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Old 10-06-2003, 01:44 PM   #31
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The way to judge a motor more accurately is by using a dyno. Plugging it in and just looking at the current draw tells you very little, if nothing. By doing this you are just looking at a no-load reading.

It all comes down to the motors efficiency, keep in mind that your only electrical connection to the armature is through the brushes coming in contact with the commutator. This presents mechanical and electrical losses, as well as all the other mechanical losses related to the shaft coming in contact with the bushings / bearings. Other things come also into play, such as armature reaction, Ldi/dt (which is why the sparks in the comm. occur), eddy or parasistic currents(the armature core is laminated instead of solid to reduce this effect), magnetic field warping in the airgap which causes neutral plane shift in the magnetic field when the motor is running. These latter things are just the nature of the beast.

Sorry for stretching this, back to efficiency. There are no two identical motors, this would be almost impossible, since manufacturing tolerances come into play. The reason why it is my opinion that just looking at the current draw on your Pulsar / Pitbull or what ever you use, is of no use, at least to me is because you need other variables that correlate to each other. For example, say a motor X draws 10A @ 3V, thatís 30W input at the motor terminals. Without an output to correlate to the input you have nothing, what is the output power?, rotor speed?, see where this is going? Efficiency = Pout / Pin.

Now lets say we have two motors, X and Y. So say motor Y draws 9A @ 3V but is 5% more efficient than motor X. Lets assume that motor X has an eff = 50% and motor Ys eff = 55%. Motor X has 30W at the terminals, this means that it will put15W at the shaft, now motor Y has 27W at the terminals, being 55% efficient it will put 14.85W at the shaft. So motor Y provides almost the same output of motor X, with 3W less and would provide 16.5W output with the same input as motor X. This difference becomes greater as the load increases, note that bushing / bearing friction, and brush friction also constitute a load. Heavier springs would increase brush-comm friction resulting in more load. Keep in mind that the motors current draw is dictated by the load, this is why the current goes up if you touch the motor shaft while running it on your Pulsar, youíre making the motor work harder. This is also why the more free the drive train the better. Efficiency is everything.

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Old 10-07-2003, 04:58 AM   #32
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Quote:
Originally posted by Joel S
The way to judge a motor more accurately is by using a dyno. Plugging it in and just looking at the current draw tells you very little, if nothing. By doing this you are just looking at a no-load reading.

It all comes down to the motors efficiency, keep in mind that your only electrical connection to the armature is through the brushes coming in contact with the commutator. This presents mechanical and electrical losses, as well as all the other mechanical losses related to the shaft coming in contact with the bushings / bearings. Other things come also into play, such as armature reaction, Ldi/dt (which is why the sparks in the comm. occur), eddy or parasistic currents(the armature core is laminated instead of solid to reduce this effect), magnetic field warping in the airgap which causes neutral plane shift in the magnetic field when the motor is running. These latter things are just the nature of the beast.

Sorry for stretching this, back to efficiency. There are no two identical motors, this would be almost impossible, since manufacturing tolerances come into play. The reason why it is my opinion that just looking at the current draw on your Pulsar / Pitbull or what ever you use, is of no use, at least to me is because you need other variables that correlate to each other. For example, say a motor X draws 10A @ 3V, thatís 30W input at the motor terminals. Without an output to correlate to the input you have nothing, what is the output power?, rotor speed?, see where this is going? Efficiency = Pout / Pin.

Now lets say we have two motors, X and Y. So say motor Y draws 9A @ 3V but is 5% more efficient than motor X. Lets assume that motor X has an eff = 50% and motor Ys eff = 55%. Motor X has 30W at the terminals, this means that it will put15W at the shaft, now motor Y has 27W at the terminals, being 55% efficient it will put 14.85W at the shaft. So motor Y provides almost the same output of motor X, with 3W less and would provide 16.5W output with the same input as motor X. This difference becomes greater as the load increases, note that bushing / bearing friction, and brush friction also constitute a load. Heavier springs would increase brush-comm friction resulting in more load. Keep in mind that the motors current draw is dictated by the load, this is why the current goes up if you touch the motor shaft while running it on your Pulsar, youíre making the motor work harder. This is also why the more free the drive train the better. Efficiency is everything.

My $0.02


thank you to all you guys who said that more amps doesnt equall a faster motor.
trust other poeple to get it wrong.
the thing is theres no identical motors [no apples are the same there ALL different.]so there is no real comparison.
only on the track.
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