Rim Inertia

[white97ex]

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

[Someguy]

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...

[fritz_269]

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.

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.

[Moppie]

help

[fritz_269]

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.

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!

[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...

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).

[ivymike1031]

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/Par...xisTheorem.cfm

[fritz_269]

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).

[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.

[fritz_269]

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.

[fritz_269]

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!

[ivymike1031]

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...

[Someguy]

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...

[Moppie]

Originally posted by Someguy


I wonder the same thing about you...

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


:smoker2:

[Cancivicd16]

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?