0
\$\begingroup\$

Im trying to build an electric wheelchair with mecanum wheels, and as you might know, each of its 4 wheels has to be independently powered for the chair to have holonomic motion.

With my desired speed to be around (brisk) walking speed only, say 1.15 m/s, and my mecanum wheels sized with a 4-inch radius, I know that I'm going to need a very low RPM (by my computations, ~110 rpm).

However, my computed power draw per wheel is around 300W, which is way large, that Im not sure if Im on the right track? My load capacity is somewhere around 136 kg, btw.

This brings me to my question: are my torque/rpm/power calculations correct?? (code-blocking it up for clarity)

per wheel:
r = wheel radius = 4 inches = 0.1016 m
circumference = 2pi(r) = 0.638 m
v = (max) running velocity = 1.25m/s
rpm = v(60/circumference) = 117.48 rpm

the chair:
u = coefficient of friction = ~0.7
N = normal force = mg
m = total mass (136 kg)
g = 9.81 m/(s^2)
vi = 0 (rest)
t = time to accelerate, say 2 s
a = acceleration = (v-vi)/t = 1.25/2 = 0.625 m/(s^2)

Finally:
Ftotal = uN + ma = m*(uN+a)
Fwheel = Ftotal/4 = 255. 248 N
Twheel = (Fwheel)(r) = 25.93 Nm

Pwheel = (Twheel)(rpm)(2pi/60) = 319 W

Is the power really supposed to be this large and my RPM that low?? Have I missed something?

Help/corrections would be very much appreciated!!

\$\endgroup\$
5
  • \$\begingroup\$ I am rather skeptical about wheelchair design with mecanum wheels, primarily because in some movement cases, they are just rubbing against the floor. Wheels are not in complete rolling motion. There is a translational component where friction is going to make the things very complicated. Surface grip is also going to play a big role. I have seen a group of robotics guys building a similar project where a person will sit on the robot. Sometimes, it worked (in a buggy way) and in most cases it didn't. I'd say - do some more research on mecanum wheel feasibility before going all in. \$\endgroup\$ Commented Apr 18, 2016 at 7:51
  • \$\begingroup\$ Just to get a sense, I'd say - do the maths for a simpler design where all wheels are rolling. Just consider forward and backward motion for 4 wheels. Compare the power input per wheel required with your mecanum wheels. Excess power is going in frictional loss (frictional force x velocity). If it's a considerable amount, best idea is to not go for it. \$\endgroup\$ Commented Apr 18, 2016 at 7:55
  • \$\begingroup\$ What about an inclination? \$\endgroup\$ Commented Apr 18, 2016 at 7:56
  • \$\begingroup\$ See this video to get a sense about the frictional rubbing part: youtube.com/watch?v=TXTo16KKm8Q \$\endgroup\$ Commented Apr 18, 2016 at 7:56
  • \$\begingroup\$ hello! sorry Ive been away for too long. thank you for the comments! @WhiskeyJack dont the rubbing concerns you have effectively just turn the free rollers of the mecanum wheels, but not the wheels themselves for they are locked? \$\endgroup\$ Commented May 22, 2016 at 11:05

1 Answer 1

-1
\$\begingroup\$

Here's a simple (frictionless) check: -

Force = mass x acceleration - with an acceleration of 0.625 m/s/s and a mass of 136 kg, that's a force of 85 newtons.

Work done in accelerating up to speed (1.25 m/s) is force x distance.

Distance covered is average speed in that 2 second time period x 2 seconds so that's 0.625 m/s x 2 seconds = 1.25 m.

Energy inputted (work done) is therefore 85 newtons x 1.25 m = 106.25 joules.

This can be converted to power by dividing by time (2 seconds) = 53.125 watts.

So it looks like something is wrong in your calculations (or mine, god forbid!).

\$\endgroup\$
2
  • \$\begingroup\$ hello! sorry ive been away too long; sigh i also figured i oversimplified my calculations, however my main confusion lies on the friction component and the (mass*acceleration) component, and whether to include them at all and what their directions are. how far should your frictionless check be for a calculation that includes friction? \$\endgroup\$ Commented May 22, 2016 at 11:10
  • \$\begingroup\$ I don't understand how I can help any more - you seem to be at 300 watts per whell and I'm at 53 watts in total. What do you see to be the main difference here? \$\endgroup\$
    – Andy aka
    Commented May 22, 2016 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.