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.

 

Hello,

Here is an example for torque calculation.
http://www.solarbotics.net/bftgu/tutorials_mech_torque.html
It uses a gearing system, but when you leave it out you will have your situation.

The link comes from the physics, mecahnics, force systems page of the EDUCYPEDIA:
http://www.educypedia.be/education/physicsmechanicsforce.htm

Greetings,
Bertus

 

Thanks Bertus.

Having read that first link, I now feel a bit silly for asking.

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 ..

 

Thanks Bertus.

Having read that first link, I now feel a bit silly for asking.

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 ..

I don't agree with this. He said 0.36 m is the circumference of the pulley and not its radius.

 

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

 

Thanks Bertus.

Having read that first link, I now feel a bit silly for asking.

Roughly it works out like this then:
F = force = 1kgf = 10N (approx)

The standard value is 9.8 newtons for a kg of mass.

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 ..

Make that 9.8 m/s^2

 

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

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?

 

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.

 

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?

 

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.