# Calculating torque/hp losses from wheel choices

1. Sep 5, 2009

### nmo

Hi guys,

I am a complete physics dunce so go easy on me, I can wrap my mind around the basic concepts but I need help with this particular proble,.

What I'm trying to do is figure out how much torque is lost or gained by changing wheels on a car.

I assume we need to ignore the front two wheels and focus merely on the drive wheels, though I'm sure there's some loses or gains to be seen from the front wheels (non drive) as well.

The first question, is how the rim size alone effects torque output. Here's what I have been thinking would be a good estimate of this:

- Assuming the tire weight is 22lbs regardless of the rim size being 17" in diameter or 18" in diameter (which is plausible according to my measurements).

- I assume that by moving the 22lbs out .5" away from the center (which I assume to be the point of force) that if T = F x D then with a 17" diameter wheel I need 15.576ft-lbs to turn the wheel and with a 18" diameter wheel I now need 16.5ft-lbs. Is this assumption correct?

Next is while it is difficult to know how the weight of the wheel is distributed, can we estimate the loss caused by a heavier wheel in general? So for example, if I have a 17" diameter wheel that weighs 21lbs and one that weighs 13.5lbs can we estimate how much less force is needed (or 'freed up') to turn the lighter wheel? I assume the distribution of weight is similar if not the same.

I'd like to have a set of formulas that I could plug values in for

tw (tire weight)
rs (rim size in diameter)
ww (wheel weight)

for two different scenarios and figure out how much torque I'm losing or gaining by changing wheels/tires.

Then I get a little confused, torque is often measured 'at the wheels' (as that's what matters).... is it wrong to assume that I can simply say that a 5ft-lb torque gain at the wheels is going to be the same throughout the power band? Meaning, does that 5ft-lbs required to turn the wheel stay constant or does it increase/decrease as the wheel speed increases? I'm assuming that the wheel is accelerating throughout the entire scenario.

Finally, I'll explain what the application is so it's clear what I'm trying to do if my above isn't clear:

MOTIVATION: I want to be able to figure out in hard numbers if there's a benefit to switching wheels for the purposes of straight-line acceleration and 'gaining' (regaining) horsepower at the wheels. It is known that lighter wheels accelerate much faster and easier, but I want to know by how much exactly given two different wheels. This is primarily to asses whether or not modifications to a vehicle are cost effect or warranted.

Thanks guys for your help, and remember that I'm a complete physics idiot so be gentle with the explanations of how to do these calculations :). I took calculus and physics in college but I admittedly haven't used it directly in years :P.

Nathan

2. Sep 5, 2009

### Stingray

Rotating components add an effective mass to the car for the purposes of straight-line acceleration. For a wheel/tire, that mass (beyond the normal one you'd find using a scale) is

$$m_{\mathrm{rot}} = I / r^2,$$

where I is the wheel/tire's moment of inertia, and r its radius. I depends on the way mass is distributed in the wheel. If it's all at the rim, it would be $m_0 r^2$, where $m_0$ is the wheel's ordinary mass. Realistically, it's going to be less than this (say 1/2). So this effect is not very significant. Rotating components closer to the engine can be much more important in lower gears due to their much higher rotational speed relative to the wheels.

3. Sep 5, 2009

### nmo

I understand from what you're saying that the components closer to the crankshaft are going to be far more important - which makes sense, since lightweight crankshaft pulleys, lightweight crankshafts and flywheels have all been used to increase response and 'power' in the past.

However, I'm not so clear about what you're saying about the wheel mass itself. I also don't think I can quite get to a point where I can somehow logically calculate the actual concrete benefit of a lighter weight wheel/tire combination - at least from what you've given me.

I am at least starting to see the rotational inertia may be very important even when we're considering a case of say a drag race where the wheel is constantly being accelerated.

Anyone else care to help the layman further understand and perhaps even calculate some meaningful numbers? :)

Thanks!

4. Sep 6, 2009

### Stingray

The practical point of my post was that if you switch to new wheels that are all (say) 10 lbs lighter, the car will act in a drag race as if it weighed about 4*(10+5) ~ 60 lbs less. So the rotational inertia of the wheels makes them act as though they weigh about 50% more than normal. A heavier car with the same engine will not accelerate as well, so you could interpret this as "lost power." It's lost by an equal proportion for all engine speeds.

5. Sep 6, 2009

### nmo

that is fairly consistent with my experience and some other sources I've seen.

However, what I am curious about is where the +5 came from in the calculation you mentioned.

I see that 4 is the number of wheels of course, 10 is the weight, but I'm a little fuzzy about where the 5 came from.

I really appreciate it :)

Thanks!

Nathan

6. Sep 7, 2009

### Stingray

The +5 came from the fact that each wheel needs to be spun up (as well as accelerated linearly). It's an estimate, and could be a little smaller or larger depending on the wheel design.

7. Sep 7, 2009

### Bob S

Hi nmo-
This an interesting and practical problem. But first, you did not provide the rolling diameter of the tire, which is important in calculating wheel RPM.
What I am going to do here is calculate the energy needed to rotate the tire at a particular velocity (I will use 60 mph) and divide by the elapsed time (I will use 4 seconds) to get accelerating power (HP) for 4 wheels.
Suppose the moment of inertia of each wheel (rim plus tire) is
I = (3/4) m R2
where the 3/4 is a coefficient related to how the mass is distributed radially, and R is the rolling radius.
The energy of one wheel rotating at a velocity of w (in radians per sec) is
E = (1/2) I w2 = (3/8) m (Rw)2

But Rw is just the velocity of the rolling wheel with radius R, so Rw= v (meters per sec).

So for one tire, the energy is E = (3/8) m v2

For a wheel (rim plus tire) mass of 35.5 pounds (16.1 Kg) and 60 mph (26.8 meters per sec)

E = (3/8) (16.1) (26.82) = 4336 Joules

If we had to accelerate four wheels from 0 to 60 mph in 4 seconds, the power would be

P = 4 x 4336 Joules/4 sec = 4336 watts = 5.8 HP, INDEPENDENT of rolling radius!

I hope this helps.

[Added note] The above calculation does not include the translational energy, which is normally included in the moving energy of the car; i.e., (1/2)M v2. Thus the TOTAL energy of the rolling tire is (3/8 + 1/2)mv2.
Thus the TOTAL power to do 0 to 60 mph in 4 sec is 13.6 HP
Bob S

Last edited: Sep 7, 2009