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


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 Jan 11 '13 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 Jan 11 '13 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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