QUERY FOR PROPELLER HEADS - Driveline Loss - What If?

[Torquemonster]

I've never quite been able to get a handle on how to calculate drivetrain losses because it seems just as soon as you adopt a formula - exceptions to it become the rule.

Most of us seem to adopt a simple percentage loss to calulate flywheel hp from rwhp or vice verser.

The percentages used vary from 20% loss for auto and AWD cars to 15% for manual trans 2 wheel drive cars - but there seems to be regular examples where Vipers may have less drivetrain loss - yet the heavy duty nature of the driveline would appear to make that illogical.

Therefore - here's a theory - I call it the "Torquemonster Had An Idea While Eating a Ginger Crunch" theory - because that's what I was doing while thinking (a dangerous pass time I know)...

What if driveline losses were a fixed power loss? That would explain why a fixed percentage does not always fit - as the percentage would be constantly changing with power variations.

This works for other power driven accessories such as alternators and A/C pumps - that take say 5hp or 15hp to drive at say 4000rpm regardless of engine power.

e.g. Say it takes a constant of 100hp to drive the Viper driveline at say 5800rpm engine speed. Therefore if the Viper made 300 flywheel hp it would put out 200rwhp, at 400 it would make 300rwhp, at 500rwhp it would make 400rwhp and so on. Therefore to gross up your rwhp dyno figures - you'd simply add 100hp.

OR - maybe there is a constant power loss - PLUS also an increasing element as more power is applied due to other losses - which I can't think of . BUT - if true there would be a fixed loss (say 60hp) plus a small % loss in addition as hp exceeds a base level.

THOUGHTS from those brighter than me? I'm drowning here out of my depth in the deep waters of my imcompetence and seek enlightment so I can resurface with my ginger crunch intact


Please "Help me!" cried the fly with the mans head.

 

[HOdbleFman]

Torquemonster, I'm not an expert on cars and I'm sure you know more about it than myself, but from solid mechanics I can tell you that your theory of increasing losses due to more power application is on the right track. While these losses may increase very little they can be calculated. Obviously as you apply more power through the drivetrain (gears and driveshaft) more stress is developed in the parts. This stress equates to increased deformation(very small) in the components (i.e. twist in the driveshaft) that ultimately results in energy release by way of heat. Since heat is power being emitted to the surroundings, you are in essence losing horsepower. Disclaimer: I do not know how substantial the amount of power that is lost this way. As for the rest of your question, I have no idea.

 

[Torquemonster]

Thanks - it makes sense. The problem with quantifying it all - how much is fixed loss and how much is proportional to power - is because you'd need a whole series of cars to be both dyno'd on the engines and then dyno'd at the wheels all on different power levels. In practice this does not happen very often - most customers choose only one form of dyno - and then there is the issue of variations between dynos - hence the difficulty.

I'll wager a total guess at it for the Viper:

60hp loss fixed plus 8% of gross hp.

Let's run that thru a few examples and see what we get....

1) 500hp at crank less 60hp fixed loss less 8% variable loss (8% of gross) = 400rwhp

2) 670hp at wheels + 60hp fixed loss then gross up by dividing sub total by 0.92 (100%-8%) = 793hp at crank

3) 850rwhp + 60hp fixed/0.92 = 990hp at crank

4) Paolo's 888rwhp corrected + 60hp/0.92 = 1030hp at crank

5) 650hp at crank - 60hp fixed less 8% of gross = 538rwhp

Now lets see how much the drivetrain loss is in each of the above theoretical cases

1) 100hp loss/500 = 20%
2) 123hp loss/793 = 15.5%
3) 140hp loss/990 = 14%
4) 142hp loss/1030 = 13.8%
5) 112hp loss/650 = 17%

Anyone think this may be a better approach than a flat % ?

- as you can see the % changes as power changes IF the ginger crunch hunch is correct

 

[Jack B]

Carcraft this month (I think) just had an article on rear wheel hp versus flywheel hp. They did actual tests on four cars On the manual transmission cars the losses were in the 17-25% range. That seems high for us since we have assumed all along that our losses were 15%. None of the cars had half-shafts either.

 

[RedGTS]

OR - maybe there is a constant power loss - PLUS also an increasing element as more power is applied due to other losses - which I can't think of . BUT - if true there would be a fixed loss (say 60hp) plus a small % loss in addition as hp exceeds a base level.

Bingo! If a car has a measured 15% driveline loss at 500 hp it will NOT have a 15% driveline loss at 800 hp--assuming rotational speeds remain the same (no gear change etc.), it will be very close to the fixed number of 75 hp measured at 500 hp, plus a small increase for the greater frictional losses associated with the extra hp.

 

[phiebert]

You are absolutely right from a physics point of view. Well maybe not on your chosen numbers but definitely on your approach.

But the change in your variable component is more related to the speed at which you turn the driveline (enertia required and friction goes up) and less to do with the amount of power you are applying to it (torque twist heat). If you are calculating the amount of loss due to the twist of your driveline you should work at NASA, not on a Viper.

So your percentage of driveline loss goes up more with RPM then with the amount of power you apply to the driveline. So I think you are better of using a smaller variable percentage. It also about acceleration. With a heavily slipped clutch, a lawn mower engine can turn your driveline at low speeds but if you want to accelerate your driveline quickly (like on a dyno) it takes a much larger amount of horsepower for enertia.

The percentage that people use on smaller horsepower cars works well because when you are talking about 200 or 250 RWHP cars a 15% number equals 30 HP or 37.5 HP, a difference of 7.5 HP (negligible). But when you apply that to 450 or 1000 HP cars you get 67HP and 150HP. That's ridiculous! Think about it. If all you did was put a blower on a Viper and didn't change the driveline at all, it isn't going to take any more horsepower to turn it except for the twist (negligible) and the enertia (slight). So you might take 2 more horse power when you apply 1000 RWHP than when you apply 450 RWHP. So using a fixed number is much more accurate than a percentage. Of course, all this changes if you are changing the flywheel and back, which changes the amount of HP required to turn it.

 

[Torquemonster]

Bingo! If a car has a measured 15% driveline loss at 500 hp it will NOT have a 15% driveline loss at 800 hp--assuming rotational speeds remain the same (no gear change etc.), it will be very close to the fixed number of 75 hp measured at 500 hp, plus a small increase for the greater frictional losses associated with the extra hp.

Cool - I wonder if anyone can come up with a fixed loss and a subsequent % loss that will be more accurate than my 60hp plus 8% guide?

 

[Miles B]

The part that is the percentage is the inertial losses, which I would estimate to be a large portion of the loss.

Think about it. If you are losing 100hp in your drivetrain, and most of that was going as heat, that is a HELL of a lot of heat. Your tranny would glow red and melt within a minute. Most is lost the same way you "lose" power into a flywheel. This will tend to be a constant percentage loss as well. The quicker you accelerate a flywheel, the more power it absorbs. The wheels, shafts, diff and tranny are all being spun up.

Because of the complexity of the system, to model it accurately you would need many factors.. some constant, some proportional to speed, some proportional to acceleration, some proportional to torque.. etc. In the end, I'd just rely on a few "known" examples, but to "know" them, you're going to have to pull the engine and put it on an engine dyno.

 

[joe117]

There must be a bunch of engine only dyno sheets around. Didn't Arrow do dyno runs of stock engines? Just compare the engine only with a drive on dyno.

I think the fixed percentage lost to driveline is just a number thrown out as a convieient, ballpark, conversion. I don't believe it is real science.

By the way, One time, I figured the heat that would be generated by the hp lost in the drivetrain. I can't remember the number right now but it's a bunch. I'm not sure I believe that much heat get's generated.

 

[Ron]

Gross > Net > RW

Since all that matters is rear wheel horsepower and torque, wouldn't it be nice if that was all the manufactures quoted. Easy to verify, easy to compare and easy to get credit for a more efficient driveline design.

 

[Larry Macedo]

Here is my take on it, which is quite simple and the percentile is thrown out the window.

Lets say your car were to dyno 465bhp and 415rwhp. The driveline loss is 50hp. What actually happens is somewhere along the translation the number becomes a percentage of loss. In this case, we'll call it 12%. What is interesting, when that percentage is used, the more power you make, the higher the driveline loss. So, lets say your making 600bhp. If you were to apply the 12% rule, it comes ot to 535.71rwhp, or a loss of 64.29hp when in all actuality it should be 550rwhp.

I guess what I'm trying to get across is that driveline loss doesn't increase when you increase your horsepower, unless you've changed gearing or driveline components.

 

[Torquemonster]

driveline loss doesn't increase when you increase your horsepower, unless you've changed gearing or driveline components.

That's what I was wondering - but Miles point about it taking more power to accelerate the drivetrain quicker kills that theory. His point is valid - it must take more power to spin up the drivetrain as fast as 1000hp could do it than it would to spin it up as fast as 450hp could.... therefore there must be some fixed hp loss PLUS an additional loss proportional to hp made...

The pattern appears clear however - the overall % lost reduces as hp increases past a certain point. Does anyone disagree with that?

 

[Larry Macedo]

The pattern appears clear however - the overall % lost reduces as hp increases past a certain point. Does anyone disagree with that?

I would totally agree with that scenario before I'd agree the % increases, Torque!

 

[Miles B]

Heh heh, if it was fixed at 50hp, the Mustang would go backwards

 

[old96er]

Some more food for thought...
It seems to me that basically the elements of driveline loss are rotational mass and friction. Friction increases proportionally to speed. Ever wonder why racers use diff coolers? All that heat of squeezing the fluid is lost power, and the faster that rear gear spins the more power you lose. Rotational mass will have the most effect during accelerating, thus the difference people feel with an aluminum flywheel. Maybe someone could switch to lighter weight synthetic lube for diff and tranny, get some carbon fiber shafts, and an aluminum flywheel. I would like to see some dyno results on that. One last comment, even though we can't change it, pinion depth effects efficiency and strength, this is why a 12bolt is more efficient but a 9inch is stronger even though the ring gear size is within a 1/4inch. Our dana super 44 is closer to the 12bolt design.

 

[Newport Viper]

Before people start modding the SRT-10. How about getting a few stock cars together at the same dyno, at the same time, to get a baseline percentage? Science no, but it is the real world.

 

[Ratech]

old96er
One last comment, even though we can't change it, pinion depth effects efficiency and strength, this is why a 12bolt is more efficient but a 9inch is stronger even though the ring gear size is within a 1/4inch. Our dana super 44 is closer to the 12bolt design.


It is not pinion depth it is pinion offset. Pinion offset for 12 bolt, Dana 44, etc. is 1.5", 9" Ford is 2.25. This changes the helix angle from 50 to 60 degrees. That causes slightly more sliding action accross the tooth face. But it also increases the contact ratio of the pinion and ring gear teeth. Frictional losses are continually increasing until the point of tooth failure. I spent 14 years as a R & D field rep. working with race teams for a US. gear manufacturer. I had engineering do a study of frictional losses. It its possible to calculate but the gathering of variables is time consuming. One last note to think about. The clutch has frictional losses also. It never fully locks up. I obtained frictional data from Borg Warner to get the effience number.
I did alot of oil weight analysis and found only a 2% difference in numbers. There are other factors on the car that have a much greater impact on friction. Tire pressure, air quality changes during the race, wheel stager, the list goes on and on.

 

[treynor]

Ah, the timeless question. A few important points --

Physics tells us that the force of friction is equal to the product coefficient of friction multiplied by the normal force between the surfaces -- that is, F=kN. In simple terms, the more the two surfaces resist movement, and the more you press on them, the more frictional resistance you get. Multiply this by speed (how fast are you moving these "sticky" surfaces) and you have a power figure.

So...
> If all you did was put a blower on a Viper and didn't
> change the driveline at all, it isn't going to take any
> more horsepower to turn it

That turns out not to be the case. A higher HP car will exert more force on the driveline components, thus causing higher friction. A simple counterexample will suffice: if you put your Viper on jackstands and put the tranny in neutral, you can turn the rear wheels by hand. You cannot produce 50 HP with your bare hands (well, unless you're Gerald ). You're able to move the driveline because the force exerted is quite small, and thus the frictional losses are quite small. So it cannot be that the driveline "absorbs" a fixed amount of HP.

As Ratech alludes, the complete answer is undoubtedly nontrivial. However, a good first-order approximation would be:

Driveline losses = intertia + friction.

Inertia is power lost to spinning up the driveline. This includes everything from flywheel to drive wheels. It is constant for a given speed change (e.g. 60 - 150 MPH on the dyno) on a given car. For the purposes of a dyno pull, this could be your elusive "constant factor"

Friction is power lost to heat. It is proportional to speed (in this case, rotational velocity) and force (in this case, torque). Classic physics says this should be a constant percentage of the total; in the real world it is undoubtedly more complex.

 

[joe117]

The rear axle oil get's very hot. Even on a street car.
The quantity of hp needed to produce this heat isn't hard to calculate.

Perhaps someone could make an estimation of how much heat is lost to the air and from that an estimate of hp loss in the rear axle assembly could be made.

To do this, a static air cool down measurement could be made, x number of degrees/min, and then a large fan could be directed on the assembly to see the change in cool down time with x mph wind over it.

This would give a data point that is probably linear. A graph could be drawn.

If we were to monitor the axle oil temp at 60 mph on a flat road, we might find that it stayed at 220 degrees. If we knew the rate that the assembly lost heat with a 60 mph wind blowing on it, we would then know how much hp it takes to keep it at a given temp.

It takes a quantity of hp to keep the oil at a certain temp given the fact that a certain amount of cooling is constantly taking place.
Whatever that given hp value is, that is the loss due to friction under those driving conditions.

Anyone have a temperature gauge monitoring their axle oil?

 

[Miles B]

Ben,
little correction here I think (but it's 3:30am here so I could be speaking out of my a$$, will have to check in the morning - the text, not my a$$). Anyway.

"It is constant for a given speed change (e.g. 60 - 150 MPH on the dyno) on a given car"

The amount of energy stored is constant for a given speed change. Rotational energy (if my brain works) is
K = 1/2 I w^2
Where I is the moment of inertia and w is the angular velocity. So for a given velocity change, yes we will have a constant change in energy. However, a more powerful engine does this velocity change in a shorter time. Same energy.. shorter time.. greater *power* absorbed by drivetrain.

The amount of power lost to providing the drivetrain with rotational kinetic energy is NOT constant relative to input power. The amount of energy is.

 

[Miles B]

Also, just a little "rule of thumb" you sometimes see in mech eng texts.. gear, belt and chain drives are REALLY efficient (well over 95%) at transmitting torque without producing heat. Heat is not where most of it is going. If you think you are losing 100hp in frictional heat, you aren't. That is a HELL of a lot of heat. Probably about the same amount of heat a Z06's engine has to get rid of through its radiator at WOT, and I don't see a water pump and big radiator in the diff and tranny.

 

[phiebert]

So...
> If all you did was put a blower on a Viper and didn't
> change the driveline at all, it isn't going to take any
> more horsepower to turn it

That turns out not to be the case. A higher HP car will exert more force on the driveline components, thus causing higher friction. A simple counterexample will suffice: if you put your Viper on jackstands and put the tranny in neutral, you can turn the rear wheels by hand. You cannot produce 50 HP with your bare hands (well, unless you're Gerald ). You're able to move the driveline because the force exerted is quite small, and thus the frictional losses are quite small. So it cannot be that the driveline "absorbs" a fixed amount of HP.

You took my comment out of context. I also said a lawnmower engine could turn your driveline. My point was that if you spin up the driveline at the same acceleration to the same speed with a 450 HP engine or a 1000 HP engine, you are using the same amount of force (losing the same amount of HP) to do so. Because the 1000 HP car can spin it up faster and because of that has to work against a greater force of enertia and friction, it will lose a little more RWHP then the 450 HP engine. But not at a constant percentage, which would mean that it takes an extra 100 HP to turn the driveline that much faster. I think that is unlikely. 20 HP maybe, but 100 HP?

 

[Miles B]

Hmm.. will it be a constant percentage.. I'm gonna have to get the old pen and paper out tomorrow morning.

One last note. You would be surprised the HUGE amount of energy that can be stored in a rotating assembly.. 50-75 pounds of shafts and gears in the tranny, driveshaft ain't light, diff gears and axles and wheels.. that's a big heavy flywheel, and believe me, the amount of power it takes to spin that up quickly would be large. I was amazed when I did the math on changing to a lighter tyre, wheel and chain on a motorcycle... the numbers were large.

 

[joe117]

Before I retired I did some work at an antenna range. The antennas were for jet aircraft and one of the sidelines was testing the protective coating on the antenna blades.

They had what they called a "rain erosion chamber"

It was enclosed in a large, round steel thing made from the end caps of a boiler. It looked like a flying saucer. It was about 15 ft in diameter.

Inside was a steel arm about 7 feet long. It would spin round and round with a small sample of the material under test mounted on the end.

They would get the thing spinning and drip controlled water drops on it to simulate the jet flying through rain.

They could get the streamlined, 70lb arm spinning so that the end was not quite mach 2.

The whole point of this is that the power source was an industrial Ford V8. It was over 500 cubes. I think it must have had 300hp.

Anyway the limiting factor was that the engine ran out of hp just before mach 2.

I thought this might be interesting. Does it have anything to do with the discussion? I don't know.

By the way, the enclosure was only about 1/2 inch steel. I always told them that if the arm ever came off, it might very well go through the enclosure and still travel a very long way.
I had figured that the energy in the 70lb arm at mach 1.9 might be about what an M1 tank gun projectile has.

 

[treynor]

Miles - you are quite correct, the energy absorbed is constant, but the power of course is not.

Phiebert - actually, what I am saying is that it IS a constant percentage for frictional losses. A 1000 HP engine will (ceteris parabis) be exerting twice the force of a 500 HP engine at the same RPM, and thus experiencing twice the frictional loss.

 

[Larry Macedo]

Miles - you are quite correct, the energy absorbed is constant, but the power of course is not.

Phiebert - actually, what I am saying is that it IS a constant percentage for frictional losses. A 1000 HP engine will (ceteris parabis) be exerting twice the force of a 500 HP engine at the same RPM, and thus experiencing twice the frictional loss.

I would have to disagree with you, Ben. What your saying is that a guy twice my size is exerting twice as much force to accomplish the same job? If anything, the task would be easier for him, thus he would expell less energy.

I guess the only way to prove this is take a 850rwhp Viper motor and strap it to an engine dyno.

 

[Torquemonster]

Wow - there are some really good posts - thanks for all your inputs.... however we are left speculating because it seems the automotive world has not got a formula that will work. Perhaps this is because it will be different for each vehicle and different for every change in driveline mass.

It will logicall take a fixed amount of power to turn the driveline at any set speed. To turn it at all could be done by hand - but to turn it at 1000rpm will require several hp and so on.

Beyond that - it will require more power to accelerate it faster to any given speed...

so - there is a basic power level needed to achieve a certain speed, and a power required to accelerate to that speed in a given time.... both factors must affect losses between engine and tires.

Let's bring it all back to the Viper. We know that a fixed flat percentage will become inaccurate as power level increases - so guessing a flat figure is not that accurate.

I proposed a flat 60hp plus 8% as producing results that seemed quite realistic....

What do you guys see as being a better "guestimate" than the 60hp + 8% - just give your best guess - to be accurate is too complex.

 

[treynor]

> What your saying is that a guy twice my size is exerting twice as much force to accomplish the same job?

... in half the time, yes. Some of this is not a matter for debate -- the laws of physics do not make an exception for vehicle drivetrains. Twice the (normal) force at the same (rotational) speed = twice the power lost to friction.

Let me provide another analogy. Suppose you have a large block of metal sitting on a concrete floor. Suppose further that you have two BeefyBoys(tm) who are competing to push the block across the room as fast as possible. They are both strong enough to get it moving, but one of them is twice as strong as the other. In fact:

Boy #1 can push the block at 5 MPH
Boy #2 can push the block at 10 MPH.

Ignoring momentum for the moment, it stands to reason that the total amount of ENERGY expended to push the block across the room is the same. After all, the block weighs the same no matter who is pushing it, and the coefficient of friction between the block and floor remains the same. However, the POWER consumed during the contest will be quite different - in fact, it will be twice as high for Boy #2, and the duration will be half as long.

If there are no arguments so far, we can then extend the analogy to match the original question...

 

[Torquemonster]

I agree with what you say Ben in princple - although the extra power to move those blocks may not be linear - it could be geometric - like a displacement hull of a ship where it takes increasingly more hp to move it faster than design speed.

 

[phiebert]

> What your saying is that a guy twice my size is exerting twice as much force to accomplish the same job?

... in half the time, yes. Some of this is not a matter for debate -- the laws of physics do not make an exception for vehicle drivetrains. Twice the (normal) force at the same (rotational) speed = twice the power lost to friction.

Two arguments this time...

1. It's not done in half the time. A 1000 HP car doesn't finish the dyno run in half the time a 500 HP car does. It probably finishes in 10 to 20% less time.
2. Friction isn't the only issue. Enertia to spin up the driveline is there too. It takes more power to spin it up faster admittedly, but I really don't think it spins it up in half the time, thereby requiring twice the force.

So my two cents, bringing this back to the Viper is that the fixed amount should be 40HP and the variable around 3%. That makes it 49HP in driveline loss on a 300 HP car. Since the formula to use a number like 15% all the time was developed for cars in the 200 to 400HP range it made sense for them. But when you apply it to a 1000HP car, you get 70HP to spin the driveline, not 150HP (using 15%). I think 70HP is more reasonable.