Calculating Change in acceleration- Tire Weight

[Floydian]

High I currently have 245/40/R17 rear tyres which weigh 26 (11.7kg) pounds each.

I am planning to change the tyres to 225/45/R17s which weigh 20 (9.1kg) pounds each.

The rims will be the same and weigh 23 (10.5kg) pounds. I do not have the ability to calculate the acceleration improvement. Could someone help with this?

 

[newbe46]

Hm... Newton's second law? F = ma ?

 

[Shannow]

Not just the linear change, need the polar moments of inertia and the mass of the vehicle...and the change in effective radius, and the current acceleration rate of the vehicle...

Too hard to work out, and probably to small to measure.

 

[Inspecktor]

Probably not worth calculating performance numbers, where you will notice a difference is ride compliance. 6 pounds is a noticeable reduction in unsprung weight.

 

[Linctex]

The driver and the car both can often stand to lose a few pounds/KGs... lighter is always faster, no matter where the weight comes off from.

Changing NOTHING else but tire size? Well, it IS calculable... but it is a Very, Very small change in real life.

 

[CR94]

That would be easy if we ignore effects of any change in effective radius, and of the changed moment of inertia of the tires. Acceleration would improve by a factor of w/(w-24)-1, where "w" is the current vehicle gross weight in pounds. That's about 0.6% quicker if, for example, the vehicle weighs 4000 pounds. The reduced moment of inertia might add roughly half that much further improvement, but we don't have enough information to calculate that.

 

[gfh77665]

Not enough information!

We need the density of the air (correlated with elevation), ambient temperature, dew point calculation, grade of the pavement driven on, amount of gas in the tank, octane rating, weight of passengers in vehicle and items in trunk, rolling resistance of the tread pattern of the tires in question, drag coefficient of the vehicle, inflation pressures, and, most importantly, is there a 7-11 Big Gulp in the beverage holder.

Input all variables into the Summit supercomputer at Oak Ridge National Laboratory, and THEN, you will have your answer.

 

[ZeeOSix]

Go to the drag strip for a before and after comparison.

 

[A_Harman]

The racer's rule of thumb: reducing rotational weight by a pound is like reducing vehicle weight by 3 pounds. So the 6 pounds saved per tire would be like saving 36 pounds total.
The improvement in acceleration would be whatever percentage of vehicle weight 36 pounds represents. So if the car weighs 3600 pounds, there would be a 1% improvement in acceleration for a given engine power output.

 

[dtownfb]

Originally Posted by ZeeOSix
Go to the drag strip for a before and after comparison.


+1

This is the only way to do it. Too many variables to calculate.

 

[spackard]

6lbs saved per tire * 4 tires * racer factor 3 = 72 racer lbs.

 

[splinter]

Irrespective of calculating formula(s), quantifiable results have demonstrated lighter wheel/tire assemblies are marginally quicker.


oldies but goodies...
http://www.nsxprime.com/forum/showt...els-Acceleration-Tests-and-their-results
https://www.hotrod.com/articles/light-vs-heavy-wheels-comparison/

 

[Al]

To complicated. Not worth the effort unless you have no life. Will be very very little.

 

[A_Harman]

Originally Posted by spackard
6lbs saved per tire * 4 tires * racer factor 3 = 72 racer lbs.


I assumed from the OP's original note that he was only changing the rear tyres.

 

[Shannow]

Originally Posted by splinter
Irrespective of calculating formula(s), quantifiable results have demonstrated lighter wheel/tire assemblies are marginally quicker.


oldies but goodies...
http://www.nsxprime.com/forum/showt...els-Acceleration-Tests-and-their-results
https://www.hotrod.com/articles/light-vs-heavy-wheels-comparison/

Of course it's quicker...the question was about calculating it...the answer was possible, but too hard, lacking critical information, and to go and measure it..and then it won't be much either..