# Math behind 5 and 4 inch flywheels

Preface:
I decided to write a post concerning the math and calculations behind the different sized wheels and number of wheels you see on the different flywheels out there.
Conclusion:
One 4" wheel requires 2.78 times the gear ratio of a single 5" wheel
Two 4" wheels only require 1.39 times the gear ratio of a single 5" wheel
Distance moved in one rotation makes a baller difference.
Calculations:
(note that I use metric)
The big 5" wheel weighs in at 160g, while the smaller 4" wheel weighs in at 90g.
The big 5" wheel has a circumference of 39.9 or about 40cm.
The 4" wheel is at 31.9 or about 32 centimeters in circumference

KE= (Mass* velocity^2)/2
Velocity = meters/ seconds
At one rotation per second, you would have a rotational velocity for each wheel equal to .4m/s for the 5" wheel, and .32 m/s for the small 4" wheel

So lets plug those values in: ( All values in standard meters, kilos, seconds)
4" wheel has a KE at one rotation per second of:
(.090*.32^2)/2=.004608 joules

5" wheel has a KE at one rotation per second of:
(.160*.4^2)/2= .0128 joules

Dividing .0128 by .004608 Gives us the ratio of the required gearing- 2.78 times the gear ratio is required to run one 4" wheel vs one 5" wheel.
This is why you commonly see people run two 4 inch wheels- it brings the required gear ratio to only 1.39 times the required ratio.

I hope this helped in terms of understanding the flywheel dynamics. If I get asked, I might explain why the KE of the launcher is so important, as well as where KE comes from.

Something is not correct here. I do not really understand the math, but I do know that it does not require 2.78 times the gear ratio for a single 4" wheel vs a single 5" wheel.

Actually, it is. This is not just speculation, I’ve tested this in the classroom, and it comes out pretty close. The only distinguishing factor is the compression on the ball, which has a partial effect as well.

A somewhat more in depth article can be found here:

The very big difference comes with the change in angular velocity- doubling the velocity of a flywheel quadruples its KE. Doubling the mass only doubles the KE.
Because the 4" wheel is 90g, and the 5" is 160g, you can already expect 90/160 or about .5625 the KE from the 4" wheel. As difference in circumference is directly proportional to velocity, and the circumference are 40:32, or 5:4 or 1.25:1, you can expect a loss of KE of 1.25^2 or factor 1.6.

2800rpm on a single 5" wheel full courts.
3600rpm on a single 4" wheel full courts.

3600rpm / 2800rpm = 1.28 =! 2.78

Your calculations are calculating the energy stored in the flywheel, not what is transfered to the ball.

However you haven’t really stated what the rpm ratio between 4 and 5" are supposed to mean.
Do you need 2.78x more rpm to store the same amount of energy or full court?
Because the later would be incorrect.

Never mind, stanley. You are right. Its angular velocity that matters, and that is in radians per second. In other words, only the masses of the two wheels should affecy their KE. So 160/90 is 1.5625 times the gear ratio is required. Still not sure why you have 1.28 times the gear ratio only…

for 1.28 I meant that a 4" wheel needs to spin around 1.28x faster than a 5" wheel.
3600rpm / 2800rpm = 1.28
nothing to do with KE of the flywheel.

Well, the ke of the wheel is the reason it needs to turn faster at all. If the angular velocity is rads per second, and not dependant on the circumference of the wheel, then should it not be a ratio of that mass the rpm has to make up? Thereby, the square root of the ratio 1.5625:1 is the effectice factor of the rpm required…
Which is 1.25:1… I concede…