Let's calculate a bit:
Torque From Gyro = (d/dt) AngularMomentumVector
Assume Gyro spinning with axle in plane parallel to the ground
AngularMomementumMagnitude = Iw
Assume wheel is a spoked wheel with all mass out OD
I = MassWheel RadiusWheel^2
Assume Gyro w is constant and robot turns at W about its center.
TorqueGyro = AngularMomementumMagnitude X W
Plug in some reasonable numbers:
8.8lbs wheel = 4kg = MassWheel
16" wheel = .200m = RadiusWheel
2400RPM = 250 Radians/Second = w
AngulareMomementumMagnitude = 4 kg X .2m X.2m X 250 Rad/Sec = 40 kg-m^2 / sec
1 Rev Per Second = 2 X PI Radians / Sec = W
TorqueGyro = 40 X 2 X Pi = 250 N-m
WOW! 250 N-m? How much is THAT?
Well, if you have a robot that is 130lbs and 28" wide, it would take 1800in-lbs to tip it (a simple static analysis 130lbs X 14 inches).
1800in-lbs = 200 N-m
So 250N-m is considerable.
Note that there are a lot of assumptions that will probably mean you will get much much less torque from your gyro than this.
- Your gyro will probably not weigh 9 lbs AND
- Your gyro will probably not be a hoop (though maybe) AND
- Your gyro will probably not be 16" in diameter AND
- Your gyro will probably not be spinning 2400RPM AND
- Your robot will probably not turn at 1 rev/sec
But if you did, there would be some significant tipping torques generated (though the dynamics would be more complex than the static analysis I did - for example in order to get that kind for torque, your wheels would have to be on the ground providing the forces needed to turn at 1 rev/sec, but of course as soon as you begin to tip, you'd loose some of that turning force -- it is a pretty complex dynamic situation to figure out).
It is something to think about.
Joe J.