Drive-wheel hp decreases as rotational inertia of brake disk increases?

[Alfred1]

I've been watching Sports Car Revolution on Speed lately. Of course most of the cars they feature are Japanese but I really like that they put everything through a dyno. The balance bling bling and performance mods pretty well.

Anyway, they are doing an Acura RSX project right now. They fitted the car with a big brake kit - you know, big rotors and big calipers. Their times around the track didn't really improve even though they were faster around the corners because they could brake later.

After dyno testing they came to the conclusion that even though the big brake kit was lighter than the stock brakes it resulted in a 7 hp loss at the wheels because of the increased rotating weight of the big rotors.

I thought it was a pretty interesting result.

Somebody asked this question on another forum and I think this is what the answer is but it’s been a while since I opened a physics book so feel free to correct me where I'm wrong.

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The power to accelerate a rotating system is given by P = (torque)*(da/dt) where a is angular position and t is time. Torque is equal to the product of the system's rotational inertia (I) and the system's angular acceleration (dw/dt) therefore this can be rewritten as

P = (I)*(dw/dt)*(da/dt)

Thinking of the brake/tire/disk system as being one solid disk and, since the rotational inertia of a solid disk is given by

1/2(M*R^2) (where M = mass and R = radius) this can be rewritten as

P = [1/2*(M*R^2)*(dw/dt)*(da/dt)] = 0.50(M*R^2)*(dw/dt)*(da/dt)

If the mass of the disk is reduced by, for example, 25% but its radius is increased by 20%, the power to accelerate the new system would be given by

P = [1/2*(0.75M)*(1.2R)^2*(dw/dt)*(da/dt) = 0.54(M*R^2)*(dw/dt)*(da/dt).


An engine will have P flywheel hp available to accelerate the brake/tire/disk system, therefore (at the same rpm)

0.50(M*R^2)*(dw/dt)*(da/dt) = 0.54(M*R^2)*(dw/dt)*(da/dt).

For a given rpm, da/dt are equal on both sides of the equation therefore

0.50(dw/dt) = 0.54(dw/t), dw/dt = (0.50/0.54)*(dw/dt) = 0.93 (dw/dt).

That is, for the flywheel power available at any given rpm, the brake/tire/disk system with the reduced mass but increased radius will have an acceleration of only about 0.93 that of the unmodified system.


The dyno roller which is accelerated by the car's tire has its own rotational inertia (I) and the power it measures is given by P = (I)*(dw/dt)*(da/dt). Since for a given rpm, its I and da/dt are the same, it will measure a lower hp reading from the lower acceleration of the modified brake/tire/disk system, i.e.,

P(1) = (I)*(dw/dt)*(da/dt) > P(2) = (I)*(0.93*dw/dt)*(da/dt).

[Dave at Pelican Parts]

Yup. The simpler way to think of it is--more mass is bad for acceleration. Mass out further from the center of rotation is worse than mass close to the center. So, given any mass reduction of rotating bits, there is a way to set up the reduced mass which is even worse for acceleration than the greater mass.

My thought when I saw the post on the other board was, "Well DUH!" But it's interesting to see that a specific brake upgrade does fall into that range--lighter but worse for acceleration.

--DD

[Alfred1]

So, by the same reasoning, reducing the mass of the flywheel (and hence its rotational inertia) actually does increase the hp at the drive wheels (if only by a little). That's the first time I really understood that.

[Dave at Pelican Parts]

Yes and no... It depends on how you measure it. A brake dyno, like a water-brake one, will not see an increase in HP with a lighter flywheel. An inertial dyno, like the Dynojet, will. I think the eddy current dynos have both a brake component and an inertia component, so they will see a smaller effect.

I've seen long discussions about the lightened flywheel contributing to HP or not... And none of the discussions wind up being that simple. At least, not simple enough for my brain!!

--DD

[Bleyseng]

Yes, and a lightened flywheel lets the engine spin up quicker which is good.
This also applies to lighter rims and tires which make the car quicker vs heavy rims. ie Fuchs vs Empi 8 spokes
I like to think of it in terms of holding something in your hands with your arms straight out. The heavier the item is the quicker you get tired or can't hold it.
Geoff

[lapuwali]

Strictly speaking, the rotational mass of the flywheel or the brakes has no effect at all on engine power. Rather, it affects how much power it takes to accelerate those objects, which has to be subtracted from the power available to accelerate the car. Note that this extra mass also has to be dealt with in deceleration. An engine with a heavy flywheel will not only be slow to rev up, but also to lose revs when you lift off the throttle.

An inertia dyno (like the DynoJet, and most other wheel-driven dynos these days) measures power by measuring how quickly the engine can accelerate a known mass. So, the extra flywheel/brake mass shows up clearly. But it will also show up on the road, so this is fine.

A brake dyno, typically attached directly to the engine, measures torque rather than power, and typically will measure this torque at a relatively fixed engine speed, so no acceleration, so no rotational mass effects.

[Alfred1]

Originally posted by lapuwali
Strictly speaking, the rotational mass of the flywheel or the brakes has no effect at all on engine power. Rather, it affects how much power it takes to accelerate those objects, which has to be subtracted from the power available to accelerate the car.

Exactly, the hp measured at the drive wheels will be less. Here's the equation for the angular acceleration dw/dt of a flywheel (solid disk) with mass M and radius R rotating at da/dt rpm and being accelerated by P hp:

dw/dt = P/[(1/2M*R^2)*(da/dt)]

At any given rpm, the angular acceleration of the flywheel is inversely proportional to its mass so, for example, if we hold all of the other values fixed, decreasing the flywheel's mass by one half would double the flywheel's angular acceleration and doubling the flywheel's mass would decrease the acceleration by one half. Obviously, if your car is only accelerating half as quickly as before, you've lost some drive wheel hp.

[Biggy72]

correct me if I'm wrong, but won't a lighter flywheel also allow the engine to reach higher rpms? I know they do, well they try to any way, lighten the fly wheels on racing mowers so they can turn higher rpms and they get there faster, but usually there's an rpm limit on the mowers set by the sanctioning body of the race.