Low RPM, high power calculations help

  • #1
Hi guys,

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??

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???

Help/corrections would be very much appreciated!!
 

Answers and Replies

  • #2
haruspex
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What does that friction coefficient represent? Sounds like it is the friction between the wheels and the floor. I would have thought you wanted the wheels rolling on the floor, not skidding across it.
 
  • #3
What does that friction coefficient represent? Sounds like it is the friction between the wheels and the floor. I would have thought you wanted the wheels rolling on the floor, not skidding across it.
yep between the wheel and the floor. i got the value from the wheels' specs. (isn't it the kinetic coefficient?)
oh, aren't i supposed to include it in the calculations as the force opposing the skid, and that which makes the wheel roll on contact?
would you mind elaborating why and what I should have done?
 
  • #4
haruspex
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yep between the wheel and the floor. i got the value from the wheels' specs. (isn't it the kinetic coefficient?)
oh, aren't i supposed to include it in the calculations as the force opposing the skid, and that which makes the wheel roll on contact?
would you mind elaborating why and what I should have done?
Static friction does no work. There is no relative motion between the surfaces, so in work = force x distance the distance is zero.
Indeed, without friction your wheelchair would never get anywhere, except downhill. When going uphill, or accelerating on the flat, friction acts in the forward direction.
See https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/ for further discussion.
 
  • #5
Static friction does no work. There is no relative motion between the surfaces, so in work = force x distance the distance is zero.
Indeed, without friction your wheelchair would never get anywhere, except downhill. When going uphill, or accelerating on the flat, friction acts in the forward direction.
See https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/ for further discussion.
thanks for the responses and the link!! however i don't seem to follow you entirely. please bear with me hehe
so what you're saying is to not include in the force/work equations the static friction, am i right? but how about if the above mentioned coefficient is the kinetic one? the one which, as you said, acts on the forward direction?
or is my other understanding of this correct, that: since the wheel is NOT skidding, there is NO kinetic friction; just static friction (which does no work)?
in summary, i shouldn't have included the term in my force equation? should it simply be F=ma then?
 
  • #6
haruspex
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thanks for the responses and the link!! however i don't seem to follow you entirely. please bear with me hehe
so what you're saying is to not include in the force/work equations the static friction, am i right? but how about if the above mentioned coefficient is the kinetic one? the one which, as you said, acts on the forward direction?
or is my other understanding of this correct, that: since the wheel is NOT skidding, there is NO kinetic friction; just static friction (which does no work)?
in summary, i shouldn't have included the term in my force equation? should it simply be F=ma then?
Yes, as long as the wheels are not skidding it is static friction and costs you nothing.
So now the question arises, why does the motor have any work to do when moving at a steady speed on the level?
At the link I posted, read the section on rolling resistance.
 
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  • #7
Yes, as long as the wheels are not skidding it is static friction and costs you nothing.
So now the question arises, why does the motor have any work to do when moving at a steady speed on the level?
At the link I posted, read the section on rolling resistance.
yep i've read it. :) however my force equation is based on the movement from rest, hence the acceleration formula from zero velocity towards maximum speed. should anything else be changed? or did i misapply a formula?
 
  • #8
haruspex
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yep i've read it. :) however my force equation is based on the movement from rest, hence the acceleration formula from zero velocity towards maximum speed. should anything else be changed? or did i misapply a formula?
Everything else looks ok. But if you just want the power, no need to find the torque. Just multiply force by speed.
 
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  • #9
Everything else looks ok. But if you just want the power, no need to find the torque. Just multiply force by speed.
oh yeah!! haha got too caught up in getting the motor ratings, i kept doing the longer cut every time.
super thanks!!! i needed the friction and rolling resistance review!
so basically. I needed just P=T(rpm)(2pi/60)= Fv = (21.29 N)(1.25 m/s) = 26.617 W of power per wheel, all this time??
Is 2.2A for 12VDC a reasonable current draw (for 4 motors) now for an electric wheelchair? :)
 
  • #10
haruspex
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Is 2.2A for 12VDC a reasonable current draw (for 4 motors) now for an electric wheelchair? :)
I haveno idea. Nothing online?
 
  • #11
I haveno idea. Nothing online?
I only see current and power ratings from posts of legit hobbyists or power wheelchair makers, unfortunately.
But at least my computations have been checked and I now have a corrected set of equations, thank you so much haruspex for guiding me through!!
 
  • #12
Baluncore
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What is the steepest hill or ramp you must climb ?
Is the motor reverse torque used to brake when descending a slope or stopping ?
 
  • #13
haruspex
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I only see current and power ratings from posts of legit hobbyists or power wheelchair makers, unfortunately.
But at least my computations have been checked and I now have a corrected set of equations, thank you so much haruspex for guiding me through!!
You are welcome. Good luck.
 
  • #14
What is the steepest hill or ramp you must climb ?
Is the motor reverse torque used to brake when descending a slope or stopping ?
Im only gonna have to reach 30 degrees max. For the braking, I've only decided on dynamic (resistive) braking on mostly flat surfaces, so unfortunately I haven't any plans on stronger brakes yet. Please do suggest/advise :)
I'm thinking about using worm gear for that, and to avoid back driving at power-off. Is it the best thing though? What do you think?
Also, how are the formulas/ratings going to change with using motor reverse torque, if it's the best method for descents?
 
  • #15
haruspex
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Im only gonna have to reach 30 degrees max.
Only? That is equivalent to an acceleration of g/2, so you will need to develop a torque of 70Nm. Of course, you don't need to be going very fast at that angle, so it's maybe not a lot of power.

Going back to the first calculation, it occurs to m that you do not need to accelerate at constant rate all the way up to maximum speed. You could instead consider the KE at maximum speed and see what power you need to reach that in 2 seconds. The real answer will be somewhere between the two. In practice, you will be torque limited at low speed and power limited at high speed.
 
  • #16
Only? That is equivalent to an acceleration of g/2, so you will need to develop a torque of 70Nm. Of course, you don't need to be going very fast at that angle, so it's maybe not a lot of power.

Going back to the first calculation, it occurs to m that you do not need to accelerate at constant rate all the way up to maximum speed. You could instead consider the KE at maximum speed and see what power you need to reach that in 2 seconds. The real answer will be somewhere between the two. In practice, you will be torque limited at low speed and power limited at high speed.
I actually dont think i have to go that steep, but i actually set it for tolerance/allowance in ratings; or is it a bit too much, from around 20degrees (actual max)?
Thank you for mentioning the need for higher torque in inclines; I actually was trying to figure out how to set the max now that I don't include the frictional component in the calculations. By "between the two" you mean between the power to reach KE at max speed, and the power for 70Nm torque, right? Meaning somewhere between 53.2W and 861W?? Also, how did you get the 70Nm value?


EDIT: i got to the 67.9Nm or 70Nm value, never mind the question on that :)
 

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