Problems sizing stepper motor

[WayneO1]

Hi
I’m new to this forum and I hope my question isn’t one that has been answered here many times before. I tried to find the answers to my questions in the archives but was unsuccessful. Over the past few days I’ve become a bit perplexed with trying to size a stepper motor to a specific application that I have. Let me start by giving a basic description of my application. I am building a fixture for an assembly line that is required to compress a spring-loaded mechanism that has about 500 lbs. of force along a path of about 3 inches until the mechanism dead stops. I then have to draw it back .375 “ (+ - .002) to accommodate assembly. The way that I am attempting to accomplish this is by using a ball lead screw linked to a stepper motor via a slip clutch. There is a linear encoder in contact with the mechanism during the last .750” of the stroke. In theory when the mechanism dead stops the motor will over ride the slip clutch the encoder will stop pulsing my PLC. The motor would them be reversed and the mechanism drawn back to the appropriate position. My confusion comes about when trying to establish my torque requirements. I am able to compress the mechanism by hand with a small torque wrench which gives a reading of somewhere from 25 to 30 inch pounds of torque. I realize that the faster I turn it the greater the torque requirement but there seems to be a radical difference between the actual torque readings that I got and the torque ratings of motors. For example as I glanced through my McMaster Carr catalog I came across a DC PM motor rated at 1.5 hp and having 54 in. lbs. of torque and claimed to have constant torque output through its entire speed range. In my Grainger catalog I saw a ¼ hp gear motor with a 10 to 1 reduction claiming 60 in. lbs. of torque. To think that I could stall either one of these motors with my little finger and an 8” torque wrench would seem to be rather foolish. Yet that’s all it takes for me to compress that spring. Why does it seem to be such a radical difference. Is 54-in. lbs. of torque on a DC PM motor the same as 54-in. lbs. of torque on a stepper motor? Torque is torque right? Maybe this is just a conceptual problem on my part. How can I compare my torque wrench reading to motor torque ratings? Any help would be greatly appreciated.

Thanks Wayne

 

[gcaudill]

You are correct, torque is torque. The torque to comress the spring will be constant over the entire speed range. The motor HP will increase as you apply that torque at higher and higher speeds.

 

[Barry1961]

A 1 horsepower motor produes 36 in-lb of torque at 1750 rpm. A 1/2 horsepower motor would be 18 in-lb. A one horsepower motor running at 875 rpm would produce 72 in-lb of torque.

torque(in-lb) x rpm
-------------------- = horsepower
63,025

If you put a gear reducer on the motor the torque will increase by the reduction and the speed will decrease by the reduction. For example the 1 horsepower 1750 rpm motor with a 10:1 reduction on it would put out 360 in-lb(or less) of torque at 175 rpm. The horsepower output will stay the same or drop due to the inefficiency of the gear reducer.

Depending on the type of reducer you can lose over 1/2 of your power just driving the gear reducer.

Remember when sizing the stepper that there is holding torque, starting torque, and running torque. The running torque will changing with the speed of motor and the holding torque will have to stop the inertia of the system.

If you can give the linear speed of the screw I can calculate horsepower ect....

Barry1961