Gearbox/reduction for CIM,RS550,RS540

We have some CIM, and RS540/550 motors left over from 2008 and 2009 kits. Interested to use them as future drive motor in the prototype platform. Normally, folks will use some gearbox to reduce the RPM of the motor to make it usable as drive motor. Have any one/team use either chain/sprocket or timing bell/pulley directly to drive the robot successfully. If yes, how and what are combination of sprocket/pulley parts numbers?


Typical gearbox ratios are around 10:1 to 16:1 for the CIM, and three times that for the smaller motors. You would need multiple stages of reduction, and the one connected directly to the motors will turn very fast! Gears work better…that’s why we use them…

k well if i understood what my teachers were tryna teach on gears…

the reason why you use the gear box is for mor torque… right?
and more torque essentially means more power

whereas if you didnt have the gear box, youd have the high speed of the motor, but no power…

is it possible to not use a gear box (for weight restraints) and still create a good amount of gear reduction?

The guys up the road from us on 612 didn’t actually use transmissions for their drive system this year, check them out in Atlanta if you get the chance. They simply used a small pulley on each CIM and a large pulley (almost the same diameter as the lunacy wheels) bolted directly to the corresponding lunacy wheel.

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.

It is possible to not use a gear box and create a good amount of torque. A gear box is just a means to an end, you can create a reduction through various means. These include gears, sprockets and chains, belts as well as less traditional means.

I recommend How Gears Work | HowStuffWorks as a good introduction to the concepts and types of gears if you are interested.

Thanks for all the input so far.

It looks like someone did it this year. From paper and commercial available part list, it is quiet easy to achieve 6:1~9:1 reduction with sprockets and timing belt pulley setup. The 6:1 is easily done at CIM’s RPM. What worry me is the 12+K loaded RPM. Will the chain jump the sprocket or the belt jump the pulley at that speed under relative heavy load. Interested to hear from someone who tried the setup, whether it is success or failure.


Here’s another plug for JVN’s design calculator. Team 234 used this tool to design their entire robot this year. I cannot push this tool enough…

I agree, this is a very handy tool and it rocks. BTW, 234 has a very smexy bot this year!:smiley: