# How to find torque reqd to move a wheel?

1. Jul 2, 2007

### dave226

Hi,
I have to find the torque required to run a wheelchair and then to match it with the electrical parameters to find motor/motor sizing, is it ok

1) if i assume some acceleration that gives me force and then finding T= F*r (N-m) where r is the radius of the wheelchair.
2) then finding Power using P=Tw and then matching these mechanical parameters with the electrical parameters for finding motors.

2. Jul 2, 2007

### Staff: Mentor

Seems like the worst case would be starting on a hill. You would calculate the force necessary to do that (including losses in the drive train), and yes, convert that into a torque using the wheel diameter.

EDIT -- Oh, and don't forget to multiply 2x or so.

3. Jul 2, 2007

### xez

Yes, you're close to being on the right track in general, though --

The acceleration doesn't give you force, the force
gives you the acceleration, though if you mean that the
accelerating force (and hence torque) is calculable knowing
the desired resultant acceleration, that's quite so.

r = must be the radius of the torqued wheel, not the chair.

F=Force=Mass * acceleration = Newtons N = kg m / s^2

T = Torque = Newtons force = N = kg m / s^2

J = Energy = Work = Force * distance = N*m

P Power = Work/time = Energy/time = J/s = Nm/s =
Force * velocity

so...
w = the hypothetical vertical velocity (speed [m/s]) of the
mass (kg) of the chair system under the conditions
that the wheel torque force (T [N]) is applying a traction
force (T [N]) on the inclined ground doing a rate of work
i.e. power (P) against gravity.

And then, yes, P_mechanical = T*w.

P_electrical > P_mechanical due to the motor inefficiency
factor and all the friction factors, so
P_electrical = K * P_mechanical, and
as a rough guesstimate K = 1.1 to 1.3.

This would be most relevant to visualize in the case of a wheel-chair
of mass (M [kg]) going up-hill at constant vertical velocity component (w [m/s])
driven by traction force (F [N] ) against a slope thus resulting in a
constant Power [Watts W] e.g. rate of work against
gravity (G = 9.81 N/kg).

The work done by moving Mass (M) [kg] rising (h) height [m meters]
against gravity { via a force F [N] = (M*G) } would be (M * G * h) [Joules],
and if that is done uniformly in time (t) [s seconds] the mechanical
power would be (M * G * h)/t.

Of course w = h/t would be your vertical velocity component
in this case; any requisite along-slope horizontal velocity
being assumed to be free of work investment.

And of course comparable analysis is applicable to the case of doing
constant power work cruising horizontally at some relatively
high constant velocity (w) against equilibrium friction forces,
though the more challenging and critical case would be going
up-hill slowly on the steepest possibly slope where the motor
torque is barely sufficient to overcome the opposing torque
due to gravity pulling the chair downward.

Please forgive any typographic / compositional errors.

4. Jul 2, 2007

### xez

Of course the other way to look at it would be:
on a slope inclined at angle (Y [radians])
relative to horizontal, you have a chair system of mass
(M [kg]) being supported by a drive-wheel of

At rest, the force of gravity applies a
down-slope directed force COMPONENT
G_F [N] = 9.81 * M * sin(Y)
to the wheel-chair system.
(the rest of the downWARD force being irrelevant since
the slope supports that force from the weight)

Since we're at rest against an applied down-slope force,
the drive wheel must apply an equal and oppositely
directed up-slope force of -G_F.

The drive-wheel is applying a force acting via the
lever arm distance of the drive-wheel's radius (r),
so the static torque needed to keep the chair from rolling
down-hill would be of magnitude | G_F * r |,
e.g.
9.81 * Mass * sin(upward grade angle)
Torque ...result given in Newtons*meters.

Please forgive any typographic / compositional errors.

5. Jul 3, 2007

### dave226

thks berkemen n xez.
xez ur formulae and guidelines are really marvelous, i want few more things to ask, though these are quite simple i know but still need some guidelines to make my logics more clear:

1)Is this torqued wheel and drive wheel is one n the same thing? is this the wheel of the wheelchair? n which factors will vary with the number of wheels of the wheelchair? as such i have 2 wheels.
2)the one thing which i asked in the first post too tat do i need to assume acceleration was because in order to find force, but u said its d force tat gives acceleration, so going once again to the basics:

F=Force=Mass * acceleration = Newtons N = kg m / s^2

T = Torque = Newtons force = N = kg m / s^2

J = Energy = Work = Force * distance = N*m

P Power = Work/time = Energy/time = J/s = Nm/s =
Force * velocity

so now i have only availability with me is that of the mass, and if i assume a velocity for my wheels of the wheelchair still around 5 qts are unknown: Force, acceleration, workdone, distance n power?

I know tat i m missing something here, or a bit confused as i m looking into my calculations from starting again. Please help me out.

Thanks!!!

6. Jul 4, 2007

### xez

You're welcome.

Yes, any wheel(s) you're powering directly or indirectly
via a motor / drive-train are the drive wheels or driven
wheels I meant.

They don't have to be the same as the other wheels of
the wheelchair if you'd like to use others. A typical
car is 2-wheel-drive with the front wheels driven and
the rear ones just rolling free.

If you have non-motor-driven wheels they'll either roll
freely or have something like a hand-brake on them.

Assuming the non-driven wheel(s) (if any) just roll
freely they don't enter into the equations relative to
force or torque etc. since they just might as well not
exist relative to any concern of moving the chair in
acceleration or motive power.

Of course non-motor driven wheels could still be used
for steering / braking (like the front wheel of a bicycle).

Or you could just use motor drive on the main support
wheels, whatever you like.

For the 'weight' or more properly mass, that's just the
total mass of everything you're trying to move.

I think if you review the previous posts and your own
design parameters you'll find that you can come up
with the answers to the quantities you reference:

force: this is determined by three things:
a) the necessary force to overcome gravity when going
up (and probably down -- with electric motor braking!)
hills. You don't have to move at all to need to supply a
force to counter gravity's pull since it takes a balancing
force just to stay in one spot and not roll down-hill
uncontrollably quickly.

b) even on a level surface force is needed to accelerate
a mass, and to continue to move against friction.
Force = Mass * Accelleration. So if you wanted to
have 1G of acceleration (whee! that's be a real rocket!),
you'd need 9.81 Newtons of Force for each kilogram of
mass you're accellerating.
Even if you're going up-hill Force = Mass * Accelleration,
except you must make sure to add up the net Force
from the Force of gravity as well as the Force your motor
produces, and their vector sum is the net force that
will act on your system and produce a
resultant acceleration proportional to the mass.

So I'm assuming you know your total maximum mass
specification, e.g. 120kg person, 120kg chair mechanism,
so total mass <= 240kg (or whatever your numbers are,
maybe mine are too high if you're designing a light
chair, maybe it'll only be 35kg, whatever).

You have some specification as to how steep of a hill
slope you require to be able to climb, so that'll give you
the force of gravity as 9.81 [N/kg] * SystemMass [kg] *
sin(slope angle above the horizontal) = Newtons Gravity
force pointing down-hill. That's the force you need just
to stay still. You need more force (F=M*a) to accelerate
to enable you to move up-hill at any speed.

Acceleration = Force/Mass.

Work = Force * Distance.

Distance = however long you're moving before you
stop or the batteries die or whatever.

Power = Work / time, or equivalently (as in the previous
post) Force * Velocity.

So if you count the radius of the drive wheel as the
distance over which your motor torque is applying a force
to the ground then your
MotorBoxTorque * DriveWheelRadius = Force applied to
pull at the circumference of the drive wheel e.g. if its
circumference is a tire pushing on the ground.

So really you can figure out most of the physics
parameters you mention if you know your Mass, Torque,
Drive Power available, Drive Power Efficiency Factor,
Hill Climbing Requirement, Battery Energy Available.

All the other physics factors are pretty much calculable
from those input parameters.

7. Jul 9, 2007

### dave226

Great thanks to XEZ, ur posts helped me out of the problems which i had.

thks once again.

8. Jul 23, 2007

### dave226

Hi just wondering if any one could provide me the way or formula for finding force required by a wheelchair having wheel of radius 'r' to climb an obstacle having height 'h'.........

any help would be of great help. Thks!!!

9. Jul 23, 2007

### xez

Imagine the wheel touching the obstacle edge as a

The obstacle is height H at its near edge.

Just as the wheel comes to the obstacle, the
obstacle touches the near edge of the wheel
at a single point somewhere along the wheel's
circumference.

The angle between the bottom of the wheel which
is on 'flat ground' and the point of the wheel that touches
the top side obstacle edge at height H is theta1.

H = R * sin(theta1)

H/R = sin(theta1)

theta1= inv_sin(H/R).

To vault the obstacle wall (which I assume to be steep
for the worst case), the wheel must proceed up at an
angle that is initially at a slope equal to (H/R)
or angle (theta1).

As before, the entire chair was to climb at that angle,
of a steady slope, the force needed would be:
9.81 * Mass * sin(theta1).

However if the chair is supported by multiple
wheels, only part of the gravity force will appear on each
wheel depending on the position and number of wheels and
the center of mass of the system, etc. In which case you
have to look at where the drive wheels, center of mass,
and obstacle vs. road positions are to figure out which
wheels apply which forces to incline and climb all/part
of the chair at what angle to figure out the actual
balance of torques and forces and angles will be.

In the worst case just assume the whole chair climbs
at theta1 as above, but obviously you may need only
a fraction of that force if you're doing things like
shifting the center of gravity and distributing the weight
of the chair over several wheels etc.

10. Aug 1, 2007

### dave226

thanks a lot xez!!! :)