On Mon, 25 Feb 2002 14:01:20 +0600 James Paul writes: > Donovan, > > > It seems to me that if your motor produces 550 Ft/Lbs of torque, > then > it would exert a force of 550 lbs at a distance of 1 foot from the > center > axis of the motor shaft. And if the free end point of the lever on > the > motor shaft were connected to a load cell, then the load cell should > report 550 lbs of force. If you shortened the lever to say 6 > inches, you > would double your torque, and therefore your force, so in that > case, the > load cell should report 1100 lbs of force. > Foot pounds of torque are not the same as foot pounds of work. The foot pounds in the "550 foot pounds per second" are foot pounds of work. If you have a cylinder with a 1 foot radius, and hang a cable on it with 550 pounds, the torque on the cylinder is 550 foot pounds. If you rotate it 1 revolution per second, the cable is pulled 2*pi*R feet per second, or about 6.28 feet per second (and the force on the cable is 550 pounds, corresponding to the weight). So, in this example, we have 550 foot pounds of torque and 6.28*550 foot pounds of work (3,456 foot pounds of work per second, or 6.28 HP). As pointed out by others, you can vary the radius of the cylinder. As the cylinder gets smaller, the torque goes up, but the power goes down (assuming the same RPM). Harold FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: http://dl.www.juno.com/get/web/. -- http://www.piclist.com hint: To leave the PICList mailto:piclist-unsubscribe-request@mitvma.mit.edu