# Gyroscopic exercise tool

1. Jul 19, 2009

### bobo105

I want to know the relationship between the resistive forces of a gyroscopic exercise tool and the revolutions per minute of the rotor.

2. Jul 19, 2009

### rcgldr

It generates from 32lbs to 35 lbs of torque when speeds of 9,000 - 13,000 RPMs are reached. I assume this means 32 ft-lbs to 35 ft lbs of torque, however the torque is zero unless you try to induce a torque, so I assume this is peak torque.

http://www.basegear.com/powerballgyro.html [Broken]

these powerballs are made primarily of metal and have twice the weight of "regular" Powerball models and are more challenging to operate since up to 250Nm of torque is produced (250Nm = 184.4 ft lb).

http://en.wikipedia.org/wiki/Powerball_(exercise_tool)

The interface between the narrow track and the ends of the gyro axis would wear out (loss of friction), resulting in an unusable powerball after a relatively short period of time. I don't know if this issue was ever fixed.

Last edited by a moderator: May 4, 2017
3. Jul 20, 2009

### bobo105

Thanks, but is there a formula that links the revs/min of the rotor of the gyroscope to the torque?

$$T = I \alpha$$

$$T = \frac{mr^2}{2} \times \frac{\omega}{t}$$

$$\omega = \frac{2 \pi R}{60}$$

$$T = \frac{mr^2 \pi R}{60t}$$

where R = revs/min, r = radius, m = mass, ω = angular velocity, t = time

So far I have found that a torque of 0.25Nm is required to rotate the gyroscope to 17000 revs/min in 1 second (with a mass of 0.45kg and a radius of 0.025m). However, I want to know, for example, how much torque is required to rotate the gyroscope at a constant 15000 revs/min.

Last edited: Jul 21, 2009