Original poster: Steve Conner <steve@xxxxxxxxxxxx>
how much is the >>braking torque when you just plug motor in and it is trying to start >>rotate rsg wheel? is it equal to the moment of inertia of wheel? >>when a wheel is achieved the full sync speed, is braking torque drops >>to zero? then we really don`t need those horse powers.
At a standing start, the braking torque is just equal to the friction of the bearings which is usually very small. The inertia doesn't really matter, since it can never stop the motor coming up to speed: it can only make it take longer. The mathematical reasoning behind this is that inertia torque is proportional to the rate of change of speed, and so is zero at any constant speed.
At full speed, the braking torque is equal to bearing friction plus "windage", the aerodynamic drag on the disc. This probably accounts for most of the torque, and manifests as air blasting out from the periphery of the disc as if it were a centrifugal blower. Or a Tesla bladeless compressor, even ;-)
Windage can be reduced by enclosing the disc in a circular housing that stops the spinning air from getting away. (if you don't believe this, put your hand over the outlet of a centrifugal blower: I promise you'll see the motor speed up.) However, the air draught is useful for electrode cooling so it shouldn't be restricted too much.
Steve Conner http://www.scopeboy.com/