Building a trommel! This trommel is for tumbling about 100 lbs of material (horse pee soaked saw dust). To envision what a trommel looks like, watch Gold Rush or Google â€˜trommelâ€™. The idea is the same, the scale is smaller! I have welded two 45 gallon drums end-to-end with holes cut in each end to insert the material. I call this the drum assembly. The drum assembly sits horizontal on four rollers (two each side of the drum assembly). On one side I use two four inch castors, on the other side I use two large pneumatic tires/wheels (actually a small hand-cart turned upside down). My hope was to drive the two large wheels with a small motor and the wheels in turn rotate the trommel (friction drive). This worked well (i.e. I could manually turn the wheels and the wheels would rotate the trommel without any slippage) UNTIL I added material to the trommel. After adding material to the trommel (Iâ€™m guessing around 100 lbs of material) the wheels slipped on the trommel. I could manually rotate the trommel by standing at one end of the trommel and rotating by hand. When I did this I noticed the material tended to cling to the inside of the trommel until the trommel had rotated about 60 degrees. Naturally, as Iâ€™m manually rotating the trommel to the 60 degree point the load is increasing. At one point I attached a pipe wrench to the rim of the trommel, which effectively increased the radius and made the trommel easier to rotate. Iâ€™m guessing Iâ€™m about 24 inches from the centreline of the trommel and exerting about twenty five pounds of pressure (i.e. if I were to hang a 25 lb weight to the handle of the pipe wrench (with handle parallel to the floor) the trommel would rotate). This indicated my plans for a â€˜friction driveâ€™ was not going to work. So now Iâ€™m considering a belt drive. Here is my problem, Iâ€™m not sure how to calculate the motor size I need. I donâ€™t want any forum member to calculate this for me, I want to know a good resource (book) that discusses how to figure this out. Here is the track Iâ€™m on, based on my above crude observation with the pipe wrench (at 24â€ radius) and 25 lb weight the torque delivered is 50 ft/lbs (24â€x25lbs = 600inch/lbs or 50ft/lbs). This takes into account drag (bearings) on the trommel and material weight in the trommel. This is the torque required not on the trommel, but 12â€ out from the trommel, how do I calculate torque required at the rim of the trommel (24â€ diameter, 12â€ radius)? After I have calculated torque required at trommel rim, now Iâ€™m looking at belt drive. I understand the speed and torque relationship between drive pulley and driven pulley. It seems that as the motor output pulley increases in size, the available torque decreases. Am I wrong about this conclusion? If Iâ€™m not wrong, how is 'delivered' torque calculated as a function of output pulley size? Thank you for your patience, again Iâ€™m looking for general information and resources. A lot of what I do on the farm is mechanical drive related and frankly, Iâ€™m pretty clueless -- but willing to learn!! Thanks!