I am here to see if anyone can help me determine the stall torque of a specific weight in Kg. I am using a 393 motor for my project which specifications say that it has a stall torque of 1.67 n-m, 14.76lbs (6.695kg) I am assuming this is the maximum load that it can move. What I want to do is find out how much torque is used when moving 730g.
I found this http://imgur.com/8jL7BOT equation on wiki and I am not sure if the s2 means speed (in this case RPM).
Am I approaching this in the wrong way or it as simple as entering the correct values into this equation?
The image you quoted is the units of what a newton-meter is.
A better way to think about torque is that it is a force * distance. If you have your .73kg load, that is about 7.1 Newtons. Now from the definition of the stall torque of the motor, it is able to handle 1.67 N-M. From this we can calculate the distance from the pivot that would be that distance by dividing the load by the torque and we get .23m.
From this we know that the motor will stall if you have your 730 gram load 23 centimeters away from the pivot.
Now with motors stall torque is not everything. As the torque increases, speed (rpm) decreases. So if your motor is 100 rpm free spinning, as soon as you add a load, it is less than this until it reaches 0 rpm when the stall torque is reached.
There is a graph somewhere on the forums of the speed vs torque but I could not find it.
If you need to move your object at a greater distance, you can use gear ratios to multiply your torque, but lose speed in the process.
Motors give you a torque, not a force. The stall torque for a 393 motor in imperial units is 14.76 inch pounds. That means that if you attach the motor to a wheel with a 1 inch radius, the motor will push on the rim of the wheel with 14.76 pounds of force. If you use a wheel with twice the radius then you will get half the force, and if you use a wheel with half the radius you will get twice as much force.
To know how much torque is used when moving 730g, you need to know value of the radius where the force (730 grams x gravity) is applied. This doesn’t depend on the motor, it’s simply (force)x(radius).
If the radius is small, then the motor will easily be able to lift 730g. If the radius is large (20+ cm) then the motor will either be really struggling or just fail to lift the load. At just under half if it’s maximum torque (so at about a 10cm radius) is where a 393 motor will be able to lift a 730g load fastest.
Motor torque Required to drive the mechanism = 0.093 Nm
That looks right to me, but be aware that your actual load must include frictional forces, and for a linear slide, those frictional forces could be significant. Also, don’t forget to consider the weight of the linear slide or whatever else might be moving, if you are driving things vertically, etc.
awesome! Do you have any idea where I can find any official documentation explaining how to work this out? It is for a university project and I need to cite the information.
If you’re talking about how to calculate torque, I would guess that any first-year physics textbook should cover that calculation somewhere. If you’re talking about the torque-speed curve for 393 motors… um… uh… I suppose the following post is as good as it gets (since the data were confirmed by Vex):
thanks for the answer. Now, someone posted that I should go higher than 7 in-lbs because i will trip the computer. Is that true and where can I get a good resource on amps? Is there some way to see how much amps Im running on the robot or do I just have to mathematically figure it out on paper? thanks
TL;DR keep motor revs above 65rpm (when sending full speed command) and you should be ok. Too big of a topic to go into detail this late in the evening, there’s lots of discussion about this topic on the forum.
It being in the same thread I thought you were the thread owner.
I am a teaching assistant for the intro to robotics class at WPI and I had a feeling that the thread owner was one of my students (my family finds the term hilarious).
It’s always nice to see more middle school teams trying to do a little more of the engineering in robotics.