Working out Motor torque

I need to calculate the required motor for a mobility scooter with the following user requirements.

The mass of the scooter with passenger = 120kg

Max Speed of scooter = 9Km/h

Time from standstill to max speed = 10s

diameter of wheel is 0.203m

The scooter can hande a slope of 5 degrees

This is my working but it doesn's seem right.

Force

F = ma +mg sin (5)

F = (120 x 0.25) + 2 x 9.8 sin (5)

F = 30 +1.7

F = 31.7 N

Circumference of wheel

= 0.63774 m

RPM

RPM = Distance per hour travelled / circumference x60

RPM = 9000 / 38.28 = 235.11 RPM

235 x 2x PI/ 60

Torque

T = 0.105 x 31.7

T = 3.32 Nm

Power (mechanical)
= T x 24.61

P(mech) = 79.2W

= 0.079 kW

An 80 W motor seems a bit low for a 120kG mobilty scooter to reach a top speed of 2.5 m/s with accleration of 0.25 m/s^2.

Would somebody be able to verify if this motor size is correct or if not where the mistake is in the calculations.

• What do “real” mobility scooters use? And some users are 120kgs plus without the scooter... Oct 13 '19 at 18:10
• It is just a scenario for a project for college and this is the mass of the scooter and user combined. Oct 13 '19 at 18:18
• You have not considered wheel or drivetrain friction. Oct 13 '19 at 19:06
• why is m=120 accelerating but m=2 climbing? Oct 13 '19 at 19:14
• I'm voting to close this question as off-topic because the stated problem is one of physics falling as purely mechanical rather than electrical engineering. Once you knew the mechanical effort required of the motor, you might conceivably have some follow question that would be on-topic as a matter of electrical engineering, but beware the requirement that questions be specific and the fact that part "shopping" questions are off topic. Oct 13 '19 at 20:18

The biggest unknown you need to determine or guess 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 value of 0.3 means rolling on wheels on a horizontal surface 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 get a intuitive feel for what the coefficient of friction should be. 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 scooter 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).

$$m_{vehicle} =$$ 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 $$=\tau_{roll}+\tau_{acceleration}$$

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

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

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

• Thanks for your answer. When working through it I work out the total torque required to be 130.93 Nm with the power 3.49 kW and peak power of 3.57kW. Oct 13 '19 at 23:02
• @danielgrisedale That feels excessive since a typical generator for an RV is 4kW. However, it could easily due to you being very conservative on estimating the rolling friction. Either that or I have an error in my math somewhere. Oct 13 '19 at 23:08
• @danielgrisedale I found two errors in my math. Froll should contain a multiplication with cos(θ), not a division. I also failed to account for how Facceleration includes a component of gravity as the incline gets steeper. I have corrected it but it should not change the answer too much near a horizontal incline. I have also neglected (and will continue to neglect) the speed-dependent losses that vary as the scooters moves faster and faster such as aerodynamic resistance and speed-dependent drive losses (too complicated for me). Oct 13 '19 at 23:48
• I think the rating of the motor is High because you have multiplied force by velocity. Should the power not be the torque x velocity in radians. this scales it down to a 342W motor which sounds sort of in the right ballpark. I may be wrong though. Oct 14 '19 at 9:26
• Aerodynamic drive losses are proportional to the square of speed for torque and the cube of speed for power. At 9 km/hr they are probably not so important, but you may be able to find drag coefficient information on the internet. The torque losses in the drive train are mostly not proportional to speed. They are proportional to the number of contact points, so a triple reduction gear box has three times the losses of a single reduction box. The type of reduction is also important. Worm gears have the most losses. Searching will likely turn up information for a manageable estimate. Oct 14 '19 at 9:28