Question About Torque and Stepping Motors

[julianwitkowski]

I'm looking for a high torque stepping motor for a project.

• Acceleration torques from 3.6 N-m to 360 N-m
  
• High output torque at low speed – continuous 10 Lb-In (1.2 N-m) to 1062 Lb-In (120 N-m)

How do I calculate much mass this can move ?

1.2 N-m = 1.2 J = Force ⋅ Distance = 1.2 kg⋅m²/s²

Does this mean it has the energy to move 1.2 kg, 1m, in 1 second?

I'd believe that, but 120 kg, 1m, in 1 second seems way too good so I'm wondering what I'm not thinking about here...

 

[matteo137]

For example, let's say that you connect to the motor axis a wheel of radius $R$. This wheel pulls a rope at which is commented your mass $M$, which you have to lift vertically by going against gravity $g$. Then $M_{\text{max}} = \dfrac{\tau}{g R}$, where $\tau$ is the output torque of the motor.

If you just have to move the mass horizontally you have to take into account the friction force of the plane, and solve $\mu M_{\text{max}} R = \tau$, where $\mu$ is for example the kinetic friction coefficient of your surface, which usually you do not know.

 

[julianwitkowski]

For example, let's say that you connect to the motor axis a wheel of radius $R$. This wheel pulls a rope at which is commented your mass $M$, which you have to lift vertically by going against gravity $g$. Then $M_{\text{max}} = \dfrac{\tau}{g R}$, where $\tau$ is the output torque of the motor.

If you just have to move the mass horizontally you have to take into account the friction force of the plane, and solve $\mu M_{\text{max}} R = \tau$, where $\mu$ is for example the kinetic friction coefficient of your surface, which usually you do not know.

Thank you very much.