DC Motor question

Hi All,
I have recently finished making a small ride-in landrover for my grandsons and have a question regarding the electric motor that drives
it. The initial 'road-test' proved excellent but was a little dissapointing when attempting gentle uphill gradients which caused to motor to stall so I just wanted to sound out some expert advise before I decide on how to solve the problem.
I am using a 24volt dc Parvalux motor rated at 200w. I am currently running the motor from a 12volt battery and the drive is geared down to run the landrover at a comfortable walking pace. My questions are as follows:
1. By applying half the rated voltage (12v instead of 24v) am I reducing the power output to 100watts, or just reducing the motor speed by half, or both?
2. Depending on the answer to Q1, if I increase the supply voltage to 24v will I need to adjust the gearing to maintain the current walking pace speed (if the motor will be running twice as fast) and will I have more power available to take reasonable uphill gradients?
I hope my questions make sense and that someone can help!
Regards, Martin.
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I do not know the motor in question but usually on DC motors the speed and power are directly proportional to the applied voltage. At 200 watts you will have about 1/4 HP available which geared down may well be sufficient for your needs. If you can run to it I would fit a 4QD controller that would give you much better control over speed.
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wrote:

It depends on what type of DC motor - series wound, shunt wound or permanent magnet field. For ALL types maximum power at 12V will be less than half the rated power.
A shunt wound machine will try to run at the same speed but the efficiency will be dreadful because of the inadequate field strength.
A series wound machine will be more tolerant of reduced voltage. The speed will be very load dependent, but at the same load it will run at a bit less than half the speed.
A permanent magnet machine will run a bit below half speed but with much better speed regulation.
By doubling the supply voltage you will more than double and possibly quadruple the output power. This should greatly improve the hill climbing. The more the than double change in straightline speed may be a problem and you would need the change the gearbox ratio or add a power controller.
Jim
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Power is squared. You reduced the power to 50 watts if I have my P=IE E=IR stuff interpreted correctly.
Wes
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steambuff wrote:

My suggestion is to find yourself a PWM controller for your motor and run at 24v. There are probably 100's to chose from, some of the robotics sites may be good for information.
Wayne....
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Thanks chaps for all your replies! It would appear to me that to get the rated 200w output from my motor I need to be running on 24v, furthermore I need to change the gear ratio (or get a pair of running shoes!) as it would seem that motor speed will increase also. I don't particularly want to go down the road (excuse the pun) of electronic speed control as it will only cost about 15 to change the gears. Zero to walking pace in 0.5 secs is adequate acceleration for a budding 6 yr old Schumaker and plenty enough for my ageing legs too! <G>
Regards, Martin.
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Speed will double if it's a permanent magnet motor, It should stay about the same if it's a shunt wound motor.
Mark Rand RTFM
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wrote:

Should work fine but the combination of more than doubled power from the motor with torque doubled again by the new gearbox ratio may give you spectacular initial acceleration.
Let us know how you get on.
Jim
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