HYEPWR

I noticed some people asking how much a reduction in wheel weight would affect a cars horsepower. So I thought I'd post up this reference for everyone. The math is not mine but "TOYGUY" from AudiWorld. Interesting results.


CAVEAT-For those that can't follow the physics involved---this calc is assuming all the weight removed is RIGHT AT THE EDGE OF THE WHEEL!! The effects of weight reduction is dependent on how close the it is from the center. In other words, you could have 2 wheels that weigh 22lbs--->one that has 80% of its weight on the edge and the other has 60%. These two IDENTICAL wheels(by overall weight) will have different effects. The calc below assumes ALL THE WEIGHT is at the edge thus giving the MAXIMUM POSSIBLE RESULT!!

QUESTION
How much does a 10lbs reduction in weight per wheel effect overall apperent vehicle weight??

ANSWER
10lbs/wheel reduction in wheel weight will have an overall apperent vehicle weight reduction of 65lbs (theorectical max of 80lbs).

In other words----1lbs saved per corner is like taking 1.65lbs off the car.


Analysis

For the vehicle in motion, the kinetic energy is given by:

Ekinetic = � * (Mtotal*V2 + 4*Iwheel*w2)

Where
Mtotal = Total vehicle mass including wheels
Iwheel = moment of inertia of a single wheel/tire combination
V = Velocity of vehicle
w = rotational velocity of the wheel (in radians/sec)

Now, V = Rtread * w where Rtread = effective radius of the tire's tread

Combining, we find:

Ekinetic = � * (Mtotal + 4*Iwheel / Rtread2) * V2

So the effective additional mass added to the vehicle due to the wheels rotation = 4*Iwheel / Rtread2

Now, Iwheel = Mwheel * Reffective2

Where
Mwheel = mass of the wheel/tire combination
Reffective = radius of gyration (which is always less than Rtread )

What is this radius of gyration (otherwise known as the radius of inertia)? It is the radius at which an infinitely thin hoop of material of identical mass would have the same moment of intertia as the body in question. It is a mathematical abstraction, but can be calculated for any object.

For example, a disk of uniform density would have an Reffective = 0.707 x Rdisk

Now, finally, the effective additional mass added to the vehicle due to the wheels rotation is:

Mdue to rotation = 4 * Mwheel * (Reffective / Rtread)2

On a per wheel basis, the EFFECTIVE TOTAL wheel mass is given by:

Mwheel, effective = Mwheel * (1 + (Reffective / Rtread)2 )

Reffective / Rtread is always less than 1 and probably somwhere around 80% by my guess. This ratio is a function of wheel and tire weight disribution.

So in this case, Mwheel, effective = Mwheel * 1.64

The absolute maximum (impossible) case would be Mwheel, effective = Mwheel * 2

E55/63 benifits

I'll use member Komp55 as an example. He just got some beautiful BBS RS-Gt wheels and he has Brembo GT brake kit. Going from stock to this combination saved him...

26.6lbs per front wheel and 23.2lbs per rear wheel. Thus, his effective weight reduction is....

26.6 * 1.65= 43.8lbs
26.6 * 1.65= 43.8lbs
23.2 * 1.65= 38.3lbs
23.2 * 1.65= 38.3lbs

effective weight savings of ~164lbs or approx 16rwhp. That should be good for ~1mph in the 1/4 mile. Not to mention that its helps TURNING/BREAKING/FUEL ECONOMY!!!!!!

Akademiks

His front wheel weighs more then rear? You wouldn't happen to know his specs on that wheel would you? It doesn't make any sense, I always thought the rears were the heaviest.

Zod

I think its the total savings, not actual weights. guessing its mostly from the rotors and calipers compared to stock ones.

Thanks for the formula...its alot to take in, but will try it out!

bobgodd

that weight calc is including brakes. the front calipers are heavier than the rears.

ah, beat me to it!

Akademiks

Ah, that would make sense, yes.

KraZy007

Are you sure caliper weight savings can be counted here? Although unsprung weight, the calipers are not part of the rotational mass, rather, they are stationary.

HYEPWR

Caliper--no.

But I believe the weight savings from the kit is *mostly* from the 2 piece Rotors--which are part of the rotational mass. Although, since it is close to the centroid---its effects are small.

The important take home here, I believe is this....

Say you have a the stock 18' rims with 265/40/18's......you switch out some 22 inch giovanni's cause youre such a PIMP.......anyway---EVEN IF You put both wheels on the scale and the stock is 32lbs and the Giovanni's are 36lbs----you say to yourself---ahh thats pretty close. The reality is that not only did you add weight but you moved the weight OUT FROM THE CENTER TO THE EDGE. This is why guys with 20's and 22's go slower at the track and can actually feel the car is slower with the big rims.....

....But they do look baller!!

MJ50

so, if u just upgrade to lightweight wheels, u gain about 5rwhp...

komp55

Your belief is somewhat wrong. Here's what I posted when I installed my Brembo GTs:

Thus, the Brembo GT front calipers contribute about 40% of the overall weight savings at each front corner since they weigh only 67% of the weight of the heavier stock calipers.

All of the math formulas are great, but all I know is that with the unsprung weight reduction achieved as a result of the Brembo GT kits and the ultra lightweight BBS forged RSGT's, the feel of the car in terms of both responsiveness to steering inputs and throttle is markedly enhanced.

HYEPWR

Sorry Komp....Im not following you?

The original question was about rotational mass and its effects on HP. The question I answered was about whether the CALIPER is part of the rotational mass and could be included in this calc---which it isnt.

Reducing Unsprung weight helps handling tremendously, as youve already pointed out---but key for here is the rotational mass which affects acceleration and hp. My statement was ;


Stock Front Rotor-----33.2
Stock Front Caliper----16.15
Total front Brakes-----49.35lbs

GT front Rotor--------25.2(8lbs difference)
GT Front Caliper-----10.8(5.35lbs difference)
total weight----------36lbs(13.35lbs difference)

So 8 out of 13.35lbs savings from the kit is from the ROTOR---which is what I suspected.

Using your explanation from the rear part of the brake instal;


So by your own account---You saved a TOTAL of 40.7lbs(13.35 per front x2 and 7lbs per back *2) and of this 40.7lbs savings--30lbs of which was due directly weight saved from ROTORS.

Just trying to understand your post...So the actual rotational weight saved in the front with the GT kit and bbs wheels is 26.3-5.35=20.95lbs?

jrcart

When I bolted on my light weight carbon fiber Dymag wheels they resulted in about 25rwhp on the dyno. Reductions in rotating mass can result in huge performance gains.

e55 baller

Jcart - welcome back! I think my Kosei's were part of the reason for my N/A 126.6 mph. Rotational mass reduction makes a big difference.

LZH

Not to mention huge reductions in braking.

c32AMG-DTM

Ridiculously fast times in your sig. - how can they possibly be N/A passes though?!?

LZH

Look at his mods:

"ASP pulley, Evo headers, Elect. cutout exh., Evo Cooling kit, K2 ECU & LSD & Hi Flow cats, AAM lowering kit, -10C fan, 160F tStat, VPR 80mm TB & CAMS, 14lb bat., H20 inj., IAT/Boost gauge, EVO rotors, AMG Lighted sills, Sprint Booster, Xpipe, custom air ducts. MT drag radials "

bobgodd

No knock on his car whatsoever, but that still isn't N/A. A N/A engine has no forced induction, not even if it came from the factory.

sack5000

Thanks Baller. That clears it up.

MRAMG1

According to Kennebell:

Tire pressure can change HP by 40 HP
Tire position on the rollers 5-6 HP
Weight of rims and tires, ONLY 5HP

At their website they list over 25 factors that can change dyno results. GOOD info for you dyno freaks. Check it out my friends.

http://www.kennebell.net/techinfo/ge...tVariables.pdf

See yeah

jikjak

so how does this apply to the evosport rotors which are 34 lbs lighter?

mekantor

Yeah, for an actual number you have to integrate density over angle and radius, which will obviously vary for every permutation of wheels and tires.

I have always thought that wheel and tire manufacturers should be reporting rotational intertia as just another number among the many specs they publish. It can be measured, or calculated from CAD if they know the densities of what they are building with.

Jon2007E63P30

Aaah, physics.

Here are a couple of physics questions.

1. You are traveling down the road in Mexico at 120MPH. How fast is your tire going when it touches the road?

2. You are driving your AMG with your 5 year old because your other car is in the shop. Your 5 year old is holding a helium filled balloon on a string because there was no way you were going to let him have an ice cream in your AMG. A boy racer Mustang KR pulls up at the light and wants to play. You look down at your kid and wonder whether he will tell you wife or not. But then you think of all the guys on the forum who would be so dissappointed that you did not defend our honor, so when the light turns green your 5 year olds head gets slammed back in the seat and the KR driver chokes on your exahust.
a. Which way did the balloon move when your kids head was thrown back?
b. How long did it take for your 5 year old to tell your wife?

TopGun32

good topic....

I say... a good set up for our cars would be light weight 18's and try to find a good compromise between a sticky tire and light weight tires.

I love my CLS55 wheels on my E, but I just hate to admit that I'm carrying a bit more weight as a trade for better looks and bit better handling.

mekantor

My physics degree is a few years old, so lets see if I still have the chops...

1. The tire has many comonents of movement: vertical, horizontal, rotational and angular (steering). Which of these are you referring to, or was it your point to trick someone due to the combination of these?

2a. Baloon moves toward the windshield
2b. Depends on the availability of icecream, but usually under 30s

Jon2007E63P30

1. No trick intended. Just velocity in the same vector as the direction of travel of the car. But given your statement, I think you know the answer.

2a. You got it
2b. funny!