The real affects of wheel weight? Weight in genera

[vtec4life]

Hello all,

I just wanted to know if an 18 inch wheel weighing 15lbs. will share similar characteristics as a 15 inch wheel weighing the same. 15lbs…

My situation: I have a 01’ GSR show car with IS300 front end conversion, complete NSX sides (into the doors), and S2k tail lights. My doors I’m sure are heavier than stock, let alone the rest of the body work being metal. To my understanding, a stock GSR weighs in at ~2,650lbs. With all the body work done to it, I have no idea what it weighs now, but for sure I want to lighten it up a little for what I’m trying to accomplish. (TOP END)

Current plans to lighten it up:- Carbon Fiber Rear hatch
- Carbon Fiber Hood
- FAL glass? windows, side panel window
- Sparcos..Brides??

My purpose: Yes it is a show car, but at the same time, I don’t intend to keep a brand new car garaged for 365 days a year. This will be my daily driver and I do want to lighten it up. I’m a top end kinda guy and that is what I’m trying to achieve is all top end power. Forced Induction? Yes. Going the whole nine yards with the engine, which in the works as we speak.

So the question is: I’m familiar with physics and know about the whole rotating mass deal; how if the mass is further away from the center, than obviously it’ll require more torque to turn it. But the main point here is….does wheel weight and size really have that much of an affect on your performance? – For the situation that I’m in, going all top end and all, wouldn’t the weight of your wheel & car play less of a role as to if you were launching from a dead stop? Thinking back to high school physics…inertia pops up in my head….I’m not too sure if this applies to what I’m asking, but isn’t it that once an object is moving, it’ll keep going until it is stopped? – In this case, once my car is moving at hwy speeds, because it is already moving, wouldn’t the weight (in general) and the size of your wheel become less important in respect of your performance?


I’ve seen 18’s weighing in at only 15lbs. and have also seen some weighing over 28lbs.!!! That’s a big difference…


- When going top end, wouldn’t you want a wider tire to grip the road?
- Would you really want small 15’s on your car at those high speeds?
- I have yet to see a production “sports car†like Corvettes, NSX, Porches, etc...roll off the assembly line with anything smaller than 17’s. Obviously to me, there is a reason for it. :unsure:


Thanks all in advance and hope to hear back from you all with some good feedback.

 

[K2e2vin]

those cars have big wheels to allow the use of large brakes and for future brake upgrades.

 

[vtec4life]

Thanks for the info: Much appreciated. Take care!

 

[D See 2]

hit us up with some pics breh

 

[asmallsol]

well there are 2 problems with those wheels. Rotational Inertia (sp? its early) is basicly the rotaional equizilent to mass.

Lets think in terms of linear units for a second. The heavier something is, the more force it takes to get it to move, lift it, ect. The equation for that is F=ma. If you apply a 100 newton force to 1 kg, it will accelerate at 100m/s. If you apply the same force to a 50kg mass, then it will accelerate at 2m/s

Now rotaional inertia takes into account more then just the mass. It also takes into account both the mass and the distance from the axis of rotaion that mass is. The general equation for a disc rotated around an axis perpendicular to its diameters, I = 1/2mr^2. m is the mass, r is the distance from the axis.

Now in rotaional terms, torque is the rotational equlivielent of force and the equation is t=I x alpha. Alpha = angular accelartion (in radians per second)

Basicly, if you apply 100 newton meter trq to a wheel that has a rotational interita or 100 kgm^2, it will have an angular acceleration of 1 rad/sec.

now the only way to increase the angualr accelartion with a set trq, is to lower the I. There are 2 ways to do this, one being more crutial then the other. (Remember I=1/2mr^2) Lets use some values.

set one. m=5kg, r=.381meters (15inches)
I=.5*5kg*(.381m)^2= 0.3629025 kg m^2

set 2 m=5kg, r=0.4572 meters (18 inches)
I=.5*5kg*(.4572m)^2 =0.5225796 kg m^2

this is an increase of 144% just by increasing the radius of 3inches.

What does this mean?

If you have 10N to apply, in set one, the alpha is going to be 27.5556107 rad/s

in set 2, the alpha is 19.1358407 rad/s

Your engine can only put out so much trq. With a wheel/tire combination that has more rotaional inertia, its going to slow down your acceleration.

 

[vtec4life]

Nice post...good thing I didn't fall asleep in physics..haha


Now that I think about it....i guess it doesn't make a night and day difference, but yet still noticable...




I've decided to go 17's. Centerline RPM's weighing at only 14lbs! But then again....it's more mass spreaded out from the center of rotation....BUT ITS ONLY 14lbs! - Shouldn't make that much of a difference from having the stock 15's on there eh?


hhmm

 

[E_SolSi]

put some 8lb 15s on there and youll see a real difference

 

[xj0hnx]

Originally posted by E_SolSi@Jun 24 2004, 11:11 AM
put some 8lb 15s on there and youll see a real difference

Huge difference.