Rim Inertia

[fritz_269]

Old PH post (written by fritz_269): This one's always good for some controversy!

IMHO, go with the header or the springs.

I still haven't figured out exactly what the Venom 400 does. None of the ads I've seen or their website explains it clearly; they just make outrageous claims with no proof. My best guess is that it re-maps the closed loop air/fuel curve. I would think that this would not be a good solution for just any engine - especially modified ones - that sort of tuning should be adjustable and done on a dyno. Each engine will want something different. Just get an Apexi VAFC and a couple hours on a dyno.

[LONG WINDED ON]

The whole debate over rim size v. acceleration makes me crazy. Reducing mass anywhere is good. Reducing rotational mass is better. Reducing rotational inertia is best. For a spinning disc (like a wheel) the inertia is proportional to the mass times the radius squared. So the weight of the steel belts in the tread of the tire is by far the highest contributor to the inertia. But no-one ever talks about tire weight!?! Lower profile tires do not necessarily have a lower weight, the sidewalls and even tread tend to be much thicker.

For rims, usually most of the mass is concentrated around the outer circumference. If, for a moment, we assume that all 15lbs of mass is around the outer edge, we get ('I' stands for Inertia):
15" rim -> I = 5.86 lb*ft^2
16" rim -> I = 6.66 lb*ft^2 (12% increase)
17" rim -> I = 7.53 lb*ft^2 (28% increase)

Now let's add in the weight of the tire.
Let's say its a 26" tire and it weighs 15 lbs. too. All of it's weight is concentrated around the outer circumference (this is a resonably accurate assumption).
26" tire -> I = 17.60 lb*ft^2

So our total inertia for each rim with the same weight tire is:
15" wheel -> I = 23.46 lb*ft^2
16" wheel -> I = 24.26 lb*ft^2 (3% increase)
17" wheel -> I = 25.13 lb*ft^2 (7% increase)

Acceleration = Torque / Inertia. So, with these wheels off the ground, it would take 7% longer to get the 17" wheel from 0 to whatever revolutions per second. (Also 7% longer to stop it with brake torque).

**edit** there is a minor error below, it's fixed a couple of posts down. Sorry! **/edit**
BUT - these wheels are not off the ground, they are accelerating a heavy car! The translational inertia of a car is just it's mass, let's say 3000lbs. The parallel axis theorem (which I won't go into here) says we can just add the inertial contributions. So:
Car w/ 15's -> I = 3023.46 lb*ft^2
Car w/ 16's -> I = 3024.26 lb*ft^2 (+0.026%)
Car w/ 17's -> I = 3025.13 lb*ft^2 (+0.055%)

Again, acceleration = torque / inertia, so if you were doing 15 seconds in the 1/4 mile before:
Car w/ 15s -> 15.000 sec
Car w/ 16s -> 15.004 sec
Car w/ 17s -> 15.008 sec

Not much difference!!!

But, let's say you got really light tires that were just 10 lbs. (instead of the 15lbs. we decided on earlier)
Car on 15" with old tires -> I = 3023.46 lb*ft^2
(15.000 sec in 1/4)
Car on 15" with light tires -> I = 2997.6 lb*ft^2
(0.9% decrease!)
(14.87 sec in 1/4)

Upshot - get lighter tires!!! Always find out tire weight when looking for new wheels. Don't worry too much about overall rim size, as long as you keep rim weight about the same.

I hope this has helped to clear up some perpetuated misunderstandings. You all know my opinion now I'll be glad to respond to comments and questions.

[LONG WINDED OFF]

PS> The one thing that light wheels do get you, regarless of where the mass lies is a lower sprung weight, which will allow your suspension to 'react' more quickly to road conditions and give you better handling. (Again - this is the rim + tire weight!)

[delsolguy]

Fritz, are you familiar with the article SCC did a while back...showing benefits of lighter, smaller wheels? Something like that. Anyway, if I recall, there was a pretty big difference when the wheels were switched. Much more than a .001 diff. I didn't read through your entire post, and I pretty much skipped to the end, so ignore me if this is a stupid post. What it seemed to me was that you were showing that a 17" won't effect acceleration in any great way compared to a 15" wheel. Basic goal of a wheel/tire combo = get lighter tires, 'cause the wheels don't really matter. Is that right?

I'll try to find the article and see exactly what they did...

[fritz_269]

I read the SCC article, and I found it very informative. I think they did an excellent job in trying to narrow down the variables. I actually went to college with Dave Coleman, the techical editor - he's a damn smart guy! The article just goes to show something that I egregariously ommitted in my above post - the real world is different from the theoretical one.

What I was riling against was that I kept hearing people say that the inertia of the various size rims what what made them slow or fast. I still stand by my conclusion - the rotational inertia of the rim has very little to do with it.

BUT - in the real world, different size contact patches, tread designs, tread compounds, sidewall constructions on different profile tires can be quite different, resulting in quite different performance. Specifically for drag racing, sidewall wrinkle is a very important parameter for getting traction, and low profile tires don't do much of it. IMHO, in the SCC drag times, it wasn't the rim size that changed the speed, it was the profile and composition of the tire!

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

I also just re-read the calcs and discovered an error! A stupid one too. When adding the rotational and translational inertia, I forgot to add all four wheels - not too many unicycles go racing. This will make the effect four times as large - but it's still a small effect, and I find my conclusions still sound. Sorry for the confusion. I'll repeat the affected calcs correctly:

**edit** These are still incorrect, I made a mistake in applying the parallel axis theorem, obfuscated by not watching my units - the (hopefully) final correction is in a post on page two - thanks to Ivymike for pointing out the error **/edit**
....
BUT - these wheels are not off the ground, they are accelerating a heavy car! The translational inertia of a car is just it's mass, let's say 3000lbs. The parallel axis theorem (which I won't go into here) says we can just add the inertial contributions. There are four wheels, so we just add four times the wheel inertia to the translational intertia (3000 lbs). So:
Car w/ 15's -> I = 3093.84 lb*ft^2
Car w/ 16's -> I = 3097.04 lb*ft^2 (+0.103%)
Car w/ 17's -> I = 3100.52 lb*ft^2 (+0.216%)

Again, acceleration = torque / inertia, so if you were doing 15 seconds in the 1/4 mile before:
Car w/ 15s -> 15.000 sec
Car w/ 16s -> 15.015 sec
Car w/ 17s -> 15.032 sec

Not very much difference!

But, let's say you got really light tires that were just 10 lbs. (instead of the 15lbs. we decided on earlier)
Car on 15" with old tires -> I = 3093.84 lb*ft^2
(15.000 sec in 1/4)
Car on 15" with light tires -> I = 3070.38 lb*ft^2
(0.8% decrease, 14.89 sec in 1/4)
....
:smoker2:

[DemonicAccord]

you're as right as any re: the 400, another fun thing it does is kill yer mileage by roughly 24-30% everyone i know that got one said yeah they offer a little more power, but always end up removing them

[delsolguy]

Yeah, judging by Technobabble, Dave Coleman strikes me as a pretty smart guy.



Now that I've read your entire post, I see what you are showing. My bad. I get it now :alien2:

Thanks for the clarification.

[Someguy]

No kidding. I traided in my Goodyear F1s for some BFG G-Force KDs. Unfortantly I don't have a scale, but the G-forces were considerably lighter and it really shows in performance. They are also very much stickier, but whenever I get the chance I'll try to find weight differences for them as well as posting 0-60 times, 1/4 times and trap speeds. Should be an interesting comparisome.

[j.fuggi]

and old tune comes to mind when i read fritz's posts....

"...if i only had a brain..."

i wish i fuckin knew half the shit you do man. but you did go to college and i'm still in high school. but you're too smart and can figure out too much@!

[ivymike1031]

Originally posted by fritz_269
There are four wheels, so we just add four times the wheel inertia to the translational intertia (3000 lbs). So:
Car w/ 15's -> I = 3093.84 lb*ft^2

Hey Fritz,

correct me if I'm wrong here, but you can't add 3000lbm + 93.84lbm.ft^2 and get 3093.84lbm.ft^2... I think you'll see what I'm saying...

[ivymike1031]

fyi, I did a quick check using (rotational kinetic energy) / (total kinetic energy), with the vehicle and wheel masses you mentioned, and found that you had overestimated the influence of the wheels considerably. Looking forward to your comments on this subject.

My comment - reducing the rotational inertia of the higher-rev'ing components, such as the crankshaft, will have a much larger impact on the vehicle acceleration. If the engine speed is 3000rpm at 55mph (higher revs conceivable), and the wheel speed is 113rpm, this means that every lbm*ft^2 at the crank will have 705 times the kinetic energy that the same inertia at the wheel will have. This is all "off the top of my head" so feel free to poke holes in it.

[praxis]

I agree with your theory...but that small difference (in 1/4 time), could be all the difference.

I would also like to state, that if all is true...one with a smaller/lightweight rim could get a lighter weight tire, as well. I would have to imagine that the lightest weight larger rim/tire combo could never be less than the lightest weight small rim/tire.

So all in all, it would be more beneficial to get the lightest/smallest, rim/tire combo. Do you agree?

[ivymike1031]

With all other things held constant, that sounds reasonable. The only complaint I have is that you'll never have "all other factors held constant." Chances are good that there will be performance considerations that make certain heavy tires a better choice than certain lighter tires, etc.

[Someguy]

Originally posted by ivymike1031


Hey Fritz,

correct me if I'm wrong here, but you can't add 3000lbm + 93.84lbm.ft^2 and get 3093.84lbm.ft^2... I think you'll see what I'm saying...

The parallel axis theorem?

My comment - reducing the rotational inertia of the higher-rev'ing components, such as the crankshaft, will have a much larger impact on the vehicle acceleration. If the engine speed is 3000rpm at 55mph (higher revs conceivable), and the wheel speed is 113rpm, this means that every lbm*ft^2 at the crank will have 705 times the kinetic energy that the same inertia at the wheel will have. This is all "off the top of my head" so feel free to poke holes in it.

705... Hmmm... Well... What crank are we talking about? What engine and car, are we talking about for that matter? I think we were talking about car "X" with power "Y" and how rim size/weight/weight distribution would effect acelleration. Let's keep the variables and knit picking at least somewhat focuesed here. If not I'll be forced to point out the spin of the Earth due to latitude, longitude, altitude, and track orientation also have messurable effects. If you disagree I have a very interesting grey block manufactured by Litton sitting on my desk that will be happy to disagree with you.

[ivymike1031]

someguy, correct me if I'm wrong, but the parallel axis theorem says the following (in rough terms):

The mass moment of inertia of an object about axis (A) given that you know its mass moment of inertia about axis (B) and that (A) is parallel to (B), is the inertia about (B) plus the mass of the object times the square of the distance from (A) to (B).

In math form:

I_A = I_B + m * r^2

where I_A is mass moment of inertia about A, I_B is mass moment of inertia about B, m is the mass of the object, and r is the radius from A to B.

It DOES allow us to simply add the wheel inertia and tire inertia, as the two objects share an axis (so r^2 is 0). It DOES NOT allow us to add rotational inertia to mass. That would be like adding apples and oranges.

for more info on PAT: http://www.efunda.com/math/areas/Par...xisTheorem.cfm

furthermore, the statement I made about the relative importance of crank inertia vs tire inertia did not depend on any particular configuration of the crankshaft, vehicle or tires. It was based simply on the ratio of kinetic energies of two same-inertia objects spinning at vastly different speeds (example rounded to nearest integers):

1 lb.ft^2 @ 3000 rpm ---> 2080 J
1 lb.ft^2 @ 113 rpm ----> 3 J

to calculate the rotational kinetic energy, simply multiply the moment of inertia by 1/2 the square of rotational velocity. Similarly, for translational kinetic energy, multiply the mass by 1/2 the square of translational velocity.

Why is the rotational kinetic energy important? The energy has to come from somewhere - and in this case, it's up to your engine to provide it. The less mass/inertia you have, the faster you go for a given amount of energy. (remember - power is the time rate of change of energy, or, in simpler terms, how fast you can turn gas into go)

[ivymike1031]

I suppose that if nobody else is going to call me on it, I should point out that 113rpm @ 55mph translates to some pretty big tires. For 26 inch diam tires, you'd get 711 rpm.

To correct the above figures:
1 lb.ft^2 @ 3000 rpm ---> 2080 J
1 lb.ft^2 @ 711 rpm ----> 117 J

relative influence: 18:1
(perhaps it's worth mentioning that the 705:1 figure is still applicable, but at about 8.5 mph, not the original 55mph)

With the faster spinning tires, I find the following:

for the 15 inch wheel case:
* vehicle KE @ 55 mph ---> 427768 J
* tire KE @ 55 mph ---> 10964 J
* tire inertia represents 2.5% of total vehicle effective inertia

for the 17 inch wheel case:
* vehicle KE @ 55 mph ---> 427768 J
* tire KE @ 55 mph ---> 11743 J
* tire inertia represents 2.7% of total vehicle effective inertia

So increasing the wheel diam (under same assumptions as initial post) results in 2/10 of 1% difference in total vehicle inertia.

If you were to remove mass from the vehicle body to compensate, you'd have to take out a meager 5.7lb.

[ivymike1031]

while I'm babbling on about this, I'd like to point out a different way to look at the inertia of the wheels+tires - as an effective translational mass. We can calculate a m_eff for the wheel+tire assy that would take into account the rotational motion of the wheels, but could still be directly added to vehicle mass for the purpose of calculating the change in vehicle acceleration:

for conservation of energy, we need the following to be true:
0.5*m_eff*v^2 = 0.5 * m * v^2 + 0.5 * I * w^2

where m_eff is effective mass of the wheel+tire assy, m is the actual mass of the wheel+tire assy, I is the mass moment of inertia of the wheel+tire assy about its rotational axis, and w is the angular velocity of the wheel + tire assy.

since w = v/r (where r is the tire outer radius), we can rewrite the above eqn and simplify to get the following:

m_eff = m + I/r^2 (remember that this is for each tire)

An interesting thing to note about this formulation is that in the initial post, all of the mass of the tire was assumed to be at the outer edge of the tire, which meant that I for the tire was m*r^2. If we plug that formula for tire inertia into the above formula, we get m_eff = 2m.
For the wheel, the same is not true, as the r for wheel inertia is less than the outer radius of the tire.

So the above formula, combined with the approximation used in the first post, shows us that every pound removed from the tire is worth twice as much acceleration as a pound removed from the body of the vehicle. (times 4 again if you do it to all tires) Mass removed from the wheel does not have quite as much of an effect, but it's still better than a 1:1 relationship.