Xevuhtess7

there's a general consensus that 1 pound of rotational weight on the wheels is about 6 pounds dead weight. i am currently taking an ap analytical physics course in highschool, so with the help of my teacher i wanted to see if its true or not. you may or may not find this interesting, but i did cause i think its cool to be able to apply what i learn in school to cars so i wanted to share. this is what i came up with:

the principals in effect:
ΣF=MA (Sum of the forces acting on an object=Mass X Acceleration)
ΣT=Iα (Sum of the torques acting on an object=Moment of Inertia X Angular acceleration)

Moment of intertia: inertia of a rotating object.

For a cylinderical object (the wheel):
I=½MR^2 M=mass of rotating object, R=radius of rotating object

now with the preliminary info squared away, we begin:

When a car accelerates, the force driving it forwards is the friction of the tires with the road. I know it sounds counter-intuitive because one would think the engine drives the car forwards, but that is an internal force within the car. The friction of the road is the external force driving the car forwards. So we know:

ΣF=MA=ƒ

From here on, let M=Mass of the car's body without wheels, and m=mass of a wheel. Put these into the equation above, and we have:

(M+4m)A=ƒ

Now, we look at the torque acting on the wheels. Torque is force X the distance of the force from the pivot point. The torque acting on the wheels is coming from the friction with the road and also from the axle, which is ultimately connected to the engine. The friction with the road is pointing in the direction of the car's motion because it is the force that causes it to move. When looking at this from a torque perspective, it is acting in the opposite direction of the wheel's rotation, so it is a negative torque. Torque from the axle is positive because it goes with the direction the wheel is rotating in. I'm going to call the torque coming from the axle E.

ΣT=Iα=E-ƒR

Put in what we know about the moment of inertia for a wheel, and multiply it by 4 because there are 4 wheels on a car. Also, notice that α, the angular acceleration, is the same as A/R (putting the angular acceleration in terms of linear acceleration) so go ahead and put that in too.

E-ƒR=4(½mR^2)(A/R)

From the force equations in the beginning, substitute for ƒ:

E-(M+4m)AR=(2mR^2)(A/R)

Simplify a bit and re-arrange to put all the stuff about mass together on one side so we can look at the big picture:

E=(M+6m)RA

Now you can see that when you change the mass of one wheel, it is multiplied 6 times. The important thing to note is that this is for the whole car. that means all four wheels. so if you lose a pound from EACH wheel, it is equivalent to losing 6 pounds off the entire car, not 6 pounds for each wheel. so no you do not save 24 pounds, as previously thought.

thanks for reading, hope you enjoyed it

rominl

iTrader: (4)

wow, cool. i skipped all the math to conclusion. if that's really the case, then it's not as bad as most of us thought huh!

tuddy

PureDrifter

iTrader: (10)

so basically, by removing 4pounds from the car (wheels) it has the effect of losing 2 pounds of "body fat". mainly because of the less mass that the engine has to rotate. this all goes on that 2 identical cars, with one having lost 40lbs from the wheels (10 each) it is equivalent to having lost 60lbs in total vehicle weight.

i think.

PureDrifter

iTrader: (10)

a couple faults i see in this.

1-you are assuming that each wheel is a solid cylinder, which we all know is not true. this will have a major impact on the moment of inertia per wheel.
2-the Radius of the wheel will have as great if not greater impact on performance than weight, and im not talking about handling advantages.oh, and im a high school student myself

XeroK00L

Yes, this is perhaps the single biggest pitfall in Xevuhtess7's assumption, since the bulk of the wheel mass is actually in the outer rim. If the entire wheel mass is in the rim, the moment of inertia would then be mR^2, not ½mR^2, and therefore the conclusion would become E=(M+8m)RA instead. So in reality the factor of the wheel mass to the body mass reduction should be closer to 8 rather than 6. In fact, the fatter the wheel, the closer to 8 the factor, since a fatter wheel has more mass on the rims than a thinner wheel.Yep, but that's why from the conclusion of E=(M+8m)RA it's clear that A (the car acceleration) is inversely proportional to R (the wheel radius). So that part is properly taken into account already. Also, E, R and A are all constants here for the subject. The only variables are M and m.And I was once a high-school student myself.

BTW, nice calculations, Xevuhtess7. We need more math people like you here on CL.

Xevuhtess7

wow im really glad you guys have taken an appreciation to this. i posted this on supraforums and scared the crap out of everyone

about the wheel and tire having different moment of inertia depending on its diameter. also taking into account that the wheel/tire combo shouldnt be assumed as a uniform cylinder. i do realize this but for the sake of simplicity i made these assumptions. also, the situation is under the condition that the wheel's diameter does not change. only the weight of it does. i can try to figure out another proof taking this into account but i think it would be really hard to prove. the results would vary for every specific case (i.e. changing from 15" to 16", 16" to 17", 16" to 18" etc.)

i can try though thanks everyone for your comments

and i found some of the supraforum guys' reactions pretty entertaining
http://www.supraforums.com/forum/sho...d.php?t=340276

LexFather

Mother of God ya'll are incredible

DaveGS4

iTrader: (2)

You had me at ΣT=Iα ...

DaveGS4

iTrader: (2)

Lol, these two are awesome

PureDrifter

iTrader: (10)

lmfao ^^.......

Xevuhtess7

yeah they had me laughing for a good half hour hehe

some good news is that i went to talk to my father's best friend, who owns a local toyota dealership as well as several others in my area. he's gonna have his master technician train me until im competent enough to be a full fledged employee im so excited! it will be awesome to have some formal instruction/experience to accompany what i do in the garage

Xevuhtess7

well if you understand ΣF=MA then ΣT=Iα shouldnt be too hard. its the same idea. the inertia of an object in linear motion is basically its mass, M. A is the linear acceleration of that object. so ΣT=Iα is the rotational version of ΣF=MA with I corresponding to M and α corresponding to A

rominl

iTrader: (4)

hahaa supra forums people are funny

but yeah, i agree, this is actually one very special and cool threads in my eyes

do we need a math forum now?

SWSC400

Thanks! good info