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?

  • $\begingroup$ Huh. That's odd. Does the Engineering Beta StackExchange not support MathJax like the Electrical Stack Exchange? $\endgroup$
    – DKNguyen
    Oct 13, 2019 at 23:45
  • $\begingroup$ It does. For some reason all your $ were escaped with backslashes, which disables their LaTeX interpretation. $\endgroup$
    – Wasabi
    Oct 14, 2019 at 2:08
  • $\begingroup$ @Wasabi Weird. I need the backslashes there for it to work on the EESE. Thanks for fixing. $\endgroup$
    – DKNguyen
    Oct 14, 2019 at 2:25
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    $\begingroup$ 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. $\endgroup$
    – Transistor
    Oct 14, 2019 at 18:49
  • 1
    $\begingroup$ @Transistor Interesting, though I am not sure why cost would be talked about any less on other engineering stack exchanges. $\endgroup$
    – DKNguyen
    Oct 14, 2019 at 18:51

1 Answer 1


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).

$ \mu_{roll}$ = coefficient of rolling friction for wheels (between 0 and 1)

  • The most important unknown that needs to be determined or estimated.

    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 $\mu_{roll}>1$ it is easier to just pick the thing and move it up rather than rolling it. 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 if it remains connected to the wheels during the test which can obfuscate the measurement for rolling friction.

    Drive train friction should technically be separate and not be lumped with the rolling friction coefficient. This decreases accuracy in inclined scenarios though where rolling friction decreases but drive friction remains constant. There are also issues where the torque through a gearbox is not reversible.

$ v=$ speed ($m \over s$)

$ \theta=$ angle of incline

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

$ g = $ acceleration of gravity $ =9.81 \frac{m}{s^2}$...on Earth. I want to know if you're using this for something that isn't.

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

$ W_{\perp vehicle} =$ normal force ($N$)$=W_{vehicle}cos(\theta)$

$r$ = radius of driven wheel ($m$)

$ N =$ drivetrain reduction ratio (i.e. gear ratio) where $N>1$ for reducing motor speed and increasing motor torque

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

  • (i.e. Number of motors that are driving at least one wheel with traction in your worst case conditions. For example, Mars Rovers have one motor per wheel so for every wheel lifted off the ground, output for a motor is lost. But difficult to actually lift a rover wheel with them being on the end of pivoting leg suspension, but much easier to do with a conventional 4 wheel x 4 motor design with fixed wheels.

    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. For example, 6-wheeled vehicle where each side has a single motor chained/belted to all three wheels on that side will maintain motor output as long as at least one wheel on each side has traction.

$\eta_{drivetrain} =$ Drivetrain efficiency (between 0 and 1 for 0% to 100%)

  • Good gearboxes can be 90%-95% efficient. Belt drives are pretty high too. 80% is a good conservative estimate in general unless you are using worm gears which have terrible efficiency. Sometimes gearbox losses are fixed in which case you just add that wattage at the very end of your calculations.

$\eta_{motor} =$ Motor efficiency (between 0 and 1 for 0% to 100%)

  • A good motor can be 80%-90% efficient or higher. 70% is a good conservative estimate. 50% for a crappy motor.

  • If you include motor efficiency, your power results will be the motor electrical input power.

  • If you do not include motor efficiency (i.e. $ \eta_{motor} = 1$), then your final power numbers will be the motor output power.

$ F_{roll} =$ force of rolling friction ($N$) (i.e. force required to maintain speed on level surface)$ = W_{\perp vehicle} \times \mu_{roll}$

$ F_{incline} =$ force required to overcome gravity on an incline (i.e. force required to maintain speed against gravity up an incline) ($N$) $=W_{vehicle}sin(\theta)$

$ a = $ desired acceleration ($m \over s^2$) $ = \frac{v}{t}$, where $t$ is the number of seconds you are willing to wait to reach your desired speed

$ F_{accelerate} = $force required to accelerate (i.e. after rolling friction and gravity have been nullified by other forces) ($N$) $ = m_{vehicle} \times a $

$\omega =$ wheel angular velocity ($rad \over s$) $ = \frac{v}{r}$

$RPS =$ wheel rotations per second $ = \frac{\omega}{2\pi}$

$RPM =$ wheel rotations per minute $ = RPS \times 60$

$ \tau_{roll} $ = combined wheel torque required to overcome rolling friction (i.e. torque required to maintain speed on a level surface) ($N \cdot m$) $ = F_{roll} \times r_{wheel}$

$ \tau_{incline} $ = combined wheel torque required to overcome gravity on an incline (i.e. torque required to maintain speed against gravity up an incline) ($N \cdot m$) $ = F_{incline} \times r_{wheel}$

$ \tau_{accelerate} = $combined wheel torque required to accelerate (i.e. after rolling friction and gravity have been nullfied by other torques) ($N \cdot m$) $ = F_{accelerate} \times r_{wheel}$

$ \tau_{constant} = $ combined wheel torque required to maintain constant speed ($N \cdot m$) $ = \tau_{roll} + \tau_{incline}$

$ \tau_{wheel}= $ total torque required ($N \cdot m$) $ = \tau_{constant} +\tau_{accelerate}$

$ \tau_{motors} = $ combined motor torque ($N \cdot m$) $ = \frac{1}{\eta_{drivetrain}} \left(\frac{\tau_{wheel}}{N}\right)$

$ \tau_{motor} = $ torque per motor ($\frac{N \cdot m}{motor}$) $ = \frac{\tau_{motors}}{n}$

NOTE: This is not strictly correct but it is about the best we can do to compensate for drivetrain friction. Technically, it should be: $ \tau_{motor} = \left(\frac{\tau_{wheel}}{N}\right) + \tau_{drivetrain}$, where $\tau_{drivetrain}$ is the drivetrain friction torque, but $\tau_{drivetrain}$ is dependent on both RPM and load torque so it is really difficult to determine.

At this point there are alternative ways to calculate the power, either using torques and angular velocities, or forces and linear velocities.

$ P_{continuous} = $ Total continuous power for all motors to maintain speed ($W$) $ = \left(F_{roll} + F_{incline}\right)\times \left(v \times \frac{1}{\eta_{motor}} \times \frac{1}{\eta_{drivetrain}}\right)$

$ = \left(\tau_{roll} + \tau_{incline}\right) \times \left(\omega \times \frac{1}{\eta_{motor}} \times \frac{1}{\eta_{drivetrain}}\right)$

NOTE: If we had a value for the drivetrain friction torque, $\tau_{drivetrain}$ we could directly calculate the power loss due to drivetrain friction via $P_{drivetrain} = \tau_{drivetrain} \times \omega$ and instead use $ P_{continuous} = \left(F_{roll} + F_{incline}\right)\times \left(v \times \frac{1}{\eta_{motor}}\right) + \left(P_{drivetrain} \times n\right)$

$= \left(\tau_{roll} + \tau_{incline}\right) \times \left(\omega \times \frac{1}{\eta_{motor}}\right) + \left(P_{drivetrain} \times n\right)$

We multiply by the number of motors because $P_{drivetrain}$ is the power loss of a single drivetrain assembly, but each motors has its own drivetrain assembly while $P_{continuous}$ is the total continuous power for all motors.

$ P_{peak} = $ Total peak power for all motors to accelerate ($W$) $= P_{continuous} + \left(F_{accelerate} \times v \times \frac{1}{\eta_{motor}} \times \frac{1}{\eta_{drivetrain}}\right)$

$ P_{continuous/motor} =$ Continuous power per motor ($W \over motor$) $ = \frac{P_{continuous}}{n}$

$ P_{peak/motor} =$ Peak power per motor ($W \over motor$) $ = \frac{P_{Peak}}{n}$

No slippage is assumed. Static friction to initially get everything moving from standstill, and speed-dependent losses such as aerodynamic resistance or speed-dependent drive-train losses have been neglected.

  • $\begingroup$ This is such a brilliant response I'll look for a way to donate you some extra points! Just what I needed $\endgroup$
    – akauppi
    Oct 16, 2019 at 19:50

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