Stephen Wille Padnos wrote: > 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. >
Actually the diameter does matter, at least for acme screws. The calculation that Steven did is correct if you neglect friction. For ballscrews, with 90% or better efficiency, you can almost neglect friction. Not so for acme screws - the majority of the torque at the motor is from friction. Here is the counter-intuitive part - a smaller diameter screw of the same pitch will be MORE efficient. Assume you have 100 lbs pushing on the nut. Steven's formula above can be used in reverse to calculate the torque on the screw from the nut. For a 100 lb load, and a 0.200" screw pitch, that torque is 100 lbs = 1600 oz, times (0.2"/(2*pi)) = 50.95 oz*in (call it 51). But now you need to add the torque due to friction. The force is equal to the load times the friction coefficient, so it will depend on the material of the nut and screw. Google says the coefficient for a lubricated bronze nut on a steel screw is about 0.16. That means our 100 lb load will result in 16 pounds (256 ounces) of tangential friction force. That force is at the radius of the screw, where screw and nut meet. If the screw is 5/8 inch in diameter, its radius is 0.312 inches, and the friction needs 256oz * 0.312in = 80 oz-in of torque to overcome it. So you need 80 oz*in to overcome friction, and 51 oz*in to move the load for a total of 131 oz*in. (You'll want a motor that can deliver at LEAST twice that amount at your maximum desired speed, for insurance against lost steps.) Now switch to a 2" diameter screw. Now the 256 oz friction force acts at a radius of 1 inch, and you need 256 oz*in of torque to overcome it. Add the 51 oz*in that actually moves the load, and you get 307 oz*in. Now you see why we like ballscrews so much. Regards, John Kasunich ------------------------------------------------------------------------- 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
