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

Question

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?

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, wheel moment of inertia, and speed-dependent losses such as aerodynamic resistance or speed-dependent drive-train losses have been neglected.

---

DC MOTOR SPECIFICATIONS:

For DC motors, often torque-speed curves are not provided. Usually only the no-load RPM and stall torque are supplied. You do not use the stall-torque to size anything because this is the torque the motor produces at zero zero RPM while it is drawing enormous amounts of current and overheating. Nor do you want to use the no-load RPM which is what the motor spins at when no torque is applied. It also goes without saying that you cannot use the no-load RPM together with the stall torque in your calculations since they do not happen at the same time.

However, not all is lost because for DC motors, the torque-speed curves are linear. As torque decreases from stall-torque to zero, the RPM increases from zero to the no-load RPM. Therefore, for the operating voltage for which the no-load RPM and stall-torque are supplied you can produce the torque-speed curve.

What will not be on this torque-speed curve that is where it is best to run the motor at. But we can use rules of thumb which are accurate enough for many applications:

• the point of maximum efficiency for permanent magnet DC motors tends to be about 1/7th the stall torque and 6/7ths the no-load RPM
• the point of maximum power is 1/2 the stall torque and 1/2 the no-load RPM

Therefore, it is best to size your motor, wheel size, and gear/belt/chain reduction ratios and so the most frequent, prolonged, and continuous mode of operation occurs at 1/7th stall torque and 6/7ths no-load RPM. Maintaining velocity, for example.

Similarly, you also probably do not want to require more than half-stall torque from your motor for modes of operation that require accelerations over a prolonged period of time. Conversely, for brief tasks where additional torque is required it may be acceptable to exceed half stall-torque. It's up to you to judge how far beyond half stall-torque is acceptable for the tasks being performed. For example, for climbing over a bump or brief obstacle you might choose that 3/4 stall torque is acceptable but too much for accelerating up to speed during normal travel. Instead you choose that to be half stall-torque instead.

Answer 2

The required tractive torque of the motor shall be equal to the tractive torque at wheel $$T_w = F_t \cdot r_w,$$ (i.e. Tractive Torque = Tractive Force x mean wheel radius). Where,

• $r_w$ is the mean wheel effective radius
• $F_t = F_r + F_g + F_d + F_{ie}$ with,
• $F_r$ the tyre rolling resistance (can be in the form of $C_{rN}$ - simplified and treated as independent of velocity)
• $F_g$ the forces due to gradient (depending on slope angle, can be positive of negative)
• $F_d$ the aerodynamic drag (as a function of air density, drag coefficient, vehicle cross sectional area, and squared of vehicle velocity)
• $F_{ie}$ = equivalent inertial force (during acceleration) - (including linear and rotational inertias, due to vehicle mass and rotating component of gear train and wheels).

So you must determine the acceleration required to reach you designed speed. And must also include overall transmission efficiency in order to get the right size of the motor.

Just to visualize for 1000 kg car and everything else constant: on flat road @ 25 km/h requires 1.0 kW power; on 5$^\circ$ gradient climbing @ 25 km/h requires 6.9 kW power; on flat road accelerating @ 1.4 m/s$^2$ (i.e. 0 - 25 km/h in 5 sec) requires 10.6 kW power.

So you really need to determine you design specifications in order to size the right motor...