On 1/23/07, Robin van Spaandonk [EMAIL PROTECTED] wrote:
If the motor is unloaded, it's own inertia should carry it past the sticky spot
if it is OU (assuming zero friction). This is because as it accelerates all the
energy is stored in the rotor as kinetic energy. If this isn't enough to get
On 1/22/07, Stephen A. Lawrence [EMAIL PROTECTED] wrote:
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.
This is the error in your reasoning. Assuming the kick is an EM pulse
generating, say, a 1
Terry Blanton wrote:
On 1/22/07, Stephen A. Lawrence [EMAIL PROTECTED] wrote:
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.
This is the error in your reasoning. Assuming the kick is an EM pulse
On 1/23/07, Stephen A. Lawrence [EMAIL PROTECTED] wrote:
I don't think you can neglect inductance, though, for a couple reasons.
Agreed; however, the point was that the energy input decreases with
increased RPM.
The inductance must be delt with as the RPM increases. If you plot
the current
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.
Now you increase the gradient by a factor
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
In reply to Terry Blanton's message of Mon, 22 Jan 2007 19:18:09 -0500:
Hi Terry,
[snip]
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
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