# Dc motors for controlling wheels of a wheelchair

I am doing a project on controlling wheels of a wheelchair. The diameter of the wheel is about 25-30 inches, the weight which the wheelchair is supposed to carry is upto 80 Kgs and it should run at a speed of 0.5-0.7 metres per second. I thought of using a 24V dc motor, what speed and Power must the dc motor provide to move a wheelchair at above specifications.

• Criss posted here : electronics.stackexchange.com/q/349100/152903 Jan 9 '18 at 20:23
• “Supposed to carry up to 80kg” - the motors must move both the weight of the human and the wheelchair itself. Jan 9 '18 at 23:15

What speed and Power must the dc motor provide to move a wheelchair at above specifications.

Wheel RPM = chair_speed / Wheel-diameter x 60
Motor speed = wheel speed/Gear_ratio.

Motor power = hard_without_more_data.
Factors include: Friction, gear train efficiency, motor efficincy, carpet-lino-concrete- ..., slope or flat, ...

Look at existing chairs.
2 x 500 Watts seems good in many cases. Lower or even much lower may be OK in your case.
In a demonstration version on a smooth hard floor with a good motor and gearbox and smooth hard wheels I'd expect that 100 Watts would probably be acceptable.
At your target speed losses from windage are minimal.

Adjust (increase) any of the following for losses due to motor efficiency (say perhaps 80-90%) and gearbox efficiency (90-95%?)

"Windage"

Power due to air resistance can be determined from the drag equation.

Drag = 0.5 x Air density x Area x Coefficient-of-drag x velocity-squared
Power = Drag x velocity.
Whence:
Power (Watts) = 0.5 x 1.2 kg/m^3 x A x 1 x V^3.
A in m^2
V in m/s
Cd = 1 = flat plate drag - about right for a wheelchair

At 1 m/s and 1 square metre area Power = 0.6 x 1 x 1 x 1^3 = 0.6 Watt!
At 2 m/s (7 kph) Power 5 Watts.
At 5 m/s (18 kph) Power = 75 Watts

ie in a "still air" environment or where minor air movement occurs power loss to windage is liable to be under 10 Watts. In external gusty wind" environments substantial power peaks may occur.

Climbing:

If the wheelchair travels on an ascending (non level) surface then extra power will be required to provide the potential energy thus achieved by the gain in height.
Energy required to move m kg upwards against gravity g a height h
E = mhg
g id about 9.8 m/s/s which can be approximated to g = 10 for this sort of working. In SI units m kg, h m, v m/s, g m/s/s:

Energy = 10.m.h Watt,seconds.
To achieve this in time t seconds
Power = E/t = 10.m.h/t.

Power required = 10.m x the height increase per second.
For a wheelchair travelling a V m/s up an incline of d degrees,
Height increase per second = v.tan(d)

Power due to slope = 10.m.V x tan(d)

For eg a 5 degree slope at 1 m/s and a 100 kg all up load
Watts = 10.m.v.tan(d) = 10.100.1.tan(5) ~= 90 Watts
ie power required to ascend a slope is non trivial above a few degrees.

For a 1 in X slope
x = 1/(tan(d))
So a 5 degree slope = 1 in 1/(tan(5)) = 1 in 11.4

• Losses from windage are minimal : have you tried pushing a wheelchair against the wind or even walking against the wind.... Jan 9 '18 at 23:21
• I guess he won't be using the wheel chair outdoors when there is a hurricane. :) Jan 10 '18 at 0:14
• @JemEripol doesn’t need to be a hurricane - a strong wind makes a significant difference... Jan 10 '18 at 7:53
• @SolarMike See addition to answer. In indoor environments loss to windage will usually be minimal. Jan 11 '18 at 4:22
• @SolarMike He adumbrates thus. If it was for cross country / Olympic / parasports / ... then he would need to have added that to the spec OR grossly underrepresented his requirement. | His description is "I am doing a project on controlling wheels of a wheelchair" and his questions are ", ... what speed and Power must the dc motor provide to move a wheelchair at above specifications" [?]. || If his need is more complex he'll need to expand his question. Responders could otherwise spend semi-infinite time proposing solutions to mostly irrelevant use cases. Jan 11 '18 at 12:45