I am trying to build a "mule": it s basically a go kart chassis with a 12v dc motor (I figured it would be easier to use). Now I need to determine how powerful should the motor be...

It is going to have 4 small wheels (15 cm in diameter - maximum), and be rear wheel driven, and it's going to do some soft kind of off roading (backyard terrain)
It is going to carry up to 150 kg at low speeds (5km/h max). Should I use a torque converter? I was thinking of using 2 cordless drills to do the job...

Do I need torque or power in this case? I'm thinking I'd need plenty of torque as it is a pretty heavy load and the project requires low speeds...

How can I determine the needed power / torque for future projects? Is there a formula (considering air friction 0).


1 Answer 1


The force required $F_{\text{r}}$ for driving a wheeled vehicle with mass $m$ is given the following formula (I will neglect slip):


The frictional force of the tires is given by $F_{\text{tire}}=c_{\text{tire}}mg.$ The dimensionless tire friction coefficient $c_{\text{tire}}$ is in general between $0.005$ and $0.010$.

Aerodynamic resistance is given by $F_{\text{aero}}=\frac{1}{2}\rho_{\text{air}}c_{\text{D}}A_{\text{ref}}v^2_{\text{rel}}.$ The dimensionless drag coefficient is between $0.28$ (good aerodynamics) and $0.80$ (bad aerodynamics e.g. for trucks). The reference area $A_{\text{ref}}$ is the projected area of the front face of the vehicle. For very small relative velocities $v_{\text{rel}}=v+v_{\text{wind}}$ aerodynamical resistance can be neglected.

The force needed for a given translative acceleration $\ddot{x}$ is given by $F_{\text{acc.}}=e_{\text{m}}m\ddot{x}$. The dimensionless coefficient $e_{\text{m}}$ is there to account that it is also necessary to accelerate the components of the motor, gearbox and so forth. In most cases it is between $1.05$ and $1.40$.

The force $F_{\text{slope}}=mg\sin{\alpha}$ is necessary to overcome a slope of $\alpha$ (in degrees, so remember to calculate $\sin{\alpha}$ in degrees and not in rad).

So in total we get:


in order to get the required power $P_{\text{r}}$ you simply multiply $F_{\text{r}}$ with the velocity $v$ of the vehicle. Note that in general, when considering the wind this velocity $v$ will not be equal to the relative velocity $v_{\text{rel}}$.

  • $\begingroup$ Re aerodynamic resistance can be ignored for slow velocities - true for the vehicle speed relative to the ground, but the wind speed can make a difference... $\endgroup$
    – Solar Mike
    Commented May 8, 2017 at 12:21
  • $\begingroup$ @SolarMike: Thank you for your comment. Edited my answer. $\endgroup$
    – MrYouMath
    Commented May 8, 2017 at 12:37
  • $\begingroup$ Like your answer anyway. $\endgroup$
    – Solar Mike
    Commented May 8, 2017 at 13:23

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