# Flywheel vehicle drive

1. Sep 25, 2010

### jimgram

I'm calculating a hypothetical flywheel/IVT/reducer/vehicle scenerio where a flywheel with a moment of inertia of 312 ft2*lb and ang. velocity of 10,000 rpm is connected to 30-in dia. driving tire of a vehicle. The vehicle is 2000 lbs. The flywheel will accelerate the vehicle 0-60 mph in 10 seconds.

I need to solve for the ending angular velocity of the flywheel. This seems simple enough based on the conservation of momentum. Initially, the vehicle has zero momentum and, in 10 seconds, it has gained momentum: p=mass*v or 176,000 ft*lb/s. This translates to angular momentum of the tire: L=p*r, so L=220,000ft2*lb/s.

The initial momentum of the flywheel is L=I*w which is 327,100 ft2*lb/s, so the ending momentum of the flywheel is equal to the initial momentum minus the momentum of the tire: Le=107,100 ft2*lb/s.

Based on this I originally assumed that I could find the ending velocity of the flywheel by using this known ending momentum: we=Le/I This gives an ending flywheel omega of 3274 rpm.

At 60 mph the tire is turning 672 rpm, so there needs to be a gear reducer between the flywheel and the tire of 4.872:1 (this is assuming that at maximum car speed, the IVT is at 1:1).

If the torque is calculated based on delta*momentum it is found that the torque to decelerate the flywheel is equal to the torque on the tire to accelerate the vehicle. But this can not be correct since there is a 4.872:1 gear box between the two.

I then found as error in my approach in that the moment of inertia changes by the inverse of the square of the ratio. Applying this, however, throws things in much more confusing results. It seems impossible to base an inertia value of the output of a gear reducer on the square of the ratio since immediately the conservation of momentum law is violated.

Any help will be greatly appreciated.

2. Sep 25, 2010

### NobodySpecial

Generally conservation of energy is simpler - at least in cases like this where you are assuming lossless transmission etc anyway

3. Sep 26, 2010

### jimgram

I think I can rephrase the problem:
Based on the mass (ms) and ending velocity (ve) of a vehicle I can calculate its momentum (p): p=ms*ve

Based on the time to accelerate (t) and tire radius (r) I can calculate the torque (tor) required on the tire: tor=p*r/t

I can aslo calculate the final angular velocity (wtre) ot the tire: wtre=ve/2*pi*r

The the linear momentum of the vehicle (p) can be converted to rotational momentum of the tire (L) by multiplying it by the tire radius: L=p*r

The delta momentum is L and this is the amount of momentum that needs to be supplied by the flywheel. Since I know the initial conditions of the flywheel (Ifw, wfwi) I can find the initial momentum of the flywheel (Lfw): Lfw=Ifw*wfw

Since I know the moment of inertia of the flywheel (I) and the flywheel begins with Lfw and then L (tire) is subtracted form it, I can find the ending flywheel momentum and therefore it's velocity: wfwe=(Lfw-L)/I

Now, when I calculate the ending ratio between the flywheel and the tire which is:
ratio=wfwe/wtre, and use that to determine the torque at the flywheel, torfw=tor/ratio , I find that this torque is significantly less than the torque calculated using the delta momentum of the flywheel.

One possible answer is that the ratio must actual vary between infinity and 'ratio' or between zero and 1/ratio (hence the IVT/fixed gear reducer combination).

If anyone has experience in this arena I would appreciate help - I think I'm at a dead end

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