RS wheel weight

[MadMaxx]

Anyone feel like weighing their front/rear wheels on an RS? Car is in the shop, trying to get ballpark total weight reduction in suspension/wheel (possible) change.

[scrming]

way to lazy myself... but I know they have to weigh A LOT... LOL

[CamaroSpike23]

you want the weight with the tires or just the rims?

[MadMaxx]

Originally Posted by CamaroSpike23

you want the weight with the tires or just the rims?

Both, I can deduct the tire weight since I have that metric.

[CamaroSpike23]

rog, give me a bit, we posted up the weights of the wheels A LOOOOOONG time ago.

[TheClassicCarKid]

I really want to say 32 pounds.
But that's just a number floating in my brain, pretty much a wild guess

[CamaroSpike23]

I've been digging..... and digging..... and digging. I cant find the thread I'm looking for. its an old one. someone asked what the weights of the wheels were, and I, being the smartass that I am, commented that they were like 32~36lbs. everyone freaked wondering where I got the info. it was a smart ass guess, then we got confirmation of the weights and they were somewhere very close to my guess. but I can not, for the life of me, find the thread

[MadMaxx]

No worries... worst case I'll pull them off tomorrow afternoon and find out for sure

[TheClassicCarKid]

Originally Posted by MadMaxx

No worries... worst case I'll pull them off tomorrow afternoon and find out for sure

If you end up doing that post up the weight on here, it's killing me now haha

[CamaroSpike23]

FOUND IT!!

WEIGHTS FOR 21" AND 20"!!!!
20x8(front) = 30lbs 13.7kg
20x9(rear) = 31.96lbs 14.5 kg

21"x8.5(front) = 31.7lbs 14.4 kg
21"x9.5(rear) = 35.71 lbs 16.2kg


tire weights 245/45/20 .. 275/40/20 .. 245/40/21 .. 275/35/21
.................. 31lbs ...... . 35lbs ...... 39.32lbs ...... 33lbs


http://www.camaro5.com/forums/showthread.php?t=11076&highlight=wheel+weight


and from Anthony at LG motorsports

I have weighed one set of stock 20" SS wheels here so far and they were 63.1 lbs each for the front with a OE tire and 68.3 lbs each for the rear with OE tire. That is ridiculous if you ask me.

http://www.camaro5.com/forums/showpo...6&postcount=20

[stovt001]

I like that nice heavy weight, because that makes up a pretty significant chunk of the car's excess weight. It shouldn't be very hard at all to get a significantly lighter package, and since it is unsprung weight, multiply those savings by 5. Wheels and tires will be the first upgrade for my Camaro, and I plan to spend good money to get the good light stuff.

[UsedTaHaveA68]

Originally Posted by stovt001

I like that nice heavy weight, because that makes up a pretty significant chunk of the car's excess weight. It shouldn't be very hard at all to get a significantly lighter package, and since it is unsprung weight, multiply those savings by 5. Wheels and tires will be the first upgrade for my Camaro, and I plan to spend good money to get the good light stuff.

Doc and I were discussing this the other day and I finally found scientific proof that 5x is probably a bit optimistic and is more like 2x.

Regardless, a set of lightweight 20" wheels and some CCM brakes would do wonders I think.

From http://www.audiworld.com/tech/wheel13.shtml

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.