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Originally Posted by fox46
By your method, (86 * 0.2) * (12 * 0.2) simplifies to 0.04 * (86 *12), the 0.04 representing 4% rather than 20%
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If you apply 12V to an RS775 that has a locked rotor, it will draw ~87 amps. This is straight ohms-law stuff. The DC resistance of the windings is 12/87 = 0.138 ohms
If you only want 20% of that current, you need to reduce the effective voltage to 20% of 12. So (0.2)(12)(0.2)(87) is the power draw at 20% of stall torque.
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My understanding is that the Jaguars control output by pulsing the power from the battery.
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Correct. It's called pulse-width modulation (PWM).
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They don't actually control the voltage or current.
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They control the effective voltage and current by turning it off and on at 15,000Hz and varying the "on" portion of the duty cycle.
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They are essentially a high speed switch which opens and closes at a specific frequency.
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Correct.
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At full power the switch is closed and supplying full power from the source.
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Correct.
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At 50% the switch is only closed 50% of the time and only 50% of the power is getting through per unit time.
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Not correct. If you reduce the voltage by 50%, the current will also be reduced by 50% (for a locked rotor which acts like a DC resistance with coil inductance), and the power will be 25%.