@Raja K You need to factor in inertia and angular acceleration, which is not easy to do at this stage.
@Coleman Benson Any software is there for drive an extra torque required to move the joints based on speed(rpm)?
@Raja K Not that weâre aware of. Programs like Simulink and Matlab, as well as ROS and others have the capability of these calculations, but youâd either need to find examples, or create the simulation yourself.
@Coleman Benson I have an additional motor which is placed under A6. It is responsible for the yaw movement of the robot. How can I calculate the torque that I need for my motor? Thank you.
@Eugene Agustin You are correct in that most robot arms have an additional actuator located in the base. The calculations are entirely different though, as well as the considerations. That actuator, assuming itâs vertical, needs to overcome friction and inertia, which is not easy to calculate without more âknownsâ. The worst case scenario for inertia is decelerating from maximum speed with the arm fully extended. Look up the foruma for inertia of a stick rotated at one end to see how it relates to torque (Torque = moment of inertia multiplied by angular acceleration).
@Coleman Benson Thank for your answer. I will search a discussion about that thank you again
i am confused in something when i put the value of link length to calculate torque ,i have to links parallel to each other
do i put the value of one link or the sum of two links ?
@mahmoud The âlink lengthâ is the distane between the axes of rotation, regardless of what is in between them.
@Coleman Benson am I right that computed torque here is the STALL TORQUE?
Do you what is larger between STALL TORQUE and RATED TORQUE? If I consider the Rated Torque what values I need to add up?
@Eugene Agustin The torque provided is simply the torque needed to maintain that position, without any acceleration. âStallâ generally indicates the motor is impeded from rotating and will still try its best, causing it to heat up very rapidly and burn. Rated torque generally means the torque which the motor can provide without issue. If you wanted a margin of error, you should consider the torques calculated here to be the torque at maximum efficiency" for the motor. Torque at max power would only likely last a few seconds.
@Coleman Benson So what is greater between stall torque and rated torque? How large is the greater between the two? BTW thank you for the explanation.
@Eugene Agustin Stall torque is the absolute max a motor can provide - to the point where the shaft cannot rotate and the motor would burn. The rated torque is what the motor can provide safely. Therefore the value for stall torque is quite a bit higher than the value for rated torque.
@Coleman Benson Thank you for your explanation. If the result of the calculated torque here is 500kg-cm. What value of rated torque of the motor should I look? Is it less than 100 or half of the result or what? (just an assumption will help). Thank you.
@Eugene Agustin Wow. For the arm to remain in place at full reach with full load, thatâs the torque your shoulder motor will need. As soon as you drop the load at full reach, it should be able to move. You should consider using linear actuators. Note that in the 500Kg-cm range, youâll need to consider industrial actuators. Remember, every gram at full reach has a huge impact on the shoulder motor.
Hello Sir, i plan to make 3DOF articulated robotic arm. I just want to know if MG996R base servo (9.4kg/cm stall torque) is enough to rotate the robotic arm (including shoulder, wrist, till end effector, and targeted object) that has weight ±211 gram. and has maximum length of 24 cm if its extended from the center of rotation to the end effector. Please reply asap, Thank You
Welcome to the community Khoukou. Lynxmotionâs RC-based AL5D arms use the Hitec 485HB in the base, which is specâd at 5.2kg-cm to 6.4Kg-cm, so yours should work provided their specs are accurate. The general idea is that horizontally, thereâs not a lot of static or dynamic resistance to motion.
Hey @cbenson,
I hope you are doing well. Thanks for this awesome tutorial. I calculated the torque required at each joint with this tutorial, I wanted to know if the same torque is applied to the link? if yes, in what direction. as I want to calculate the total deformation of the link in Ansys.
Thanks once again
@JJenn Welcome to the RobotShop Community. The links between joints can be treated as beams subject to force as opposed to torque. You will need to sum the weight acting at the end of each beam (and to be accurate, the weight of the beam itself at its center of mass) to calculate the deformation.
Uh-oh hello, there is few thing that i dont quite understand when i calculate the joint by myself i use gravity in the formula and pretty much get a triple digit of torque when in this calculator i only get 2 digit how does this calculator work exactly? just like in the tutorial? which mean not counting for gravity?
@RIzami Welcome to the RobotShop Community. If youâre using a force as opposed to a mass, youâre factoring in gravity, For example Kg is a mass, while Newtons ( 1N = 9.81 x 1Kg) is a force. It will depend on what value you want for the torque (ex. Kg-cm vs. Nm).