Help with torque and power required please

In summary, the conversation discusses the specifications and considerations for building a skateboard-like device powered by an electric skateboard system. This includes determining the gear sizes, motor sizes and specs, calculating torque and power needed, and accounting for factors such as friction, gradient, and weight distribution. The maximum motor output suggested is 378 watts or 1/2 horsepower.
  • #1
Kevj999
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I am looking to build a skateborad like device that i want to power with electric skateboard-like system. However I only want it to go 1.1m/s using 60mm wheels moving a 100kg load for 12 km. I am looking for help determining gear sizes, motor sizes and specs, really anything. I have seen so much assistance through this forum. Thanks so much!
 
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  • #2
The torque needed will depend on how steep the hills are. It will take more torque when climbing hills than to overcome friction when traveling on the flat. You need to find the torque needed to overcome friction, while climbing the steepest hill.
What grade is the steepest hill ?
Motor or wheel angular velocity is measured in radians per second which is 2 * Pi * RPM / 60.
From that the power needed will be the torque multiplied by the angular velocity.
watts = Newton*metres * radians/second
 
  • #3
I think i would want to calculate for a max 30% grade.
 
  • #4
The maximum force you can apply to your skateboard is determined by the wheel-road friction force ##\mu N##.

I don't think your friction coefficient ##\mu## will be higher than 0.7. If you are powering 2 of 4 wheels, then the normal force ##N## is about half the supported weight, i.e. ##50\ kg \times 9.81\ m/s^2 = 490.5\ N##.

So the maximum force your skateboard can produce is ##0.7 \times 490.5\ N = 343.35\ N##. At the velocity you want, you then need ##1.1\ m/s \times 343.35\ N = 378\ W## of power or about ½ hp. That is what your maximum motor output should be (or less).

At 1.1 m/s, your wheel will revolve at ##\frac{1.1\ m/s}{0.030\ m}\times\frac{30}{\pi}\frac{rpm}{\frac{rad}{s}} = 350\ rpm##. So, for the gearing, whatever rpm your motor is, the gear ratio ##GR## will be ##GR = \frac{rpm_{motor}}{350}##.
 
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  • #5
Note: It doesn't appear that "@ jack action" has accounted for the 30% grade in the above solution.
 
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  • #6
JBA said:
Note: It doesn't appear that "@ jack action" has accounted for the 30% grade in the above solution.
It doesn't matter. It represents the maximum friction force that the skateboard can produce. Whether it can or cannot climb a 30% grade with that force is another characteristic that can be evaluated separately. If it cannot, then there is nothing you can do about it (except increase friction or weight distribution).
 
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  • #7
If you climb a 30% grade at 1.1 m/sec you are rising at 0.33 m/sec.
Your total? mass is specified as 100 kg. The change in potential energy per second will be m·g·h.
100 * 9.8 * 0.33 = 323.4 joules per second = 325 watts, or about half a HP.
That ignores frictional losses in the system.
 

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