Patrick, > 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).
This confuses me a bit. If the diameter of the screw is 0.5" the circumference is pi*0.5 = 1.57" which gives 1" linear travel per 5 rotations (7.85") so a 1::7.85 ratio. > 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. Nope, most likely the 200 oz.in is the hold torque. This means that at a full step, with full current, the motor will be able to withstand up to 200 oz.in When the motors start moving, torque is less. How much depends on the motors, kind of stepper controller used and the speed. > 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? * 7.85 * efficiency. Ball screws have a high efficiency and since the calculation is a rough estimation anyway I use 100%. I did a simple measurement with a weighbeam on the current leadscrews and determined the force needed to move the leadscrew and selected my motors based on this. Replacing the old leadscrews by ballscrews would even reduce the torque needed but I decided to stay on the safe side. Hope this helps a bit. Regards, Rob ------------------------------------------------------------------------- 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
