Dyno question from other board

[Alfred1]

Someone asked this question (below) on the other board and I initially believed that his friend was correct that the hp loss on a given chassis dyno would be constant and that the chassis hp measure would reflect this and not be an approximately constant percentage of the flywhell hp figure. I thought about it more and now I think that he is basically right and his friend is wrong. Does anybody here know one way or the other?

This is my reasoning:

To simplify things, I will consider hp at only one rpm, say 5000, and that the chassis dyno runs are both done in second gear so that the rotational inertia of the drivetrain is the same for both runs. Since the power of a rotating system is given by

Power = (rotational inertia) * (angular acceleration) * (angular speed)

angular acceleration = Power / [(rotational inertia) * (angular speed)]

Now, if you know that the engine produces, say 37%, more flywheel hp at 5000 rpm with nitrous injection, the new flywheel hp value is 1.37*F (where F is the initial flywheel hp value) and since the rotational inertia and angular speed are the same for the two chassis dyno runs, the new angular acceleration of the dyno drum is 1.37 times that of the first drum angular acceleration resulting in a chassis dyno hp reading of 1.37 times that of the initial chassis dyno hp reading. The chassis dyno hp value has increased by the same factor that the flywheel hp value has. Since the initial flywheel hp value is given by F, the second flywheel hp value is 1.37*F. Similarly, if the initial chassis dyno hp value is given as C, then the second chassis dyno hp value will be 1.37*C. Clearly, C/F = 1.37C/1.37F . That is, the ratio (or percentage) of the initial chassis dyno hp value to initial flywheel hp value is the same as the percentage of the second chassis dyno hp value to the second flywheel hp value even though the flywheel hp value had increased by 37%.

A friend and I got into an argument regarding dyno numbers.

Instead of trying to explain in words, I will just show.

Him:

200 Crank HP
160 Wheel HP

40 HP loss

If you make that motor 400 crank HP then the dyno will say 400 minus 40 hp.
Even at 800 HP, it would still be minus 40 HP.

I say:
200 Crank HP
160 Wheel HP

20% Loss

So, if 400 crank HP, then it would be 400 minus 20%, therefore, 80 HP loss to the wheels.

These are just numbers. The loss is disputable between the crank and wheels, yada yada yada.

But, just based off that information, who is right?

I need it to be proven. Mathmatically / Scientifically preferred. Not just dyno sheets as there are too many factors that can come into play and what not.

Or am I wrong? I still swear that I am right, but who knows now. Oh, someone here will know

[silver2.0]

Your over my head, but it just makes sense to me that a percentage of loss would make more sense.

[Alfred1]

I guess to really simplify what I was trying to explain above is that the drivetrain/dyno-drum combination can be thought of as one large flywheel and that in a fixed gear this "flywheel's" rotational inertia is the same from one dyno run to another. Increasing the engine hp at the crank by a certain factor will increase the acceleration of this "flywheel" by the same factor and therfore the chassis dyno hp reading by the same factor. The result is that the ratio of chassis dyno hp / crankshaft hp will remain constant.

[Alfred1]

Ok, I wanted to get a concrete answer to this question so I emailed a physics professor and here is what he said:

----------------


Hi Alfred,
Replies inserted into the body of your text below:

----- Original Message -----

Sent: Sunday, May 09, 2004 10:19 AM
Subject: Re: Moments of inertia


| Thank you for taking the time to answer my question.
| Can I ask you a couple more questions and then I won't
| bother you again?
|
| I know that it's only by experiment that we know that
| forces obey the laws of vector addition so is the
| equation for the effective moment of inertia that you
| sent me also shown to be true only by experiment or
| can it be proven mathematically?

REPLY: The basic laws of physics are based on experiment, including vector addition and moments of inertia. I derived the formula that I sent you from these basic laws.


|
| Also, someone on a bulletin board that I'm a member of
| asked if when a car with known crankshaft horsepower
| is placed on a chassis dynamometer to measure the
| drive-wheel hp, is the hp loss through the drive-train
| a constant value or will the chassis hp measured
| always be a constant percentage of the crankshaft hp.
| I'm not a physicist or engineer so the best
| explanation that I could come up with was that if the
| car performs two back-to-back chassis dynamometer runs
| with the transmission in the same gear for each run,
| then the drive-train and chassis dyno drum system
| will have a certain moment of inertia that is the same
| for each run and if the hp is measured each time at
| the same rpm (say 5000), then the equation
|
| Angular acceleration = Power / [(moment of inertia) *
| (angular speed)]
|
| would show that increasing the crankshaft hp by a
| certain factor would increase the angular acceleration
| of the chassis dynamometer drum by the same factor
| (since the moment of inertia and speed remain constant
| for the two runs) and therefore the chassis dyno hp
| reading would always be (disregarding friction) a
| constant percentage of the crankshaft hp.
|
| Am I on the right track with this answer?

REPLY: I understand your answer and it may be correct. I would
prefer, however, to answer along the following lines. In most cases, friction is proportional to the velocity (or rotational velocity in the case of gears); e.g. wind-resistant on a moving car increases with the speed of the car. Thus, if speed of the crankshaft were unchanged when the power is increased
(e.g. starting to go up a hill, but the car is kept at constant speed
by pressing on the accelerator), then the friction should be unchanged and the "percentage of the crankshaft hp" should decrease; i.e. more efficient transfer of power.

Bob.