# Calculating torque requirements ..

Discussion in 'Physics' started by JakesSA, Sep 3, 2009.

1. ### JakesSA Thread Starter Member

Mar 20, 2009
15
0
Hi

What is the formula for a rough estimation of torque requirements for an electric motor?

I want to be able to move say 1 kg over 0.36 meter in 1 second? Please let me know if there are any other variables required for the calculation.

Much Appreciated

PS: There are no complicated mechanics involved, think of a motor shaft directly attached to a pully, .36 meters in circumference, driving a belt with 1kg weight attached.

Last edited: Sep 3, 2009

Apr 5, 2008
19,911
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3. ### JakesSA Thread Starter Member

Mar 20, 2009
15
0
Thanks Bertus.

Roughly it works out like this then:
F = force = 1kgf = 10N (approx)
d = distance = 0.36 meters
T = Torque

T = Fd
T = 10N x 0.36m
T = 3.6Nm

Easy as pie.

EDIT:

PS: Force is calculated like this:
F = mass * acceleration
I just took 10 m/s as acceleration ..

Last edited: Sep 3, 2009
4. ### mik3 Senior Member

Feb 4, 2008
4,846
70
I don't agree with this. He said 0.36 m is the circumference of the pulley and not its radius.

5. ### mik3 Senior Member

Feb 4, 2008
4,846
70
circumference=C=2*pi*r

Thus,

r=C/(2*pi)=0.36/(2*pi)=0.057 m

Thus the torque required to lift the 1 Kg 0.36 m (one turn of the pulley) is

T=F*r=10*0.057=0.57 Nm

6. ### Mark44 Well-Known Member

Nov 26, 2007
626
1
The standard value is 9.8 newtons for a kg of mass.
Make that 9.8 m/s^2

7. ### JakesSA Thread Starter Member

Mar 20, 2009
15
0
Err .. but the circumference is 0.36m, surely that is the distance travelled to complete one revolution?

Also, is force ≠mass * acceleration? If so, how can a standard value be derived?

8. ### mik3 Senior Member

Feb 4, 2008
4,846
70
Torque=Force*distance from the point of movement

Thus you need the radius of the pulley and not the circumference.

The torque I calculated is when the 1 Kg is moving at a constant speed. When you accelerate it you need more complicated calculations due to its inertia which result in the need of higher torque.

9. ### JakesSA Thread Starter Member

Mar 20, 2009
15
0
Ai .. now I really wish I paid more attention in high school physics ..

Insofar the radius is concerned I might just mention that the setup is not a mass suspended on a belt hanging over a pulley. Rather think of a belt running over two pulleys with a motor connected directly to one of the pulleys. A weight is then attached to the belt and moved horizontally. For example, many electric sliding doors work on this principle.
Having said that, does the radius vs. circumference in the distance calculation still apply?

Last edited: Sep 4, 2009
10. ### mik3 Senior Member

Feb 4, 2008
4,846
70
Yes, it still applies.

Have a look here:

http://en.wikipedia.org/wiki/Torque

In this case, you need to calculate the net horizontal force and then calculate the required torque. This horizontal force depends on how tight the belt is, how much it bends when you put the weight on it etc.