What are the formulas for SCARA manipulator where no torque is required from actuator to carry the weight of the links, load, or itself; only turning them around.
@Michael Unfortunately the formula for the SCARA robots is beyond the scope of this article, but we can see there are articles published about exactly that. Look for SCARA Robot kinematics.
hi, how would you calculate the torque for an hexapod robot with 2 DOF?
@Bam This should help: https://www.robotshop.com/blog/en/robot-leg-torque-tutorial-3587
Fix one of the three joints in the leg.
If I’m understanding this calculator correctly, then the “A” value should be the mass (weight) of the actual servo at a given joint, and “M” value is the mass (weight) of the materials used to create the linkage between the pivot points. Should I include this “A” value separately at each joint if I already include its’ mass (weight) as part of the value of the “M”?
@Robb In these calculations, the weight of the connecting linkage / mechanical structure “M” is at a different location than the actuator “A” and therefore impacts the calculations / torque differently. More info here: https://www.robotshop.com/blog/en/robot-arm-torque-tutorial-7152
Did you use Jacobian matrix for this
@Abhimanyu No - simple torque balance at the “worse case scenario” (but without inertia or friction).
how to servo motor cection and caculation for Robotics
Sorry ,
How to servo motor selection And Caculation Formulas.
@Munees This tutorials shows the formulas: https://www.robotshop.com/blog/en/robot-arm-torque-tutorial-7152
my worst case torque is 36 , i have 13 kg cm servo…will it do the work sir??
I am working on robotic manipulator(articulated 6 DOF) and i want to apply gravity and inertial compensation for which i have to control the torque of DC motor ,so do you any method to do that or any tutorial.
@amesh Unfortunately we do not reply during weekends. If you provide only 1/3 of the required torque, you won’t be able to move the arm much past vertical; 13Kg-cm won’t work with the input parameters you provided. You will need to decrease the payload and/or decrease the weight of each joint and/or decrease the length of each joint (or choose a more powerful actuator).
@Awais Mughal Excellent question indeed - the equations used here do not factor in inertia, nor do they include torque control. You will need to update the “worst case” scenario and include a downward acceleration for all links and actuators, as well as the load. We currently do not have time to do this, but will certainly consider it for future articles.
So to lift a 300 lb person with a forearm of 15" weighing 10lbs would require an elbow servo with 5271Nm of torque?
@JM Since you only indicate one length, we assume you mean shoulder servo? You would need 300lbs*15in/12 = 375lb-ft = 508Nm This is why normally rotational actuators are not used for heavy loads - you might consider a linear actuator.
I would like to ask you how to calculate the torque if we use the newtons law F=mg & T=F . L so how can i find the torque as mass is not given meaning that when i go to to the workshop to implement the design
@Nasr You’ll need to estimate the masses of each joint and actuator in order to estimate the torque. If not, over-estimate the torque needed per joint and you will simply need to live with whatever weight the arm supports at the end.