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I am confused with the torque ratings of servo motors, or I guess it is general while choosing any motor.

For example a servo's torque is rated as 10kg-cm, so does that mean it can handle a 100N force from 1 cm distance to its center or 50N force from 2cm distance?

Am i interpreting it right?

If it is right, then consider a motor mounted to the wall from its back and a rope is wrapped to its tip(the rotating part, i don't what it is called); then how should I calculate the maximum weight the motor will be able to lift upwards?

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Your comment is correct. However this value is max. torque value not nominal value.

If your engine is 10 kg-cm, we can simply say:

$$F = m * a$$

$$F = 10 kg * 9.80665 m/s^2$$

$$F = 98.0665 N$$

This value indicates the maximum force you can apply.

If we examine the torque and forces simply:

$$τ = r * F$$

If the distance increases, the force value to be applied decreases, so the load on the motor decreases. It is theoretically true that there is a linear relationship here. In practice, it tries to approach this line, but this is directly related to your engine quality.

In general, it would be more correct to give the following answer. You can also tow a load that you connect directly to the shaft of the motor. However, it may not be correct to use it at maximum torque for this. It doesn't mean it don't work, but the longer you stay at maximum torque, the more you can damage the motor itself or the gears. Therefore, I think that staying 10 percent below the specified value is sufficient as an operating range. Of course, we cannot verify this without the engine's datasheet.

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  • $\begingroup$ On Robotics we are fortunate enough to have MathJax support enabled, allowing you to easily create subscripts, superscripts, fractions, square roots, greek letters and more. This allows you to add both inline and block element mathematical expressions in robotics questions and answers. For a quick tutorial, take a look at How can I format mathematical expressions here, using MathJax? $\endgroup$
    – Ben
    Commented Jun 28, 2022 at 13:54
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Yes, you are right. 10kg-cm = 100N force from 1 cm distance. You can use a servo motor with nominal torque force 10kg-cm to lift up 10 kg-cm, but you will most likely damage gears first. enter image description here

For example, there is a dependency chart of rotation speed (degree) vs load (kg), for KST Servo x10 v2.0 (9.5 kg, 7.4V, 1.2A) https://www.kstsz.com/kstsz_Product_2042526799.html, so you can estimate lifetime for your servo's motor.

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  • $\begingroup$ Why using the motor at "nominal" values damages it? What should be the safety margin then? $\endgroup$
    – muyustan
    Commented Oct 10, 2019 at 7:15

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