# finding the right torque and motors

I am trying to find the right motors for the following case:

I want to have a total of two motors on opposite sides of a board. The board will have a maximum weight of 150kg (330 lbs) and needs to have a maximum speed of 20km/h (12mph). Time for the motors to reach this spead is trivial, but anywhere between 5-10 seconds looks about right.

Note that those numbers can be exceeded but I need to be able to go at least this fast at peak RPM.

I am trying to find the Torque the motors will need in order to move this weight and any other specification such as Voltage and Current, in order to find the right motors.

The diameter of the wheel should be anywhere between 10 - 20 cm (4 - 8 in)

I also don't care if the motor is brushed or brushless. I would prefer to have a brushed ones since I value cost more then lifespan.

What I tried:

Horse power = kg.m/s = 150kg/2 * 20km/h = 5400kg.m/s = 72HP (sounds unreal)

which then by the use of online calculator was converted to Torque to be equal to

T = 2.57 kg.m^2/s^2 (sounds ok? or is it too small?)

[first to imperial HP = 71, then to imperial T = 18.6 - only calculators I found were using imperial]

In this scenario RPM was chosen to be 20000 since I saw a motor that had that much and since raise in RPM means decrease in torque I figured the bigger the better in order to not get a crazy high torque which I expected provided that I have 72HP

Another approach:

Downwards force due to the weight = m.g = 150kg.g

its split between two wheels so F at each wheel is half of that assumming weight falls in the center

F1=F2 = 75kg.g where F1 and F2 are forces at each wheel.

Torque = F1.d where d is the radius of the wheel (in m)

T = 75kg.g.0.075m = 55,2 kg.m^2/s^2

Those two numbers are very differnet though.

What should I do?

• Torque is related to acceleration, so we need to know either the required acceleration, or time required to get to required speed. Feb 12, 2021 at 2:02
• You should also include a diagram of your arrangement including any gearing or pulleys. You need to account for losses if any are significant.
– K H
Feb 12, 2021 at 2:37
• @KH weight of those are accounted for in the 150kg with a bit over so they wouldn't have any impact
– Miro
Feb 12, 2021 at 8:32
• @Mattman944 it is given. I want to be able to reach 20km/h at peak rpm, any normal time to reach that is fine. Since its a DC motor, am i not reaching this spead almost instantly anyway? ( for a chosen rpm)
– Miro
Feb 12, 2021 at 8:38
• What about losses? You need to fully describe your system for someone to really guess at a solution to help you. It takes more power and a more powerful motor to accelerate things more quickly. 1 second, 5 seconds or 1 minute are all "normal" acceleration times depending on the application. It's necessary to know how it will be arranged and the efficiency of any attached gearing systems.
– K H
Feb 12, 2021 at 8:51

Energy = 0.5mv^2 (I'll use SI units here) so to reach 20km/h = 5.55m/s you'll need to provide 0.5 * 1500N * 5.55^2 = 23kJ. If you want to achieve that in 10 seconds you'll need 2.3kW output. Intuitively that makes sense, as a 2HP moped will return about that level of performance with slightly less mass and power. A 10cm diameter wheel has a circumference of 31.4cm, so at the midpoint of the acceleration 2.775m/s it will be rotating at 8.84 revolutions per second = 530rpm. That's irrelevant but interesting. The propulsive force when accelerating is given by F=ma where m is 150kg and a is 0.555m/s^2 so you'll need 82N of force. With a wheel radius of 0.05m that's 4.16Nm of torque.

• A good start for the OP. You haven't considered efficiency, a good mechanical system can have an efficiency of 90+%, a poor one can be much less. Feb 12, 2021 at 13:33
• Torque probably isn't constant for all motor speeds, another thing to consider. For the next step, I would use numerical methods via an Excel spreadsheet. Put the motor torque speed curve into Excel and break the problem into time steps. Feb 12, 2021 at 13:40
• Both valid comments. Given that the design requirements are quite broad, it’s hopefully a worthwhile starting point.
– Frog
Feb 12, 2021 at 19:16

I chased down my university textbook and found this induction motor torque formula:

T=(3 x 9.55 x Pt)/Ns


Pt is rotor power not motor power, Ns is motor rated speed.To calculate Pt, need to find stator resistance, total leakage reactance and angle between them.