I would like to design a robotic arm to hold a weight X at length Y (in my case I want to hold X=2.5 lbs at Y = 4 inches). Starting out simply, I would like try building an arm with a gripper plus one servo joint.

[Servo Joint] ----- Y ------ [Gripper]

When designing an arm, would I want to say that the gripper has to have enough torque to hold the desired weight (e.g. 2.5 lbs) at a minimal distance (however long the fingers are) then design the servo joint to bear the weight of the gripper + the load?

I would like to be able to hold the object at full extension


2 Answers 2


You have the right idea, just be sure to design for the servo to bear the moment force (aka torque) generated by the load at Y = 4 inches from the joint, not the 2.5 pounds of what you're trying to hold.

$\tau = r*F*\sin(\theta)$


  • r is the displacement (your 4 inch arm)
  • F is the magnitude of the force (2.5 pounds + the gripper)
  • Theta is the angle between the force vector (gravity, pointing down) and the lever arm

You also want to account for the torque exerted by the weight of the 4-inch arm itself. The displacement you use to calculate this is not 4 inches, but the distance from the servo to the center of mass of the arm (probably 2 inches).

  • $\begingroup$ so in the worst case I would need torque = 4*(F_gripper + X) * sin(pi/2) + [2 * F_arm * sin(pi/2)] (where X is object weight 2.5lbs and F_arm is weight of the arm)? $\endgroup$
    – Jon
    Commented Jan 11, 2013 at 16:51
  • 2
    $\begingroup$ Yep, that's your operating torque. Add a little overhead for safety and you're good to go. $\endgroup$
    – Joe Baker
    Commented Jan 11, 2013 at 22:27

this page gives a nice overview of joint torque requirements for robot arms. and even provides a simple calculator. http://www.societyofrobots.com/robot_arm_tutorial.shtml


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