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I have heard of the effect of rotational mass on acceleration for along time but I didnt really know how much the effect was. So I decided to try it in real world.
For the sake of the test, I used the same piece of road, same ambient temperature, same opposite car (my uncle's vette). For all races, we would roll it 85 kmh all the up to 260 kmh. Each test was repeated 3 times for accuracy.

First test was with my stock tires ( continintsl contisportcontact in stock oem sizes front and rear)
We ended up head by head (side by side) at the end of the race.

Second race was with my stock tires as well but I added a 30% smaller underdrive pulley
I beated him by 3 cars length.

Now is the shicking news

Third race was with goodyeares eagle f1 asymmetric 2 in stock oem sizes front and rear. The goodyears are about 5 lbs total heavier than my stock setup.
My uncles had me by 2.5 cars length.

So I couldnt belive it and decided to put back my stock tires and try it again.

So I reinstalled my stock continental tires back on
I again beated him by 3 cars length.

Im not sure if this new to you guys or not, but it was definitely new to me as far as real world results. So I decided to share with you guys.
I never thought that changing only a brand of a tire (that is a little heavier) and not the size; would cost me alittle more than 5 cars length.

So I ended up selling my goodyears and placed an order of a set of lighter tires.
 

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no news to me

I have bought 96-tires and 64-wheels from Tire Rack since 2000 (back then the founder was personally installing tires on my car ).

I may be stupid for spending that kind of money (the movers refused to move them all during my relocation), but I like to experiment, and be able to mount the optimal set of tires for it's intended purpose. Thus I think I have gotten to know a lot about tires

My conclusion is that the smallest and lightest wheels that fit over the brakes, the lightest tires preferably 4% under stock diameter, and of course the stickiest compound that you can buy will beat all 20" 21" wheel/tire combos, in real road (not smooth race track) driving conditions in terms of handling, acceleration performance, and ride (less unsprung weight) and most important of all being able to hold the back end in bumpy corners.

I grin when ever I see a signature list of mods that combine a light weight fly wheel with heavy 21" wheels and tires.

Has any one found a nice looking 18" wheel that fits the V-10 M cars?

It may have to be a Custom order OZ wheel to fit over the brakes.


It is bit more complicated to calculate a mass equivalent (static) that can be compared to the rotating mass's inertial resistance to acceleration.

I have searched a few calculations and analysis, but none are to my satisfaction.

Some mention that when accelerating 1-lb at the tire it is equivalent to 6-lb in the trunk, others say it is a 1:10 ratio thus 20-lbs (total in this case) at the tire is like 200-lbs in the trunk which I believe applies better to the rate of acceleration the M5 is capable of.

The faster a car accelerates the more HP is required to overcome the inertial effects of the rotating masses (tires, wheels, rotors, axles, crank, flywheel etc.) This is a fact.

It takes more power to accelerate a flywheel or tire from 1000-rpm to 2000-rpm in 5-sec than the same rotating mass accelerated in 10-sec

The Drum based dyno's measure the time a drum of known angular inertia takes to be accelerated from a starting rpm to an ending rpm.


The HP calculation is based on the work done over time based on the difference in stored flywheel energy at start of run vs. the stored flywheel energy achieved at end of run.

The quicker the drum is accelerated to the higher stored energy state the more HP was required from the engine.

Thus mass equivalent rules or ratio's that describe effects of tire weights can only be applied to cars of similar weight and acceleration capability (a slow accelerating Corolla will barely notice the difference between a 25-lb tire and a 20-lb tire while 1100 lb F1 car will show a substantial reduction in actual time to distance).
 

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