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I built a basic wooden go-cart and I want to convert it to electric power, and my question is how to calculate the motor. I searched here and apart from this 24V 250W motors I can't see any similar questions. The go-cart is not intended to be high performance. Details are:

  • It has 4 wheels which rotate independently on two axles, the front axle is for steering and rear axle fixed.
  • It's intended for paved or mostly-smooth and mostly-level surfaces
  • For simplicity I planned one motor driving one rear wheel, or two identical motors driving two wheels.
  • The wheels are 23cm diameter (0.72m circumference) and I realize gearing is needed between the wheels and the motor.
  • Total mass of go-cart, passenger and motor <= 50 kg
  • Maximum speed of say 4 m/s, go-cart doesn't need to accelerate faster than 2 m/s/s
  • The max speed and acceleration are very approximate and I don't mind if they are greatly different from the above. I based the max speed on the speed of a running person
  • Powered by 12v or 24v lead-acid batteries
  • Motor to be controlled by controller or even a large rheostat https://surplussales.com/Potentiometers/Rheostats/PotsRheost-1.html

So my approach is to calculate the kinetic energy (KE = 1/2 M V^2) at different speeds to see what power is needed:

  • Speed of 1 m/s = 2.2 mph (equivalent to walking). KE = 25 x 1^2 = 25 Joules
  • Speed of 2 m/s = 4.5 mph (fast walk). KE = 25 x 2^2 = 100 Joules
  • Speed of 3 m/s = 6.7 mph (jogging?). KE = 25 x 3^2 = 225 Joules
  • Speed of 4 m/s = 9 mph (running). KE = 25 x 4^2 = 400 Joules
  • To acquire a velocity of 2 m/s in 1 second = 100 Joules in 1 second (requires 100 watts)
  • To acquire a velocity of 4 m/s in 2 second = 400 Joules in 2 second (requires 200 watts)
  • Power = V x A, so assuming a 200 watt motor, this needs 200/12 = 17A from the battery.
  • This website https://www.powerstream.com/battery-capacity-calculations.htm says (step 3) if we drain a lead-acid battery in 1 hour we get about half of the rated capacity. So to get 17A for 1 hour would need a 12 volt 35 A/hr battery
  • When we know the required current (eg 17A above) this would be the minimum current handling capacity of the motor controller or variable resistor

I realize this method is very approximate, the KE isn't dissipated in 1 second, and the calculations don't allow for friction, gearing, motor or other losses. So my question is: how easiest to calculate the required motor for the go-cart as described?

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Oct 7 '20 at 20:08
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You can use ramp medel and required climb-up velocity. For example you need to climb 15-degree ramp at 4m/s and you cart is about 100kg. So it equivalent lifting up 100kg mass at speed 4sin(15 deg) = about 1.04 m/4 so you required mgh/t or mg*v power. Then you got 100 * 9.81 * 1.04 = 1020.24 watt of mechanical power. (don't forget to consider efficiency of motor to convert electrical energy to mechanical energy).

If you planned to use DC motor you need to know torque-speed curve of DC motor enter image description here (Image credit: National Instruments Corporation) At 50% of noload-speed, the motor will provide maximum power, so you can design gear reduction to make motor turn at 50% of no-load speed when climbing the ramp. There are no absolute way to select motor. there are to many unknown factor such as air resistance or mechanical friction. this calculation will give some guild line and you need to do some experiment to verify and adjust some setting.

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