distance:  Notice that the Ferrari never catches up! (although the Vetts lead diminishes after the 1/4 mile is done.) Vette wins (in this case) by 0.2 seconds in the 1/4 mile and in the previous graph the Vette wins by 0.25 seconds 0-60. This excersise was performed without any air resistance. The 0-60 times look a little fast by 0.2 seconds, and the 1/4 mile times a little fast by about 0.4 seconds. Enjoy | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 494 Registered: 4-2002 |
I thought it might be fun to compare the Ferrari high reving engine to the American V8 big iron. I choose a Corvette with an Ls1 and compared to an F355. (I leave it an an excersize to the reader to do the same for the Modena and Z06.) The first graph shows both engine curves plotted on the same graph and at the same scale.  Notice that the LS1 torque curve basically is a straight line downward from peak torque. This is a sure sign of breathing issues, but this also give it that big rounded HP peak. The second chart shows the available thrust at the rear wheels in all gears for both cars.  Notice that the Torque curve of the Vette has a BIG advantage getting out of the hole, and that the Ferrari has more thrust at some speeds. Notice that if the Ferrari owner wants to keep up, he can always rev up the engine to 5000 and feed the power through the clutch, while the vette driver just revs to 2300 and uses the clutch normally. At $200 a clutch this is a loosing proposition for the (unsponsored) Ferrari driver. This was all confusing, so I decided to do some numerical integration to answer the question as to why cubes win. This final graph shows the time to speed graphs of a 3200 lb F355 and a 3200 LB Vette. I penalized each shift by 0.7 seconds.  This really doesn't address the issue much better than the car with the big cubes gets out of the hole faster and then maintains its advantage. Notice that the Ferrari looses 0-60 by a realistic 0.2 secand is again tieds at 100 MPH in velocity--it still has distance to make up. From hear on up the Ferrari has an advantage in that the 100-top speed range is given 3 gears while the Vette ony gets 2. Finally, notice that when the Ferrari "makes its move" on the Vette, it takes all of its advantage right in the middle of the power band, not at peak TQ, nor at peak HP. Here we have a 350 HP American V8 with a dificiency of a whole gear beating a 380 HP high reving engine, both pulling the same weight. | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 163 Registered: 10-2001 |
"Rich, I'm not talking about rear wheel rotation, gearing isn't necessarily one to one. I'm talking about revolutions of the flywheel as function of time. " And your position is that to go from 4750 revolutions of the flywheel per minute to 5250 per minute will take longer in seconds than going from 8000 to 8500 in a car that has its torque peak at 5000rpm? Hop in your car and go time it - then maybe you or someone can use the math to show why it wasn't true, instead of using the math first, ha! Use a low gear (1st or 2nd) to lessen aerodynamic drag on the 8000 to 8500 run. Anyone who has ever timed their own car or has even watched motorcycle or auto racing where they show you the tach on the screen will know that the cars blow through the first 8k or so in a blink of an eye and only when it gets towards the redline does it take longer for an additional 500rpm or whatever. Yes, some of this is an aerodynamic issue (though not so much on bikes) but most of it is that the engine just isn't putting out as much torque. Or is the real world misleading? | ||
| PSk (Psk)
Member Username: Psk Post Number: 372 Registered: 11-2002 |
Okay the acceleration debate is over, lets now discuss why manufacturers and race car makers are interested in horsepower figures. If we look at the formula: P = T * rpm/5252 We can see that a motor that produces high power produces 2 things, good torque at high rpm and can rev to high rpm. Thus a motor that produces good power and can rev allows more flexibility with gear ratios (and we have already proven how much difference these make, and note the fact that trucks need something like 20 gears due to the very small torque producing rev range ...). Also a motor that can rev and can produce good torque at high rpm shows that it can breath and that the makers are clever buggers , thus credit to the them (and Ferrari). With F1 we have a CC limit of 3 litres thus the engineers are designing engines to rev very high because the peek force produced on the piston can only be so good using a piston based engine and this is a factor of bore and valve area. Thus they have optimised this ... thus by making the engine rev and breath very well they can get a huge advantage on our simple power formula and thus as they have gears (refer to the 360 Modena example) acceleration overall is improved as the higher you can rev in the lower gear the more mechanical advantage you gain and the better off you are when you change into the next gear. Thus what should magazines quote? Well I think they should show a torque graph so you can see how much usable torque you have to play with ... power is really a bragging right. Lets face it we have American muscle cars with around the same Hp as Ferraris but top out at 130 mph ... why is this?, because their usable rev range is low and thus their gear ratios have to be compromised. If you look at Jim's Lola and GT40 you can see that these engines can make cars go very fast ... but going shopping with a first gear ratio that can do nearly 100mph (if not faster) would wear a normal cheap clutch out real fast and waste petrol ... hence the wonderful Mustang/Corvette/Camaro/? is compromised for street use, and thus they are acceleration machines only. Pete ps: I think I will go and have a look at another thread , hehe | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 499 Registered: 2-2003 |
Rich, I'm not talking about rear wheel rotation, gearing isn't necessarily one to one. I'm talking about revolutions of the flywheel as function of time. I'm also not talking about distance covered, I'm talking about force applied as a function of time. Finally, for those of you who are talking about graphs of acceleration remember that aerodynamic resistance increases as the square of speed, e.g., resistance at 100mph is four times that at 50mph. Mitch, you're the fellow who said that engineers figure out where stuff breaks and then put in a cam to put power near that redline for "know nothing" customers. If your understanding of automotive engineering is this puerile, why waste time replying to these threads? I'm asking where you got "thrust" not what it might be. I'm familiar with Force N down on the tire, and force F, the friction that tire can exert on the road, but I'm saying what is thrust, e.g., how are you deriving your value. Finally, Mitch, you gave us the knowledge that at 7,000rpm, no wear occurs. I'm not sure you're really grasping any of this, despite your pretty graphs. If I'm wrong, I'm wrong, but I doubt you have the knowledge to know either way. Seems that Dr. Selevan approved my equations, and I'm thinking he might know best. | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 162 Registered: 10-2001 |
EFWUN, the only way time enters the equation is that in, say , 1 second, of course the rear wheels will be rotated twice as many times at 8000 rpms as they are at 4000 rpms and the car will be going twice as fast. But what we are talking about here is the cars ability to accelerate - not its ability to cover a certain distance in a certain time. I still maintain that a car will gain 10mph much more quickly near its torque peak rpm than it would take at its hp peak rpm. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 493 Registered: 4-2002 |
Just to clarify: 1) Instantaneous acceleration is maximized at peak TQ in any gear for that gear (my graphs agree with this statement) 2) Total acceleration is maximized when one shifts above peak HP* (my graphs also agree with this statement) *gearing has to be carefully choosen to maximize the area under the HP curve for this maximum total acceleration | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 492 Registered: 4-2002 |
EFWUN: Thrust is the tractive force at the contact patch of the (driven rear) tires. It is what actualy moves the car. This force is engine torque times transmision ratio times differential ratio divided by the rolling radius of the tire. However, I spent all of 3 minutes on google: search on 'peak acceleration RPM'. On the first google page, there are 10 articles: http://www.teutonic.ca/articles/techCorner/hpTorque/ In particular, in the 7th paragraph they state: "A car accelerates hardest (fastest) at it's torque peak." In http://www.renthal.com/website/resources/guide.htm "The torque peak is the point at which the bike has maximum acceleration," How many times do you need to be shown that you are wrong before you figure it out? (37, 99, 642)? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 498 Registered: 2-2003 |
Mitch, I've no idea where you're getting the term thrust, or the data you used to create that "chart." Please try to understand that the only force a dyno can measure is torque, while the only calculation of the power that produces is horsepower or KW. In any given second that your car is accelerating, the motor is rotating twice as quickly when you're at power peak, and therefore delivering nearly twice the accelerative force over that discrete time interval. If you want to pull out a tree stump, that is, strain against static resistance, then you'd want to be at torque peak, because "time" doesn't enter the equation. However, if you want to accelerate, that is, create a rate of change relative to a discrete interval of time, then you need peak power, because over that discrete interval (1 sec, or 1/1000 sec), the motor is producing more power because it is rotating twice as fast while producing nearly the same torque. I.E., at 8,000rpm, the motor rotates 2.186 times per second, while at 4,000rpm, the motor rotates 1.09 times per second. If we agree that the motor makes 275lb-ft of torque at 4,000, and 240lb-ft of torque at 8,000, and we agree on an imaginary value of 1 as a distance for every rotation, then we can clearly see that 275 x 1 x 1.09revs equals 299.75 units of our force per second. Similarly, 240lb-ft x 1 x 2.186 equals 524.64 of our "units" of force per second. Over any given second (or fraction thereof), at 4,000rpm the motor is imparting 299.75 of our units of force to accelerate the car, while at 8,000rpm, the motor is imparting 524.64 units of our force over the same time interval to accelerate the car. There are obviously other variables, and as I said, "our units" of force are a simplification, but there is proof that higher revs impart more force per unit time, and create a more rapid change of velocity as a function of time. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 490 Registered: 4-2002 |
As someone what actually does drive an F355, I can attest that the acceleration near redline is a triffling less than the acceleration near peak TQ. Just like in the thrust chart I provided. One of the things that continues to amaze me with this F355 engine is how much faster my blood pressure goes up that in my other car with the same power to weight ratio and a conventional American V8 engine. Its the 'sound' that convinces me that acceleration is harder near peak HP than at peak TQ. But the car actually does accelerate faster at peak TQ in each and every gear. My blood pressure reading falsely indicates otherwise. If you go back to the HP and thrust charts I posted, you will see that the torque curve is more like a plateau in the Ferrari engine. You have pretty much the same TQ over the entire power band. If you look at the curves for a typical american V8 (doesn't matter which one) you will see that after peak TQ the graph goes down at a constant rate as HP peaks then diminishes. The power band of the American V8 has more TQ at the low RPM band than in the high RPM band. This gives a decreasing TQ with increaseing RPM power band. This gives a big wide fat lazy HP curve (rather than the straight up to max power curve of the Ferrari). Based on the shapes of these curves, one would expect that the american sports car would gain an advantage just after each shift (accelerating faster) and suffer some setback just before the subsequent shift. Low and behold, this is just what one sees when drag racing big iron agains simmilarly powered little iron. | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 161 Registered: 10-2001 |
"If you've driven a 360 for instance, can you really maintain that the car accelerates harder at 4,750rpm than at 8,000?? " I haven't driven a 360 so can someone please grab their favorite passenger at lunch time and do an experiment? Get in 2nd gear and time how long it takes the engine to go from 4500 to 5000rpm (if you can hit the stopwatch buttons that fast) and then do it again in 2nd gear from 8000 to 8500rpm. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 497 Registered: 2-2003 |
Gentleman, as a favor to me, please read this carefullly, without personal animosity. *LC*, your argument is flawed because you leave out time. A motor making 300lb-ft at 4,000rpm produces power as a function of that torque expressed as a function of time. Similarly so does the same motor making 250lb-ft of torque at 8,000rpm. HOWEVER, as the motor causes acceleration (a rate of change over a specified interval of time) the motor at 8,000rpm is producing more power (e.g., 250lb-ft X rotational speed per time) than the same motor making 300lb-ft X angular speed over the same time, because Rotational SPEED is doubled, there are twice as many rotations occuring for any given interval of time. Over any fraction of time you choose, the motor spinning twice as fast (8,000rpm)is supplying 250lb-ft times 2X, while the motor at torque peak is supplying 300lb-ft times X, and thus, over any given interval of time, the motor at peak power is delivering nearly twice the force per unit time, and the car accelerates harder. If you've driven a 360 for instance, can you really maintain that the car accelerates harder at 4,750rpm than at 8,000?? Your argument works when you're talking about overcoming a static resistance, yes, at torque peak, the motor will rotate against higher resistance than it will at power peak, but acceleration occurs as a rate of change over intervals of time, and the motor spinning twice as fast produces 2X the (torque times rotational distance)minus the small reduction in torque value. *Victor, a couple of dumb mistakes in posts fired off without much thought don't really disable my argument, which at this point is pretty well thought out. I'd welcome the opportunity to talk at any time!! | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1093 Registered: 4-2002 |
OK, OK, you've convinced me that max accel corresponds to max torque. What wasn't obvious to me (and possibly others) is why you would rev way beyond the torque peak - Answer: When you shift up, the lesser torque multiplication of the next gear screws you out of a bunch of torque, and you're better off staying in the lower gear, even if you're way past the torque peak (all within reason, of course.) What I don't understand now is why are horsepower figures quoted, if it's torque that delivers the goods? It's been illustrated with several examples here that, *because of the extra torque multiplication of a lower gear*, revving past the torque peak, and revving past the hp peak all the way to redline is optimum. So what good does it do to quote horsepower? (Note that I understand that you can rev so high that everything peters out, but the specific Ferrari examples given here showed that staying in the lower gear as long as possible - to redline - is better. Probably because the Ferrari torque curve stays reasonably high at high rpm and doesn't take a sharp nosedive.) | ||
| Ben Cannon (Artherd)
Member Username: Artherd Post Number: 255 Registered: 6-2002 |
Hi Jim, just came back to this, and I thought I'd answer your questions to the best of my ability First- Airplane engines are OLD OLD OLD tech, they have not changed much if at all (like, literally, same part #s, no new alloys, etc.) since the 1950s. They are also rebuilt more often than they have to be, because when an F-car motor goes bad, you wait for a tow truck. When an airplane motor goes bad, your family shops for coffins... That said, nothing is as simple as looking at it with only tow paramaters (time near max power/load and wear/overhaul.) At least, it does not tell you the whole story. Modern engines are so complex, and have so many processes intersection that happen at the same time (ie, rev harder and ring sealing increases, but oil film size decreases and vibration increases which can come closer to causeing contact... I guess I don't really have a point, other than, these aren't tractors, and driving one at 8/10ths will let it live a fairly long, and HAPPY life Best! Ben | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 160 Registered: 10-2001 |
It's hard to tell what you guys are arguing about now, ha! If Lawrence is simply arguing that a car will accelerate faster around its torque peak than way up at redline than I don't think anyone can seriously doubt that. Speed is directly proportional to RPM so you can just go check it for yourself. If your torque peak is at, say, 5000, have someone time how long it takes your car (on the road) to go from 4750 to 5250rpm at full throttle. Then, if redline is, say, 9000, (drive in the *same gear* and) have them time how long it takes to go from 8500 to 9000. Even with the increased efficiency of high rpms, I think you'll find the car will cover the 500rpms down by its torque peak much quicker than up at its redline (or hp peak). That doesn't mean you should shift just after you pass the torque peak though. Because at redline in 2nd gear, your car may be making more rear wheel torque (engine torque times gear ratio) than it can make at torque peak in 3rd gear (which has a much lower ratio). The reason racing teams want their cars to rev higher and higher is (among other things) that it allows higher speed to be reached in a particular gear. Otherwise the driver will have to shift to the next gear and lose the gear ratio advantage provided by the previous gear (I'm sure gear ratios are top secret but let's say it goes from like 1.5 in one gear to say 1.2 in the next: by having to shift, the driver who shifts loses that .3 of torque to the wheels compared to another car that could continue on in the previous gear at that given road speed). | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 159 Registered: 10-2001 |
oops | ||
| James Glickenhaus (Napolis)
Member Username: Napolis Post Number: 976 Registered: 10-2002 |
Not Moi... | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2801 Registered: 6-2001 |
I saw a blue maser coupe in armonk once when i was delivering some packages. I pulled over in the entrance to some estates, and apparently this guy wanted to get in, but not before i finished looking at a map to see where i had to go next . | ||
| James Glickenhaus (Napolis)
Member Username: Napolis Post Number: 974 Registered: 10-2002 |
Not yesterday. Best | ||
| TomD (Tifosi)
Advanced Member Username: Tifosi Post Number: 3214 Registered: 9-2001 |
Jim, that sounds like fun BTW, I saw a blue maser coupe up on 9w yesterday - was that you | ||
| James Glickenhaus (Napolis)
Member Username: Napolis Post Number: 972 Registered: 10-2002 |
The only thing I know is that after a long winter this sunday I'm going to start up my MK-IV, warm it up carefully, plant my right foot until it reaches 7000rpm,feel the torque that that 427 puts out, and | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2800 Registered: 6-2001 |
EFWUN, where do F1 cars make all their torque? wayyy up high because of their small displacement. The faster the engine spins the more air is can take in in a given time. | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 8 Registered: 1-2003 |
EFWUN Lbft of torque and Nm of torque are different wasy of expressing the same thing. Just as lb of force and N of force are different ways of measuring the same thing. They just use different units of mass and accelaration. In the linear system mass and weight are not the same thing so this needs to be corrected by the force of gravity. This is shown in Lawrences post. When a force moves a mass over distance this becomes work when you work over a period of time this becomes power. Look at it this way, if you walk a mile and run a mile you burn the same number of kcal. The amount of work performed in each case is the same. When you run though you use more power because you are doing the work at a faster rate. You are going faster. I have ordered Carl Lopez's book on the physics of racing hopefully this will add some insight. Efwun, you write excellent prose and make strong arguments for your point of view. I have followed your posts in this forum and others. Unfortunately you have been mistaken before. Earlier in this thread your were mistaken on the piston speeds in a 180degree V8 and about the numeber of power strokes per crank revolution in such an engine. You obviosly are well educated, and well read. Your grasp of technical matters however, is not up to the quality of your prose. There is nothing like a spirited debate to get the cobwebs out of ones head! It might be fun to talk in person to have other discussions. | ||
| TomD (Tifosi)
Advanced Member Username: Tifosi Post Number: 3210 Registered: 9-2001 |
man, I have not checked this thread in a while - what the hell is going on, is this a science class | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 496 Registered: 2-2003 |
Gentlemen: No one has actually disproven the equations I've posted, including Victor (you've made the actual point more clear, torque becomes POWER in the instant that there is movement because torque in foot pounds is not Newtons of force in Kg/meters/sec. **TimN**, I think you must publish your findings and run, don't walk to F/1, where you can immediately put a stop to the waste of hundreds of millions of dollars in research to raise rev limits (pneumatic valves, unobtanium parts etc) in search of horsepower! Simply tell them their cars accelerate faster at torque peak than at peak horsepower!! While you're there, you may want to tell them about **Mitch's** conspiracy theory, so they can sack those silly engineers who have wasted hundreds of millions of dollars researching higher rev limits!! Perhaps Mitch can also find evidence withheld by all the manufacturers about those 400mpg carbs!! The answer was clear all along, just use a low-revving, stump-pulling motor, and Forget all that high tech stuff!! Come on, fellas, work through the equations and realize torque is resistance to movement or pull against stasis, while horsepower is the calculation of the force produced by angular momentum in the instant that torque causes motion. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 570 Registered: 4-2002 |
This is my final attempt to convince some of you that car achieves maximum acceleration at rpm where engine torque is maximum providing you are comparing acceleration in the same gear. I am using figures for my 328 for torque, power, gear ratios, and speed in gears. Derivation does not consider friction and drag on vehicle. What I am demonstrating is that I can compute the acceleration of the vehicle from either torque or power standpoints and get the same result. One method is based on power and change in kinetic energy while the other is based on torque and Newton's second law. Either way, the result is the same. Car accelerates faster at peak torque than it does at peak power providing you are in the same gear. Power is rate of change of kinetic energy � first derivative of it with time To save typing I�ll just put result. P = M * v * dv/dt = m * v * A M is mass, v is velocity, and A is acceleration Therefore and substituting W/g for mass A = P * g / (W * v) My 328 puts out 260 hp at 7000 rpm. There are 550 ft-lbs/sec in 1 horsepower. 1000 rpm is 19.8 miles per hour so 7000 rpm is 203.28 feet/second in fifth gear. My 328 weighs 3170 according to the manual. From the acceleration formula derived from above A = 260*550*32.2/(3170*203.38) = 7.147 ft/second ** 2 That is acceleration at maximum power (7000 rpm) in fifth gear. Now let�s get at it from the torque standpoint to demonstrate equivalence of my method. The final drive is 3.6111 (5th gear) according to owner�s manual. I need to compute radius of wheel based on car�s speed at rpm and final drive. Speed = omega * R where omega is rotational velocity of wheel and R is radius from ground. Omega = 7000 * 2 * pi/60 = 203.0 radians/second R = speed/omega = 203.38/203.0 = 1.0014 ft Engine torque at 7000 rpm computed from power output T = 260 * 550/(7000 * 2 * pi/60) = 195.08 ft-lbs Force pushing car ahead including gear multiplier F = T/R = 195.08 * 3.6111/1.0014 = 703.47 lb From Newton�s second law, a = F/m = F * g / W A = 703.47 * 32.2 / 3170 = 7.146 ft/second **2 Please notice I got the same answer both ways. One was a torque approach; the other is the power approach. Now, let�s compute the acceleration at 5500 rpm where there is maximum torque. This will be done from the torque standpoint first. Torque is 213.5 ft lbs according to owner�s manual. F = T/R = 213.5 * 3.6111/1.1104 = 769.9 lbs A = F * g / W A = 769.9 * 32.2/3170 = 7.82 ft/second **2 Let�s do it by the power formula Speed of vehicle = 5.5 * 19.8 = 108.9 mph = 159.72 ft/sec A = P * g/(W * v) Power = Torque * rpm = 213.5 * 5500 * 2 * pi/60 = 122967 ft-lbs/sec = 223.58 hp A = 122967 * 32.2/(3170 * 159.72) = 7.82 ft/second **2 Notice the same result is obtained by either approach. The car accelerates faster at maximum engine torque than at maximum power. | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 141 Registered: 1-2003 |
If this thread goes over 200 posts, I am going to print it and mail it to the Smithsonian! | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 489 Registered: 6-2002 |
Matt - for clarification. You attach a string to a spoke near the rim, and with the bicycle balanced, you pull from what direction? From the back or from the front? Jim S. | ||
| PSk (Psk)
Member Username: Psk Post Number: 370 Registered: 11-2002 |
Matt, I seriously think you have very little to do with physics as every post you have made is full of primary school physics errors:
Der, this point would actually be at whatever maximum rpm you can get out of this gear so could be passed maximum power rpm too. How old are you?, how come you understand the force moves the car at the rear wheels and yet keep talking about peek power. You keep proposing incorrectly that the maximum rear axle force is at force is at peak Hp, and yet I (and others, including Mitch's excellent graphs) have mathematically proved that the maximum rear axle force is ofcourse provided at maximum engine torque. Maximum velocity and acceleration have very little to do with one another, for example: Your maximum speed in a gear might be 100mph, but it could take all day on a very long straight road to get there ... thus not much torque required to gain that speed, but heaps to accelerate this mass. This example completely blows holes in your incorrect theory, so please stop confusing EFWUN Your example regarding first gear and effort on the brake pedal ... again the most force required will be at maximum engine torque because this is when the engine is providing maximum forward force ... the problem with this invalid example is that the car will have gained momentum (increasing the faster you go) and thus the brake force also has to counter act this. This has nothing to do with our acceleration debate. Rich, Thanks for getting back, and yes I concurr and did work this out with the Modena example. My initial confusing was the comment regarding acceleration, and as we have proved over and over this occurs at peek torque ... but over all acceleration including gear changes has to be calculated per car, etc. Matt and others, In the end lets go back to the start of the rear axle force and what pushes a car along. In the combustion chamber a certain amount of fuel and air is sucked in and then ignited by the spark plug. This combustion pushes the piston down and exerts a force on the crankshaft. This force tries to rotate the crankshaft and thus we have torque ... this torque works it way past the clutch and is multiplied through the gearbox and diff and out to the rear wheels. If this force is greater than the opposing forces from rolling and air resistance and anything else the car will ACCELERATE. If it is the same the car will continue to move forward at a constant rate or velocity. So Matt confuse everybody as much as you like and add to the post count, but you like EFWUN and many others are confusing velocity and terminal speed with the incredibly simple concept of acceleration. Again Torque for acceleration, and power for top speed in a constant gear. Mitch's graphs prove this as do the magazine graphs that James mentions. Cheers Pete | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 7 Registered: 1-2003 |
EFWUN I am afraid you are mistaken when you state that there is no time component in torque as there certainly is. In the USA we are stuck with an anachronistic system of weights and measures known as the linear system. This uses lbs feet and gallons. Newtons second law to which you are using in your posts has to use SI units to make sense F(newtons)=mass(kg)xacceleration(m/sec^2) newtons have the units of kgm/sec^2 In the linear system we have a problem with the defenition of forceand mass. We use lbs to measure both but they are very different. This is where there is some confussion. A 170 lb man will weigh much less an the moon because there is less gravity. A 100kg man will have the same mass on both the moon and on the earth. We have come to define force lbs as the force that is exerted on a mass by the force of gravity on the surface of the earth. This force is an accelaration force directed downward at 32ft/sec^2. Hidden in the lbft units of torque is this unit of acceleration An easy way to understand this is to look at the power ratings of any European car. They wil quote torque ratings as ftlb and as Nm. If you go down to Sears and look at torque wrenches you will see that they are all calibrated in both lbft and Nm. You will agree that feet and meters both have similar units that is length. Newtons and lb(force) must also have similar untis. As a matter of fact there is a simple conversion factor. lb(force)=Newtonx0.2248 The units for newton are kgm/sec^2 As you can see lb(force) has to have units of mass times distance devided by time squared or mass times acceleration. This proves that torque has a time component in it. Which means horsepower is what you buy but torque is what you drive As an aside take a look at this months issue of Car and Dirver. They compare a Jag S type R, MB E55 AMG, Bmw M5, And an Audi RS6. They talk about the E55's torque reserves and the instant torque of the RS 6 that hammer the car to a 0 to 60 time of 4.4secs. Prof. auto testers and authors seem to think that torque rules too. | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 158 Registered: 10-2001 |
BTW, Pete, of course your acceleration will slow as you rev higher, i.e. you were accelerating the fastest in 1st gear as you went past the 4600 rpm torque peak. You are accelerating much more slowly by comprison up at 7000k. But, you are accelerating faster at 7000k in 1st gear than you will be able to in 2nd gear (regardless of rpm), and this is all that matters. So you press on until that is no longer true. | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 157 Registered: 10-2001 |
Man, I've got to come back here more often. PSK Pete was trying to figure out what I was talking about and the logic behind my statement that normally revving well past torque peak provides best acceleration and I was no where to be seen. I think some people have stepped forward to explain it but basically, just remember that what gets to your wheels goes through the transmission and differential. Let's use the little Dino 246 and ZF gearbox as used in my budget Fiat as an example. Peak torque is 164 at 4500rpm. If I rev to 8000 in 1st gear, it is only making 100 ft.lbs. of torque. However, multiply this by the 1st gear ratio of 2.991 to get 299. This amount is higher than 2nd gear can produce at any rpm - even at the peak 4600 rpm (since 164 x 1.763 gear ratio is only 289). Therefore, the driver that wants to be putting the most torque to the road stays in 1st gear until 8000 (and lands in 2nd at 4700rpm). He makes the same calculations for the difference between 2nd and 3rd gear and determines that at 7000rpm he is making only 126 ft. lbs. but this multiplied by the 1.763 gear ratio of 2nd gear is more torque than 3rd gear will produce at any rpm and so he shifts just after 7000, landing at about 5200 rpm in 3rd. And so on. Does that make sense? Just go ahead and double (or triple) all those output numbers for V12 and V10 cars, ha! | ||
| arthur chambers (Art355)
Intermediate Member Username: Art355 Post Number: 1256 Registered: 6-2001 |
Maybe we should restrict the math majors to only 2 posts per day? Just kidding. Art | ||
| matt (Matthewmag)
New member Username: Matthewmag Post Number: 31 Registered: 11-2002 |
Wow - another 20 odd posts and the argument is still going on. Not only am I wondering if people would change their opinions if Ron Dennis or Frank Williams posted here but I'm trying to think of something really simple that will put an end to this. Firstly, the points of agreement again. The only thing which propels a car is torque at the driving wheels. Maximum torque at the driving wheels will make your car accelerate fastest beacuse it gives rise to the maximum force acting on the car in a forward direction. F=MA ......... maximise the force and you maximise the acceleration. So like in any Physics problem, if you want to demonstrate a theory (albeit a pretty well established one - thanks Sir Isaac) you need an experiment. Here is one that you can perform very easily. Warm your car up and drive it along in first gear at peak torque. Now hold the rpm constant by using left foot braking while your right foot is firmly planted on the floor. Make a note of the force required to keep the rpm constant. Let the brakes cool off a little and now repeat the experiment at max power rpm. Note the force required to hold the rpm constant this time. Tip: you do not need to insert your bathroom scales between your left foot and the brake pedal to appreciate the difference in force required. Since the force applied to the brake pedal is proportional to the torque at the wheels (within limits this is true) (and to point out the blindingly obvious, the torque is of course in the reverse direction to the torque at the driving wheels from the engine.) Now why would the result of this experiment support the theory from classical mechanics if peak axle torque occurs at peak engine torque rpm? Don't forget that you can resolve all the forces into a single forward force provided by reaction at the road to the axle torque and a single rearward force provided by the addition of the reaction force from the reverse torque of the brakes. Some more simple questions for anyone struggling: Why do cars reach maximum speed at maximum power rpm and not maximum torque rpm? Why did Hans's Subaru Justy CVT hold peak hp rpm when trying to achieve maximum acceleration? Or did those incompetent Subaru engineers win the world rally championship by accident? Why why why????? We're all agreed that maximum axle torque provides maximum forward force and that maximum forward force is needed for top speed...... Just like maximum forward force is needed for maximum acceleration. So why the crazy disagreement? Like Pete says it's just physics. So true, although if you read my post carefully, you might have taken the hint that I have more than a passing connection with the subject..... BTW...... has anyone thought about that bicycle problem in my last post? I didn't just throw that in to be facetious - if you think about it, it demonstrates the difference between a torque and a force acting on a wheel. And has some bearing on this discussion. James - following your point about our desire to keep engines running cool, and your declared love of thermodynamics, something that might interest you is the "over-expansion engine" it is something that I thought was very exciting when I heard about it. It is kind of unusual and has an expansion ratio much greater (3x ??) than its compression ratio. It over expands the gasses after combustion to provide its own cooling! Sounds totally wrong because it's doing work to keep itself cool but from a thermodynamic point of view it works just as well. In practice they only work well at full throttle. Their advantage in the end was light weight and simplicity because they need no cooling system - not even a fan. It only ever found use as a field pump! I can't tell you how much torque or power it produced : ) or at what rpm it reached its top speed. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 487 Registered: 4-2002 |
Manufactures determine red line based on where things start to break. They then choose a cam designed to put peak power fairly close to redline. This gives them the most HP they can afford to deliver to a 'know nothing' customer. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 495 Registered: 2-2003 |
To repost the explanation; Horsepower is the expression of torque in the instant that any motion occurs. Torque is not a separate force; it does not exist as rotational power: Torque is instantaneous rotational force (e.g., static force). Horsepower is the work that force can perform over time, e.g., Work = Force x 2 x Pi x radians, ergo; 1 lb x 2 x 3.14etc x 1 ft = 2 x 3.14 lb-ft = 6.283 lb-ft Then; Power = Work / time = 6.283 lb-ft / min 1 hp is defined as 550 lb-ft / s = 33,000 lb-ft / min as measured So, if 1 lb-ft of torque is applied in one minute (1 rpm) = (6.283 lb-ft / min) / (33,000 lb-ft / min) = 1 / 5252 of 1 hp. Thus; we can see that there are 5252 lb-ft of torque per minute in one Horsepower. Or...if we multiply torque in lb-ft by the RPM this torque is produced at, then divide by 5252, we get the power output at that RPM. THEREFORE: the peak POWER output is where ever the manufacturer determines on the dyno, e.g., 8,200rpm for the 355, while the peak TORQUE output is not necessarily the same, NOR is it the place at which the motor makes peak POWER!! The motor makes more POWER at peak horsepower than it does at peak torque, power being defined as units of force/time. Manufacturers determine redlines as a function of where peak POWER exists on the rev curve, and things like valve float etc. All vehicles with conventional transmissions accelerate at maximum g at peak power, not torque. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 494 Registered: 2-2003 |
Just the way I typed it in. Trust me, I understand torque is foot pounds, but thanks for the pot shot, Sean. | ||
| Sean F (Agracer)
Junior Member Username: Agracer Post Number: 83 Registered: 2-2003 |
lbf = pounds-feet, and is generally used when refering to torque. ft/lb = feet divided by pounds which is ... well nothing as far as I know. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 493 Registered: 2-2003 |
Tim, you've gotta understand that in paraphrasing a mathematical expression, I wrote lb/ft as a way of saying foot pounds of torque. Attacking the units doesn't show your mastery of the fundamental principles, it shows your rather mean spirited focus on argument rather than discussion. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 492 Registered: 2-2003 |
Tim, what I'm saying is that there is no time component in torque. I'm not using lb/ft or vice versa as "divided by." That's the same reason I wrote quotation marks around "multiplied" because I know that you'll seize upon that. Please read Matt's post, and the link within it, and then read my posts with an open mind. If you think clearly about it, instead of looking for something to pick on, you'll see that horsepower IS torque expressed as a function of rotation and time. | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2792 Registered: 6-2001 |
EFWUN, it does make a difference. You are defining toruqe as feet divided by force, when it is supposed to be feet multiplied by force. How canb you say that it doesnt make a difference how you write it? thats like saying 5x5 is the same as 5/5. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 489 Registered: 2-2003 |
LC, your proposal about the derivatives of acceleration graphs in R & T makes sense if you understand that the redline (where R & T and all magazines shift) is actually past peak horsepower in most cars. Ie, the Modena peaks at 8,200rpm, but revs to 8,500, where "power" really is tailing off. Please review Matt's post and mine with an open mind, and the idea of a giant diesel formula one car as the ultimate expression of your proposal. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 488 Registered: 2-2003 |
LC, you're wrong. Plain and simple, just reversing foot pounds and pound feet doesn't make any difference whatever. My last three posts are the definitive answer to this question, derived after a lot of thought. Just because you don't understand them, and have to resort to telling me the Units are wrong because I wrote pound feet v. foot pounds, doesn't make your position "tenable." Stop arguing for the sake of arguing, and really read Matt's post, Dr. Selevan's post, and mine. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 569 Registered: 4-2002 |
Quote: The reason motors rev well past this optimal breathing point, is because the actual "power" being produced is a function of measured torque "multiplied" by the rotational speed of the flyheel as a function of time. 275lb/ft of torque at 4,750rpm results in 245units of this "power" or HP. At well past optimal efficiency, say 8,200rpm, torque may have dropped to 240lb/ft, but "Power" is up to 395 units because the measured torque is "multiplied" by the rotational speed of the flywheel as a function of time. Unquote Your units are incorrect. Torque is ft-lbs, not ft/lb which is foot per pound. You consistently make this error. You are also trying to defend an untenable position. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1089 Registered: 4-2002 |
James: Just a couple of comments - Quote: If an engine could run with wide open throttle plates, with ideal gas flow, with a load that is modulated to maintain a mechanically achievable RPM (no valve float, etc.), then ideal efficiency would be achieved. Sounds like a continuously variable transmission to me. Unquote. The engine you describe sounds like a diesel. No throttle plate. RE: CVT - the old Subaru Justy had one. It basically, in normal driving, would try to lug the engine, thus forcing the driver to open the throttle plate a bit more than he probably otherwise would - more efficiency, better mileage. At full throttle, however, the CVT would stick the tach at the HP peak. Was weird to drive, as the engine would stay at one rpm and you didn't hear it rev as the car accelerated. Of course, it is a little hard to use the terms "accelerate" and "Subaru Justy" in one paragraph. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 487 Registered: 2-2003 |
Dr. Selevan is precisely correct. A motor is optimally efficient at the point where gas speed as determined by valve lift and duration, as well as piston speed, result in optimal filling of the cylinders. At that point, the twisting ability produced by the explosions of gas is optimal, and that is torque peak. However, torque is not doing work, because it has no distance (angular) or time components. The reason motors rev well past this optimal breathing point, is because the actual "power" being produced is a function of measured torque "multiplied" by the rotational speed of the flyheel as a function of time. 275lb/ft of torque at 4,750rpm results in 245units of this "power" or HP. At well past optimal efficiency, say 8,200rpm, torque may have dropped to 240lb/ft, but "Power" is up to 395 units because the measured torque is "multiplied" by the rotational speed of the flywheel as a function of time. Dynos measure torque, but the calculated number, horsepower, is the actual "power" produced. It is this power (torque "multiplied" by rotational speed as a function of time) that accelerates the Modena, or an F-150 or any other vehicle. Therefore, motors make maximum power at peak horsepower (department of redundancy department) and cars accelerate fastest at peak horsepower. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 486 Registered: 6-2002 |
Not that this thread desires another post, but an off-line communication with EFWUN stimulated a tangential thought. An internal combustion engine is most efficient at wide-open throttle. Efficiency here is defined as work per unit of energy consumed. Our car engines are nothing more than heat pumps. They convert expanding gas into mechanical energy. Heat expands the gas. Thus, the more heat one can pump through an engine in a unit of time, the more work that can be performed. This equates to pumping as much air:fuel (14:1) per unit time as possible. This is accomplished through larger cylinder displacement or higher flow rate (RPM). There are no other choices. Limitations to this ideal for a fixed displacement are 1) gas flow (intake and exhaust valve area and duration - valves and cams) and 2) RPM. Maximum efficiency comes about because the same amount of internal friction is present whether running at partial cylinder pressure or full pressure. If an engine could run with wide open throttle plates, with ideal gas flow, with a load that is modulated to maintain a mechanically achievable RPM (no valve float, etc.), then ideal efficiency would be achieved. Sounds like a continuously variable transmission to me. One other thought - our desire to keep our engines running "cool" is somewhat self-defeating. In the limit, if the engine block is cooled to ambient, then all the heat of combustion would go to heating the surrounding atmosphere, and none would be used to expand gas. We actually would, from an thermodynamic perspective, prefer that the cylinder wall achieve a temperature equal to combustion temperature, and stay there. In this way there is no heat flow to the atmosphere, and all the work goes into expanding gas. I loved thermodynamics. Jim S. | ||
| Dave Wapinski (Davewapinski)
Member Username: Davewapinski Post Number: 531 Registered: 8-2001 |
I have not had time to read through all this (but will), but I had a few questions. I agree that pushing a car is needed at some point to properly break in an engine and to keep it running properly. Of course too much heat can hurt an engine, but has Ferrari designed their engines to better handle the heat due to being first a racing design? Some people select a gear to keep the RPMs up even while driving around the block. Some people have problems with clutches, others do not. Does selecting a gear to keep the RPMs up while driving around the block affect the clutch or any other part? Sometimes endurance is important. Does keeping the RPMs up while driving around the block or running it fast on the road at high RPMs drastically decrease MPG? On a piston aircraft engine, generally the best compromise between power and MPG (or endurance) is around 75 % power. Some planes around 65 % power. Above this point, fuel consumation increases drastically with minor increases in speed. Does this hold true with Ferraris? I assume RPMs are a good indication of power in a Ferrari. Assume redline at 6000 RPM. 75 % of 6000 is 4,500 RPM. Assuming good radar detection and that it can be done safely, on a trip is running a Ferrari at 75 % or about 4,500 RPM (with whatever is the resulting speed) the best set of compromises? | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 485 Registered: 2-2003 |
Fellas, here's an example I just thought out; hopefully a cogent one. A dyno (called a "brake"), uses resistance to measure one variable without regard to time; Torque. However, horsepower is the calculated expression of that torque over any unit of time. So, while the instantaneous (actually without a unit of time) twisting force of the motor may peak at "torque" peak, say 4,000rpm for easy reference, that cannot be translated without reference to time, ergo, we measure "horsepower." The motor makes say 300lb/ft of torque at 4,000, say, and only 250lb/ft at 8,000rpm. However, the only way we can evaluate that instantaneous force is to add a time component, and this is horsepower. So, the motor is spinning twice as fast while delivering slightly less torque, but over any division of time, (or rotation, because we cannot measure rotation without translating instantaneous torque into a unit containing time) the actual "power" is greater at 8,000rpm. Cars accelerate as a function of the area under their horsepower curves, so a motor with excellent torque has a larger area under its power curve than a motor with minimal torque and a skyhigh redline to maximize its power. The smaller the area under the power curve, the more gears you need to keep the car on the upper reaches where it is making power. However, understand that torque is just a measured quantity from which we calculate horsepower to evaluate its ability to do work as a function of time, and because the flywheel is spinning twice as fast, and the torque measured by the "brake" is only perhaps 20% lower, the motor makes more "power" per unit of time, and therefore the car accelerates hardest. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 483 Registered: 2-2003 |
Charts (even good ones like Mitch's) don't explain everything. Torque peak (4,750 or so) is the point at which the Modena will best pull against a tree stump. Horsepower peak is the point at which most power is being delivered per unit of time to the flywheel and the rear wheels. Yes, torque peak is where the motor makes most "pull", say 275lb/ft at 4,750rpm. However, while torque is less at 8,200rpm, (say 245 from PSK's example, 30lb/ft down from peak), the motor is spinning nearly twice as quickly. Therefore, because torque is the static pulling power, we need to know that the horsepower at 4,750 is actually around 245hp, because over any unit of time, the flywheel is spinning half as quickly as at horsepower peak. Over any unit of time, no matter how small, the motor is delivering more "power" at horsepower peak, by definition. Looking at the formula one example, if "torque" were an arbiter of overall acceleration, someone would have a CVT and a 3,000rpm motor with 1,000lb/ft of torque. Nevertheless, no one does, and in fact, motors are designed to reach 18,000rpm in the search for Horsepower, not the ephemeral notion of torque. Horsepower IS the expression of torque over time and rotational speed, so the analysis must come down to how much horsepower is produced, and yield the result that where horsepower peaks, so too does acceleration. | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 130 Registered: 1-2003 |
I am tempted to fly all of you guys somewhere to get together in the same room to roll this one around! I have never seen such talented gray matter at work.....Can I hire a few of you? | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1085 Registered: 4-2002 |
"Stay in a lower gear until torque falls into a canyon." Crude, but representative of your charts. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 483 Registered: 4-2002 |
I took some time and drew up a passable tourque curve for a Ferrari F355, and used excel to do the rest of the math. The first figure shows the torque curve and HP curve appropriate for an F355 Engine:  You will notice that the point of maximum Torque (256 lb-ft*5800 RPM) fixes one specification point. You will also notice that converting peak HP (380 HP at 8250 RPMs) back to torque at the RPM of peak HP fixes the other end of the tourque plateau. I diddled around with the torque curve beyond these points to get a passably accurate touque curve for the F355 engine. The next chart shows the thrust available in any gear at any speed:  Notice peak acceleration occurs at the RPMs of peak torque in 1st gear. Also, note that there is more acceleration in a lower gear at any RPM than in a higher gear at THE SAME SPEED. Replotting the HP curve on the scale of the thrust axis gives the following chart:  Notice that the point where one should be in the higher gear is exactly where the HP curves of two gears intercect! | ||
| PSk (Psk)
Member Username: Psk Post Number: 369 Registered: 11-2002 |
Matt,
The engine will be producing more torque but due to the 7th gear increase over the 6th gear the final torque multiplication will mean that less torque is applied to the rear wheel(s) ... unless the ratio change is so minor and the change puts the 7th gear at peek engine torque. Thus the comment that:
Is unfortunately wrong. As you will see by my 360 Modena example the torque at the rear wheels is always maximum at peek engine torque ... but that the gearing ratio increase drops the rear wheel torque more. So maximum force and torque at the rear wheels is always at peek torque, no matter how you slice the bread ... and the Mustang guy did not dispute this. You will notice in the 360 Modena example how the torque at the rear wheels reduces as we go up the gearbox ... this is why acceleration slows down, and also why your super high 7th gear example will not accelerate the car (ie. the torque and thus force available at the rear wheels will not be enough to counteract the equal opposite forces provided by rolling and air resistance, etc. which the car had reached when it topped out in 6th gear). Thus again it is all about torque, because torque pushes the car along, and the amount available at the wheels controls how much acceleration you have under your foot. The 360 Modena and my previous F = m.a example prove this completely. Do not get me wrong, Power is extremely important otherwise we would all be puttering around like trucks (lorries), as the ability to produce torque over a wide rev range provides the flexibility to make use of suitable gear ratios. (Note: If you put a diesel truck motor in a Formula 1, it would accelerate extremely quickly over a VERY short rev range and then the petrol engined car would piss all over it ... I have shared the track with a racing Truck and it out accelerated my Alfa Sud with 200 hp that I classic raced way back then ... but then it hit a wall (aerodynamics and NO power ) ... and zooming past I went ) It is just Physics, and torque multiplication. Pete | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 482 Registered: 2-2003 |
Tim, a course in vector analysis of forces e.g., finding the angle from N (or whatever) of a force, and then using sin cosine and tangent to analyze the total force is generally not applicable to this discussion. This discussion relates to torque only as it applies to rotation of a shaft. I understand you're using your vectorial analysis to compute the force at the contact patch by multiplying that force by the moment arm (the radius of the tire). Nevertheless, the argument is far more fundamental than that. Torque is the instantaneous ability of the motor to pull, or twist as an expression of weight and distance. Horsepower is that same force as a function of time. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 481 Registered: 2-2003 |
Tim, I've tried to keep personal animosity out of my posts. Where do you get off telling me I don't know what a vector product is? Or that I should leave this discussion to people (ostensibly you) who "understand this stuff?" If you can't understand that horsepower is the expression of torque at rpm, while torque is the expression of a static twisting force, then say so. Vectorial analysis has nothing to do with an understanding of why cars accelerate. Your personal attacks are unwelcome and unwarranted. I'll admit it is a long time since I had my four semesters of physics on my way to a degree in Bio Chemistry from U Penn. Leaving seconds squared out of an equation made me look silly as well. Nevertheless, I have an excellent grasp of a concept that is apparently foreign to you, as well as a grasp of fundamental courtesy that you obviously lack. Your facts are wrong, so you make personal attacks. I find that offensive. | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2784 Registered: 6-2001 |
i didnt say that a motor produced two types of forces. you do not know what a cross product (or vector product) is. you have to stop pretending you know what you are talking about. just because you think something is right, doesnt mean it is. Neither torque, nor power are forces, and inever said they were. you probably understood what i said incorrectly. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 480 Registered: 2-2003 |
Perfect post, Matt. Read the link in Matt's post, Tim et al, and then scratch your heads a bit. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 479 Registered: 2-2003 |
Tim, you're the fellow who wrote "where x is a cross product." If you'd like to persist in thinking that a motor can somehow produce two different types of forces, e.g., torque and horsepower, go ahead. If you'd like to believe that your car accelerates harder at 6,000rpm than at 8,200rpm, please be my guest! No problem. There's no real point in attempting to make sense to you. And if you would like to make fun of my attempt to simplify this so that you (Tim) would understand it, please go ahead. | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2783 Registered: 6-2001 |
The fact that EFWUN doesnt even know simple units of physical quantities pretty much discredits all of this arguments. I see no point in farther argument with him, PSK. | ||
| matt (Matthewmag)
New member Username: Matthewmag Post Number: 30 Registered: 11-2002 |
Hey guys, I know it's rude to butt in on a chat right at the thick end of 120 posts but this has been keeping me amused because some folks here are steadfastedly arguing opinions on physics when there is not opinion to be argued only fact. Following this thread, there are a number of misconceptions but the big one that a few people seem to make is looking at this as a "static" problem when in fact it is a dynamic one. Like any such problem, it becomes clearer if you draw a diagram of all the forces involved. One thing that everyone here agrees on is that it is TORQUE AT THE DRIVING WHEELS which provides the force to move a car forward (or more precisely, it is the equal and opposite reaction of the road on the tyre due to friction.) So if you maximise that force, you maximise acceleration - it's as simple as that. I use the word force here carefully. Yes, it is torque at the wheel, but the actual force (exerted on the car by the road) which pushes the car forward is acting in a straight line. The question "is that force maximum at peak torque or peak power?" is easily answered and I am amazed that this argument has gone so far when it is a fact of physics....... Rather than answering it myself though, this is a reasonably well written explanation by a guy called Dan Jones who is into Mustangs. It's written for the layman - there isn't much that can be argued about it. It's about ten times shorter than I could have written it. The author has done a good job to make it concise without too many loose terms open to interpretation. http://www.mustangsandmore.com/ubb/DanJonesTorqueVsHP.html It's worth stating that in physics problems one of the fastest ways to understand the problem is to consider extreme examples. Consider the car moving at top speed - it is not accelerating or decelerating. It is a good approximation that all of the resistance to motion is wind resistance. In any case, the car is in a state of dynamic equilibrium because the forward forces (provided by the reaction at the road due only to torque at the rear wheels - there is NO other driving force)are equalled by the force of wind resistance...... You might have observed that maximum speed occurs at approx. max power rpm..... Now imagine you had a super high 7th gear that when engaged would drop engine rpm to peak torque..... What would happen? Surely if the engine is producing more torque, you'll get more torque at the rear wheels. Errr, NO. You wouldn't go faster and the reason is simple - maximum torque at the rear wheels does not correspond to maximum engine torque. It might not seem logical to some people but it is true. This is why a car has gears - to provide maximum torque at the wheels over a reasonable speed range. There are fundamentally only 2 factors which affect a car's acceleration rate - power and weight (well, ok mass for the really fussy). That's why car manafacturers love to quote power to weight ratios (and never torque to weight ratios). That's why a car like a Westfield Hyabusa is so quick (lots of power, little weight, not much torque.) It's why Ferrari don't fit deisel engines to accelerate faster and it's why formula one cars rev so high..... Now if you're still not convinced, or you found that first site a little easy going you can check this out. If you have matlab you can carry out the simulations for yourself. http://www.clarkdomain.com/academic_projects/docs/ea_opt.pdf An interesting point that some have touched on is that depending on the car you have, actual acceleration at max power may be less than at peak torque in the same gear. WHAT? After I've said all that? The reason is that because wind resistance increases with the square of velocity, you can have the situation where increasing wind resistance has a greater effect than increasing engine power output - this is particularly obvious on motorbikes. But if you change up a gear and try accelerating at peak torque, you will not accelerate so quickly. I don't feel that I need to give my qualifications to add to this argument because it is accepted fact by all automotive engineers and any competent physicist and there is quite possibly a more highly qualified physicist than myself on this board. I can say that trivial problems like this can sometimes give even the most qualified professors a hard time - sometimes the answers seem counter-intuitive but often you're just missing something trivial and obvious. Now if you really want to give yourself a headache, consider the following - a bicycle is carefully balanced in its normal upright position, a light string is attached to one of the spokes of the rear wheel near the rim, just forward of the contact patch of the tyre with the ground. The string is gently pulled. Which way does the bicycle move? This problem is not only counter intuitive to some, it also demonstrates the power of drawing a simple force diagram. The correct answer is fairly easy to reach if you draw the diagram. Also, you can check it out for yourself very easily........ and it's a lot safer than reaching for an extra gear while testing your car's top speed. | ||
| PSk (Psk)
Member Username: Psk Post Number: 368 Registered: 11-2002 |
EFWUN, Acceleration units are m/s^2 or feet/s^2 for Americans. Pete | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1081 Registered: 4-2002 |
Wow, Lawrence. I have to admit your "look at the magazine graph" argument is pretty convincing. Even simple for people like me to understand. While I had always assumed that the graph curved because of increasing aero drag, this wouldn't be the case in the first couple of gears where such drag is minimal. I was going to come up with an argument based upon our local oval track, which used to be 1/4 mile. As the track was wide, and the straights short, the racing line was closer to a circle than an oval. Thus the speed didn't vary that much, with the cars coming out of the corners at typically 6500rpm and accelerating to 7500, both of which are well above the torque peak of the engines in use. Why not gear them so that they ran closer to the torque peak? - I was going to ask. Then I answered my own question. The change in gearing that would necessitate would reduce the torque to the wheels because of less torque multiplication. i.e. you turn 7500rpm with 3:1 gears, or 5000rpm with 2:1 gears. Even tho engine torque may be more at 5000, the 3:1 gearing multiplies the available torque at 7500, giving a better result. It's an interesting result - VERY simplistically it means that gearing (torque multiplication) wins the battle. I suppose that in this limited example, where there is no gear shifting, you wouldn't want to run past the HP peak, as then the torque would be falling off very fast, faster than the gearing advantage would "compensate" for. (bad choice of words, but that's what came to mind) | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2782 Registered: 6-2001 |
EFWUN, the unit force is not lb/ft/sec. it is lbft/sec^2, and forces can do work. if forces couldnt do work, then how could anyhting get done?? power is not work either. The equation for work is W=Fd. the term power and work is not interchangeable. note that work is not power. power is W/t. you shoud let the people who know what they are talking about explain this thread, and not try to explain things with invalid arguments. despite what you think, its not me thats making the stuff up as i go. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 482 Registered: 4-2002 |
"It is readily apparent that when you get to the shift point, the acceleration rate is falling off." Agree "The acceleration of the car is the instantaneous slope of this curve." agree "It lessens as you go up in the band." agree; However, one changes fron one gear to the next based on which gear provides the greater thrust after the transmission effects are taken into account (differential and tire size are constant). See Carroll Smith: "Tune to win". Chapter on gear ratio selection | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 481 Registered: 4-2002 |
Summary: a) in any particular gear your _instantaneous_ acceleration is fastest at peak TQ b) given a choice between gears, you will accelerate faster in the gear where HP is higher | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 568 Registered: 4-2002 |
I cannot believe some of you people who still think car accelerates faster at peak horsepower. Here is simple demonstration that everyone should be able to understand. Look at a graphical acceleration test of a sports car from say 0-100. Road and Track did some in their April 1997 issue. Look on page 75 where they show tests of a porsche, ferrari, acura, corvette. Note that all the graphs have a negative second derivative between gear shifts during acceleration. They are graphing speed versus time and show the shift points. Please note that the lines are not straight between shift points. The slope of the curves lessens as you get near the gear change. The slope of this curve is acceleration. The fact that the sign of the second derivative is negative implies that acceleration is on the decrease. Now why is this so I ask rhetorically. It is because each is reaching its redline where there is maximum power produced. The steeper portion of the segment is at the low end where torque is maximum. The acceleration of the car decreases as you pass the point of maximum torque that is lower in the band. It is readily apparent that when you get to the shift point, the acceleration rate is falling off. The acceleration of the car is the instantaneous slope of this curve. It lessens as you go up in the band. Just draw a tangent line at the lower portion of the band. The line always rises more than the graph because the line on the graph is losing its steepness - negative second derivative, less acceleration as you stray from the maximum torque engine speed. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 478 Registered: 2-2003 |
Pete, I think you're probably a good egg, but I'm not "persisting", I'm RIGHT. Motors don't produce two types of force, torque is not WORK! it is a static force. Look at it this way, F=MA, right? You're trying to assign your own values to F, in lb/ft. Yet, the value of F is already set in units of lb/ft/sec!! E.g., Mass (in lbs) x Acceleration (in feet per second), equals F, in lbs/ft PER SECOND, or lb/ft/sec. You cannot do away with the necessary unit of time! Torque, measured in lb/ft, is a STATIC force, it cannot do any "Work." Horsepower, is "Work", and is capable of accelerating a car. In any given gear, a car (given mass in lbs) accelerates at its maximum acceleration (in feet/sec) at maximum F, which is ONLY exists as pound-feet per second. F=MA. Plain and simple. | ||
| PSk (Psk)
Member Username: Psk Post Number: 366 Registered: 11-2002 |
EFWUN, Whatever makes you happy. I have proved in my calculations via working out the force applied to the road that it is maximum at peak torque (in a constant gear) and yet you persist. What I have also proved is that OVERALL a 360 Modena will accelerate faster (if gear changes are involved) by revving to peek power ... this is due to the torque in the lower gear being higher than the torque in the next gear, NOT because HP is higher. Torque is a force that simply rotates around an axis ... remember a dyno can only measure Torque, not power! Anywhere I will let this rest with myself agreeing to disagree with your power versus acceleration comments, and that is cool ... what a boring world it would be if we all thought the same Cheers Pete | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 477 Registered: 2-2003 |
From Pete's manipulation of the equations, we get 248hp for the Modena at 4,750rpm, or peak torque. Conmensurate with that, we get 246lb/ft of torque at peak Horsepower of 395. Understand that torque is a static, non-moving force, and that horsepower is what results when that force moves, and you understand that cars accelerate hardest at peak POWER, not torque. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 476 Registered: 2-2003 |
Hans, that's because what everyone here is missing is that "Torque" is a STATIC force, to find what's really going on you need to add the time component. In other words, the STATIC rotational "force" is torque, with may peak (as per PSK's erudite analysis) at 4,750rpm for the Modena. Nevertheless, what Pete is missing is that you're really putting to the rear wheels is POWER, and that must have the time/movement component. In other words, torque is the "force" required to "Hold" the motor stationery, a quasi imaginary number, while POWER is the Force that results in the INSTANT that the motor rotates, even a thousandth of an inch. The reason your car accelerates fastest at peak POWER, is because the acceleration is coming from a rotating force. At peak torque, the POWER is actually a bit lower than it is at peak power. | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2780 Registered: 6-2001 |
Hans, its not torque that you need to overcome drag, it is power. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1078 Registered: 4-2002 |
James: I'm still confused by the HP .vs. Torque issue. I realize that HP has a time component, as in your example of going 150mph - a small engine via torque multiplication could come up with enough torque to overcome the drag, but not fast enough. Ergo, not enough power. However, I still wonder if HP has some relation to acceleration. As per your example, it takes a fair amount of HP to go 150mph, almost entirely in the form of fighting aero drag (over time!). However, let's remove the aero drag, and let the only thing holding the car back be it's mass or, probably more correctly, the inertia of it's mass. (I'm not an engineer, nor do I play one on TV, so excuse me if the terminology isn't totally correct.) What I'm getting at, is that at 150mph, it's the aero drag holding the car back, using the HP. If you remove the aero drag, what's to prevent the car from accerating at an infinite rate? It's inertia. So, I don't quite understand how aero drag and inertia are so conceptually different *to the loading of the engine* that you have max accel at the torque peak, and max capability to fight aero drag at HP peak. Also, I still can't reconcile why a g-meter showed more acceleration at the HP peak than at the torque peak when I experimented with this several years ago on an old Corvette. (And, PSK, the consumer quality G-meters measure to a claimed 3% accuracy, with a repeatability much better than that - ?1/2%-1%? I don't recall off hand.) My previous experimentation was very brief, so I'd like to play with it again, but my son has his G-Tech 200 miles away in Portland, and my GT4 is in tiny bits all over my mechanic's shop! | ||
| Rob Schermerhorn (Rexrcr)
Member Username: Rexrcr Post Number: 485 Registered: 11-2002 |
Psk, that's the ticket. Your analysis is exactly how I determine theoretical shift points and for gearing when running a race car who's ratios can be changed. To all: Nice discussion, very well informed group we have here! | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 483 Registered: 6-2002 |
And we return to P. Thomas's correct observation - the final arbiter is the Dyno. PSK - Nice analysis. The pessimist observes that the glass is half empty. The optimist, that it is half full. The engineer simply observes that the glass is too large. Jim S. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 567 Registered: 4-2002 |
"But it is still 100% correct that in a particular gear the maximum acceleration will occur at peek torque ... as proven again here." ...PSk Thank you PSk for taking the time and effort to put numbers onto the argument. Some need numbers to be convinced. What you have so well written should prove the point. I have seen arguments over torque and power crop up elsewhere. Some opinions are so engrained and so strong that to suggest they are wrong gets personal. It is almost like an argument over religion. | ||
| PSk (Psk)
Member Username: Psk Post Number: 363 Registered: 11-2002 |
Okay for those that need real numbers, I have obtained figures for the 360 Modena: Max Hp - 395.1 Hp @ 8500 rpm Max Tq - 275.0 ft lbs @ 4750 rpm Rear Tyres - 275/40 ZR 18 (thus approx 670mm dia.) Weight - 1390 kg Ratios: 1st 3.29:1 including diff (4.44:1) = 14.6076:1 overall. 2nd 2.16:1 thus 9.5904:1 overall. 3rd 1.61:1 thus 7.1484:1 overall. 4th 1.27:1 thus 5.6388:1 overall. 5th 1.03:1 thus 4.5732:1 overall. 6th 0.85:1 thus 3.774:1 overall. Right torque at peek power: Tq = Hp/(rpm/5252) = 395.1/(8500/5252) = 244.13 ft lbs at engine. Thus lets work out how much force is applied by the rear wheel(s) to the ground at peak torque and peek horsepower per gear: 1st: @ 8500 rpm we have 244.1 ft lbs x 14.6076 = 3566.085 ft lbs. Thus using T = F.d (and it is this simple Tim, no dot products. T at ft lbs or Nm is simply a force in lbs or N times a distance in feet or Metres) Thus F = T/d and d is the radius of the wheel (1.11 ft or 338.6 mm) = 3566.1/1.11 = 3210.1 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 14.6076 = 4017.1 fb lbs. Thus the force applied by the wheel is = 4017.1/1.11 = 3616.1 lbs to the road via the wheel. 2nd: @ 8500 rpm we have 244.1 ft lbs x 9.5904 = 2341.3 ft lbs. Thus F = T/d = 2341.3/1.11 = 2107.5 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 9.5904 = 2637.4 fb lbs. Thus the force applied by the wheel is = 2637.4/1.11 = 2374.1 lbs to the road via the wheel. Thus while maximum force is applied to the road at peek torque, due to the large ratio increase revving to maximum rpm in first does apply equate to maximum acceleration, ie. a = F/m. Lets look at 3rd gear: @ 8500 rpm we have 244.1 ft lbs x 7.1484 = 1745.1 ft lbs. Thus F = T/d = 1745.1/1.11 = 1570.9 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 7.1484 = 1965.8 fb lbs. Thus the force applied by the wheel is = 1965.8/1.11 = 1769.5 lbs to the road via the wheel. Thus again due to the large increase in ratio, maximum rpm in 2nd equates to more force on the road than 3rd at peek torque ... but it is getting closer. 4th gear: @ 8500 rpm we have 244.1 ft lbs x 5.6388 = 1376.6 ft lbs. Thus F = T/d = 1376.6/1.11 = 1239.2 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 5.6388 = 1550.67 fb lbs. Thus the force applied by the wheel is = 1550.67/1.11 = 1395.9 lbs to the road via the wheel. 5th gear: @ 8500 rpm we have 244.1 ft lbs x 4.5732 = 1116.4 ft lbs. Thus F = T/d = 1116.4/1.11 = 1004.9 lbs to the road via wheel. @ 4750 rpm (peek torque) we have 275 ft lbs x 4.5732 = 1257.63 fb lbs. Thus the force applied by the wheel is = 1257.63/1.11 = 1132.1 lbs to the road via the wheel. As you will notice the higher the gear the closer the peek torque of the higher gear is getting to the same force as the lower gear at peek Hp. I won't do 6th, as I have already proved thanks to Ferraris choice of ratios that I am wrong ... when their gearbox is involved. But it is still 100% correct that in a particular gear the maximum acceleration will occur at peek torque ... as proven again here. Interesting ... isn't it. Thus I have debated both directions now, and will concurr that revving your Ferrari to maximum power is the way to go, thanks to good torque at maximum rpm due to the excellent breathing of the high-performance Ferrari engine. This would definitely not be the case on my people mover, and many other vehicles. Pete | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 482 Registered: 6-2002 |
Hans - EFWUN defined these disparate quantities well when he said: -------------------------------------------------- "Torque is instantaneous rotational force. Horsepower is the work that force can perform over time" -------------------------------------------------- Translating to English - torque is rotational force. Work is defined as a force applied for a distance. Power is the rate that work can be done. At 150 mph, the car has to push a lot of air aside. The drive wheels do this by rotating (force at the tire contact surface) at a rate that equals 150 mph (power). If they do it for 1 mile, then that much work has been done. Horsepower, a measure of power, is the metric that is used to determine how much force can be applied to the wheels and at what rate. For example, if geared 1:100, a small engine could generate enough force to push aside an opposing force equal to air resistance at 150 mph. However, the small engine could not push aside the air when turning the wheels at the rate equal to 150 mph - it can do the work, but at a slower pace. Another example - the Ferrari engine could climb a 1 mile 30 degree ramp in 60 seconds (example). The smaller engine with 1:100 gearing could climb the 30 degree ramp, but it would take 300 seconds. The work done (lifting the weight of the car to the same altitude) is the same. The rate that it was done is different. Horsepower by any other name. Jim S. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1073 Registered: 4-2002 |
Hopefully you engineering types won't gloss over this question. If torque is almighty, why are horsepower figures quoted? What relevance do they have? If torque is the only figure of interest in putting the 'juice' to the wheels and accelerating the car down the road (not in terms of velocity, but in delta-velocity - acceleration), of what relevance or use is horsepower? Why do the manufacturers quote horsepower? Of what use is it? Why is it more widely quoted, both in advertising circles, and around the racetrack, than torque? And, no, I'm not trying to be a smart-ass devils advocate, I'm seriously wanting to know. | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2773 Registered: 6-2001 |
Yeah, torque is a vector product. the link to the applet i posted does a nice job of demonstrating it. While i understand that the discussion is heating up, i have more knowledge on the subject than a couple others on this page. i just finished up a matter in motion physcis course. with what i learned in that class, combined with what i learned back in highschol physcis a few years ago, i feel i have a very solid understanding of what is being discussed. obviously not the same understanding as you, or other guys who have their engineering degree already do. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 481 Registered: 6-2002 |
Tim N - you are correct concerning cross product - although I do not believe it has application to this discussion of torque. Do yourself a favor - stay out of the kitchen - it's getting hot. Jim S. | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2772 Registered: 6-2001 |
being an EE major, surely you would know about cross products (ie. acceleration of charged particles in an electric field), and that they are not just a figment of my imagination to make EWFUN look wrong. he seems so sure that my math (cross products aka. vector products)is impossible. perhaps he is the one who fails to grasp fundamental concepts of physcis.
Do you still think my defined terms are meaningless and my mathematics impossible? | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 480 Registered: 6-2002 |
EFWUN - for my own edification - I would appreciate clarification of the discussion between you and Lawrence C. In one post, Lawrence states: -------------------------------------------------- "Peak engine torque produces peak rear axle torque in any gear. Peak rear axle torque puts the most force on the road surface. Putting more force on the road surface accelerates the car at a higher rate. Force on road surface equals axle torque divided by radius of wheel. More axle torque means more road force" You responded with: -------------------------------------------------- "LC, you are failing to grasp a fundamental concept of physics. "Force on road surface equals axle torque divided by radius of wheel." Wrong. Actually, the longer the torque arm, e.g., a wrench, or a wheel radius, the greater the torque, because torque is Multiplied by the torque arm not divided. Is it easier to tighten bolts with a 4 inch wrench, or with a foot long "breaker" bar??" -------------------------------------------------- I responded with: -------------------------------------------------- "EFWUN - I believe Lawrence Coppari has it correct. He is working backwards from torque, to find linear force at the tire/road patch. Given (T) torque at the center of the wheel, the force necessary to stop it from turning, at a distance equal to the radius (R) of the tire will be F = T/R. This comes from the simple relationship of Torque = force x distance from center of rotation (assuming right-angle application of force)." -------------------------------------------------- You responded with: -------------------------------------------------- "This is a waste of time. Physics is a science, not a discussion of opinions, there is no such force as "translational" force, you can't make this stuff up as you go!" -------------------------------------------------- I feel your pain. Help me out here. I believe, (and I certainly do not wish to put words in anyone's mouth) that L.C. was translating torque at the wheel to a force vector pushing the car forward (acting through friction at the tire patch/road interface). Don't hit me, please. I only spent 5 years in engineering school, having to take physics courses and mechanical engineering courses all along the way (major was electrical - but all my electives were in mechanical). Jim S. | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 6 Registered: 1-2003 |
Horsepower cannot be multiplied by a transmission otherwise we could use a transmission to achieve 200mph speeds in cars with 150 hp. A transmision can only multiply torque | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 478 Registered: 4-2002 |
Ok; now I think I see through this mess: PSK is talking about _instantaneous_ acceleration in a single gear. And indeed, peak TQ corresponds to peak _instantaneous_ acceleration in that gear. However, I still maintain that a car will achieve a higher _instantaneous_ acceleration at peak HP RPMs with a gear with higher multiplying ratio than at peak TQ and a lower multiplying ratio. This is what the common man means as acceleration (not _instantaneous_ acceleration). | ||
| Timothy Guay (Timguay)
New member Username: Timguay Post Number: 25 Registered: 2-2003 |
I love the physical laws. I love my Ferrari. I think I should just buy an automatic so I don't have to make these decisions, my wife usually makes them for me! | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 5 Registered: 1-2003 |
Imagine a pillar stickng out of the ground. Attached to the pillar is a one inch bolt that cannot be turned. Now place a socket wrench with a one foot long breaker bar on it so that it is parallel to the ground. Place a fifty pound weight on the end of the bar. You are now applying fifty pounds of FORCE to the breaker bar and the bolt is feeling 50 lbft of TORQUE. In a car if the rear axle torque is 400 lbft and the distance from the center of the axle to the contact patch of the tire then there will be 400 lbs of force pushing the car forward to produce acceleration. The maximum force available to accelerate a car forward in any given gear will therefore be at the torque peak of the engine. The issue of when to shift to a higher gear while accelarating comes down to where you are in the torque curve af the engine. Lets say your engine has a perfectly flat torque all the way to red line. It produces 300 lbft of torque. Second gear has an overall gear ratio of 3:1. There will be 900 lbft of torque at the rear wheels to accelerate the car all the way to red line. If you shift the car to third gear with an overall gear ratio of 2:1 there will now be 600 lbft of torque to accelerate the car. This car will accelerate best if shifted at redline. Now take an engine with the same gear ratios but a diferent torque curve. This engine makes 350 lbft 2000 rpm below redline 300 lbft 1000 rpm below redline and 200 lbft at redline. In second gear this translates to 1050 lbft, 900 lbft and 600lbft of torque at the rear axle. In third gear there would be 700 lbft, 600 lbft and 400 lbft of torque at those same rpm in this engine it is clearly better to shift to third gear well below redline since there will be more torque availble at the rear wheels in third gear at lower rpms than in second gear at redline. The exact shift points for different cars will depend on the torque band and gear ratios of the car but the concept will hold that the greatest acceleration will occur when maximum torque is applied to the rear wheels | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2766 Registered: 6-2001 |
i didnt say anything about a "translational force" (but there is a such thing as translational motion!) i said that the equation F=ma is an equation for translational motion. Infact, F=ma isnt even the real way of writing newton's 2nd law is written as F=dP/dt. maybe you, EWFUN should open a textbook and read it. when you apply a torwue to a body, its acceleration isnt describd as f=ma, because it is not accelerationg in a line, but in a circle, this is known as angular acceleration, which is defined as t=Ia, where t is torque applied, I is moment of inertia (remember, we cant pretend that this is a point mass, but rather a continous body), and a is angular acceleration (a is supposed to be alpha, not "a"), so then from that we see t/I=a. Also, you can tell that Torque is not a force though unit analysis. The unit for force is kgm/s^2, also known as a newton, while torque's unit is Nm. is Nm the same thing as N? clearly not. now, to tell you about "cross products". Force is a vector. you cant treat vector quantities the same as scalar quantities. heres an applett that you can hopefully understand. http://physics.syr.edu/courses/java-suite/crosspro.html hopefully mark or Rob Schermerhorn can weigh in and back me up. i understand how what i said in this post, and my last one, is confusing and seems impossible, but it isnt. honestly further arguing will go nowhere, but if you feel the need to contest what i am saying, i will be happy to direct other members of this board here to explain it to you better than i can. | ||
| PSk (Psk)
Member Username: Psk Post Number: 360 Registered: 11-2002 |
Wrong,
As James has said way before you guys are confusing velocity and acceleration, you are also confusing torque multiplication through the gearbox. Just remember if we did not have gears the car would accelerate fastest at peak torque and past that it will slow in ACCELERATION but still gain velocity until peek velocity is obtained for that gearing. CVTs will prove this point. Anyway I rest my case, and it has been very interesting. BTW I am not confusing angular acceleration in my formulas. The acceleration of a crank shaft would be rotational but as you calculate Torque via a Force at 90 degrees to the centre of rotation (even if it is a spanner or wrench) then the acceleration is rotational ... and the acceleration is still maximum at maximum force and thus Torque. This also applies exactly the same at the wheel. A wheel is just the same as a wrench, except the nut is turning the wrench and pushing your hand ... which ofcourse is the road. Try putting a wheel brace on a wheel and running the engine (actually please don't ... ) but you will feel the torque similar to the road. Again there are many real world variables, like friction, wind resistance and traction but maximum acceleration will occur when the driving wheel has maximum torque ... no matter which gear as we are talking torque at the wheel. Thus you may need to do some extra calculations to work out due to friction, and other losses where exactly maximum torque is applied to the driving wheel in each gear ... but I bet is at maximum engine torque (because it cannot get greater than that). I would love to prove this theory via a G meter, but it would have to be very, very accurate because we might be talking very small differences ... Cheers all. Pete ps: Tim, Torque is a force applied around an axis, thus a rotational force. Plain and simple. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 566 Registered: 4-2002 |
"Peak engine torque produces peak rear axle torque in any gear." I am talking about same gear acceleration. Sure, lower engine torque in a lower gear can make a car accelerate faster. The argument is for the same gear with max torque versus max power. That is what we were discussing I assumed. If you are allowing for different gears, then bets are off because less engine torque can produce more axle torque with lower gearing. And that is why Ferraris are fast. They have relatively low torque engines with a very broad band of rpm. The tranny multiplies meager torque and applies it to the rear wheels making it a fast car. And by the way, drag is proportional to the square of velocity not cube. Power consumed by drag is cubic. Power is drag times velocity. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 476 Registered: 4-2002 |
Ok, here is a (somewhat realistic) senario: Take an engine that produces 255 lb-ft of TQ at 5600 RPMs (this is equivalent to 266 HP); and produces 380 HP at 8250 RPMs (241 lb-ft). {Anyone student of Ferraris can see that this is the F355 engine. But the engine is not important here, the number are!} The ratio between peak TQ RPMs and peak HP RPMs is 1.47--this is almost precisely the ratio between 1st (3.07) and second (2.16) gear (1.42:1). If you believe PSK, then the F355 accelerates faster in second at 5600 RPMs than it accelerates in 1st at 8250! {{A pause to allow the readers to sop laughing and catch their breaths}} In reality, 255 lb-ft of TQ at 2.16:1 has 550 lb-ft of driveshaft TQ, while 241 lb-ft of TQ at 3.07:1 has 740 lb-ft at the driveshaft. Guess which accelerates faster. I am willing to take my F355 to the drag strip and accelerate in 1st at 8250 if someone else is willing to get their F355s doors blown off in second at 5600! You see, the Ferrari TQ curve is almost flat, flat enough that although the peak TQ is a long RPM distance from PEAK HP RPM, yet the actual TQ falls very slowly. So slowly that there is ALWAYS more thrust at the rear wheels in a lower gear at higher revs than in a higher gear at lower revs. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 473 Registered: 2-2003 |
TimN, I'm inclined to think you're just kidding; a "cross product?" Open a physics book, and turn to a discussion of angular momentum. Your defined terms are meaningless, and your mathematics impossible. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 472 Registered: 2-2003 |
This is a waste of time. Physics is a science, not a discussion of opinions, there is no such force as "translational" force, you can't make this stuff up as you go! However many angels can dance on the head of a pin (to quote Jim G.), has nothing to do with scientific fact. Peak acceleration occurs at peak POWER, plain and simple. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 478 Registered: 6-2002 |
EFWUN - I believe Lawrence Coppari has it correct. He is working backwards from torque, to find linear force at the tire/road patch. Given (T) torque at the center of the wheel, the force necessary to stop it from turning, at a distance equal to the radius (R) of the tire will be F = T/R. This comes from the simple relationship of Torque = force x distance from center of rotation (assuming right-angle application of force). Jim S. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1067 Registered: 4-2002 |
Tim: If "maximum acceleration occurs at maximum Torque", then why do I get a higher reading with a g-meter at max HP? | ||
| Tim N (Timn88)
Advanced Member Username: Timn88 Post Number: 2764 Registered: 6-2001 |
actually your definition of torque is wrong, its t=r x F where x represents a cross product. the resultant torque is perpendicular to the rotation and perpendicular to rhe force acting on the moment arm (or maybe im thinking of angular frequency, but im pretty sure im not) i guess for a circular body where the angle between r and f is 90 (pulley, wheen on ground, etc), then you could think if it as force times moment arm length.
even though you are right with the last sentence, your proof is wrong. you are mixing formulas for translational motion and circular motion. the formulars are analagous, but you cant plug in a translational motion quantity into an equation for rotational motion. one last thing pretty much everyone here thinks, and is wrong about, is that torque is a force. IT IS NOT a force. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1065 Registered: 4-2002 |
Lawrence: Try putting a G-Tech or equivalent in the car. You'll find highest G's at the HP peak. You'll need to test in a lower gear, as aero drag screws up the results at high speed. (i.e. drag increases with the cube of velocity, so at speed in a higher gear, the drag will be significantly higher at the HP peak than at the torque peak. Test needs to be done at speeds where aero drag is relatively insignificant.) | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 470 Registered: 2-2003 |
LC, you are failing to grasp a fundamental concept of physics. "Force on road surface equals axle torque divided by radius of wheel." Wrong. Actually, the longer the torque arm, e.g., a wrench, or a wheel radius, the greater the torque, because torque is Multiplied by the torque arm not divided. Is it easier to tighten bolts with a 4 inch wrench, or with a foot long "breaker" bar?? Next, understand something, Torque is an instantaneous concept, it is NOT related to rotational power, but is only the force existing statically. POWER, is that torque over time, e.g., 1 sec, or 1/100000th of a sec. That is why motors are rated for torque AND horsepower. As explained mathematically below, horsepower is the expression of rotational torque. Not an opinion, a law of physics!! | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 565 Registered: 4-2002 |
Peak engine torque produces peak rear axle torque in any gear. Peak rear axle torque puts the most force on the road surface. Putting more force on the road surface accelerates the car at a higher rate. Force on road surface equals axle torque divided by radius of wheel. More axle torque means more road force. Peak torque occurs somewhere below peak power. On my 328, peak power is at 7800. Peak torque is at 5500 rpm. Car accelerates the greatest at peak engine torque because that is where the force on the road is greatest. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 467 Registered: 2-2003 |
Torque is instantaneous rotational force. Horsepower is the work that force can perform over time, e.g., Work = Force x 2 x Pi x radians, ergo; 1 lb x 2 x 3.14etc x 1 ft = 2 x 3.14 lb-ft = 6.283 lb-ft Then; Power = Work / time = 6.283 lb-ft / min 1 hp is defined as 550 lb-ft / s = 33,000 lb-ft / min as measured So, if 1 lb-ft of torque is applied in one minute (1 rpm) = (6.283 lb-ft / min) / (33,000 lb-ft / min) = 1 / 5252 of 1 hp. Thus; we can see that there are 5252 lb/ft of torque per minute in one Horsepower. Or...if we multiply torque in lb-ft by the RPM this torque is produced at, then divide by 5252, we get the power output at that RPM. THEREFORE: the peak POWER output is where ever the manufacturer determines on the dyno, e.g., 8,200rpm for the 355, while the peak TORQUE output is not necessarily the same, NOR is it the place at which the motor makes peak POWER!! The motor makes more POWER at peak horsepower than it does at peak torque, power being defined as units of force/time. Manufacturers determine redlines as a function of where peak POWER exists on the rev curve, and things like valve float etc. All vehicles with conventional transmissions accelerate at maximum g at peak power, not torque. Plain physics, opinions not necessary! Stop the madness! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 473 Registered: 4-2002 |
PSK: what you are forgetting is the the gearbox is involved in the computation of maximum acceleration. So given a gear (any gear) maximum acceleration in THAT gear occurs at peak TQ, however, if there is a lower gear (than the one you are in) and if the HP available in that lower gear is greater then the HP at peak TQ, then you will accelerate faster in the lower gear with lower TQ because the gearing makes up for the TQ. In effect, the gearing (1.2X in my example) overcomes the TQ deficit. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 476 Registered: 6-2002 |
PSK - -------------------------------------------------- "Thus in conclusion Torque is required for acceleration, and Power is required for speed. This makes sense as Torque is obtained by the ability for combustion to be turned into a rotational force, and Power is the engines ability to keep producing Torque at higher rpms." -------------------------------------------------- I could not have said it better. Stated in a more poetic form - Top speed is the stalemate between horsepower and wind resistance. (Qualifiers include horsepower at the wheels) Jim S. | ||
| Bart Boonacker (Sharky666)
Member Username: Sharky666 Post Number: 266 Registered: 9-2002 |
Psk, you forgot reason #3 to rev past maximum Torque, creating joy in the form of sound I just finished this at school, max. Torque, Power etc. etc., couldn't explain it easier then you just did. | ||
| PSk (Psk)
Member Username: Psk Post Number: 359 Registered: 11-2002 |
Okay driving home again and I remembered a simple formula that will mathematically prove that maximum acceleration occurs at peek torque: T = F.d where T = Torque, F = Force and d = Distance. Distance would be the crankshaft radius in an engines case and the Force would be provided by combustion, thus producing Torque. F = m.a where m = Mass (rotating mass such as crankshaft and piston, etc.) and a = Acceleration. Thus putting the two formulas together we get: T = m.a.d as Mass and the Distance are constant for a given engine then the formula can be rearranged, and the only variables are Acceleration and Torque: a = T/(m.d) Thus maximum acceleration occurs at maximum Torque. Now you say what about at the wheels. Well that torque is multiplied through the gearbox and differential and becomes a Torque at the wheel, which again abides by the T = F.d calculation, except that Distance would be the radius of the wheel and tyre. Thus the most torque at the wheel corresponds to the most Force and using F = m.a and Mass this time would be the mass of the car means that the larger the Force the faster the mass (car) will accelerate. Thus in conclusion Torque is required for acceleration, and Power is required for speed. This makes sense as Torque is obtained by the ability for combustion to be turned into a rotational force, and Power is the engines ability to keep producing Torque at higher rpms. Thus there are only 2 reasons why you rev past maximum Torque: 1. To enable an upward gear change to set the engine revs at maximum point once in the next highest gear. 2. To max out the speed in that gear. This also proves my point that maximum torque no matter where on the rev range will provide the most acceleration, as I stated in an earlier post. End of story Pete ps: James has also being trying to say this with graphs, etc. and this is what I have been trying to say all along. | ||
| PSk (Psk)
Member Username: Psk Post Number: 357 Registered: 11-2002 |
Peter,
I just cannot leave this alone, please read Jame's post again, Peter I think you missed something :
and my posts as this was what I was talking about. But I give up as I was being theoretical (not just talking about Ferraris ... we all know that you have to rev a Ferrari) as is James and Rich. Acceleration is what it is all about ... revving until the cows comes home only slighty increases velocity until you reach peek velocity for that gear. Again it is F = m.a, force eventually peeks out at the peek force that your engine can produce in that given gear for that car of given mass. But to go on revving that far would also be stupid as the force would have been increasing in slower increments over time as the power durve topped off. Look at James graphs ... Remember an engine (even an F1 engine) will not keep on producing torque and power no matter how high you rev it and it is not just due to cam profiles and timing and ignition and valve springs it also due to port shape and flow ability and harmonics and exhaust shape and lengths, etc. It is all one big complicated equation. Throwing a car on a dyno will help produce figures for your car and if you tested in each gear and followed up with calcs you will find peek up change points ... but then you go to a track and find that due to track surface or corner characteristics that it might be faster to change up slightly earlier say to optomise traction ... Mitch, still think that changing so you end up at peek torque in the next gear is the best as this would corresponding to maximum engine acceleration and thus car acceleration. After all again power is just a factor of torque and all an engine really produces is torque (ie. a turning force) it just varies at different rpm. And if you can keep on producing torque at high rpm then you make large power, but actually your combustion is not as efficent as you torque has dropped off and it is only the ability to built revs that keeps the acceleration going. Peek torque is where the very best combustion and efficiency occurs and the engine is producing the maximum force possible on those pistons. but ofcourse we are not interested in efficiency we just want POWER One day we will all be driving CVTs and those transmissions (if we give the car full throttle) will maintain the engine at peek torque during acceleration ... and we will be accelerating not only fast but also efficiently, so you will be able to smile at the greenies as you wizz by into the hairpin Pete | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 123 Registered: 1-2003 |
Mitch....Your post made a great deal of sense to me. I do not believe that anyone mentioned tourque curves and the fact that when you 'land' in the next gear, you should be in the optimum area of tourque for that gear. With this in mind, my '77 308 had 60K miles on it with a fresh major. Any comments out there on the optimum shift speeds ( 2 to 3, 3 to 4th, 4th to 5th? Again, great replies to this thread! | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 470 Registered: 4-2002 |
Lets say you have an engine with a red line at 9000, peak power at 8000 and a peak torque at 6000. Now consider that you are in third gear with a ratio of 1.2:1 and shifting upwards to forth gear at 1:1 1) If you shift at peak power (8000) you end up at 6666 RPMs (above peak TQ). 2) If you shift at red line (9000) you end up at 7500 RPMs (almost peak HP) 3) If you shift to meet peak TQ after shifting, you shift at 7200 (under peak power) I suggest that none of the above answers are correct! In this particular case, you should shift at an RPM that equally brackets the RPMs of peak HP (8000). This means you shift at 8763 RPMs and end up at 7302 RPMs. This RPM band has more power (HP) under the curve than any of the first three answeres. More importantly, the used RPM band depends on the gearing! Now back to Ferraris: Modern Ferraris have a TQ curve and HP curve such that red line is not far enough beyond the peak HP that you should shift at redline for maximum acceleration. If you 'got more revs' and the HP curve doesn't 'die to fast' then the extra revs will increase acceleration. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 475 Registered: 6-2002 |
P. Thomas - you risk confusing us with facts. It is more fun to debate hypotheticals. Dyno? Dyno? We don't need no stinking Dyno! EVERYONE knows the answer to this discussion, we simply can't agree on what it is. Your excellent input added fuel to my fire. What we really wish to maximize is velocity. Thus, as velocity is the integral of acceleration (that is, the area under the curve), a given acceleration for a period of time can be matched by one-half the acceleration for twice the time, or twice the acceleration for one-half the time. Sounds simple. However, as the shape of the torque curve is complex, it is not an easy extrapolation to intuitively recognize which design results in the maximum area. I have taken a stab at a graphical presentation. Of interest is that the torque curve is identical to the acceleration curve, as mass is constant. The velocity, as noted on the graph, does not decrease, but increases more slowly. It is the area under the acceleration curve that is of most interest, and can be maximized by providing more acceleration (torque), or lengthening the period of influence of the force. By increasing the torque, one shortens the time necessary to go from 4,800 RPM to 8,000 RPM (for example), and in the limit, approaches an impulse function (infinite area). S1 represents the slope of the velocity curve at peak acceleration, while S2 represents the slope of velocity on the down slope of acceleration. Note that the velocity is still increasing, merely at a slower rate. The shaded area is the area between peak torque and redline, the area of interest. We must turn to the Celestial Design Committee to really understand this stuff. Jim S.  | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 456 Registered: 2-2003 |
Listen, this isn't a question of opinions. To maximize acceleration, you shift at the motor's maximum revs (e.g., the redline, in a 355 8,500, in a 550, 7,600, etc. ) The manufacturers figure the gear spacing from gear to gear with maximum acceleration at redline in mind, so that changing up at 8,500 rpm in a 355 places the motor at a relatively optimal point on the ascending rev range in the next gear. Also, there are two peaks. Torque, and Horsepower. A 355 for instance may make peak torque at 6,000rpm, but it makes peak horsepower at 8,200rpm. The redline is at 8,500rpm for valvetrain safety and other reasons, and shifting the car at that point is what the manufacturer had in mind when calculating their gear spacing. | ||
| P. Thomas (Ferrari_fanatic)
New member Username: Ferrari_fanatic Post Number: 7 Registered: 4-2003 |
You guys are killing me. I think that there are several points to be argued here: 1) Does revving an engine past the designed redline yield more acceleration? Depends. On what you ask? Several things: First and foremost, where did the engineers design the peak of the power band? Has anyone EVER seen a dyno graph?? This would answer your question. Next add a transmission, and if you redlined, then shifted (putting under the power band) you would surely lose acceleration. But here is my question: Don't you think that the F Engineers computed peak torque/HP curves and correlated precise transmission gearing with the engine power curve? Second, where is the redline set?? Electronically controlled (ie ingnition, or fuel cut out)??The tach?? If the engineer cut spark or fuel before hitting the peak power curve then yes, revving your engine beyond redline would yield more power. But if the peak power was below the eltronic cutout (or fuel cutout for that matter) then no, continual revving past "Redline" would yield only nominal gains. Guys, I think we need to define "Redline". The next question probably churning in your brain is what determines a power band? Several things, but mainly cam timing. The higher the lift, and longer duration the more your powerband will be pushed into the "higher" RPM range. Does higher RPMs cause more wear and tear? Depends on what part you are talking about. Pistons and rings? Yes! Quantify how much more, who knows. Does lugging an engine below its torque curve stress an engine? Yes. Once the cam timing has been determined then other factors follow. ie intake and exhaust systems, cylinder head porting, VALVE SPRING TENSION: Why you ask? What good is it to have an engine designed to rev to 10,000 RPM if the valves float at 9,000?? My point is that ALL F engines are highly MATCHED machines. The only way to see if the engine produces more torque/HP is to bypass the rev limiter and put it on a DYNO. PERIOD. | ||
| PSk (Psk)
Member Username: Psk Post Number: 353 Registered: 11-2002 |
Well said James, refering to:
Exactly and revving any further is a waste of time (ie. 3000 rpm above red line) as acceleration, but not velocity, will slow. Thats is definitely all for me, and James has put it so well ... Thank you all for such an interesting topic Pete | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 474 Registered: 6-2002 |
Perhaps this is from the Department of Redundancy Department, but I am not sure, after reading these educated posts, that an essential point has been made. While peak torque might be achieved at some RPM less than redline, torque (force) is produced to redline and beyond. Thus, the mass (car) will continue to accelerate beyond peak torque. Let us be careful not to confuse acceleration and velocity. The car will continue to gain speed at RPM greater than peak torque, albeit at a decreasing rate. Shifting above peak torque (redline), when mated with a transmission designed for this power-curve, will allow the next gear to be engaged at PRECISELY peak torque, thereby providing maximum acceleration in the next higher gear. Mathematically, it is the sum of time versus torque that equates to performance (the integral of torque over time - area under the curve - well said Mitch Alsup). Thus, graphically, shifting at redline and falling to peak torque (with the properly designed gear ratios) provides the maximum area under the curve. This is the reason that Ford engineers, when suffering clutch slippage in the 1969 Daytona 24 hour, asked their drivers to shift ABOVE redline, thereby avoiding engaging the next gear at peak torque, and solving the clutch slippage problem. Perhaps this has all been said. I just missed it. Jim S. | ||
| PSk (Psk)
Member Username: Psk Post Number: 352 Registered: 11-2002 |
Okay I thought about this some more driving to work and will conceed that Rich is right many in many of his points:
Agree that the engine that is making decent torque at higher revs is making more power. Area under graph. Do not agree that you can just keep on revving engines higher and thus expect to continue making more power. Eventually you reach a breathing inefficiency of the engine and power drops off dramatically ... if the engine can survive the mechanical abuse of such high revs. This is due to cam and ignition timing and porting abilities, etc.
Agree.
Agree, as long as the power drop is not so significant that you loose more time revving out than changing to the next gear. Most engines power drops off very quickly so 3000 rpm past peek power will definitely not be producing good power ... so as long as it is producing more power than in the next gear including ratio advantages (ie. lower ratio is easier to push with due to mechanical advantage), I agree. I have never driven a car that is still pushing good 3000 rpm above peek power ...
Agree, again based on my above point. I still think you all have missed my point, which was revving massively past peek power is a waste of time. I agree with all of you that if your motor is still making torque then you should keep revving as it will be making more power. Thus as long as the torque drop off is not worse than the power gain due to revs then keep on it. Again in the real world there would be motors that produce huge torque at low rpm that would out accelerate a high revving engine, but this is probably because the torque advantage is too great for a meaningful comparison. Please remember that you have to be producing torque to be making more power, yes it will drop off due to the engines inability to continue to make efficient combusion due to breathing difficulties, etc. but it still has to be making some torque and it has maintain a level high enough so that x the rpm/5252 ratio is more than at lower rpm. Take my people mover for example 3000 rpm past peek power and a bus would be able to out accelerate me as it would have absolutely no power ... and holes in its block  Again I am not saying not to rev as you have to keep in the power band but 3000 rpm past peek power is the bit I cannot understand, or though I understand the theorical ratio difference point that Rich made. Pete ps: Making an engine rev to 18000 rpm is not just about making it strong enough. Thus no matter how strong I made my people mover engine it will never rev to 18000 rpm, never ever. It will get to a certain revs and not be able to produce enough combustion to create enough force to balance out the friction and thus accelerate the engine internals ... thus it will just sit there and rev its heart out with the throttle wide open. No longer will it have enough power left to accelerate the rest of the car and thus acceleration has stopped in that gear ... based on F = m.a and we have constant mass and now constant Force we are no longer accelerating. So if you rev to this point you are an idiot as maximum acceleration is what it is all about. That is my point. 3000 rpm past peek power would be that point for most engines if not all ... so I do not agree that in first gear revving 3000 rpm past red line would correspond to the best car acceleration. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1062 Registered: 4-2002 |
Psk: What kind of blows a hole in your argument is discussed here: http://www.ferrarichat.com/discus/messages/112/228318.html?1049665691 A 308 QV has a torque peak at 5000-5500rpm. But I virtually guarantee that shifting at anything below 7k will result in reduced performance. Not that I do that, mind you, but for different reasons. | ||
| PSk (Psk)
Member Username: Psk Post Number: 351 Registered: 11-2002 |
I actually think we are all saying the same thing, ie. rev to the designers intended maximum rpm. I just was commenting about Rich's comment of revving his Dino to 8000 rpm ... where I am sure that it will have stopped making torque and thus power would have fallen off and the car slowed. Now theorically due to ratios, etc. that is where he should change ... but he has to make the engine breathe at that rpm, which a standard 246 Dino would not do ... thus no point abusing the motor is there. Engines do not just keep on making torque and power the higher you rev them. Even if it was mechanically possible to keep on revving your Ferrari motor to F1 like revs I can assure you that it would be producing very little torque and thus power because the valves and cam timing, lift etc. would not allow the engine to breathe at that rpm. While the power formula: P = Tq * rpm/5252 shows that as the rpm increases the power goes up (which is true) if the torque drops down fast enough so will the power and thus so will the cars acceleration ... otherwise we have utopia (sp ?). This is why F1 engines rev so high to make the big power but they have to work very hard to make the engines keep producing torque so that they do keep making power. Without torque you have NO power ... so it is not just revs. Pete | ||
| PSk (Psk)
Member Username: Psk Post Number: 350 Registered: 11-2002 |
Efwun,
That is what I am saying, or trying to. Thus changing up WAY OVER the red line makes no sense to me. Thus if your car peeks at 3000 rpm (slack car and all ) then change at 3000 rpm. If your car peeks at 18000 rpm then change then. I own a Mazda people mover that peeks at around 4500 rpm (okay you can laugh, but it gets the family around and me to work for very little gas) but it car rev to 5500 rpm. The point I am trying to make is that above 4500 rpm it has no torque because it has stopped making torque and the torque curve is diving very fast ... thus I change up (if in a hurry) at around 4700 because there is absolutely no point revving it right out. Now when I used to race, my car made torque all the way up to 9000 rpm thus I would rev it all day long to 9000 rpm. Thus it is all about where your peek torque is and where it stops making torque (If your engine stops making torque then by the formula that was provided the power starts to drop ... as power is torque divided by 5252 rpm, thus torque going down so is power ... now if torque does not drop too fast then you can keep revving, but usually you hit a wall) , and if your car makes peek torque at 3000 rpm and none above it then it will definitely not be making torque above 5252 (which is where power and torque meet) and thus the area under the graph will not be greater because your engine is not a revver. Thus please read what I am saying it is all about WHERE your engine peeks and works not just that it can rev. Pete ps: If it can rev and still makes good torque then ofcourse you have to rev and yes the calculation for power does work. ps: Think of a tractor or truck engine that DOES not rev ... no point revving that thing to the sky falls down is there? ps: Mitch my debate with Rich was assuming that both cars had the same gear ratios to prove that the engine with the MOST torque will win all things being equal. Obviously gear ratios are required to make up for torque deficiencies on low torque motors, or motors with a short power range. Again missed my point. More torque means more area under the graph and good bye low torque motor if GEAR RATIOS ARE EQUAL, etc. You guys need to think outside the Ferrari square. I am not saying do not rev your Ferrari ... read my other posts ... a Ferraris peek torque is very high (and they keep making torque up high because they can breathe) so you have to rev it to make it work. | ||
| Timothy Guay (Timguay)
New member Username: Timguay Post Number: 21 Registered: 2-2003 |
All right dudes, Feel your heart right now? Does it feel the same way at 7,000 rpm? I didn't think so. C'mon lets have some fun, that's what we're doing on the computer right now instead of driving. As I speak we are expecting 6-10" of snow here in Ct. New England can suck. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 444 Registered: 2-2003 |
Pete: Trust me that if you want to maximize acceleration, you need to change up at the redline, not somewhere else. The task in gearing a racer is to find ratios that allow you to be on the torque curve at the apex, and on upshift drop your revs from redline somewhere near your torque peak in the next gear. Road cars like the 355 have close ratios so that the rev drop doesn't step you outside of the range. If you're not changing up at 8,500rpm, you're not going to have enough revs on the clock to pull your hardest in the next gear. Simple fact. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 466 Registered: 4-2002 |
PSK is simply wrong about torque. Torque at RPM is HP (HP = TQ * RPM /5252) The car with the most area under the HP curve will accelerate faster. This area begins at the lowest RPMs after a shift and ends at the next shift. PSK would be right except there is this thing called a transmission and a rear axel ratio. A car with a 6000 RPM red line has to have 33% more torque than a car with a 8000 red line just to break even! American Muscle cars typically torque curve that gives a broad power curve with big cubes. Ferrari has a slowly rising torque curve that gives a great top end rush and just enough cubes to get the job done.* The American engines with big cubes beat the Ferrari only because the ratio of engine size (cubes) is greater than the ratio of red line RPMs. * getting the job done can be debated all day long..... | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 814 Registered: 10-2001 |
The S2000 is strictly an autocross car. It's the first car in years that gives Europas serious competition, which incedently are awful track cars too in stock form. | ||
| PSk (Psk)
Member Username: Psk Post Number: 342 Registered: 11-2002 |
Ken
I am so confused now that I might go and have a beer. I am not trying to say that at all. I am trying to say that shifting at slightly higher than peek torque is what it is all about, and that a motor with higher torque, no matter where in the rev range will smoke a lower torque motor. Obviously you have to change slightly above peek torque so you are close to peek torque in the next highest gear as you want maximum acceleration. I definitely do not rev the balls off my motors just for the fun of it ... no point unless the motor is pushing hard. Anyway I'll drop it now as I think there are too many wires crossed ... pity as I really did want to chat with Rich a bit more and understand what he was getting at. An interesting off this topic comment is that the Honda S2000 seems to be hopeless on the track (in Aussie production car racing anyway) which does not make sense to me as 9000rpm and light weight should work ... but it could be the driver I guess. Maybe Rich you could send me an email and explain your thoughts ??? Pete | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 118 Registered: 1-2003 |
Tim, No, the prop heads are not getting to me! I leave my beanie in the sedan when I get home! All I wanted to understand is where the edge of the envalope is between running the engine at its peak Vs premature wear. There is SO much great info on this thread, obviously from talented individuals like yourself. Pete | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 813 Registered: 10-2001 |
I think Tim means that if revving high and then shifting floats your boat, don't sweat it if you're not at the exact right shift point. | ||
| PSk (Psk)
Member Username: Psk Post Number: 338 Registered: 11-2002 |
Tim, Sorry don't understand your post either. I consider myself to be a tech head, and have a Mechanical Engineering certificate (nearly 20 years ago though) and a Degree in Computer Science (with a minor in mathematics), but while I understand that the car that produces the most torque at higher rpm will be able to handle revs better and this will help with the next ratio change. A car with that much torque advantage at lower revs will still accelerate faster as they have more energy to push the car along. And yes I do understand the force balancing equation ... etc. Many years ago we used to have a NZ race car built with an aeroplane engine that only revved to (I do not know exactly) 3000 rpm but produced huge torque, and this car used to sound so funny in amoungst the high revving normal race cars ... but it was a race winner. Thus Rich's argument makes no sense unless your engine continues to produce good torque at the higher rpm ... which just does not happen in the real world. Thus if you revved his Dino to 8500 rpm the torque (and thus power) would have dropped off completely and the car acceleration would have slowed. Rich calculations are theorical and based on a linear torque curve which is only possible in theory, and over certain rev ranges. That is my point ... maths may say that that is possible and the correct change up points but torque does not keep building until whenever you like ... if it does please send me that engine I'm not wound up or anything ... could not give a toss , but maybe I should sit down and do the maths to see where he was coming from ... but I have a life and little kids If Rich's comment was true then the aeroplane powered car would not have been competitive, and it was and gear ratios were free here, so the high revving cars could have (and did) choose what ever ratios they like to make the most of their engines. Thus higher torque wins on the track ... not higher at whatever rpm, just the most overall. Pete | ||
| Timothy Guay (Timguay)
New member Username: Timguay Post Number: 15 Registered: 2-2003 |
Peter, don't let these tech heads (sorry I am one too) get you down. There is a physical ramification to everything. For every action there is an equal and opposite reaction. Mass can never be destroyed, it can only change form. Every time I get on the chat lines I get pumped about driving (since I obviously am not behind the wheel). Whatever sounds good to you is the correct way to run, whether for pleasure/excitement or for preservation. Have fun Tim Guay. | ||
| PSk (Psk)
Member Username: Psk Post Number: 334 Registered: 11-2002 |
Rich, I don't agree with your comment, about the car that makes the highest torque at higher rpm will accelerate the fastest. The car that makes the most torque will accelerate faster. Thus lets say that both cars have exactly the same gear ratios (and same weight, tyre sizes, etc.), and (using your examples) one makes 400 ft. lbs. at 3000 rpm (car A) and the other makes 300 ft. lbs. at 5000 rpm (car B). Thus on the drag strip the car with more torque will accelerate faster as it is torque that turns the wheels. If both drivers change at peek torque then car A will change earlier and also be ahead on the road as (remember the gear ratios are the same) car A has more torque pushing it down the road and thus will be accelerating faster. Thus in the end you are an idiot if you rev WAY past peek torque as the acceleration will actually slow (unless you have a turbo engine and need to rev to a certain point to get the next gear revs right ... otherwise you will drop off boost), and this is why large capacity engines rule drag racing as they make HUGE torque. Now I need to explain myself a little here. If your engine has come on cam and has a nice flat torque curve and thus is still making decent torque numbers then keep revving, but if your engine starts a torque dive then change up. So you could have an engine that torque peeks (by a small amount) at a lower rpm that still keeps making good torque at the higher rpm and then your point is completely valid ... but this is not normal, usually the torque figures climb until peek torque and then drop off fairly suddenly. My last race engine once on cam (at 5500 rpm) had a wonderfully flat torque curve all the way up to over 9000 rpm (where my shallow pockets limited any higher rpm) so I would rev it to 8500-9000 all race long. This is because the torque did not drop off. Now and engine that produces peek torque at 3000 rpm (car A) will definitely drop off before you get up to the Honda S2000 peek rpm as that is a low revving v8 type torque peek ... An F1 engine may make similar torque (?) at 18000 rpm but it would have had its carbon body blown off by the lower revving torque producer ... with same gear ratios. This is why American iron, or muscle cars can still blow a Ferrari away (some) in a straight line ... it is all torque. Thus a Honda S2000, a wonderful car that it is, will not be as fast of the line as a many other cars because you have to wait for the engine to spool up to the high revs before it can produce enough torque to push it along at a decent rate. Now if you do a full race start and hold the engine at peek revs and dump the clutch then maybe you will have a chance ... but again that American V8 with huge torque at ONLY 3000 rpm will be gone ... because he can also dump at peek revs and HAS MORE torque to accelerate his car with (assuming equal traction, etc.). To assume that drivers are stupid enough to continue revving their engines WAY past peek revs is only showing that the driver does not understanding that it is the torque that is accelerating the car ... and yes when you hook up to the next gear the torque will drop but the engine with the most torque at that rpm will accelerate up to peek torque the fastest ... thus again the American hugh torque motor wins on the drag strip. Now in the real world they do not have the same gear and differential ratios, so the American hugh torque motor is handicapped by long ratios and unnecessary weight ... this is the only reason the Honda has a chance Pete ps: It is not power that accelerates a car it is torque, and again same gear ratios the motor that produces the most torque will win. Maybe I have missed understood your post, but you have to compare with equal ratios and tyre sizes, etc. to compare the torque turning the wheels ... again most pushes harder. There is a saying in motor racing that goes something like power is nice but torque wins the race | ||
| Bob Cowart (Bob_cowart)
New member Username: Bob_cowart Post Number: 5 Registered: 3-2003 |
Yes and the engine that develops its torque at a higher rpm can take advantage of various gearing strategies to maximize its acceleration. Which brings me to a question for owners of the F355 and F360. Do any of you know what the torque curve looks like for these 2 machines? - Thanks in advance. Bob | ||
| rich (Dino2400)
Junior Member Username: Dino2400 Post Number: 154 Registered: 10-2001 |
Interesting topic. Also remember that the car that makes more torque at higher rpms will always be faster. If a certain car makes 400ft. lbs. at 3000rpm but makes only 100ft.lbs at 5000rpm, the driver of that car will have shifted to the next gear. Say another car only makes 300ft.lbs at it's peak, but that peak is 5000rpm and it's still making 200ft.lbs at 8000rpm. This guy will not shift as early and will spend much more time in the lower gears. He will thereby accelerate faster. Why? Because power isn't about what the engine is putting out - it's about what makes it to the wheels and this means gear ratios come into play. Some people think shifting at their listed torque or power peak leads to the fastest acceleration but in reality you should only shift if you car is going to be putting more power to the road at the resulting rpm in the next gear. Usually this point is significantly past (like 3k rpm past) the torque peak on a car and declines by perhaps 500rpm per gear since the ratios are getting closer. For example, on the 246 dino engine, torque peak comes at something like 4600rpm but shift points would be 8500, 8000, 7500, 7000. That's the math anyway. In reality, it would scare the crap out of me to rev 1st gear that high, ha! But that particular engine should easily handle 9000 and you can drive it all day at 7000 and still get many many miles out of it (and then rebuild it for $5k or so: there is something to be said for simple vintage engines, ha!) What was a better engine for its technological time, the dino v6 of 35 years ago or the honda s2000 engine today in 2003? My vote is the dino of course. What is a better engine today, honda's best engine or ferrari's best engine? Ask Villeneuve, ha! | ||
| Bob Cowart (Bob_cowart)
New member Username: Bob_cowart Post Number: 4 Registered: 3-2003 |
Greetings Chris, We share the same enthusiasm for our cars. All 3 cars discussed, have engines that require high revs in order to explore their available power. The M3 develops 333hp at 8000 rpms, while the F360 develops its peak power at 8500 rpms. There's nothing quite like these 3 engines and I again confess my bias towards engines that rev well into the upper ranges. I live close to the mountains and 50% of my driving is on mountains roads where the engine is constantly between 6000 - 8100 rpms. So far so good and the engine begs to be driven in that rpm range. Reving your car to 9000 rpms must be a pretty heady experience. - regardless, these cars were meant to be driven. Those who hold back are really denying themselves the essense of these fine machines. It's their respective engines that seperates them from the competition!! happy motoring- Bob | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 34 Registered: 1-2003 |
Great point Bob. Low/Mid RPM torque is clearly lacking in the S2000 compared to some other cars, and this can be an issue for other people as well as yourself. The S2000 was made to be driven at high RPMs, and I happen to find it a fun all around package to drive. It has a small 2 liter 326 pound engine, and to make power you must make it scream at high RPMs. However, it was indeed engineered to do this, which is why I brought it up in this high RPM thread. Watching the S2000 tach hit 9000 RPMs always makes me smile, especially knowing that Honda engineering intended this for its engine. The F360 is clearly in a much more exclusive class, and is more refined than the S2000 in every way I can think of. However, before I made up my mind to buy an S2000 Roadster I did test drive the BMW M Roadster (and also the Boxster and Corvette convertible). I am happy to agree that the M Roadster I drove had a very impressive and powerful engine, however as a complete driving package I was a little disappointed and personally ranked it slightly lower than these 2-seater convertibles -- in terms of overall driving feel to me. Test drives are really important though, since the best choice for one person can be quite different than the best choice for another. All of these cars are exciting! Kind Regards, Chris | ||
| Bob Cowart (Bob_cowart)
New member Username: Bob_cowart Post Number: 3 Registered: 3-2003 |
There's no doubt that Honda's VTEC is an engineering accomplishment. After all 120hp/liter is amazing: however; there are other important aspects to consider when rating these cars in terms of technology, drivability and finally the grin factor. Torque is an important aspect of that equation. Consider these facts: The F360, an engineering masterpiece, cranks out 275 ft. lbs. of torque at a very usable 4750 rpms, or 76.3 ft.lbs. of torque/liter of displacement. Each lb. of tq. must pull 11.96 lbs of weight. The BMW M3, another engineering masterpiece cranks out 262lbs of torque at 4900 rpms, or 81.8 ft.lbs. of torque/liter of displacement. Each pound of tq. is pulling 12.99 lbs. of weight. The S2000 by comparison cranks out 153lbs. of torque at a very high 7500 rpms, or 76.5 lbs. of torque/liter of displacement. Each pound of torque must pull 18.3lbs. of weight. All this to say that all 3 cars are at the fore front, but the S2000 needs to be driven in the upper rpm range at all times to fully explore the fun factor. This in my HO gets tiring after a while and makes it less drivable under normal conditions. A proper balance of torque at usable rpms and HP with a smaller displacement high reving engine is a blissful experience. The F360 and M3 have it in spades. | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 33 Registered: 1-2003 |
The NSX engine is another fine example of an engine designed for high RPM operation. Clearly, Honda engineering is delivering the best of both worlds in the NSX and S2000 (high RPM performance and general streetability) with their unique VTEC approach. Furthermore, the S2000 engine is the ONLY naturally aspirated production engine that generates 120 bhp per liter. It is an engineering marvel that this has been accomplished together with Honda reliability, and at such a low price point. Having said this I still do love to drive my 308 GTB QV. The exhaust note is wonderful, and the general feel of the car is outstanding in all regards. I never cease to be amazed when I consider that it is 20 year old technology, as it was so far ahead of its time. Kind Regards, Chris | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 807 Registered: 10-2001 |
Pete, my Brother In Law has an NSX and I love it a lot even though it's not got the soul of a Ferrari or my ancient Lotus. You said you thought the VTEC was a waste on it, yet it has been discussed that the low end streetability of an agressive cam is mitigated by the VTEC system. Why do you think the NSX would still have its docile low RPM performence with the agressive cam setting it has if it didn't have the VTEC? I would think there would be a compromise tune like other higher RPM cars that will cost you both low end torque and high end HP. With the VTEC, you get more of both at the same time. | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 26 Registered: 1-2003 |
Good points Pete. I believe the difference is that the longer valve cycle duration and higher valve lift needed for high performance high RPM operation are really to compensate for high RPM breathing limitations in the intake and exhaust paths. At low RPMs these limitations are not typically apparent, and an aggressive cam profile (long duration and high lift) is actually unwanted since it results in erratic and high RPM idling, as well as poor low RPM drivability. This in the end hurts efficiency and emissions in an engine operating throughout the RPM range. On a racing engine, it is easy to merely toss out low RPM operating characteristics and slip in a very aggressive (long duration high lift) cam optimized for a specific high RPM power band; turn up the idle speed and instruct the driver to keep up the RPMs. However, a larger compromise always needs to be made with respect to just how aggressive the valve operating cycle can be made when the same duration/lift is used throughout the RPM range. Yes, mechanically advancing/retarding the valve timing through various mechanisms has been used by every manufacturer for about half a century. This is great, but is different than varying the cam profile (lobe shape/slope) -- especially with respect to the duration cycle. I do agree with you that Honda engineering has been much more effective in motorcycle racing, than in F1. Kind Regards, Chris | ||
| PSk (Psk)
Member Username: Psk Post Number: 332 Registered: 11-2002 |
Hmmm, I wouldn't go to the end of the earth batting for the VTEC system. In my opinion a bit of a gimmic, and for NORMAL road cars a waste of time. Note I exclude the S2000 here as it is a sportscar and 9000 rpm makes sense here, but in you people mover going shopping no need. Variable valve timing by rotating the camshaft hydraulically as BMW and most European and Japanese manufactures do, fills in torque holes and allows for more agressive cam profiles ... thus doing what Honda do with little of the complication. Now the Honda idea could probably do more, but I wonder why they do not vary the profiles at lower and more accessible rpms? BTW: The Honda NSX comes with VTEC but the engineers responsible for that car did not want to use it as it was not required, ie. the cam profile was already very agressive ... but the advertising men and women of Honda made sure it was included. Thus very much of a gimmic with that car. Even some of their motorbikes use this technology so they have obviously got the reliability sussed ... all very interesting and I give Honda credit for lateral thinking. But I'll stick with the more simplistic variable valve timing and yep I'm sure the BMW M3 goes very well and gives a Japanese engine a run for its money power/cc, etc. BUT while I am a Italian and European car enthusiast I own a Japanese people mover as the family car as the Japanese have really got the manufacture of the motor vehicle sussed and they are so dependable. The biggest credit I give them is with their electrics ... they always work and lovely design of the plugs, etc. Even Bosch is not that good. Just my two pennies worth ... Hondas biggest mistake at the moment is NOT going ahead with it's own F1 car and getting involved with slackers BAR ... Pete | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 25 Registered: 1-2003 |
Bretm: Don't get me wrong, I was referencing your nerd flame to where it comes from. Chris | ||
| BretM (Bretm)
Advanced Member Username: Bretm Post Number: 3356 Registered: 2-2001 |
Because the Germans and Italians aren't using constantly variable cam timing as opposed to the on/off VTEC system, which is better, I wonder, a cam that optimizes timing at every rpm or a cam that turns on at a set point and back off again at that set point... And apparently my living in NJ has something to do with all this as you seem to bring it up in your posts, you bringing that up obviously adds more validity to what you say, because I'm sure when people see you make comments like that they must think to themself how intelligent you are. | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 23 Registered: 1-2003 |
Bretm: How correct you are. Honda's VTEC variable cam timing changes the cam profile for the higher RPM range. Imagine, how useful more valve lift and longer duration within the higher RPM band could be in other engines -- yet with the ability to switch right back to more sedate duration/lift for the low RPM range. Not only is varable cam timing an exciting technology, but it has also been proven reliable by Honda. Don't worry, when I used to live in north eastern Bergen county I would just go inside my sealed air conditioned home and breathe deeply so I could think more clearly, once I had cleared those fumes. Not a problem for me now... Regards, Chris | ||
| BretM (Bretm)
Advanced Member Username: Bretm Post Number: 3355 Registered: 2-2001 |
Yes, you got me, it must be those smokestacks, what was I thinking. The Japanese do infact lead the auto industry, it has been the Germans and Italians stealing Jap technology all along... | ||
| Mike Charness (Mcharness)
Member Username: Mcharness Post Number: 412 Registered: 8-2002 |
Chris: My pleasure... glad you like it. Maybe one of these days someone will make several from that design and sell them, and send me royalties! | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 22 Registered: 1-2003 |
BretM: You must be overcome by the fumes from those Jersey smokestacks... Our QVs unfortunately only put forth around 80 bhp per liter, and weigh a whopping 600 pounds more than a Honda S2000. Don't be surprised when you watch a higher horsepower, lighter weight S2000 hurry past you around a racetrack. Mike: Thanks for posting that detailed article about the fire extinguisher bracket. I quickly welded one up out of smaller steel bar stock for my 308 QV's halon extinguisher. I painted it satin black, and it looks great. Hopefully I won't need to use it... Kind Regards, Chris | ||
| BretM (Bretm)
Advanced Member Username: Bretm Post Number: 3352 Registered: 2-2001 |
Top or not, the Japanese tree strikes me as more of a bush. Like being the coolest of the nerds... | ||
| PSk (Psk)
Member Username: Psk Post Number: 330 Registered: 11-2002 |
Honda are definitely the top of the Japanese tree, or though the Toyota 4AGE 1600cc engine was a beauty ... Pete | ||
| Mike Charness (Mcharness)
Member Username: Mcharness Post Number: 409 Registered: 8-2002 |
I've driven an S2000. Great car for the money -- and it SCREAMS at 9000 RPM! | ||
| Chris F. (Chrisfromri)
New member Username: Chrisfromri Post Number: 21 Registered: 1-2003 |
Very Interesting thread! I started reading this thread with particular interest towards my own 308 QV. Having said this, I should further comment that one of my cars happens to be a 2002 Honda S2000 Roadster. Its engine is a naturally aspirated 2 liter inline 4 cylinder that develops 240 bhp (120 bhp per liter naturally aspirated). Valve timing varies for higher performance duration/lift within the 6000-9000 RPM band (redline is 9000 RPM). Honda engineers designed this engine to run, and breathe best, at high RPMs. The car's short throw close ratio 6 speed gearbox was also designed to allow the driver to keep the RPMs high. This, in a 2800 pound 50/50 weight balanced Honda. To be sure, the generalizations made in this thread towards Honda engines clearly are inaccurate when it comes to this Honda engine. I could go further to suggest that as much as I do appreciate and enjoy my own Ferrari, only Honda engineering delivers this level of engine performance and efficiency (120 bhp per liter naturally aspirated, Low Emission Vehicle classification, good fuel efficiency, very high value) in a fun to drive, good handling sports car. Before you reach for your flame thrower consider value: my S2000 cost about $30K in 2002, and my 308 QV cost about $60K in 1984. Kind Regards, Chris | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 448 Registered: 4-2002 |
7000 is strictly for the F355/F360 engines. This has to do with the engine architecture, the build materials, and maintanence schedule. Architecture = {180d V8, 3.03" stroke, DOHC, main bearing size, throw bearing size, 100 PSI oil pressure} Materials = {forged billet crankshaft, nikisil cylinders, ceramic coated pistons, forged titanium con rods, 5 tiny titanium valves per cylinder, direct valve actuation, dry sump w/ heat exchanger} Also note: a quick zing up to red line hardly moves the oil temp, so the engine is still operating in the quadradic wear region, its only when you flog the snot out of the car that the cubic wear region takes hold as the oil gets hot. | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 804 Registered: 10-2001 |
Mitch, that was an amazingly detailed post. Is 7000 RPM the 'no wear' threshold of all engines or just a specific Ferrari engine? How was this information obtained? Why don't car manufacturers have the red line at 7000 RPM and design the engine within that perameter? BTW, my Lotus Twin Cam engine redline is 6800; maybe Colin wasn't too dumb huh? | ||
| PSk (Psk)
Member Username: Psk Post Number: 320 Registered: 11-2002 |
No worries EFWUN, will be interested in what you find ... very interesting stuff to chat about and sure beats work The written word can be a bast*rd when it comes to explaining things ... especially when we are busy and trying to chat at the same time Pete | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 378 Registered: 2-2003 |
Pete, I may have gotten this wrong, although not from any lack of understanding of motors or anything else to do with cars. I'm not saying a 180 crank turns it into a 2-stroke, where there is a firing event every stroke of the piston. I understand the 4-stroke cycle pretty well. I'll post again when I've done my research and probably found the I'm wrong, or at least not explaining my idea correctly. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 446 Registered: 4-2002 |
There are several RPM ranges in a typical engine, and each range has different wear characteristics. Many modern road car engines have more wear related to getting started in the morning than from their entire life driving down the roads! But back to (warmed up) Ferrari engines! From idle to about 150% idle, the engine may be underlubricated, and wear is greater than when slightly higher RPMs are used, and/or when large throttle openings are used in the low RPM band. Idle is set where wear is low, and the engine does not overheat, and fuel consumption is acceptable. From the point where the engine is fully lubricated until stresses start to overcome basic friction, wear is basically linear and basically zero*. Modern oils and engine materials and surface treatments provide very hard, low friction, long wearing surfaces. This range extends upwards to about 80% Red line or around 7000 RPMs for an F355/F360 engine. In this range, cylinder and ring wear are more a function of cylinder pressure which is approximately equal to throttle opening than of RPMs. Other engine wear components are basically fully supported by a nice thick oil film preventing anything to anything-else contact. As stress becomes greater than friction (above 7000 RPMs), different mechanisms begin to change which parts in the engine are wearing. In this region (7000 to 8000 RPMs), wear is basically quadradic with respect to RPMs. But even here the rest of the car will wear out before the engine, and the difference in wear of 8000 RPMs is only about 30% greater than that of 7000 RPMs which was basically zero anyway. At even greater stress levels (8000+) the engine cooling adds another bit to the exponent of wear and wear becomes cubic with respect to RPMs. From 8000 to about 10,000 RPMs wear continues to accelerate, the oil get hotter and supports the components les and less well, seals are consumed by friction. Valve guides wear due to the geometry of actuation not being dead center of the valve stem and the related side/slide loads. Wear accelerates at a rapid pace. Above 10,000 RPMs you enter the terminal region where parts are stressed beyond their physical capabilties: con rods break, wrist pins distort, rings flutter, cylinders become 'unround', crankshafts have horible harmonic distortions, valves overheat their oil film in the guides, valve springs float, and the engine takes itself apart in short order. Many call this region the region of exponential wear, however, it is the region of stress accumulation until failure. *engine wear is basically zero, however ancillary wear continues unabated. | ||
| PSk (Psk)
Member Username: Psk Post Number: 317 Registered: 11-2002 |
EFWUN, As you so nicely put in the airbag thread 'I am not medically inclined' and well mate, you do not understand the theory of a 4-stroke engine. Please read the others posts above your last and do some research. BTW: I have read (but have no medical degree, etc. to justify) that whiplash us caused by the head going forward and then bouncing back and damaging the neck on the backward movement ... hence the head rest. I believe this does not matter whether you get hit in the back or have a front impact or whatever. My sister suffers from whiplash after being a passenger in a car that rolled down a hill ... with out headrests. Anyway, that is off topic for this thread, but getting back to the subject. PISTON SPEED IS ONLY DETERMINED BY THE RPM OF THE CRANK AND THE STROKE. A 180 degree crank does not turn an engine into a 2 stroke ... think about the cam timing and everything just would not work. Please let this die and accept that you have got it wrong and have just learnt something, as I have in the airbag thread. After all one of the great things about this site is what you can learn by listening to (er, reading from) others ... and accepting that you do not know everything Cheers Pete | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 4 Registered: 1-2003 |
Thank you for explaining the mechanics that make a 180 degree V8 have better power characteristics that a 360 degree V8. I knew that the 360 degree cranks have better balance and are therefore smoother. I couldn't quite understand the better breathing characteristics of a flat crank untill you explained it in such a concise manner | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 441 Registered: 4-2002 |
Re: 180 degree crank versus 360 degree crank. F355/F360 have a 180 degree crank. In this arrangement each cylinder bank is a fully balanced 4 cylinder engine with two pistons moving up and down in unison, while the other two move down and up in unison. For end to end shake reasons the inner pistons are paried and the outer pistons are paired--just like a 4-banger. With the pistons moving thusly, the inlet phasing is strictly 180 degrees, and the exhaust is strictly 180 degrees. The bank firing order is LRLRLRLR. This give perfect breating and allows each intake ane exhaust bank to resonate and at some RPM levels achieve more than 100% volumetric efficiency. Chevy, Ford, DC all used 360 degree cranks. This causes the firing order to be LRLRRRLL (ford) or LRRLRLLR (chevy). The doubled up LL and RR overload the exhaust, and require x-pipe arrangements to dampen the burble. The double pulses make it impossible to extract added volumetric efficiency with resonant intake tracks. However, the 360 degree crank engines are smoother. At any given point in time on one bank, one piston is accelering downward, one accelerating upward, one decelerating upward, and one decelerating downward. The 'total' amount of shaking is just over 1/4 that of the 180 degree crank engine. This is a pure trade off between engine shaking and power output. | ||
| Lawrence Coppari (Lawrence)
Member Username: Lawrence Post Number: 558 Registered: 4-2002 |
These are 4 cycle engines. Each cylinder fires every other crank rotation. That's it. You do not get 2X explosions per crank rotation with a single plane crank. If you did, you'd have to have a 2 cycle engine - big chainsaw. It does not matter about the configuration of the crank. Each cylinder fires every other rotation. Period. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 351 Registered: 2-2003 |
Without antagonism, PSK, I'm going to go back and research this before making a pronouncement, but it is currently my belief that the 2-Four Cylinder function of the 180degree crank results in twice the number of explosions per crank revolution as, for example, a chevy. Exhaust pitch relates to the speed of the exhaust pulse (e.g. the scream of a BDA when the exhaust pulses go supersonic) and the size of that pulse, the throb of a Z-06 at idle represents half the exhaust pulses per any specific time period than does the hum of a 355. I will research carefully and post when I'm sure. | ||
| victor v villarreal (Vvvmd)
New member Username: Vvvmd Post Number: 3 Registered: 1-2003 |
Piston speed is a function of stroke and engine rpm. The actual timing of the crank in the engine does not affect piston speed. the reason Ferrari engines use a 180 degree crank is that it makes eshaust tuning easier in V8 engines. At one time NASCAR racers were using 180 degree headers that were very complicated actualy running under the engine to join cylinders from the right side of the engine to the left in V8s with 90 degree cranks. This was to optimise the exhaust pulses to improve exhaust flow out of the engine and increase power. | ||
| PSk (Psk)
Member Username: Psk Post Number: 315 Registered: 11-2002 |
Ken, Yep bad choice ... just randomly selected any manufacturer that came to my head. Possibly all those marques make better engines than Ferraris, I was just trying to point out that they design them for different purposes, er, for family cars. Pete | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 465 Registered: 6-2002 |
Ben Cannon, I concur with your opinion ("Drive the car, please,") but am a confused concerning your comment "Engine's rings seal with pressure from combustion. If they don't see significant throttle opening, they may not function properly at all." I note that in your profile you fly a propeller driven aircraft. Why is it that an engine overhaul must be done every 2,000 (or something close) hours in an aircraft, while an automobile engine does not have to be overhauled for some time beyond that? At an average of 30 miles per hour (for those that have "computers" in their car, check the average mph over a 2-week period - I suspect it will be somewhat less than 30 mph) this equates to 60,000 miles. When I asked this question to Jim Chapman (previously referred to on another thread), an aircraft engineer during and after the 2nd War, his response was that one does not drive an automobile engine at near maximum thrust continuously. Between idling, trailing throttle when decelerating, low-speed motoring (0-60 mph), etc., the automobile engine is loafing. Consequently, wear and tear on rings, cylinder liners, piston thrust services, cam lobes, timing chains, main bearings, etc. are accentuated in an aircraft engine because of the constant load. There is the obvious observation that the consequences of engine failure in an aircraft are qualitatively different than in a car, but nonetheless, the hours chosen before TBO (time before overhaul) reflect a balance of cost and safety. Clearly, aircraft manufacturers specify their TBO based on the nature of thrust used to fly an aircraft (maximum at takeoff - near maximum to cruise). While not a pilot, I slept in a Motel 6. I'm not sure I understand the compression thing, though. Jim S. | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 798 Registered: 10-2001 |
I question if Ferrari engines are 'better' than a Honda engine. Honda engines are very well made to very close, but not racing, tolerences. After all, they don't leak! I say that they're in a different state of tune. Honda engines are tuned to operate at lower RPMS and also have the torque curve at lower RPM's than a Ferrari engine. Same with my Lotus: it doesn't do a lot before 3000 RPM's. But if you tweaked my Accord engine (albeit with computer changes to match cam and timing changes) I would bet you'd get 8000 RPM and even though it's a 4 banger, it could have pretty respectable HP (it's 130 as is). As it is, both are excellent engines designed for two different purposes. | ||
| PSk (Psk)
Member Username: Psk Post Number: 311 Registered: 11-2002 |
EFWUN, Yes, after reading your post twice, yes you are right (but not the comment about twice the piston speed), but the only reason a Ferrari v8 sounds like a high revving 4 cylinder is that we have twice the number of cylinders firing compared to the 4. A Ferrari v8 still has to obey the laws of a 4-stroke engine, and that is to fire every second revolution for each cylinder. Thus the length of the crank stroke is the only thing that affects piston speed ... unless ofcourse the engine breathing is so pathetic that it cannot produce the required combustion to power to those high rpms. Piston speed could be exactly the same as a 4 cylinder and the v8 would still sound like it is revving harder ... but it is actually not. Please remember a Ferrari engine is the same in theory as all the other 4 stroke engines, i.e. Toyotas, Hondas, Chevs and Ford, etc. they just use better components designed for high performance not high life and fuel economy ... no magic involved. Again I stick by my previous comments regarding firing order ... the only difference a 180 degree crank makes. Pete | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 347 Registered: 2-2003 |
PSK, think about what you wrote. The 180 degree crank makes the car sound like two (2) higher revving four cylinders . . . why? Could it be that the motor sounds like that because the piston speed is double??, and because you've actually got two motors with four cylinders each firing every crank revolution?? I'll give you a hint, yes. | ||
| Ken (Allyn)
Member Username: Allyn Post Number: 797 Registered: 10-2001 |
I'm a firm believer that an Italian Tune Up is a very good thing for car and driver both. | ||
| Steven R. Rochlin (Enjoythemusic)
Member Username: Enjoythemusic Post Number: 296 Registered: 4-2002 |
Ben, >>>Drive the cars, please.<<< Amen :-) Enjoy the Drive, Steven R. Rochlin | ||
| Ben Cannon (Artherd)
Junior Member Username: Artherd Post Number: 244 Registered: 6-2002 |
The curve is not only not liniar, but is not always hold true at all. Sure, 7grand will last longer than bouncing off 8750, but running at 2grand all day long can actually DESTROY the motor faster. Engine's rings seal with pressure from combustion. If they don't see significant throttle opening, they may not function properly at all. Drive the cars, please. Best! Ben. | ||
| PSk (Psk)
Member Username: Psk Post Number: 307 Registered: 11-2002 |
Tommy, You might have a point there, regarding the old leaded fuel ... as I have not worked in this industry for a few years and leaded fuel was around when we used to do this. The only counter point to that is that I would have thought that the Alfas, as they were new and young cars, would have run on unleaded fuel ... but cannot remember. Yes it mainly was to clear carbon build up, but other things happen like grit in the fuel and other stuff playing with injectors, and carbs, etc. and also sticking linkages. In the end as somebody else said, are you saving (?, I do not think this is saving the car at all) for the next owner and thus missing out on the fun? I do not thrash or abuse my cars, but I can guarantee that I DRIVE my cars and my running costs are minimal, and I smile alot when driving Pete | ||
| Dr Tommy Cosgrove (Vwalfa4re)
Member Username: Vwalfa4re Post Number: 989 Registered: 5-2001 |
I always thought the old "Italian tune up" only applied to cars that ran on leaded gas. Something about burning off build-up of the lead additive. | ||
| Greg V. (512tr17teeth)
Junior Member Username: 512tr17teeth Post Number: 60 Registered: 3-2003 |
hey PSK, does this rule apply to old saabs, particularly the SPG turbo. I'm just wondering, because I have a high mileaged one. | ||
| PSk (Psk)
Member Username: Psk Post Number: 304 Registered: 11-2002 |
Efwun,
The degree of the crankshaft has nothing what so ever to do with the piston speed. It is simply the stroke length that affects piston speed. The 180 degree crank just affects firing order and thus sound. American and Aussie v8's do not have 180 degree cranks and thus you get that v8 burble sound, while Italian and most race v8 engines have 180 degree cranks as they are stronger and probably light. This makes the engine effectively 2 inline 4 cylinders bolted together and thus they sound like a higher revving 4 cylinder ... and much better to many. A Rover 3.5 V8 with a 180 degree crank sounds absolutely fantastic ... Regarding the original conversation about saving their Ferrari motor by not revving her ... well please remember that the motor WILL go off tune if not revved ocassionally, as it will carbon up and effectively sludge up and then one day you will try and rev it and it will not play the game. This will be because the combustion chamber, ports and valves will have so much carbon hardened and attached that the ports have been reduced and the engine will not be able to breath. If you think that is bullsh!t then go and buy a one owner hardly driven or carefully driven car and open its engine. It will be very caked up ... the most common NORMAL engine job is to decoke it and get rid of all this carbon build-up ... I know I used to do them for my fathers garage ... and we used to give these engines an almighty big thrash to try and clean them out BEFORE we had to open them to save the owner money ... sometimes this worked and you would dump carbon and exhaust crap all over the road, and then the engines would run very, very much better. This is why the Italian tune-up works, and is not a joke. All engines need to be USED properly ocassionally to get good mileage out of them ... thus with an engine like a Ferrari that will go out of tune so easily if you do not give them a good work out once a week then you are asking to end up with a car that spends most of its time getting unnecessary little tune ups every month. Naturally this depends on how much you drive the car, but I would give her a full rev every time you go for a long drive myself. This is the same for Alfa Romeos, especially the 33 Boxers that wifes used to buy (and some men too) to go puttering to the shops with ... used to call my fathers garage every so many months complaining of misfires, etc. What they would have a heart attack about is that the mechanics used to love these cars as they would be given the instruction to go and give them a huge thrash before they would do the tune ... and these owners thought that their cars were pampered all the time ... btw this was at an official Alfa Romeo dealership. So when you take your ever so babied Ferrari in for another tune up, and the mechanics laugh behind your back, because they all know that you do not drive your Ferrari like it is a performance car ... they jump in and thrash your lovely car around their favourite test track before they even think of starting the tune up ... it just has to be done and it is soooo much fun ... otherwise you will be back within a week complaining about misfires, etc. So for gods sack, it is a mechanical machine, use it as intended and if you want to buy a historic musuem piece to cherish and make last forever buy a painting ... not something that has been designed to be used. All that engineering wasted ... might as well have bought a Toyota. Pete | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 462 Registered: 6-2002 |
Dave - yes, it was your coat tails that I jumped on. If it seems that I can see so far, it is because I am standing on the shoulders of giants (Newton, of course). EFWUN, the operative word is "significantly". For those of us who earned our money the "old fashioned way," the formula {fun=revolutions/dollars} is merely an approximation of the general theorem that {fun=miles/dollars}. As in any fraction, the greater the denominator, the less the quotient. My pleasure no longer (important phrase, as it implies that it once did) comes from racing to the mechanic, but is now derived from appreciation of what things can do - they don't have to do it to bring me the pleasure. Knowing that a number of very intelligent Italian engineers spent many hours designing a fine engine, capable of wonderful speed, power, and sound, is sufficient for me. Sure, the occasional casting aside of logic and pursuing with reckless abandon the music and mechanics of fluid sloshing around in my bladder with forces unintended by the Celestial Design Committee is exhilarating, but this is quickly tempered by the sound of coins slipping out of my pocket on to the floor, a metaphor representing far greater visceral pain. No - I appreciate the chassis design and mechanical engineering for what they are. I do not have to beat it to death to enjoy it. When I wear a perpetual calendar mechanical watch that can display day, date, time, moon phase, month, leap year, day of the month, am/pm, and adjust (mechanically) for leap year for the next 400 years, I do not have to live that long to appreciate what some watch maker (engineer) accomplished. No, just knowing that it will work that way is sufficient. Jim S. | ||
| Dave328GTB (Hardtop)
Member Username: Hardtop Post Number: 484 Registered: 1-2002 |
James, Thanks for saying what I was saying with such eloquence. Efwun, Actually I would bet the engine run at 3500 will last more than twice as long as the one run at 7000. However, I'm not about to donate an engine for the the experiment. Dave | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 326 Registered: 2-2003 |
James, the point is not particular to Ferrari engines (other than V-12 with inherently lower piston speeds). The issue is that because of the evolution of piston wall metals, ring metals, and piston materials themselves, wear no longer tracks linearly with piston speed. There are elements of wear (ring vibration, piston "slap" etc) that are just not issues below certain piston speeds. In other words, a motor run nearly continuously at 3,500 rpm will not last twice as long as a motor run at 7,000 rpm, particularly in a car with a rev limit of 8,500rpm. Ferrari has not repealed the laws of physics, and the 355 motor is not a necessarily low piston speed motor (it has, after all, a 180 degree crank). However, running the motor at 4,500rpm on the highway instead of at 3,000rpm won't wear it out significantly faster. If you constantly run at large throttle openings and shift at the redline (e.g., track use) that is entirely different, and not what I believe this fellow was asking. | ||
| James Selevan (Jselevan)
Member Username: Jselevan Post Number: 461 Registered: 6-2002 |
So that those less analytic don't become confused, there seems to be two plots to this thread. One, the original, asked about the consequences of driving frequently at high RPM vis-a-vis engine life. The other, the subplot, espouses the virtue of Ferrari engines, their immunity to high RPMs, the fun of driving and listening to high-reved Ferrari engines, etc. Back to the potential for confusion. Unless the laws of physics have been repealed (by some left wing die-in group), engine life is directly (not mathematically linear) related to piston travel (in feet) per mile of car travel. Less we forget, internal combustion engines, basically air pumps, operate continuously with metal-against-metal. Piston thrust surfaces, rings, cams, shims, bearings, chains, belts, seals, bushings, etc., all have finite lives, and the more they turn and rub, the shorter the life. Thus, please, someone let me drive in 5th gear as early (in the shift sequence) as possible, and as long as possible (without lugging the engine, of course). Let us not confuse fun with longevity. In order of likely failure, connecting rods are WAY down on the list. If a connecting rod fails, there was likely a metallurgical flaw. No, engines more often die a slow and expensive death. Oil leaks from cam-end seals or main crank seals (remember all those unnecessary rotations driving around in 2nd or 3rd gear?). Chains, with all those wonderful moving parts, stretch with consequent cam timing affects and subsequent rough idle. Compression drops because of valves POUNDING a few extra times into valve seats that would have really appreciated a shift to a higher gear. Mitch Alsup gave us an eloquent description of what a valve spring must feel like when getting hit up-side the head with a cam lobe. Compression is also compromised as rings rub, like fingernails on the blackboard, one too many times, and pistons lose their shape from sandpapering themselves against the thrust surface of the cylinder. Let's consider those centrifugal advance weights, pivoting on a brass axle, in those old distributors, and how higher revs slap those weights out as far as they will go. No, brother, give me 5th gear. Oh, but I've forgotten that water pump, and its unnecessary rotations against a seal and bushing, and, oh yes, those throttle plates opening and closing, smooth as they may be aided by tiny little cute ball bearings, or perhaps a simple metal-on-metal bushing. It's 5th gear for me, oh brother, it�s 5th gear for me. Jim S. | ||
| Timothy Guay (Timguay)
New member Username: Timguay Post Number: 8 Registered: 2-2003 |
Ya Baby, Keep it at a c-hair under red while under a full powerslide (with sticky traction), when she hooks up and snaps back to centerline mass you'll drop a couple hundred rpm's, then get ready to do it all over again! God these are the best! The noise (Tubi's), the heart pounding, the Jap crap in the rear view! C'mon you know it's true! By the way, just got new Goodyear Eagle F-1's in 245's. Don't let the rest of us down, put it to the mat, face it and embrace it! Tim. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 433 Registered: 4-2002 |
As long as we are talking about various stressed components in a car; consider valve springs. As long as the follower is touching the cam, the spring is a device with very little movement going on. It goes down and pushes back on the follower, it goes up and lifts the valve. The center of the spring remains 1/2 the distance from the seat to the follower with tiny amounts of vibration. The instant the follower leaves the surface of the cam, the spring reverts back to a harmonic device (binggggg) and starts to vibrate. As the balistic trajectory of the valve reconnects with the cam (thump) a wave of undamped energy is sent down the spring. This wave bounces off the seat and travels back up the spring, where it bounces off the follower, and repeats in an undamped fashion. Just a couple of seconds of valve float, and the springs have self heated to the point where they are loosing tension--forever--toast as they say. | ||
| Ben Cannon (Artherd)
Junior Member Username: Artherd Post Number: 239 Registered: 6-2002 |
Yes, conrods are VERY strong in compression, but not so strong in elongation (where the lower caps are in use and stressed now.) There's a reason an un-muffled F-car's overrun sounds so savage :D) PS: As to running at higher revs, these cars are built for it. Nothing lasts forever in an automobile, rev it, and rebuild/replace the motor later. running 5,000 versus 7,000 on a F355 (rev cutout 8750) isn't going to hurt anything at all. Take each shift to the redline. Enzo would have wanted it that way. Best! Ben. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 310 Registered: 2-2003 |
Mitch is absolutely correct. The moment of highest stress (other than being wildly overrevved) on connecting rods and piston crowns is on the overrun into a corner. Lots of drag racers slam their clutch in the instant they pass the line, because lift-off is the moment when the entire thing's gonna go out the tailpipe. My posts relate to decreasing ring life, guide life and valve life by running at high revs, but Alsup's point is correct. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1012 Registered: 4-2002 |
Mitch: I've always wondered about that. At our local circle track, the motors always let go on decel going into the corner. In our shop, we currently have 2 trucks with blown motors that happened while going downhill at high, but not excessive revs. | ||
| Mitch Alsup (Mitch_alsup)
Member Username: Mitch_alsup Post Number: 430 Registered: 4-2002 |
It might surprise a bunch of people, but the maximum stress on the connecting rod occurs when the engine is at red line (or other max RPM) and the throttle is closed! And this stress occurs on the transition between the exhaust stroke and the intake stroke--not the power stroke. On the compression stroke, as the piston decelerates towards TDC, the compressed mixture pushes on the piston and the connecting rod, reducing the tension in the connecting rod. On the exhaust stroke as the piston decelerates toward TDC and the throttle closed, the vacuum in the intake pulls on the piston and increases the tension in the connecting rod. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 308 Registered: 2-2003 |
David, when you get paid as little as we did to drive those things, every little bit helped!! Also, knowing the car and the motor inside out was helpful in understanding where to find tenths. Helping the mechanics made them that much more willing to work late for you when you shunted! It was not an always glamorous life; even GV got his hands dirty occassionally. Basically the only guys who flew in and drove were the richie-richs like KC, JW et al. Talk soon. | ||
| David Stoeppelwerth (Racerdj)
Junior Member Username: Racerdj Post Number: 75 Registered: 1-2003 |
Efwun-Please keep all those stories coming!! I doubt that he is intending to running in a narrow RPM band that we did (7500-9500).Your command of the mechanics of the Formula Atlantics far exceeds mine. We just got use to letting the crew's earn their wages. WE just Flew in and drove-The most fun IMO. | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 306 Registered: 2-2003 |
David, I understand what you're saying, and you're pretty much correct. I just meant, I don't think this guy, driving around a gear or so lower than necessary, is using his motor on the same destructive exponential equation as what Jennings was describing. (P.S., our piston to wall tolerances were larger (8 thou or so) than a street motor, at least then) Also, as you may know, those older BDA's ran at 13+ to 1 compression ratios, and had O-ringed heads to try to keep all that banging inside where it belonged!! They were pretty reliable, but after 3-4 hours, you'd be 300rpm down and yelling for a change!! | ||
| Steven R. Rochlin (Enjoythemusic)
Member Username: Enjoythemusic Post Number: 286 Registered: 4-2002 |
David, The reminds me of my mechanical watch friends... If you want the unit to NEVER degrade per se, NEVER use it and keep it in a hermetically sealed enclosure. Things may degrade in that environment to some extent though. Naturally how you treat/maintain/use/lube it will dictate the lifespan of a mechanical device. Of course it would be rather hard to *truly enjoy* a mechanical watch or car that is hermetically sealed IMHO. Spin the fluids, have fun :-) Enjoy the Drive, Steven R. Rochlin | ||
| David Stoeppelwerth (Racerdj)
Junior Member Username: Racerdj Post Number: 73 Registered: 1-2003 |
The only point I was trying to make was the higher the RPM'S will equate to some degree loss of engine life thru engine wear or heat. I was not inferring that the RPM'S example of the Cosworth's back then equates to today's Ferrari engine life. The motors from Jennings did not have to be broken due to the close tolerances he used in the rebuild's. I would like to think Ferrari motors are bullit proof today. According to my dealership, occasionally the current motors sometimes go bad. | ||
| Hans E. Hansen (4re_gt4)
Intermediate Member Username: 4re_gt4 Post Number: 1006 Registered: 4-2002 |
EFWUN may have hit it on the head, in that the original question may have been about cruising with the engine 'singing' a bit, but not totally maxed out. The old 308 V8s don't have enough power to pull a greased string out of a hen's ass below 3000rpm, and barely enough to do so at 4000. So most of us drive around town with the tach parked at roughly 3000 as a matter of course. I asked my mechanic at what rpm to drive it and he said, "You do what you want, but I drive them at 4000 because I want something to happen when I step on the gas pedal." (That was Carlo Durante, just in case some of you Seattle area folks are wondering.) | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 97 Registered: 1-2003 |
I sincerely thank all that replied to this thread. A lot of great expertise given on this subject. I now feel very comfortable at pushing the RPM's higher than I normally would, but still being under what is perfectly tolerable. Cheers, Peter | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 302 Registered: 2-2003 |
I'm pretty sure that when this guy is talking about running at high rpms as a choice, he's talking about leaving it a gear low, e.g., cruise at higher rpm than needed. You're right about running at high rpm, particularly under the load of large throttle openings. My only point was that the rev limit in something like a 355 is set somewhat conservatively, and running around at 6,000 versus 4,000 isn't going to halve engine life. I agree that it will reduce engine life somewhat. | ||
| Dave328GTB (Hardtop)
Member Username: Hardtop Post Number: 482 Registered: 1-2002 |
All other factors being equal, engine life will be determined by loads on the internals and engine hours. Running at high R's increases both. Yes, Ferrari motors are designed to take high R's with short strokes, etc. Nevertheless, any motor that is run at high R's all the time won't last as long. Hard running also increases heat as any track runner knows and that ain't good for exteneded life either. Dave | ||
| EFWUN (Efwun)
Member Username: Efwun Post Number: 299 Registered: 2-2003 |
David, I'd take a mild exception to your comments. When we spun the BDAs to 10,000 or so (only Rosberg ran 10,500, and he regularly holed blocks!!), we were using comparatively ancient technology as to rings and materials in general. The problem is that increasing piston speed dramatically increased wear. (and one weekend was GREAT if you could get the motor to last that long!! Some guys got 4 hours!!) A modern Ferrari motor, (I've gotta believe) uses far more sophisticated ring materials as well as silumin liners. I'm not sure that increased wear with increased RPM necessarily follow the same disastrous equation we used to experience. Finally, I'm pretty sure that the piston speeds in the BDAs were significantly farther into the red zone for the materials of the day, than the same piston speeds in an V-8 Ferrari with 180 degree crank are to today's materials. In other words, 8,500rpm in a 355 doesn't stretch the envelope like 10K on a BDA; the comparison might be more apt if you spun the 355 to 10K also (except, uh, BANG!!). | ||
| David Stoeppelwerth (Racerdj)
Junior Member Username: Racerdj Post Number: 70 Registered: 1-2003 |
When we were racing Vintage Formula Atlantics with BBD Cosworth engines, Steve Jennings (our motor rebuilder in California) gave up these choices. Run at 9000 rpms and send me your motor after 1500 miles, 9500 rpms 1200 miles, 9800 rpms 1000 miles and then up to 10500 rpms (where the Pros back in the 1970's were running them and the motor needed to be rebuilt after every weekend! I think one might start to see a pattern developing here!! | ||
| Eric (Vette)
New member Username: Vette Post Number: 17 Registered: 3-2003 |
..F-cars are creatures that live at higher rpms. fyi, when you take your car to the dyno, you dont measure horsepower, rather you measure torque. horsepower is actually derived from an equation, which is torque divided by 5,250 times rpm. because of that, in general, as long as your car's engine is built well enough to take the high rpms(yours is..) and can feed the engine properly at high rpms (again, yours is.. its designed to be) then the higher the rpm your car will turn the more power the car will make. an example would be a F-1 car that turns 16k rpm. anyway, your F-car was designed to be a track car.. in being so, it has a fairly large powerband located high in the rpm range.all thats pretty much common knowledge. from what ive learned, yes, rpm does wear your engine faster than lower rpm will. but what the hell.. its a ferrari! thats how its supposed to be driven! | ||
| Ming Cheng (Onlinesys)
Member Username: Onlinesys Post Number: 260 Registered: 5-2002 |
I don't think you could degrade or ruin the engine life as long as the engine is warmed up with the proper engine oil. I was told that you should drive the F-car at high RMP from time to time to make the engine healthy as it was meant to build for. | ||
| Peter Sedlak (Peters)
Junior Member Username: Peters Post Number: 96 Registered: 1-2003 |
You read chat about pushing the redline while driving FCars. If one drives normally at high RMP, would this not degrade engine life (take my 308 as an example car)? Is there a benefit to the engine while driving at higher RPM's or is it just the 'feel' of the engine at higher turns that makes us feel good when driving? |