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  • Building a Trommel, help with calculations!

    Discussion in 'Calculations' started by twebb, May 4, 2013.

    1. twebb

      twebb New Member

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      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 ‘trommel’. 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 (I’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 I’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. I’m guessing I’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 ‘friction drive’ was not going to work. So now I’m considering a belt drive. Here is my problem, I’m not sure how to calculate the motor size I need. I don’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 I’m on, based on my above crude observation with the pipe wrench (at 24†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 12†out from the trommel, how do I calculate torque required at the rim of the trommel (24†diameter, 12†radius)?

      After I have calculated torque required at trommel rim, now I’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 I’m not wrong, how is 'delivered' torque calculated as a function of output pulley size?

      Thank you for your patience, again I’m looking for general information and resources. A lot of what I do on the farm is mechanical drive related and frankly, I’m pretty clueless -- but willing to learn!!

      Thanks!
       
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    3. paul.wang

      paul.wang Member

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      Hi Twebb,I recommend to you a book "Practical Drive Engineering – Project Planning of Drives", of which "8 Calculation Example: Travel Drive" estimationis useful for you.You can find him, from Google or Baidu.
       
    4. twebb

      twebb New Member

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      Thanks for the resource

      Paul,

      I downloaded the document and took a quick peek, I'm expecting this to be very helpful. THANK YOU!!

      twebb
       
    5. ChrisW

      ChrisW Well-Known Member

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      Establishing the worst case scenario for a trommel drive is straight forward. Consider the drum filled evenly to the centreline with your material then rotate it 90 degrees. Calculate the contained mass and the distance to the centre of gravity and you have the max torque. Any greater fill will reduce the required torque so this is an effective method and gives you some safety margin as it can never be achieved in practice.

      Another value you need to calculate is the maximum rotational speed. If the peripheral centripetal acceleration is more than one g the material will not cataract and you will not have any screening action, it will stick to the drum walls.

      A useful project, good luck.
       

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