Rim Inertia

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!!! :eek:

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.
:cool:
[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!)

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...

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:

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

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.

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.

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@!

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...

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.

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?

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.

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. :)

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/ParallelAxisTheorem.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)

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.

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.

ok...weight of honda factory steel wheel 18 lbs. lets say that for all practical purposes...liek fritz said..the stock tires are 15 lbs....33lbs total...now lets say we get some Rays engineering volk racing TE37 drag wheels (by the way ....anyone know the cost on those???) weighing in at a beefy 8.25 lbs...(given thats a 13 " wheel) so we will say we bump to some 16's and we will say they are 12lbs. 6lbs under stock. then we get some lo profile tires to match our factory tire circumference and height. (wouldn't want to mess up the speedo would we) and the lo pro's lets say weight 13 lbs. what type of difference would we be looking at (anyone feel free to make adjustments to my numbers to make them practical) total weight of wheel tire combo, 25lbs that is an 8lb drop....per corner. cutting total weight by 32 lbs. (we could even go with some racing hart 17's weighing in at a beefy 13.5 lbs) oh, wheel weight numbers are coming from turbo, june 2001. anyway what i am getting at. is, in theory the total inertia would depend on the wheel and tire's put together correct? therefore the ultimate goal would be (given you want to keep factory tire circumferneces, so not to mess with the speedo, while trying to attain maximum performance) to reduce TOTAL rim and tire weight by as much as possible correct. with the numbers i proposed 33lbs stock and 25lbs with the racing wheels you are looking at a 24% decrease in rotational mass per tire correct? (8 divided by 33....or loss divided by initial?) this would offer the best gains correct? i need to cut this off...just my .02

Originally posted by ivymike1031

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.



Ya know, when I sit down and do this a completely different (and much messier) way I end up with the same conclusion for the hoop model...

If we use the hollow cylinder model and assume the mass of the tires is constant for different profiles then we get something like:

KE(tire)=.25M(R1^2 + R2^2)*(v/r2)^2 + 1/2Mv^2

plus if we add the rim as a hoop (ignore the disk portion of the rim):

KE(tot) = KE(tire) + 1/2(MR1^2)(v/r2)^2 + 1/2Mv^2

I'd imagine the mass of our rim (hoop) is going to be roughly proportional to its circumference so it will be proportional to its radius. I think we can also assume that the weight of the tread section of each of our various profile tires weighs approximently the same....

So what do we conclude from all this? Given our assumptions, as far as acelleration is concerned mass of the tire and rim approach twice the significance of mass of non-rotating parts as the profile goes down, although the rim much slowly? So little wheels for you drag racers. Handling is a whole other topic though. :)

Interesting... I hope I didn't just assume too much though all of this...

Note: I tried to post this Monday, but it wouldn't go through :( I've just gotten back to my 'puter and will try to catch up on the rest of the discussion. :D

Originally posted by ivymike1031
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...
Good call, I didn't show my units. But I think it may still be correct! I based that addition on the "parallel axis theorem" which states that "The combined effects of translation of the center of mass and rotation about an axis through the center of mass are quivalent to a pure rotation with the same angular speed about an axis through the point of contact of a rolling body." (from "Physics" by Halliday & Resnick) It's basically a shift to a rotational reference frame, so everything ends up in rotational units.

Kinetic energy should work out:
Kt = 4 wheels * 1/2 Icm * Omega^2 + 1/2 M * Vcm^2

where Kt = total kinetic energy
Icm = rotational inertia about the center of mass (the axle)
Omega = angular speed
M = mass
Vcm = forward velocity at the center of mass (i.e. the axle, i.e. the velocity of the car itself)

I gotta split now, but I'll double check my calcs more throughly tomorrow. :)

:confused: :confused: :confused: :confused: help

OK, I think I've got it. IvyMike is right, I was careless with my units, and thus missed a factor which will make the effect of the wheels/tires far less than I previously posted!

Skipping the parallel axis theorem, I'll simply derive an 'effective' rotational inertia for the non-rotational mass of the car. To make things easier for the moment, let's split the car into quarters, and only deal with one wheel and the vehicle mass on that wheel.

KEt = 1/2 Iw w^2 + 1/2 Mv v^2
where
KEt = total kinetic energy
Iw = rotational inertial of the wheel/tire (kg*m^2)
w = angular velocity of the wheel/tire (rad/s)
Mv = mass of the vehicle excluding the wheel/tire (kg)
v = forward velocity of the vehicle (m/s)

We start by relating the forward velocity of the car (v) to the rotational velocity of the wheel (w):
v(w) = w * r (rotational speed * circumference / 2 pi)
where r is the outside radius of the tire.

Now substitute into the KEt equation
KEt = 1/2 Iw w^2 + 1/2 Mv (r w)^2

Now we can see clear to make up a term to call the effective rotational inertia of the non-rotational mass:
Ieff = Mv r^2
And we see that the units work out correctely to kg*m^2

And we can reduce the KEt equation to:
KEt = 1/2 (Iw + Ieff) w^2

So when I just added the mass of the car to the rotational inertia of the wheel/tire, I was off. It should have been larger by a factor of r^2 - about 17% in this case. :eek:

Note: by symmetry, it's easy to reverse the above and see that you can solve in terms of v as well:
Meff = Iw / r^2) ('effective' mass of the wheel in terms of forward velocity) (kg*m^2 / m^2 = kg)
and then KEt = 1/2 (Meff + Mv) v^2
And of course, I just noticed that this is nearly identical to what Ivymike showed just a couple of posts ago! :)

So, to once again retrofit my numbers, I'll reprint the end of my original post with corrections:

...
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. We can modify the translational inertia to an 'effective' rotational inertia (see directly above) by multiplying the translational inertia by the tire radius squared. This will allow us to add and subtract the inertial contributions of the wheel, tire, and vehicle mass as they are now all in the same units. For a radius of 1/2 * 26" = 13" = 1.08 ft, that factor is 1.174 ft^2. So the effective rotational inertia of the translational mass is 1.174 ft^2 * 3000 lbs = 3521 lbs*ft^2.

There are four wheels, so we just add four times the wheel inertia to the effective rotational inertia of the vehicle. So:
Car w/ 15's -> I = 3614.8 lb*ft^2
Car w/ 16's -> I = 3618.0 lb*ft^2 (+0.089%)
Car w/ 17's -> I = 3621.5 lb*ft^2 (+0.186%)

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 -> 15.013 sec
Car w/ 17s -> 15.028 sec

Not very much difference!
...

thanks for the corrections ivymike!
:cool:

Don't forget that the wheels are translating in addition to rotating... If you leave out either the translating or the rotating component of the wheel & tire inertia, you'll cut its effect in two...

This, of course, won't make a difference in your comparison between various wheel sizes, because they were assumed to have the same total mass, but I think it's worth pointing out (for the case where you go with a low-mass wheel).

Originally posted by fritz_269
I based that addition on the "parallel axis theorem" which states that "The combined effects of translation of the center of mass and rotation about an axis through the center of mass are quivalent to a pure rotation with the same angular speed about an axis through the point of contact of a rolling body." (from "Physics" by Halliday & Resnick) It's basically a shift to a rotational reference frame, so everything ends up in rotational units.


The thing you may have missed is that the center of rotation will no longer be the center of rotation of the wheel - it will be a parallel axis that passes through the point of contact ("...about an axis through the point of contact..."). This means that you not only have the inertia of the wheel about its axis, but also a parallel axis term m*r^2, where the distance from the wheel central axis to the new axis, "r," is the radius of the wheel, and "m" is the mass of the wheel assy.

Example: if the inertia of the wheel assy about its central axis is Iw1, then the inertia of the wheel about the "contact" axis Iw2 (parallel, through point of contact) is found as follows:

Iw2 = Iw1 + m*r^2

This gives us a total kinetic energy of:

KE_total = Iw2 * 0.5 w^2 or
KE_total = 0.5 w^2 * (Iw1 + m*r^2)

Let's compare this formulation to the version I showed previously:

KE_total = KE_rotation + KE_translation

KE_rotation = Iw1 * 0.5 w^2

KE_translation = m * 0.5 v^2

substituting v = w*r (tangential velocity is rotational velocity times radius) gives
KE_translation = m * 0.5 * w^2 * r^2

substituting both into the eqn for KE_total gives
KE_total = (Iw1 * 0.5 w^2) + (m * 0.5 *w^2 * r^2)

if you collect the w^2s and 0.5s, you get the following:
KE_total = 0.5 * w^2 * (Iw1 + m * r^2)

which is the same thing that you'd get by using what you called the parallel axis theorem. The parallel axis theorem that I'm used to using (which I used earlier in this post) is described here:
http://www.efunda.com/math/areas/ParallelAxisTheorem.cfm

Originally posted by ivymike1031
Don't forget that the wheels are translating in addition to rotating... If you leave out either the translating or the rotating component of the wheel & tire inertia, you'll cut its effect in two... I just made the assumption that the total wheel/tire mass was included in the overall vehicle mass (3000 lbs in my example).

that's fair enough, but then you have to remember to subtract a portion of the vehicle mass if/when you go to lighter wheels. I just figured that keeping them completely separate would better illustrate the relative magnitude of the two, and the relative significance of reducing the wheel mass&inertia.

Where do you find the time, Mike? - I had just pulled the physics book down off the shelf and figured out the exact same thing. You keep beating me to the punch. :)

If Ip is the Inertia through the point of contact (i.e. the instaneous axis of rotation) then the parallel axis theorem tells us that
Ip = Icm + m r^2
Where Icm is the rotational inertia of the wheel through the axle,
m is the mass of the translating body, and
r is the distance from the axle to the point of contact.

And simply, KE = 1/2 Ip w^2.

I simply tried to add Icm + m, forgetting the r^2. (fortunately, I'd just happened to have picked r=1.08, so my error wasn't too large! ;) )

Thanks for the catch, it's nice to be kept honest. :cool:

Originally posted by ivymike1031
that's fair enough, but then you have to remember to subtract a portion of the vehicle mass if/when you go to lighter wheels. I just figured that keeping them completely separate would better illustrate the relative magnitude of the two, and the relative significance of reducing the wheel mass&inertia. True, but most people weigh their car with the wheels on! :)

Originally posted by fritz_269
Where do you find the time, Mike?

heheeh... it doesn't take nearly as much time when you use this stuff on a regular basis. I have to use P.A.T. at least once a month at work...

Originally posted by ivymike1031


heheeh... it doesn't take nearly as much time when you use this stuff on a regular basis. I have to use P.A.T. at least once a month at work...

I was wondering about that. :) I was at work when I first replied, and had to pull my physics book out when I got home.

Originally posted by Fritz

Where do you find the time, Mike?


I wonder the same thing about you... :)

Originally posted by Someguy


I wonder the same thing about you... :)


I just wonder about all 3 of you........


:smoker2:

After all of that, did anyone take into consideration the fact that a stock rim and tire has its mass closer to the center, where as the 16" rim with low pro tires has much more mass spinning farther out from the center of the wheel? From what I have heard this is the single most influencing factor, not just the weight. Does this make any sence?

Cancivicd16, did you read any of the posts before responding? Several of the calcs dealt ONLY with the effect you are referring to. Take, for example, the very first post in the thread: 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)

So for wheels with exactly the same weight, inertia increases with wheel radius, or as you say "a stock rim and tire has its mass closer to the center, where as the 16" rim with low pro tires has much more mass spinning farther out from the center of the wheel"

all i gotta say is there is deffinatly a difference between lighter wheels and heavier wheels. i just put a set of rota slipstreams (16x7) that weigh 13.5 pounds as compared to the stock lude base model rims which are like 20lbs and are 16x6.5 and i loose traction when i mash the gas while driving in the rain and it never happened before. oh and its the stock tires on the new rims too, yeah i know they suck but im not throwing away perfectly good tires.