Hello again, [Machine-related, but off topic, so I'm really looking for someone to steer me toward a good reference on the subject, or to communicate off-list.]
I'm thinking about replacing my Z-axis leadscrew with a larger diameter one that will have less backlash. I'm trying to figure out how to calculate how much (linear) force a certain stepper will provide with the new screw. The screw's pitch is such that one turn gives 0.2" of travel, or 5 turns per inch. That would seem to give it a mechanical advantage of 5, but what has me a little bit confused is the translation of a torque (from the motor) to a linear force (by the nut). I believe it is correct to say that a 200 oz.-in. motor, for example, can exert a force of 200 oz. (12.5 lb) at a distance of one inch from the center of its shaft. Since the leadscrew has a mechanical advantage of 5 and is 0.5" in diameter, does that mean that I should expect (200 oz. / 0.5") * 5 = 2000 ounces (or 125 pounds) of push? thanks again, Pat .................... Patrick Ferrick Town of Webb School Main Street Old Forge, NY 13420 (315) 369-3222 (315) 369-6216 .................... ------------------------------------------------------------------------- This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse0120000070mrt/direct/01/ _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
