Terry Blanton wrote:
Let's suppose you have a circular magnetic gradient of 1 gauss per
degree which delivers 1 Nm of torque to a rotor. Further suppose
that, at the discontinuity of the gradient, you kick the rotor of a
motor past the "sticky spot" with a mechanical force.
Could you use a flywheel to provide the "kick"?
Now you increase the gradient by a factor of 100. Windage and bearing
resistance do not change. How long and hard must you kick?
Well, if the gradient is 100 times stronger, then the "wall" you need to
climb to get past the sticky spot would presumably be 100 times higher,
so you'd need 100 times the torque to get it over the "bump".
So I would think, anyway.
Of course you gain 100 times as much energy going "down the hill" before
you get to the bump, too.
Do not forget to consider that the revolutions of the motor are what
deliver work and the kick is shorter with faster rotation.
True; but the energy it takes to get over the bump doesn't depend on the
speed of rotation. It's the same whether you do it fast or slow.
Similarly, the energy you get out going "down" the gradient before you
get to the bump will be the same, no matter how fast the motor is turning.
Power goes as torque * rpm, because the energy per _revolution_ depends
only on the torque -- it's something like 2*pi*tau. Net energy out
during 1 minute is the energy produced per revolution, times the number
of revolutions turned in a minute.
Terry
