Say I had an object with 4 motors/wheels attached (in a fairly standard arrangement).

I need to calculate the amount of torque required from the motors to be able to move the object of x kilograms consistently (without skipping any steps) at a velocity of y, travelling up a slope of angle z.

I'm guessing this would also depend on factors like the grip of the tyre and such?


1 Answer 1


This is a standard dynamics problem. Let's use this figure I drew:

Corrected car going up incline.

Some definitions:

$$ \begin{align} m & \mbox{, the mass of the vehicle in kg.} \\ \mu_{\mbox{rolling}} & \mbox{, the rolling friction coefficient of your tires.} \\ \theta & \mbox{, the incline of the plane in radians.}\\ g & \mbox{, the gravitational constant, 9.81 } m/s^2 \\ r_{\mbox{tire}} & \mbox{, the radius of the tire in m.} \end{align} $$

The normal force of the vehicle against the incline is going to be $$ f_{\mbox{normal}} = mg \cos{\theta} \\ $$ The normal force is going to play into your frictional force in that $$ f_{\mbox{friction}} = \mu_{\mbox{rolling}} f_{\mbox{normal}} \\ $$

The force opposing the vehicle, trying to cause it to roll down the slope, is $$ f_{\mbox{opposing}} = m g \sin{\theta} \\ $$

Now, keep in mind that the opposing force and the frictional forces are both at the point of contact between the wheel and the surface. Also note that frictional coefficients are constant for a given force regardless of contact surface area, so as long as all four tires are equal and evenly loaded, you don't need to worry about the per-tire information yet.

So, you have your forces to overcome; opposing and friction. Now you need to convert those to torques:

$$ \tau = f r_{\mbox{tire}} \sin{\phi} $$

Where here I've used $\phi$ to help distinguish between the applied force angle and the incline angle $\theta$. Anyways, the force is being applied at a right angle, $\sin{90} = 1$, so you can leave that term off and you're left with

$$ \begin{align} \tau_{\mbox{ losses}} &= (f_{\mbox{opposing}} + f_{\mbox{rolling friction}}) r_{\mbox{tire}} \\ &= (mg( \sin{\theta}+\cos{\theta} \mu_{\mbox{rolling}})) r_{\mbox{tire}} \\ \end{align} $$

Now your speed comes into play, with the formula:

$$ P_{\mbox{rotating}} = \tau_{\mbox{applied}} \omega \\ $$ where $P_{\mbox{rotating}}$ is the rotational power output in Watts, $\tau_{\mbox{applied}}$ is the torque applied in Nm, and $\omega$ is the rotational velocity in rad/s. Your vehicle's linear speed, in $\frac{m}{s}$, is given by:

$$ v = r_{\mbox{tire}} \omega \\ $$ so $$ \omega = \frac{v}{r_{\mbox{tire}}} \\ $$

Putting it all together:

$$ \begin{align} P_{\mbox{rotating}} &= (mg (\sin{\theta} + \cos{\theta} \mu_{\mbox{rolling}})) r_{\mbox{tire}} \frac{v}{r_{\mbox{tire}}} \\ &= v m g(\sin{\theta} +\cos{\theta} \mu_{\mbox{rolling}})) \\ \end{align} $$

Now, finally, this is the rotating power required to move the entire vehicle. At this point, however many (equally sized and loaded) tires you use needs to contribute a corresponding fraction of this power; 4 tires = each one needs to deliver at least 1/4 of the power given above.

  • $\begingroup$ @Bending Unit 22 - thanks for the edit. I'm still new to Latex. $\endgroup$
    – Chuck
    Aug 4, 2015 at 13:34
  • $\begingroup$ This is a nice reference on math meta $\endgroup$ Aug 4, 2015 at 20:19
  • $\begingroup$ sinθ changes to cosθ at the end here, is this a mistake, or am I missing something? Protating=(mgsinθ(1+μrolling))rtirevrtire=vmg(sinθ+cosθμrolling)) $\endgroup$ Aug 23, 2015 at 10:00
  • $\begingroup$ @Alec Dorrington - you are correct. I revised the equation. Sorry I normally post from my phone so the equations can get a little muddled on my end. $\endgroup$
    – Chuck
    Aug 23, 2015 at 12:59

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