I've been looking for a way to size an electric motor for a vehicle that could be assumed as a box-shaped load with a mass of 60 kg, up a 10° incline, at 1.6 m/s, using continuous tracks with 1 motor per side, but for some reason while searching for the answer I found different formulas for this and all of them gave me different answers.

What is the actual formula I should use to find the power required to move this load? As detailed as possible since it's for a practical application not only a theoretical question.

The last one I found is:

$$P = \frac{F_t\ v}{\eta}$$
Where $F_t = Rolling\;Resistance\; +\; Gradient\; Resistance\; + Air\; Resistance$
and $\eta = Transmission\; Efficiency$.

Is this the correct formula I should be using? Am I missing something? Should I consider Air Resistance since the vehicle is not moving too fast? I noticed both rolling resistance and gradient resistance have a friction coefficient, do I use the same one on both equations?


2 Answers 2


At 1.6[m/s] the Air resistance is negligible so I would dismiss that term.

However, a word of caution: Rolling resistance Force is not the same as friction force if you are using wheels.

I would point you to the following page for a rough estimation of Rolling resistance.

The other factors that might come into play are negligible compared to what you have outlined (namely gradient and Rolling resistance).

If you use tires (which I expect with a load of 60 kg), the equation should have a form like:

$$ P = (c\cdot W\cos(\alpha)+ W\cdot sin \alpha)v$$

which can be approximated by the following (as you mentioned in your comment):

$$ P = (c\cdot W + W\cdot sin \alpha)v$$

  • $\begingroup$ When calculating the gradient resistance, should I actually use a friction coefficient? I thought that maybe using tires that it wouldn't make that much of a difference, though tracks might. So my equation in this case becomes (withut air resistance): $$ P = (cW+ Wsin \alpha)v $$ $\endgroup$ Commented Oct 2, 2020 at 10:29
  • $\begingroup$ Oh $c$ is the rolling resistance coefficient used for calculating the rolling resistance as it says on the link you provided before. The equation I posted would be the sum of the rolling resistance and the gradient resistance times velocity. I noticed your fixed equation looks a lot like the gradient resistance with the friction coefficient. $\endgroup$ Commented Oct 2, 2020 at 11:55
  • $\begingroup$ I updated the post, basically incorporating your comments and my reply, while stressing that this is for pneumatic tires. However, I am curious as to where you got your initial equation with the Transmissability Ratio (I've encounter this term sometimes in Gear boxes or Vibration Isolation problems, so I am unsure of the context its being used). If you'd be so kind to provide a reference, I might get a better understanding of where you are coming from. $\endgroup$
    – NMech
    Commented Oct 2, 2020 at 12:05
  • $\begingroup$ Yes, of course. I actually got the equation from the same site you linked, I didn't notice that the page for the rolling resistance was linked at it that's why I missed it until you linked it. Just so I can understand, if I use the rolling resistance coefficient, it replaces the friction coefficient from gradient resistance? $\endgroup$ Commented Oct 2, 2020 at 12:07
  • $\begingroup$ I might be getting confused but it's best to ask than not. But would this be the correct equation? ($c$ is the rolling resistance coefficient and $\mu$ is the friction coefficient)$$P = (F_r + F_g)v = (cW + W(sin\alpha + \mu cos\alpha)))v$$ $\endgroup$ Commented Oct 2, 2020 at 12:16

The friction coeficient is defined by the surfaces in contact, if they don’t change then it won’t.

So tires on tarmac has a friction coeficient but that changes with rain, snow or spilt diesel...


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