Patrick Ferrick wrote: >[snip] >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? > > The leadscrew diameter isn't important for the force calculation.
The motor can output 200 ounces force at a one inch radius, as you said. For one revolution, the 200 ounces of force is applied over 2*pi inches of travel, but produces 0.2 inches of table motion. Just divide: 200 * (2*pi*r) / (0.2) = force in ounces =2000*pi, or about 6280 ounces force (close enough to 400 pounds). The units are ounces * (inches) / (inches), which is ounces since the lengths cancel out. - Steve ------------------------------------------------------------------------- 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
