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Unread 03-04-2009, 13:54
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Re: Gearbox/reduction for CIM,RS550,RS540

Power stays the same, since power is torque times speed. You need gear (or belt or chain) reduction to reduce the speed, as well as to increase torque.


The first step is to figure out how fast you want the wheels to turn. A typical top speed for a robot is about ten feet per second. With eight inch wheels, which have a two foot circumference, you'd need the wheels to turn five turns per second (or 5 x 60 = 300 revolutions per minute) at full motor speed. Full speed of the CIM is about 5000 rpm, so you need a reduction of 5000/300 = 16.6:1

Look at the belts and pulleys you have, as well as the chains and sprockets. Figure out how many teeth are on the smallest ones, and how many are on the biggest ones, for each type of belt or chain. Divide the smallest by the largest, to get the maximum reduction for that type of drive system. For example, if you have a 36 tooth sprocket and a 15 tooth sprocket, you can get 30/15 = 2:1 recuction. If you have a 16 tooth pulley and a 56 tooth pulley, you can get 56/16 = 3.5:1 reduction.

You could connect a belt drive reduction directly to the motor, then add a chain reduction, and you multiply the ratios to get the total reduction. So if you used a 3.5:1 belt reduction, then a 2:1 chain reduction you would get 7:1 total reduction. Adding another 2:1 chain reduction would get it to 14:1 which would probably work pretty well. It will be a big, bulky drive, but it could work. Or buy some more sprockets, such as a 9 tooth and a 36 tooth, and you could get a 4:1 chain reduction.

Last edited by MrForbes : 03-04-2009 at 13:56.