I was bored in my Organic Chem. lecture so I did some calculations on taking mass off the wheels and flywheel and equivilency to taking mass of the car. These are the equations I came up with:
For every pound off the flywheel is like taking:
=(Final Drive)^2 x (Gear Ratio)^2 x (R*)^2 x (1/2) + 1
off the car.
R*= Ratio of "Flywheel Diameter/Wheel&Tire Diameter"
For Example: In first gear with the FS Flywheel (13 pounds lighter) and stock wheels.
(3.82)^2 x (3.67)^2 x (10in*/20.03in)^2 x (1/2) + 1 = 24.48
This means for every 1 pound off the flywheel it's like taking 24.48 pound off the car, so since our FS flywheel is 13 pounds light, it's like we took 318.25 pounds off the car in first gear!!!
*I don't actually know the diameter of the Flywheel I just guessed it's 10 inches.
I then calculated the same thing for wheels/tire combo.
First I approximated a wheel with it's entire mass on the outer rolling radius. This gave me a ratio of 2.
Then I approximated the wheel/tire as a disc of mass and this gave me the ratio of 1.5.
So, basically taking one pound off a wheel is the same as taking between 2 - 1.5 pounds off the entire car or about:
1.75 pounds per pound off the wheel
Notice how radius of the wheel has no effect on this because a larger wheel also means you aren't spinning the wheel as much to go the same speed. However a larger wheel will effect you drive ratio and the above flywheel equation, slowing you down in those ways.
In case any of you were wondering, yes organic chem. is boring as hell!
![Image]()
If any of you want to see my actual derivations of these equations, I'll post them. It's just some basic calculus and physics.
If anyone has any questions or comments [I have been known to be wrong] feel free to ask.
![Image]()
PS. For the first equation I approximated the flywheel as a disk of consistant density.
PPS. Losing wieght this way is even better than taking wieght off the car because you don't lose the additional traction that comes with real wieght.
[ 03-15-2002: Message edited by: Generic Focus ]</p>
For every pound off the flywheel is like taking:
=(Final Drive)^2 x (Gear Ratio)^2 x (R*)^2 x (1/2) + 1
off the car.
R*= Ratio of "Flywheel Diameter/Wheel&Tire Diameter"
For Example: In first gear with the FS Flywheel (13 pounds lighter) and stock wheels.
(3.82)^2 x (3.67)^2 x (10in*/20.03in)^2 x (1/2) + 1 = 24.48
This means for every 1 pound off the flywheel it's like taking 24.48 pound off the car, so since our FS flywheel is 13 pounds light, it's like we took 318.25 pounds off the car in first gear!!!
*I don't actually know the diameter of the Flywheel I just guessed it's 10 inches.
I then calculated the same thing for wheels/tire combo.
First I approximated a wheel with it's entire mass on the outer rolling radius. This gave me a ratio of 2.
Then I approximated the wheel/tire as a disc of mass and this gave me the ratio of 1.5.
So, basically taking one pound off a wheel is the same as taking between 2 - 1.5 pounds off the entire car or about:
1.75 pounds per pound off the wheel
Notice how radius of the wheel has no effect on this because a larger wheel also means you aren't spinning the wheel as much to go the same speed. However a larger wheel will effect you drive ratio and the above flywheel equation, slowing you down in those ways.
In case any of you were wondering, yes organic chem. is boring as hell!
If any of you want to see my actual derivations of these equations, I'll post them. It's just some basic calculus and physics.
If anyone has any questions or comments [I have been known to be wrong] feel free to ask.
PS. For the first equation I approximated the flywheel as a disk of consistant density.
PPS. Losing wieght this way is even better than taking wieght off the car because you don't lose the additional traction that comes with real wieght.
[ 03-15-2002: Message edited by: Generic Focus ]</p>
