JACKK

I have narrowed my wheel choices down to 2 wheels, and one wheel is 8lbs lighter than the other.
Would the performance of the car be the same for the following.............
20lb wheel with 32lbs of junk in my trunk
-and-
28lb wheel with no cargo
????

JACKK

Don't correct my spelling - Just advise if you know answer

Former A6 4.2 sport

No, it would not.

Adding 1 lb. of unsprung weight, is the same as adding about 10 lbs. in the trunk or cabine, etc.

In case of your wheels: it would be 320 lbs, not 32 lbs, in your trunk.

JACKK

Thanks for the info - Lighter it is

2gen

just curious, do u have a simple explanation of the physics behind this?

Bollinger

The wheels are rotational mass, so they are not the same as something in the trunk. The "unsprung" comment was misleading though, since it is only rotating mass that matters.

320lbs seems far too high. I say this because quattro has far more rotational mass than any 2WD system and it seems to do quite well in racing. In Speedvision GT they make them add 150lbs of penalty weight above the extra mass of quattro and it still can beat them all.

Do you have the formulas behind this math?

Former A6 4.2 sport

Simple test would be to run 20lbs wheels with about 320lbs worth of additional people in the car and then the same car with 28lbs wheels and those 320lbs removed. The theory is that under the same conditions - road, weather, etc - both tests of the same car will exhibit similar cornering, acceleration and breaking results - suggesting the negative effect of the wheel weight on handling and so on.

Over the years I tried many different wheels in my cars, and noticed a noticable negative effect of heavier wheels on acceleration and breaking. No G-tech hard numbers, just the "feel of my pants".

Former A6 4.2 sport

I am not very race knowlegable, but it seems to make sence, at least to me, that 4WD cars have sagnificant traction advantage and no wheel speen unlike most powerfull 2WD cars, hence requiring a weight penalty in Gt racing.

2.7Tango

If by performance, you mean acceleration or braking on a smooth surface, then the wheel weight can make a difference, but never more than 2x. So, unless the heavier wheel has a smaller or equal rotational inertia (I think this is the term we used way way back in physics) than the the lighter one, the impact of an additional 32 lbs of wheel weight would be somewhere between 32 and 64 lbs of unrotating weight.

If by performance, you mean the ability of the suspension to keep the tires planed on the road surface, then lighter is better and this can help cornering, braking, and acceleration, but it is not so easy to quantify. This is where I would expect the biggest benefits of a lighter wheel to come in. However, I have heard people discuss rules of thumb like 10x.

If by performance, you are talking about the wheel's ability to maintain its structural integrity through the punishment it receives from bumps and potholes in the road, then lighter is not necessarily better.

I think the second two issues are the ones that will probably have the biggest impact. I am not sure if you are doing the kind of driving that will make the 8lbs less unsprung weight be noticable, but 8lbs is a big chunk of something that probably weighs in the 15-35 lb category anyway, so you want to make sure that weight savings isn't buying lighter unsprung weight at the cost of durability. Of course, it is possbile the lighter wheel is also more durable.

Ming 2.7T

Give it a read and let us know what you think. I think,,,, it came to the conclusion of about 1.7 times??? Your an engr, so look it over and then try to explain in "simpleton" ( my language!!!).
Hope this helps some.

Effect of Wheel Mass on Acceleration Eduardo Martinez 2001

What is the effect of a change in wheel/tire mass? Several people have quoted factors of x6 or x8 to determine the equivalent non-rotating mass variation. That is, an increase of 10lbs/wheel could be as detrimental to performance as carrying Arnold Schwarzenegger and Scooby Doo as passengers.

I decided to refresh my Physics and came up with the following conclusion: from the point of view of acceleration, an increase of X in wheel or tire mass is no worse than an increase of 2X in passenger mass. Not 6x, not 8x, just 2x worst case. This is why.

At any given speed/gear combination there is maximum torque T available at the wheels. The torque does two things: (1) opposes the rolling resistance and the aerodynamic drag and (2) accelerates the car. The equation that describes the equilibrium is:

T = I*u + M*r*a + D*r

*: denotes multiplication
T: torque at the wheels
I: total moment of inertia of the rotating parts
u: angular acceleration of the rotating parts
M: total mass of the vehicle
r: external radius of the tires
a: linear acceleration
D: total drag and rolling resistance

The first term, I*u, is the torque used in making the wheels rotate faster. The second, M*r*a, is the torque used to accelerate everything (wheels included) in the direction of movement. The third is the torque used to cancel the drag and rolling resistance.

Another way of writing the equation above is:

T/r = I*u/r + M*a + D

Now the terms are forces. The left side is the force available at the contact patch.

If the tires are not slipping, the angular acceleration and the linear acceleration are related in the following way:

a = u*r

Replacing in the equation and moving things around, we get

T/r - D = (I/r^2 + M)*a

^: denotes exponentiation (r^2 means "r squared")

We can say that the left side is the force available for acceleration. Such force accelerates an "non-rotating equivalent mass" E,

E = I/r^2 + M

Now suppose that we increase the mass of the wheels and tires by an amount X. Both the total mass and the moment of inertia will increase; let's call the new mass M' and the new moment of inertia I'. Obviously,

M' = M + X

What about I'? Well, the moment of inertia of a "punctual mass" [m] (a mass concentrated in a point) at a distance [r] from the axis of rotation is [m*r^2]. That is,
the moment of inertia depends critically on the distance between the mass and the axis of rotation.

In a real wheel+tire combination the mass is distributed in different amounts at different
distances from the center. In order to compute the total moment of inertia we would need to know the mass distribution and use integral calculus. We can do a simpler thing though. We can make the pessimistic assumption that all of the mass increment is
located on the periphery of the tire, that is, at a distance [r] from the center. This assumption is pessimistic because in a real wheel some of the mass will be located closer that [r] and will contribute less to the total momentum (it is not too pessimistic though: most of the mass is located pretty far from the center, if not at the periphery). So now we can compute I',

I' = I + X*r^2

The new "equivalent mass" is,

E' = I'/r^2 + M' = I/r^2 + M + 2*X = E + 2*X

In other words, from the acceleration point of view, the equivalent non-rotating mass increment corresponding to an increment X in rotating mass is - at worst - 2X.

NOTE: After doing some approximations and assumptions about mass distribution in a typical wheel+tire combo, I believe that 1.7X is a better approximation. A 10lb/wheel mass increase would not hurt acceleration worse than carrying RinTinTin.