Rotational Mass & Acceleration - A Case Study

[Mach V Dan]

Shaving one lb off a tire on a 12in wheel is way different from shaving one lb off a tire on a 22in dub. Its easy to see it can't be '8 lbs' in both cases.

 

Er, no, it is exactly the same. The PROPORTIONAL change will be less (one pound off a ten-pound wheel compared to one pound off a 100-pound wheel...) But the effect of removing one pound of tire tread from each tire (at the exact outside of the tire diameter) will be EXACTLY equivalent to eight pounds off the car, regardless of wheel diameter.

 

--Dan

Mach V

FastWRX.com

[ChE Outback XT]

You need to use calculus.

 

Yes, you do. Let me take a stab at it... (if you don't want to read the whole spiel, I'll just say that his original simplifications aren't too far off)

 

I've attached a worksheet I created in Mathcad to test an approach to solving this problem. Here is my general approach:

 

1. Determine the energy the car has at 60mph

2. Determine the average power used to get the car to 60mph

3. Use this power to accelerate a car with lighter rims to 60mph

 

Details of the approach:

 

1. The energy of the car is the sum of the kinetic energy of the car itself (plus a 200lb person in my example) moving straight forward and the rotational kinetic energy of each wheel. The kinetic energy of the wheels is a function of their moments of inertia. I assumed that the wheels and tires are not uniform discs, but have their mass concentrated near their perimeters. To be precise, I assumed their center of mass along the radius is located 3/4 of the way from the hub to the rim for the wheel and 3/4 of the way from the rim to the tread for the tire. I then used the definition of the moment of inertia (yes, the calculus version) to determine it for each wheel. The formula works out to 1/2*Mass*(inner Radius^2+outer Radius^2) where the inner and outer radii are equidistant from the radial center of mass (the 3/4 I talked about). The total moment of inertia is the sum of the tire's and the wheel's.

 

2. I assumed an LGT can get to 60 in 5.6s. This gives an average power of roughly 130hp. (This makes sense as we have a peak of around 206whp and when accelerating you spend most of your time outside the peak of the power curve.) I also did a trial using drag data posted on this site of roughly 14.1 seconds to get to 97mph. This produces an average of about 131 hp.

 

3. I recalculated the moment of inertia for lighter wheels as specified at the beginning of this thread. This lowers the total energy of the car at 60mph, meaning the same power can better accelerate the car. How much better...?

 

RESULTS

 

Yes, it's true that lighter wheels help, but not much. In my calculation the 0-60 time went from 5.6s to 5.59s. (Equivalent to roughly 0.25 whp) This is pretty close to the original estimate of a 0.08s improvement. The problem with the approximation of 1 wheel pound being equivalent to 6 static pounds is that the wheel pound's influence is dependent on wheel size.

 

This was posted earlier and makes the point:

 

Originally Posted by bemani:

Shaving one lb off a tire on a 12in wheel is way different from shaving one lb off a tire on a 22in dub. Its easy to see it can't be '8 lbs' in both cases."

 

"Er, no, it is exactly the same. The PROPORTIONAL change will be less (one pound off a ten-pound wheel compared to one pound off a 100-pound wheel...) But the effect of removing one pound of tire tread from each tire (at the exact outside of the tire diameter) will be EXACTLY equivalent to eight pounds off the car, regardless of wheel diameter.

 

--Dan

Mach V

FastWRX.com

 

The equation for the moment of inertia of a uniform disc is 1/2 Mass * Radius(squared). That means the difference cause by removing 1 pound from a 12 inch wheel will be 0.125 lb-ft^2 while it will be 0.420 lb-ft^2 on a 22 inch wheel. The difference in moment of inertia is offset because the larger wheel spins slower at a given speed, but it is still different...

 

Assume each wheel is fitted with a tire that has a 3inch sidewall and we remove that one pound from each wheel:

The car with 12 inchers would save 0.001642 hp as it accelerates to 60mph.

The car with 22 inchers would save 0.002280 hp as it accelerates to 60mph.

 

One pound doesn't make much differnce, but it does affect the 22 inch wheel at least 1.4 times as much as the 12 inch rim. The "EXACTLY 8 pounds" argument doesn't hold.

 

I hope this helps clear things up. Let me know if you have any questions on my methodology.

 

http://www.shootskiers.com/LGTtires.jpg

[Mach V Dan]

The "EXACTLY 8 pounds" argument doesn't hold.

 

Sure it does, if you assume (as I did) that you somehow magically removed a pound of TIRE TREAD at the exact perimeter of the tire. Since we're talking about wheel weight, it would be less...Dave Coleman used a rough figure of 1.5 instead of two, figuring that the mass removed was roughly halfway between perimeter and center of the rotating assembly. (Or in between so the effect was halfway. Whatever.)

 

Anyway, I agree, Outback XT, that the effect on the car is negligible. If we assume two pounds (I know, I know...overstating...) of equivalent weight reduction for every pound of wheel weight, and we lighten the wheels by four pounds (super-light fifteen-pound wheels!), that's still only 4 pounds x 4 wheels x 2 multiplier = 32 pounds difference. So we go from power/weight ratio of 250/3500 (wagon w/driver) = .0714 to 250/3532 = .0708, a change of 0.8%. Hardly noticeable. And assuming a more realistic 1.5 multiplier, and a more normal weight reduction of only two pounds/wheel, we're down to something like 0.3% change in power/weight.

 

So, it probably won't make enough difference for you to feel. On the other hand, a lot of little changes to the car add up. So if you're building an all-out race car, get the lightest wheels, and the lightest everything else, and pretty soon you'll have a very light car.

 

--Dan

Mach V

FastWRX.com

[ChE Outback XT]

Sure it does, if you assume (as I did) that you somehow magically removed a pound of TIRE TREAD at the exact perimeter of the tire. Since we're talking about wheel weight, it would be less...Dave Coleman used a rough figure of 1.5 instead of two, figuring that the mass removed was roughly halfway between perimeter and center of the rotating assembly. (Or in between so the effect was halfway. Whatever.)

FastWRX.com

 

I think we missed each other in the middle. Are you talking about the same overall sized tire with different sized wheels in the middle, or totally different tire sizes?

[Mach V Dan]

Are you talking about the same overall sized tire with different sized wheels in the middle, or totally different tire sizes?

 

The original quote I was repsonding to was talking about TIRE weight, and I was using the extremely simplistic assumption that the weight was basically at the very outside perimeter of the rotating wheel/tire. Hey, they're low-profiles...I'll just assume they're all-the-way-low-profile. My point was basically that if you are talking TIRE weight, and disregarding actual thickness of the tire, that a pound off ANY diameter tire is going to be equivalent in terms of the affect on acceleration. So (again, ignoring tire sidewall thickness -- I'm basically assuming tires have no sidewalls), one pound off a 17" tire = one pound off a 12" tire = one pound off a 22" tire. No difference. Again, this was more a theoretical assertion than anything else. I don't know any cars with 12" diameter tires these days.

 

I see your point about the 12" vs. the 17" wheel. Assuming the same outer TIRE diameter, the 17" wheel will obviously have mass farther out.

 

On the other hand, moving from a 17" to 18" wheel is going to have (again) a negligible change in the mass distribution of the wheel/tire combo. Yes, it'll be some change, but it'll be small.

 

These effects are often wildly overstated online, and in many articles. Outback XT, your math looks right to me, and it verifies that wheel weight in the magnitudes we usually see is not going to change the car's performance envelope to a signficiant degree.

 

--Dan

Mach V

FastWRX.com

[praedet]

There are some problems with your equations and basic assumptions Outback XT, but as stated above, the results are close enough that it doesn't matter...

 

Ted

[firedawgs]

Here is a easier way to loose lbs. from your car.

 

 

 

GO ON A DIET YOU FATTY. HA HA

[iyamdman]

It looks like the cheapest mod you can do for your car, is buy lighter and better tires than the stock RE92's when they wear out at 15k.

 

So what would be lighter AND better?

 

If you did go up a size, to like an 18" x 7.5 - your rim might be slightly heavier or the same, but if you use a lighter tire - you shouldn't see any acceleration performance change.

 

In theory, handling should improve slightly. Is this correct?

[FiftyTwoEighty]

And then of course, there is the matter of whether you are using standing start acceleration as the only performance standard. If so, then a wheel/tire combination with less outer rotational mass would be in order but if you were wanting to improve top end performance or even something like 70mph to 100mph performance, then a wheel/tire combo with more outer rotational mass may be in order so that you would be able to use the inerta more advantageously. I would like to qualify this statement by saying that I have no proof, scientific or otherwise, for the aforementioned statement. Just seems reasonable to me.

 

Common sense combined with your own performance goals are probably more important than any rule of thumb per se'. If you wanted to get really anal about it all, you could also say that offset could also play a significant role in how well the vehicle is able to put power down to the ground through the drive axles.

 

Naturally, that is only one very small part of how your wheels/tires are going to change the behavior/performance of the car. As others have already stated, there is also the matter of traction, road surface, air pressure, barometric pressure, etc.... Those sticky tires that you have may perform wonders on the track but you can also expect that to show up on the dyno. It cracks me up when people oooh and ahhh over dyno plots that show a difference of 5 hp for this or that. It has gotten to the point where proof is required for every little change that you can make to your car. I'm not saying that its a complete waste of time but from my experience, I can only say that nothing is exact when you are talking about a mass produced vehicle.

[ChE Outback XT]

...but if you were wanting to improve top end performance or even something like 70mph to 100mph performance, then a wheel/tire combo with more outer rotational mass may be in order so that you would be able to use the inerta more advantageously. I would like to qualify this statement by saying that I have no proof, scientific or otherwise, for the aforementioned statement. Just seems reasonable to me.

I don't think that's really the case. Rotational weight basically acts like any other weight on the car (except rotational mass has a greater effect per pound)... slower acceleration, lower breaking performance. In the case of wheels and tires, weight can also affect suspension performance.

[2005garnetGT]

Yes, you do. Let me take a stab at it... (if you don't want to read the whole spiel, I'll just say that his original simplifications aren't too far off)

 

I've attached a worksheet I created in Mathcad to test an approach to solving this problem. Here is my general approach:

 

1. Determine the energy the car has at 60mph

2. Determine the average power used to get the car to 60mph

3. Use this power to accelerate a car with lighter rims to 60mph

 

Details of the approach:

 

1. The energy of the car is the sum of the kinetic energy of the car itself (plus a 200lb person in my example) moving straight forward and the rotational kinetic energy of each wheel. The kinetic energy of the wheels is a function of their moments of inertia. I assumed that the wheels and tires are not uniform discs, but have their mass concentrated near their perimeters. To be precise, I assumed their center of mass along the radius is located 3/4 of the way from the hub to the rim for the wheel and 3/4 of the way from the rim to the tread for the tire. I then used the definition of the moment of inertia (yes, the calculus version) to determine it for each wheel. The formula works out to 1/2*Mass*(inner Radius^2+outer Radius^2) where the inner and outer radii are equidistant from the radial center of mass (the 3/4 I talked about). The total moment of inertia is the sum of the tire's and the wheel's.

 

2. I assumed an LGT can get to 60 in 5.6s. This gives an average power of roughly 130hp. (This makes sense as we have a peak of around 206whp and when accelerating you spend most of your time outside the peak of the power curve.) I also did a trial using drag data posted on this site of roughly 14.1 seconds to get to 97mph. This produces an average of about 131 hp.

 

3. I recalculated the moment of inertia for lighter wheels as specified at the beginning of this thread. This lowers the total energy of the car at 60mph, meaning the same power can better accelerate the car. How much better...?

 

RESULTS

 

Yes, it's true that lighter wheels help, but not much. In my calculation the 0-60 time went from 5.6s to 5.59s. (Equivalent to roughly 0.25 whp) This is pretty close to the original estimate of a 0.08s improvement. The problem with the approximation of 1 wheel pound being equivalent to 6 static pounds is that the wheel pound's influence is dependent on wheel size.

 

This was posted earlier and makes the point:

 

 

 

The equation for the moment of inertia of a uniform disc is 1/2 Mass * Radius(squared). That means the difference cause by removing 1 pound from a 12 inch wheel will be 0.125 lb-ft^2 while it will be 0.420 lb-ft^2 on a 22 inch wheel. The difference in moment of inertia is offset because the larger wheel spins slower at a given speed, but it is still different...

 

Assume each wheel is fitted with a tire that has a 3inch sidewall and we remove that one pound from each wheel:

The car with 12 inchers would save 0.001642 hp as it accelerates to 60mph.

The car with 22 inchers would save 0.002280 hp as it accelerates to 60mph.

 

One pound doesn't make much differnce, but it does affect the 22 inch wheel at least 1.4 times as much as the 12 inch rim. The "EXACTLY 8 pounds" argument doesn't hold.

 

I hope this helps clear things up. Let me know if you have any questions on my methodology.

 

http://www.shootskiers.com/LGTtires.jpg

 

i see no integrals or derivatives, i see no calculus here, only physics

[ChE Outback XT]

i see no integrals or derivatives, i see no calculus here, only physics

Garnet- the formulas for determining the moments of inertia, "I", are the solutions to the integral definition of the moment of inertia for a disc or ring. I just didn't do the derivation. Most physics equations you see (in algebraic form) is actually a solution to a differential equation. Even high school physics uses a lot of calculus, you just don't see the original forms of each equation.

 

Either way, the point of my little mathematical masterbation above was just to show that it's tough to lose enough weight in the wheels to affect acceleration. Suspension performance is probably affected much more. (As a friend of mine pointed out, lighter wheels/tires are often accompanied by bigger breaks, which pretty much makes it a wash anyway.)