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Can anyone remind of the correct way of working out the volts amps and watts on a DC motor?

I know it is something like:

\$ V=I \times R \$

\$ I=\dfrac{V}{R} \$

\$ W=\dfrac{V}{I} \$

or something like that...

So, basically, to get the answer of one you multiply or divide the other two? I forget what order it goes in.

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What I suppose I really need to work out basically how I can get what size motor I have on my scooter when it has no labelling on whatsoever. Is there a way of measuring the power of the motor i.e horse power watts etc what is the formula for that. It uses(the scooter that is) 2 12v 33Ah batteries? so we need to know is it power and the watts as we have the voltage im guessing is the 2 batteries? –  Stuart Nov 11 '12 at 10:22
    
Note that your battery capacity does not give you much of a clue as to how much current the motor draws. All you know is that the motor is probably running on either 12V or 24V. If the 12V batteries are connected in series then the motor probably operates on 24V. Current draw of the motor will change depending upon load. You can not directly infer the motor current from the battery capacity unless you did a complicated set of experiments. So in the end you'll have to measure the motor current with an ammeter or put a small value resistor in series and measure its voltage drop. –  Michael Karas Nov 11 '12 at 13:15
    
Is it an electric scooter? –  damien Nov 16 '12 at 9:45

2 Answers 2

The power of a motor is the product of its speed and torque. The power output is greatest at about half way between the unloaded speed (maximum speed, no torque) and the stalled state (maximum torque, no speed).

The output power in watts is about (torque) x (rpm) / 9.57.

The efficiency of a motor η = Shaft Power Out/Electric Power In

Ref:http://www.me.ua.edu/me416/LECTURE%20MATERIALS/MotorEffic&PF-CM5.pdf

(vector) Input Power = V·I and Loss = I²R

If you can measure real power V*I but cannot measure HP, assume 87~90% for η

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Is your motor functioning? Do you have any data belonging to the scooter model? (If yes, you may be able to find the motor used, and access the datasheet). There are TWO major powers that are of concern with motors. THE FIRST is the input electrical power. This power, also known as Pin (because it's the input power), is: P = VI, where P is in Watts, if V is volts and I is in Amperes. The OTHER power is the motor output MECHANICAL power (that is, the mechanical rotational power the motor exerts). This is: P = T*w, where T is torque (in N.m (Newton-Meters)), and w is angular velocity (in radians/sec). P is then in Watts. If you're American, you may want to use RPM and H.P. In this case, use an online converter, or look at what torque/rotational units you need to use. The two powers are not the same. The MECHANICAL power will always be LESS than the input electrical power (due to the second law of thermodynamics), and the ratio of the two: n = Pout/Pin, where the same units are used for P, and Pout is mechanical power. n will change depending on the loading (how difficult it is to turn whatever's attached to it) and the voltage applied to the motor. But assume 24V, if there seems to be no other voltage control in your scooter.

If you do some simple experiments, find the speed-torque curve of your motor (assuming it still runs). Do this by finding the torque (applied load) that makes the motor stop moving. This is one data point. Next, find another data point between the load and the speed. Since the relationship between the torque applied to the motor and the speed of the motor is ideally linear, you can use this characteristic to find the LINEAR relationship between the torque applied, and the speed of the motor. If you like, make a line of best fit. Since the OUTPUT power of the motor is the torque times the rotational velocity. You can generate an output power parabola as well. From all this data, should definitely find a motor (given a motor's spec. sheet provides motor curve info) that matches your application. Just make sure the current of the new motor isn't too much higher.

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