Thanks, yeah its definitely a torque, but I have no idea what it relates to and how much i need. I was thinking of maybe taking a motor out of a cordless drill
Kevin
Kevin
You said you do not comprehend torque and have no idea what it means to you when building equipment.
I repeat what Mildrone said - shame on your instructor.
OK so here we go.
Force X distance = work ie HP Kw whatever
Torque is a force on a cylindrical object to make it turn.
Tie a piece of rope to a 2' diameter wheel (radius = 1 foot) so it will not slip. Mount wheel above floor and tie a 100 pound weight to it. Put a torque wrench on wheel shaft and apply force to wrench. A 100 lb force you put on wrench at one foot will raise the load. The weight applies if you want to call it that "counter torque" better yet think of it as torque imposed by the load.
IF you keep twisting the wrench and applying 100 lb force to wrench you will keep raising the weight.
Work is proportional to time so the faster you go the more work you accomplish ie feet raised per unit time
NOT how much sweat you are producing.
HP = 550 pounds raised one foot in one second.
HP (work) = torque X RPM / 5252
This example is a constant torque load
it will always impose 100 ft lb torque. Your conveyer is a constant torque load provided the number and size of your spindles is constant.
HOMEWORK
get an inch pounds torque wrench and measure the torque of the drive pulley shaft.
You can also get a piece of heavy string or wire and wrap around shaft and then hang weights.
YY pounds weight X shaft diameter / 2 X 12 gives you foot pounds.
RPM of drive sheave = desired speed of conveyer (fpm) / sheave circumferance.
Now plug this value into HP = torque X RPM / 5252
and you will get motor HP
We will do the gearbox next lesson.
You may argue but I am only going to do the programming. I will save you the Peter Nachtway et al lecture
- programming cannot overcome poor machine design.
If you cannot distinguish poor vs good machine design you are going to have problems.
Dan Bentler