# Drivetrain mass increase skews torque output?

1. Jan 14, 2007

### Sleeper

I have a car with a switchable transfercase that will allow me to choose between running FWD and AWD. I'll be putting the car on the dyno shortly, and I was just theorizing drivetrain output differences. Obviously HP numbers should drop roughly 15% because of the increase of mass on the drivetrain going to AWD, but I was wondering how torque output would be affected? I cannot see rotational force being affected greatly? Though it must since HP will be down roughly 10-15%, torque should be the same right?

2. Jan 15, 2007

### brewnog

Firstly (practically), you might have trouble finding a 4WD rolling road!

Secondly, you need to be very clear about what you define as "horsepower" and "torque". A typical rolling road will measure torque and speed at the wheels, and use a crude calculation (based on a pure guess at drivetrain efficiency) to estimate the engine torque and power.

Now. Your horsepower figures at the wheels for 4WD will not drop as you say, due to an increase of drivetrain mass. They'll drop due to a reduction in transmission efficiency, because of the need for more gears, bearings, lubrication etc. Since, at a given speed, power is directly proportional to torque, your torque at the wheels will drop by the same factor. (In fact, the resulting reduction in torque will limit the top speed, giving you your power decrease.)

Bear in mind though, the maximum torque and power your engine will develop is regardless of the number of driven wheels.

3. Jan 15, 2007

### Sleeper

Thanks for the reply. So the transmission efficiency drops when a greater load is placed on it, i.e. more drivetrain mass? The transmission uses the same gears regardless and spins the output shaft as normal, just the transfercase now spins because of a locking pin being engaged, which controls whether or not the driveshaft spins.

4. Jan 16, 2007

### brewnog

No. The efficiency is nothing to do with the rotating mass. A system rotating at constant speed with no opposing forces will require zero torque to drive it, no matter how heavy it is.

The efficiency drops when you're trying to drive power through more gearboxes, differentials, clutches and bearings than before, as you are when you engage four wheel drive.

5. Jan 17, 2007

### Sleeper

So its related to friction? More parts turning = greater friction or transfer of HP output to efficiency loss through the friction of having to turn more parts?

6. Jan 18, 2007

### brewnog

Yes, friction is a pretty big part of drivetrain inefficiency (bearings in particular). You also lose quite a bit to heat through gears and lubricating oil, which is essentially near enough all from friction anyway.

7. Jan 27, 2007

### venik

:uhh:

excuse me you dont know what your talking about so dont give advice to anyone about drivetrains. i have a modded vr-4 but for learning purposes lets use the stock #'s, it's all-wheel drive, and the engine is benchmarked at 320 crank hp. When my car was stock it put down ~245 awhp avg with 3 pulls made. With no engine modifications and the same crank, i put on a set of lightweight rims and re-dynoed at an avg of 260awhp. Please tell me your reasoning with this friction theory of your's. In your wonder land i would still dyno 245awhp if i put rims made of lead & concrete blocks on my car.

"A system rotating at constant speed with no opposing forces will require zero torque to drive it, no matter how heavy it is." :rofl:

this is true, but we don't live in a frictionless vacuum. Nor do we benchmark wheel horsepower without opposing forces. Also I don't know if you've ever seen a dyno operator do his job, but it isn't done at a constant speed. The machine gets it's values by acceleration of mass heres an equation for you...
F = mass x acceleration
In our little experiment the Force is essentially a constant, so when we lower the mass, guess what happens, the acceleration increases proportionally. The machine doesnt know you're drivetrains dtl%, but it doesnt care, all that matters in a car's power is whats it's putting to the ground. The machine has heavy rollers, and it knows the mass of those. The computer knows 3 things, speed, constant mass of the rollers, and time. It uses delta V and delta T to find the acceleration and plugs it into that formula I gave you to find the force, then it converts the force to power using P = F/v to find watts, then it divides the watts by 746 to find your wheel horsepower. There is no guessing involved.

You seem to grasp the idea of inertia, an object in motion wants to stay in motion, but a more massive object wants to stay in motion more than a small one. More rotational mass in the drivetrain means its going to be harder to speed up and slow down, but thats what we want in drivetrains... to speed up and slow down with ease, so we get rid of the rotational mass.

"Secondly, you need to be very clear about what you define as "horsepower" and "torque". A typical rolling road will measure torque and speed at the wheels, and use a crude calculation (based on a pure guess at drivetrain efficiency) to estimate the engine torque and power."

nope all it gathers from the roller's is the speed, and it has a stopwatch timing your acceleration. The only guessing here is you guessing how a dyno and drivetrain operate. Secondly dynos don't estimate engine power, they estimate wheel power.

"Now. Your horsepower figures at the wheels for 4WD will not drop as you say, due to an increase of drivetrain mass. They'll drop due to a reduction in transmission efficiency, because of the need for more gears, bearings, lubrication etc. Since, at a given speed, power is directly proportional to torque, your torque at the wheels will drop by the same factor. (In fact, the resulting reduction in torque will limit the top speed, giving you your power decrease.)"

reduction in torque will limit your top speed giving you a power decrease? i think its the other way around buddy, the power decrease will limit your top speed. Anyways thats besides the point, top speed on a dyno for any car is all about gear ratio's and redline. Biggest factor in the top speed of a car is wind resistance and drag, you don't have any on a dyno.

/end rant

8. Jan 27, 2007

### brewnog

Great. You've been here 2 minutes and are ready not only to insult me, but to presume what knowledge and experience I have.

Changing the mass of drivetrain components has absolutely no effect on the power developed. Of course it will increase acceleration of the vehicle, and reduce the time seen on your dyno run, but it certainly doesn't increase the horsepower, regardless of where it's measured.

You certainly would. It would just take ages to get there.

Here's an equation for you:

Power = force * velocity

THIS is how your power is calculated. The torque is measured in exactly the same way as it is on an engine dynamometer.

Indeed, this is all true. However, changing your rotational mass does not affect the power developed, which was the point I was making. I'm not denying that benefits are to be gained by reducing masses throughout the drivetrain. The point is that these benefits aren't in the form of power.

This is just not true. A rolling road measures torque. At least, all the reputable ones I've ever dealt with do. What kind of rolling road are you using, which has no provision for applying any load to the cars' wheels other than the rollers' inertia?

The torque and speed at the rollers is measured, the power at the wheels is calculated, and the power at the flywheel is estimated.

No. This is incorrect. You do not suffer any brake power decrease as a result of having more drivetrain resistances. You have additional opposing torques in the drivetrain which reduce the power seen at the wheels.

Venik, if you decide to reply to my posts in future, please would you consider being as polite and respectful with your words as other members around here are? Thank you.

Last edited: Jan 27, 2007
9. Jan 27, 2007

### FredGarvin

Most of the dynos that I have worked on did indeed measure torque. I figured that is the information most people are looking for when they get on a dyno. However, I did have rolls that we used for vehicle testing at assembly plants at the end of the line. Those did not measure torque. They were simply rolling masses that allowed the vehicles to run in place while we checked other powertrain systems. Torque and power measurements were not necessary.

10. Jan 27, 2007

### venik

Did you skip over the part about my lightweight rims freeing up DTL to the wheels? It's a fact and there are millions of Dyno charts to prove it. no it doesnt increase hp, it decreases the amount that is lost in moving an extra 100s lbs of rotating mass.

Ive fought similiar arguements such as is DTL a direct percentage or is it a fixed value. But this is seriously rediculous to think that DTL doesnt even exist.

there is no way to "measure" torque on a drum spinning at 1000 rpms, you can only calculate it using the power equation. if there there was we would still use the calculating method because it'd still be much cheaper and just as effective if not, more effective.

I know what your thinking, and im not saying we get HP out of nothing, just that there is power wasted turning all that mass. Picture 2 bars 2 feet long each, they rest on a fulcrum 2 inches from the edge of both of them. Heres the catch, one bar weighs 50 lbs, the other weighs 5 lbs, are they both going to be just as easy to lift? thats almost just like how a drivetrain works except they use gears instead of levers, lots of heavy gears. Really all a gear is, is a series of levers that make a circular shape right?

11. Jan 28, 2007

### FredGarvin

That is not true. Our small dyno measures torque and calculates power. To be exact, it measures the counter-torque required. There are also many, MANY styles of in-line torquemeters available on the market. I just ran a test on a high speed pump that used an in-line torque meter that was good up to 50,000 rpm. Measuring torque directly is nothing new.

12. Jan 28, 2007

### Stingray

In terms of actual car behavior, drivetrain inertia acts like an effective gear-dependent ("static") mass resisting straightline motion. It is not particularly useful to think of it as reducing the effective power produced by your engine. The power difference measured by a dyno after changing the mass of some drivetrain component (and nothing else) is actually irrelevant. Different dyno designs will give you very different results. None of these measurements will be relevant for anything.

It might be useful to explain this using an example. Say that your wheels weigh 20 kg each (including tires). Say that they have moments of inertia ~1 kg-m^2, which will be typical. If the tire radius is ~30 cm, the car will accelerate as if each tire had a mass of [20+1/(0.3)^2] ~ 30 kg.

Now think about how much power looks like it's lost to the rotational inertia of the wheels and tires if the effective mass concept is not used. It is possible to ask this at low speeds. The problem is that it depends on how much your car weighs and how much power its engine produces. You get a much more widely-applicable answer if you just stick to thinking about effective masses instead.

Last edited: Jan 28, 2007
13. Jan 28, 2007

### venik

how much the car weighs is irrelevant on a dyno, anyways my buddy did 2 sets of dyno pulls on the same state-of-the-art dyno, one with a CFDS, and one with a stock DS. his max power was avg 302awhp with a stock DS, and avg 310 awhp with the carbon fiber DS. 8 awhp was freed up, with 15 lbs in the DT lost and this is with DS which have a very small radius. LW rims have a much better effect. No matter what you say and what ever formulas you bring up and try to prove me wrong with, like i said before, it's been proven before and proven again that LW drivetrain parts are going to put more hp to the wheels.

14. Jan 28, 2007

### Stingray

Congratulations on continuing to ignore what knowledgeable people are telling you. Stop learning engineering from car magazines.

I'll try this once more. I agree that dynos give you different results before and after changing weights of rotating components. What I'm telling you is that the numbers you get from a dyno are not precisely "hp to the wheels" in a useful sense. If you just want bragging rights, then fine. But the measured power difference in your friend's car will not quantitatively predict any changes in that car's performance.

If dyno measurements took place at constant speed, rotational inertia would not affect measurements at all. Most chassis dynos you're likely to find do not take steady-state measurements. They allow the wheels to accelerate. When that happens, some power from the engine acts against the rollers and some acts to accelerate the drivetrain. So you'd now see a loss in your measurements. But that loss depends entirely on how quickly the dyno allows the rollers to accellerate. On most dynos, that rate is not adjustable, and is whatever it is. So you end up with a loss which has no physical relevance. It is completely arbitrary. Allow the wheels to spin up more quickly and your friend will see an even larger difference. But if that number meant anything, it shouldn't depend on minor changes in the way it was measured.

Again, that does not mean that a lighter drivetrain is irrelevant. I am only saying that good dynos strive to remove those effects from their measurements because they muddle up what you're really looking for. Cheaper machines often ignore this problem or apply a "correction factor" which assumes that all cars have identical drivetrain inertias. You can probably imagine how good an assumption that is...

15. Jan 29, 2007

### venik

I'm studying petroleum engineering at UCSB one of the top engineering schools in california. I'm not ignoring anything you say, you are right in everything you say but you are looking over key elements. I'm aware of the constant speed dynos and those aren't made to measure power they are made to fine tune specific rpm points, yes they tell you the power you are yeilding but not at a constant speed. At a constant speed you can't accelerate, which is essential to find the power. Ask me which one i would like my car to be tuned on, I'd take the constant speed dyno. Those dynos are meant for tuning and they will do it good by finding the AFR and timing that you make the most power at in each rpm point, thats really important, but there are only a handful of people operating those for commercial use in the US. Anyways back to the arguement...

"But the measured power difference in your friend's car will not quantitatively predict any changes in that car's performance."

If it's increasing the acceleration of the drum, it increases the acceleration of the car. to calculate power acceleration NEEDS to be in the equation, Power = MASSxACCELERATIONxVELOCITY. Where the mass is the mass of the car, yes LW parts will affect the mass of the car, but even adding that much mass into your trunk, you will still accelerate faster than before you added the LW part.

This was copy and pasted from a dyno operator.

"How does a chassis dyno measure Horsepower?
from the HP = T*RPM/5252 equation.
A chassis dyno (no load on the drum) measures the instantaneous drum speed, and time during a pull. From both, it can calculate drum acceleration. Knowing the rotational inertia of the drum, it can calculate the applied torque. Then finally, from the calculated applied torque, and the speed of the drum, it can calculate HP applied to the drum using the above formula"

16. Jan 29, 2007

### AlephZero

Hmm... so when the car is going at constant speed (zero acceleration) it doesn't need any power? That's strange, I just tried driving 10 miles down an empty road with the engine switched off, but it didn't work for some reason.

Assuming that what you measure on a rolling road will read directly across to real performance is very poor physics, whether you learned it "from one of the top engineering schools in california" or wherever.

17. Jan 29, 2007

### Stingray

Yes, constant-speed dynos are often advertised as being good for AFR tuning. But they're also optimal (if properly used) for measuring power. Have you ever heard of brake hp? The dyno applies a brake to the rollers until they stop accelerating. They then measure the torque applied by that brake. If you apply torque to a rotating drum, you are expending power. No acceleration is necessary.

Since you are an engineering student, you should be able to understand the following derivation which shows (a) no acceleration is necessary and (b) it should be avoided. Say that the dyno rollers are drums of radius $R$ and moment of inertia $I$ with brake torque $T$ acting on them. Say that a car's wheels have radius $r$, and are being driven with torque $\tau$. Let the effective moment-of-inertia of the drivetrain be $i$ (measured at the wheels). This heavily weights components which sit behind gears.

Anyway, ignore rolling resistance losses. They're easy to bring back in, but irrelevant for what I'm talking about. The the balance of power in the system now reads
$$\tau \omega - T \Omega = I \frac{d \Omega}{d t} + i \frac{d \omega}{dt} ~.$$
$\omega$ is the angular velocity of the wheel and $\Omega$ is the angular velocity of the drum. If the tires aren't slipping much, these are obviously related by $\omega = (R/r) \Omega$. It's also reasonable to call $\tau \omega$ the power $P$ you want to measure. So the previous equation turns into
$$P - T \Omega = \left[ (R/r) i + I \right] \frac{d \Omega}{dt}$$
You should now see that an appropriate $T$ will cause the right-hand-side to vanish. If the dyno can measure the brake torque (some can), it can measure power without acceleration.

Now think about what happens if $T=0$. This is the cheap inertia dyno you've been talking about. These don't measure torque directly, but instead (as you've said) measure $d\Omega/dt$. So they think that a car is producing power
$$P_{\rm{dyno}} = I \frac{d \Omega}{dt} ~.$$
But the power equation shows that
$$P/P_{\rm{dyno}} = 1 + (R/r) (i/I) ~.$$
So the power you measure is always off by some fraction depending on the design of the dyno and your drivetrain inertia. That fraction is not physically relevant.

There are many other issues with interpreting chassis dyno results as well. But they're not the focus of this thread, so I've ignored them.

Last edited: Jan 29, 2007
18. Jan 29, 2007

### Q_Goest

Hi venik,
As the others are trying to explain, the drivetrain with a lower inertia doesn't change the engine power. It changes the amount of power stored in the drive train during acceleration. The increase in measured power during acceleration may very well differ from the measured power under constant load, but that difference is dependant on how much power is being stored in the drive train during acceleration. If one measures power at constant RPM, drivetrain inertia will have no affect on the measurement. If measured during acceleration, it will drop from this constant RPM measurement by the amount of power stored by drivetrain inertia.

You are correct in suggesting that by reducing drivetrain inertia, more of the engine power is transfered into accelerating the vehicle. Note however, that the difference between the power going into accelerating the vehicle and the power produced by the engine is going into things like drivetrain inertia and also frictional losses. Note also that the drivetrain inertia is stored energy just as a flywheel stores energy, it is not lost as heat like frictional loads are.

19. Jan 29, 2007

### Staff: Mentor

I'm not a car guy, but I have a pointed question that may help clarify things: how easy/difficult is it to know what the total moment of inertia of the drivetrain is? Is seems to me that the simplest way to find it would be to calculate it based on the torque applied to a dyno (of known moment of inertia) under steady acceleration. If you don't know it and use an uncontrolled accelerating dyno, you haven't learned anything useful - all you learned is how your car performs on that dyno and measuring the same car on a different dyno may give you completely different results.

And once you have the moment of inertia, it should be left-out of the measurements if at all possible because, as others have said, it gets in the way of measuring actual engine performance. And once you have it, it is an easy matter to calculate the effect on acceleration of, for example, changing to lighter rims.

[mentor hat] Venic, I don't know how you act toward people you meet for the first time in person, but on this site we are a little nitpicky about applying basic human courtesy to the conduct in the forums. If you start with a neutral tone, you may find that we have some remarkably knowledgeable people here who can be an excellent resource. [/mentor hat]

Last edited: Jan 29, 2007
20. Jan 29, 2007

### venik

we dont want to find out the engine's horsepower on a dyno, most importantly we want to TUNE the car on a dyno, but as far as numbers go, hp to the wheels is more important than hp at the flywheel. As i've stated before we can increase what it's putting to the ground without changing the engine's power and by decreasing frictional and/or inertia losses. For frictional losses we can coat the drivetrain with very smooth material, and for inertia losses, we use carbon fiber, or other lightweight/strong materials. Still frictional loss is no competition for drivetrain loss, unless you're rev limiter is at 8000 rpm, I think you are regurgitating what you heard about engine dynos, mostly frictional loss there because there is almost no mass between the crank and the flywheel.

"If you don't know it and use an uncontrolled accelerating dyno, you haven't learned anything useful - all you learned is how your car performs on that dyno and measuring the same car on a different dyno may give you completely different results."

I've noticed that switching from one dyno to another, I may dyno 500 on one, and 560 on another. But i don't go to a dyno to find my horsepower, I go there to get a professional to tune it, and I go to the same one every time. Yes if i went to another one i would get different numbers, but as long as I redyno on the same dyno and my number's increase, my power at the wheels is increasing.

"Hmm... so when the car is going at constant speed (zero acceleration) it doesn't need any power? That's strange, I just tried driving 10 miles down an empty road with the engine switched off, but it didn't work for some reason."

in a vacuum, frictionless space it would be possible without power, but you're working against air resistance, gravity, etc.

"Assuming that what you measure on a rolling road will read directly across to real performance is very poor physics, whether you learned it "from one of the top engineering schools in california" or wherever."

As i said before, if it increases the acceleration of the drum, it will increase the acceleration of the car, and acceleration takes power. Nobody said it will read directly to real performance, however if the dyno tells me I'm gaining power consistently, then I'm doing something right.

"So the power you measure is always off by some fraction depending on the design of the dyno and your drivetrain inertia. That fraction is not physically relevant."

On a roller dyno, the drive train should be completely excluded when you are trying to find what you are putting to the wheels, when you go out back on the road you are taking that drivetrain with you so there is no sense in calculating what your horsepower would be without it. having said that, who cares? If i wanted to find out exactly how much it is increasing my power to the wheel's maybe i would care, but if the power is increasing on the same dyno, I got my answer and i can use it for whatever proving i want regardless if they are exact numbers, they increased the average hp, that is the point I'm trying to make.

It seems like everyone is flaming me for off-topic things, so i think my work here is done, sorry to piss everyone off especially you physic's majors. Drive train inertia = lost power correct?

Last edited: Jan 29, 2007