# How can I calculate the power and torque required for the motor on a wheeled robot/vehicle?

How can I calculate the power and torque required for the motor on a wheeled robot or vehicle if a particular acceleration or movement up an incline is required?

• Huh. That's odd. Does the Engineering Beta StackExchange not support MathJax like the Electrical Stack Exchange? – DKNguyen Oct 13 '19 at 23:45
• It does. For some reason all your $ were escaped with backslashes, which disables their LaTeX interpretation. – Wasabi Oct 14 '19 at 2:08 • @Wasabi Weird. I need the backslashes there for it to work on the EESE. Thanks for fixing. – DKNguyen Oct 14 '19 at 2:25 • On EE.SE the inline MathJax uses the \$ syntax whereas on many of the others it is just \$. It makes it easy to discuss cost on EE.SE. – Transistor Oct 14 '19 at 18:49
• @Transistor Interesting, though I am not sure why cost would be talked about any less on other engineering stack exchanges. – DKNguyen Oct 14 '19 at 18:51

The biggest unknown that needs to be determined or estimated is the coefficient of rolling friction.

For reference, a coefficient for rolling friction of 0.3 is already very high and is for something like for soft wheels (which deform) on a dirt road (which isn't flat or hard) where it is light enough it won't sink in. Most of the time it should be more like 0.1 to 0.2 with it being lowest on smooth hard surfaces with smooth hard wheels.

A coefficient of friction of 0.3 means rolling on wheels takes 30% the force of just lifting it up. Ideally you want it to be zero. At >1 it is easier to just pick the thing up rather than roll. Knowing this definition should help you an intuitive feel for things so a value can be estimated. You will have to estimate or measure this most important value or conservatively guess a worst case.

If measuring rolling friction (by pushing or pulling the vehicle to determine the fraction of the weight that must be applied to slowly budge the scooter on a horizontal surface) be aware of drive train losses (i.e. gearbox friction) if it is present and linked to the wheels during the test which can obfuscate the measurement. This should technically be included in drive efficiency term and not the rolling friction coefficient but may be easier to measure it along with rolling friction and just lump it all together as rolling friction. This decreases accuracy in inclined scenarios though where roling friction decreases but drive friction remains constant. If doing a lumped measurement and the gearbox was present but not the motor you can include motor efficiency separately using the drive efficiency term while leaving gearbox friction lumped in with rolling friction.

I laid everything out so you should only need to read it from top to bottom and look backwards for variables, never forward. I also tried to lay it out so hopefully you know where everything is coming from (as long as you have a basic understanding of power, torque, force, and friction...maybe even if you don't).

$$n =$$ minimum number of motors engaged with ground

(i.e. Mars Rovers have one motor per wheel so lifted wheels do not contribute to propulsion. Other robots may have multiple wheels ganged to a single per motor via belt/chain/treads, so as long as one of those wheels is in contact, the motor still propels the robot)

$$m_{vehicle} =$$ mass of vehicle (kg)

$$g =$$ acceleration of gravity $$=9.81m/s^2$$

$$W_{vehicle} =$$ weight of vehicle (N) $$=m_{vehicle}\times g$$

$$v=$$ speed (m/s)

$$\mu_{roll}$$ = coefficient of rolling friction for wheels

$$\theta=$$ angle of incline

$$\eta =$$ drive efficiency (between 0 and 1 for 0% to 100%). Use 1 if you need output power (or in calculations of required output like torque). Use actual efficiency if you need input power

$$F_{roll} =$$ force of rolling friction $$=W_{\perp vehicle} \times \mu_{roll}=W_{vehicle}cos(\theta)\times \mu_{roll}$$

$$a =$$ desired acceleration $$(m/s^2)$$

$$F_{acceleration} =$$ ADDITIONAL force required to accelerate $$=m_{vehicle} \times [a + gsin(\theta)]$$

$$r_{wheel}$$ = radius of driven wheel (m)

$$\tau_{roll}$$ = torque required to overcome rolling friction (i.e. to maintain constant speed)$$= F_{roll} \times r_{wheel}$$

$$\tau_{acceleration} =$$ ADDITIONAL torque required to accelerate $$=F_{acceleration} \times r_{wheel}$$

$$\tau_{total}=$$ total torque required to accelerate $$=\tau_{roll}+\tau_{acceleration}$$

$$\tau_{total/motor} =$$ Total torque per motor $$= \frac{\tau_{Total}}{n}$$

$$P_{continuous} =$$ Continuous power to maintain speed $$= F_{roll} \times v \times \frac{1}{\eta}$$

$$P_{peak} =$$ Peak power to accelerate $$= [F_{roll} + F_{acceleration}] \times v \times \frac{1}{\eta}$$

$$P_{continuous/motor} =$$ Continuous power per motor $$= \frac{P_{continuous}}{n}$$

$$P_{peak/motor} =$$ Peak power per motor $$= \frac{P_{Peak}}{n}$$

Speed-dependent losses such as aerodynamic resistance or speed-dependent drive-train losses have been neglected.

• This is such a brilliant response I'll look for a way to donate you some extra points! Just what I needed – akauppi Oct 16 '19 at 19:50