A traditional HPG/HPM has a current running radially from the axis to the rim (centrifugal current) or radially from the rim to the axis (centripetal current) according to the direction of rotation and the magnetic field direction. The brushes at the axle and the rim are both stationary in the traditional setup to form a stationary external circuit.
In the image below, we have a rotating brush that rotates with the disc and magnet, while there is a stationary brush circuit. The stationary brush circuit is our motor and the rotating brush circuit is our generator. The generator won't oppose the motor, and the motor won't oppose the generator. They will work together. The green arrows are showing the current flow.
There is a slight error in the image below, but if I correct the error, then it will lead to a lot of confusion. The only way to correct the error without the confusion, is to draw it in 3D, and my 3D drawing abilities are very poor. So, instead of drawing a poor 3D illustration, I will build a prototype and take a video. I'll have the prototype finished today.
GB
It's easier to highlight the emfs involved. Below I tried to highlight all 3. You could say that 1 and 2 cancel but then you are stuck with 3, so I don't directly see the generating part generating.
Quote from: broli on June 06, 2010, 07:36:35 PM
It's easier to highlight the emfs involved. Below I tried to highlight all 3. You could say that 1 and 2 cancel but then you are stuck with 3, so I don't directly see the generating part generating.
How can you say the generator part won't be generating? Do you agree the stationary circuit has an EMF that is opposite to the rotating circuit? If you add an additional EMF, which is the stationary brush circuit on the right half (EMF #4), then you can see how it will generate. The stationary brush circuit will have an EMF that is opposite to the EMF generated on the disc, so it will be generating a voltage/current without opposition to the motor. I'll be releasing the video soon.
GB
Video to aid in the visualization, http://www.youtube.com/watch?v=XpluXYcoA2k
@Broli: I really appreciate you looking at this. When the rotating brush is on the left side, it is easy to make the mistake of saying the rotating brush circuit and the rotating disc will have the same EMF and will cancel since they are both rotating together. But, don't forget there is a stationary circuit on the right side with an EMF pointing in the opposite direction that is part of the electrical path for the rotating circuit, which will allow the induced EMF on the disc to flow through the rotating brush to the stationary brush circuit.
The green arrows in my illustration shows the direction of the current and I believe it is correct. The current should flow diametrically across the rotating disc, then flow diametrically in the opposite direction through the rotating brush circuit and stationary brush circuit, then back to the disc.
GB
@Broli:
I added a fourth EMF on the stationary brush circuit to your modified drawing. Please let me know why you ignored or overlooked this part of the circuit (I think we will both agree there is an EMF in that part of the circuit, so why ignore it when it comes to #3?). If it was being ran as a generator, then this stationary brush circuit would have an EMF as indicated in the below image, which is opposite to the EMF generated on the disc in #1. I may have screwed up showing the correct direction of the EMF in the image, but you should understand what I am trying to say by looking at the green arrows.
Thanks,
GB
GB how is it that a stationary wire piece has an emf induced? This is basic hp theory :P . In a motor the stationary part causes the torque whereas the moving part causes the back emf. In a generator it's the reverse. So how can the stationary part be inducing an emf?
Quote from: broli on June 07, 2010, 03:16:17 AM
GB how is it that a stationary wire piece has an emf induced? This is basic hp theory :P . In a motor the stationary part causes the torque whereas the moving part causes the back emf. In a generator it's the reverse. So how can the stationary part be inducing an emf?
Induced or not induced, it has an EMF in that part of the circuit. Are you telling me the stationary circuit has no voltage or current flowing through it, or there is no charge separation in order to have an EMF? Fact is, there is an equal and opposite EMF there, regardless if it is stationary.
GB
Quote from: gravityblock on June 07, 2010, 03:26:42 AM
Induced or not induced, it has an EMF in that part of the circuit. Are you telling me the stationary circuit has no voltage or current flowing through it, or there is no charge separation in order to have an EMF?
GB
Induced means generated by homopolar action, I'm saying a stationary circuit part cannot have an induced emf. Emf occurs when charge is moving, this movement forces an attraction or repulsion from the surface current of the magnet this force in turn causes emf which makes current flow. Stationary circuit parts however can torque the magnet. That is only if one part of the circuit is attached to the magnet, if it wasn't the torque on the magnet cancels. This is basic stuff gb.
Quote from: broli on June 07, 2010, 03:30:17 AM
Induced means generated by homopolar action, I'm saying a stationary circuit part cannot have an induced emf. Emf occurs when charge is moving, this movement forces an attraction or repulsion from the surface current of the magnet this force in turn causes emf which makes current flow. Stationary circuit parts however can torque the magnet. That is only if one part of the circuit is attached to the magnet, if it wasn't the torque on the magnet cancels. This is basic stuff gb.
I know what induced means. I don't care if the current in the stationary external circuit is induced or not. It has an EMF to cause the current to flow from the rotating brush circuit. Voltage and current is flowing through this stationary circuit, right? Then there must be an EMF there, regardless if it was induced on the disc, it is now pointing in the opposite direction in the stationary circuit.
GB