drivetrain losses as a percentage???????

It sounds like another example of a guy that knows what he knows, and thinks he knows a lot more that he actually doesn't.

Did you take high school physics? Do you remember friction? The basic equation there is:
F = mu * N
Where F is the force to overcome friction (how much is lost)
mu is the coefficient of friction (this is a constant for rubbing two surfaces together based on what they are; i.e. ice on ice or a cardboard box on concrete will have two very different values)
and N is the normal force pushing the two surfaces together. The ice cube sitting on the hockey rink will have a low normal force, whereas the refrigerator in the cardboard box will have a high one. In those two examples, it will be the weight of the object.

In the case of your drivetrain, you've got the same two gears with the same lubricant between them, so the mu will be essentially the same. However, with a lot more torque going through them, the normal force will be much higher, by the percentage increase in torque! Thus the force lost to friction will be higher by that percentage.

If you didn't take, or don't remember a thing from high school physics, then all of the shortcuts I took will make me look bad to the brains on the forum, but I hope that was simple enough for the non-technical to relate to.

 

It all depends on how you are MEASURING "horsepower".

If you measure at a steady state - holding the engine at an RPM - and measure the torque
using a load cell or cells - the losses are in one form.

If you measure as the engine accelerates - the rotational inertia of the drivetrain induces
"losses" as a percentage of the power transmitted.

Simply because - given a "given" drivetrain - with a "given" inertia - more power will spin the
drivetrain faster - which inertia will oppose more.

To distill - most common chassis dynos are of the accelerative type.

They allow the engine to spin up either at a rate determined by the mass/inertia of a big
drum (dynojet) or determined by the resistance of a retarder (mustang/etc).

These dynos will show losses as a percentage of the engine power.

(well, technically - there are two components - one fixed, and one percentage)

Jim

 

^^^ Good post.

 

All depends on how many twists and turns the power is making on its way to the wheels, and how many accessory drives there are. Frictional power lossess will be less if there's just a flywheel output leading to a couple gears, R&P, couple CVs and one camshaft. Now, if you go adding 3X8 drop gears, more take offs for distributors/lots of accessories, 4 camshafts, etc....well, you you see where that's going....

 

The accepted number for A US car with a front engine and 4 spd manual trans in 4th gear (traditionally this locks the intup and output shafts together) is 15%.

The accepted number for 308/328 with its extra gears in the drop gear case and input/output shafts that do not lock together is 17%-18%. This comes from some testing done at Norwood years ago I think. Other ferrari models would be in between these 2 numbers.

 

I am surprised by these numbers. I always thought that the hypoid gears in a traditional differential were extremely inefficient. That is one of the reasons that front wheel drive is an immediate gas mileage improvement: no 90 degree turn for the power.

Anybody know why the Norwood numbers don't agree with this?

 

Nope and no milage improvement for FWD. Front drive works well on small light cars to keep weight on the drive wheels and it's also overall cheaper to build which also fits small light car design goals normally.

 

In a typical American car, 4th gear is 1:1 and does not even use a gear. So all the transmission has to do is hold the input and output shafts in a line. Here, there is one gear in the drive line (the differential gear).

In a F345/F355 there are a minimum of 4 gears in the driveline {Drop gear, 90 degree gear, ratio gear, output gear}. This is a tradeoff, it trades off energy for center of gravity. The energy transfer is not as good as the American arrangement, but the ability to drop the engine to the limits of the dry sump helps the CoG.

For example, the Vette LS7 engine has a dry sump, but the engine is not lowered in the car because the flywheel remains as large as in the non dry sumped models.

To certain extent, the argument goes somethign like
A) the direct 4th gear make the car look good on the dyno
B) the lower CoG makes the car look good on the track.

Ferrari has been doing this since about 365 GTB time. So, you will never get the low driveline loss of the american muscle cars (and live axles are better than independent suspensions too) and look good on the dyno; you will look good on the track where CoG really comes into play.

 

As far as losses, there are two ways to determine them.

(We are assuming here, that the "type" of chassis dyno is fixed)

1) Remove the engine, dyno it using a proper engine dyno - replace engine in chassis
and dyno - calculate losses.

2) Dyno a number of STOCK vehicles of the same type, and using the manufacturers published
horsepower numbers, calculate the "loss" between the average of the measured RWHP and
the published HP numbers.

#1 is more accurate than #2, but #2 usually suffices.

Remember: LOSS (%) = 100 * (1-(wheel/engine))

 

A) I suspect that Ferrari measures its HP on a brake dyno with no acceleration going on. This gives the highest possible number (for marketing purposes). You may take this as being optimistic wrt HP.

Also note since Ferrari engines have higher RPM limits, this necessarily puts more rotational inertia in play while accelerating than in comparison to engines with lesser RPM limits. So while accelerating, the Ferrari engines look somewhat proportionally worse. This is equivalent to the "you can't make up for lack of TQ with higher RPMs at equal HP by throwing gears in the transmission" argument.

B) I think a tuner or dyno shop should just let the dyno numbers and graphs do the talking by themselves. Informing the customers as to procedures, limits of accuracy, and any strange weather (or gasolinie,...) than may be interacting the the dyno runs performed.

 

For example, the Vette LS7 engine has a dry sump, but the engine is not lowered in the car because the flywheel remains as large as in the non dry sumped models."

Wrong they use a torque tube straight to trans, this whole system creates lower CG.

"

 

Yes they use a touqure tube, but that has basically no affect on CG.

 

Didnt lower the engine, have a look.

Mitch, even if 4th you are still rotating every gear through the thick gear oil in the musclecar box.
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Agreed, but it is the gears with power crossing over in the teeth that create the majority of the drag induced losses, the others are just frothing up the oil.

But, at operating temperature, the oil in the tranny is no thicker than the oil in the crank case. They both operate around 10 cSt. That is, the 75W90 gear oil has the same kinematic viscosity as the 10W-40 engine oil when both are around 200dF; and both are near 10cSt.

 

What Mitch said.
Direct drive is important for efficiency.
Interestingly, it's disappearing from transmissions. Not used that much on the EPA test, I suppose!

 

And speaking of rotational inertia, the power that is needed to increase the rotational kinetic energy of all that is spinning in a given amount of time increases with the square of the rotational speed.

 

Which is one reason bigger engine displacement in a smaller package is so advantageous (i.e. Vette LS7). The 2K RPM advantage of (say) F458 (9K RPMs) versus LS& at 7K RPMs comes with a 65% rotational inertia disadvantage--which is also why Ferrari has to work so hard on lightening all the little driveline pieces.....)