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How to estimate effect of wheel/tire weight on acceleration
Ok everyone, after much thought, I think I have come up with a simple method to estimate the change in straight line acceleration after changing your wheel/tire size or weight:
First of all, in the calculations to be shown below, I am going to assume that the wheel is a cylindrical thin shell and I am going to neglect the weight contained within the wheel spokes. Also, I am going to assume that the tire is a thick-walled cylindrical tube. So, keep these assumptions in mind.
The main mathematical formula to use is the following:
torque = moment of inertia * angular acceleration
We know what the peak torque of the engine is from the dyno charts. We need to compute the moment of inertia of the old wheel/tire combination and compare that with that of the new wheel/tire combination to see how much the angular acceleration will change with the new setup.
Now,
total moment of inertia = moment of inertia of the wheel + moment of inertia of the tire
moment of inertia of the wheel = (wheel mass) * [ (wheel radius)^2 ]
moment of inertia of the tire = (1/2) * (tire mass) * [ (tire inner radius)^2 + (tire outer radius)^2 ]
In these equations, ^2 denotes a squared quantity.
Note the factor of 1/2 which appears in the tire moment of inertia. The factor 1/2 comes in here because the tire cannot be approximated as a thin shell, unlike the wheel.
Now, let's consider an example:
Stock R56 17" crown-spoke wheels with 205/45/R17 Dunlop performance run-flats:
When you plug in these numbers into the above equations, you get:
total moment of inertia of this wheel/tire combination = 4085.5125 lb.inch^2
Let's say we want to replace this combination with 18" light OZ Ultraleggeras with 205/40/R18 Pirelli PZero performance run-flats:
new wheel weight = 17 pounds
new wheel radius = 9 inches
new tire weight = 22 pounds
new tire inner radius = 9 inches
new tire outer radius = 12.2 inches
new total moment of inertia = 3905.24 lb.inch^2
So, as you can see, the total moment of inertia of the new combination is slightly less than that of the old setup. The angular acceleration is inversely proportional to the moment of inertia. So, with the same torque applied, the new setup will have about 4.6% larger acceleration than the old setup.
Of course, this new example setup includes a different tire whose grip may also affect the straight line acceleration.
Anyways, I hope this method will be useful to someone out there!
__________________ 2010 Redline Time Attack Enthusiast FWD Class Winner at Sebring (Thanks to RMW!)
07 Laser Blue R56 MCS with White Top, JCW Aero/Engine/Suspension/Brake Kits, Forge Intercooler, RMW Tune, Factory LSD, Vorshlag Front Camber Plates, JCW Strut Brace, Hotchkis Competition Rear Sway Bar
Your on the right track but those equations are for uniform bodies. The tires have most of their mass way out at the tread so the difference in moment will not change much between the different tire sizes. A good assumption for the moment of these is to subtract an inch from the OD and just use the MR^2 formula.
The wheels are a bit more complicated. I'm not really sure how to estimate where the center of mass is of a wheel cross section. Your assumptions on the wheel need a little work. Your basically assuming both wheels are exactly the same except one is 1" larger than the other (this is fine to do since you don't have any real way of measuring the actual). You are also assuming that all the wheel mass is at the outside edge of the wheel. This is not right. You need to come up with another value, say 2/3 or 3/4 of the total diameter of the wheel. This should give a better estimate of the total moment of the wheel/tire.
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'02 MCS, Sport, Premium, Webb Sportbox RAF, 15% pulley, Milltek cat back, H-Sport comp rear swaybar, Powerflex engine and gearbox inserts, Stewart Warner boost and oil temp gauges in E-Pod, Optima Red Top 35, Whalen knob, M7 front grill, Hella FF200 fog lights on Out Motoring brackets, MCAW
Thanks for the comment. I realize the assumptions I made are a bit crude.
No problem. It's good to see your getting an idea of how it works.
ED955S, you can't directly relate moment to acceleration. If you did, you would have infinite acceleration if your wheels didn't weigh anything. You still have to take into account the mass of the car. By finding the moment of the tire/wheel, you can determine the torque required to accelerate it. This is done by multiplying the moment with the angular acceleration. This means if you have no acceleration, there is no torque needed to change its velocity. As the acceleration increases, the effect of lower moment of inertia comes into play. Still, this is just going to tell you the difference in torque to spin the wheels. Say you do all the calculations and come up with 5 ft*lbs. This means that the engine now has an additional 5 ft*lbs of torque to accelerate the car rather than rotate the wheel and tire.
This is why a wheel/tire change can influence dyno results if the dyno is inertial. True measurements of engine output should be done at constant rpm steps instead of a sweep throughout the rpm range.
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'02 MCS, Sport, Premium, Webb Sportbox RAF, 15% pulley, Milltek cat back, H-Sport comp rear swaybar, Powerflex engine and gearbox inserts, Stewart Warner boost and oil temp gauges in E-Pod, Optima Red Top 35, Whalen knob, M7 front grill, Hella FF200 fog lights on Out Motoring brackets, MCAW
where: IR = inertia resistance [N] m = car mass + equivalent mass of rotating parts [kg] a = car acceleration [m/s2], (from 0 to 100 km/h in: 6 s (4.63 m/s2), 18 s (1.543 m/s2)) mcar = car mass [kg] meq = equivalent mass of rotating parts [kg]
= [ Iw (1/rw)2 + Ip [FONT=Symbol]h[/FONT]f (if /rw)2 + Ie [FONT=Symbol]h[/FONT]t (if ig / rw)2]
where: Iw = polar moment of inertia of wheels and axles ≈ 2.7 [kg m2] Ip = polar moment of inertia of propeller shaft ≈ 0.05 [kg m2] Ie = polar moment of inertia of engine ≈ 0.2 [kg/m2] + polar moment of inertia of flywheel and clutch ≈ 0.5 [kg m2] [FONT=Symbol]h[/FONT]f = mechanical efficiency of final drive [FONT=Symbol]h[/FONT]t = mechanical efficiency of transmission system ([FONT=Symbol]h[/FONT]g x [FONT=Symbol]h[/FONT]f) ig = gearbox reduction ratio [ig1 or ig2 or ………….] if = final drive reduction ratio rw = tire radius [m]
* ( + ), (+) with the car in acceleration. {tractive resistance}
(-) with the car in deceleration. {tractive effort}
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2009 R56S, solid BRG, manual, JCW suspension and strut bar, Pirelli PZero, JCW air filter, IE fixed camber plates.
That formula should take everything into account. It's just that you have to find the equivalent mass of all the rotating parts in the entire drivetrain. Since everything is the same except wheel and tires, you could drop everything else out of the equation so IR= the polar moment of inertia of the the wheels and tires. That's what we're trying to calculate above.
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'02 MCS, Sport, Premium, Webb Sportbox RAF, 15% pulley, Milltek cat back, H-Sport comp rear swaybar, Powerflex engine and gearbox inserts, Stewart Warner boost and oil temp gauges in E-Pod, Optima Red Top 35, Whalen knob, M7 front grill, Hella FF200 fog lights on Out Motoring brackets, MCAW
