Gear Reduction

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Posted by admin | Posted in Parts And Stuff | Posted on 15-09-2010

Tags: audio, camera, gear, gear reduction, gear reduction box, gear reduction calculator, gear reduction motor, gear reduction starter, photography, shopping

Gear Reduction

how to figure out gear reduction ratios?

hello! I am building a ball mill. it is a container of minerals and metal balls that spins at about 50-100 rpm. the metal balls roll to the top and them crashing down on to the mineral being crushed.

I am following the plans that are at this website

http://tutorialtub.com/info/building-a-ball-mill-for-grinding-chemicals

so I need some help figuring out the ratios for the motor. the motor spins at 1100 rpm. any help would be greatly appreciated.

Looking at the link, this chap is driving an axle that rotates the drum by friction. So first, you need to know the RPM of this axle. You can find this by measuring your drum diameter, and the diameter of the axle (he has a rubber hose on it, making it larger in dia. but grippier).
If, say, the drum is 6" dia., and the axle 2", the ratio is 3:1.
So the axle would need to rotate 3 times faster than the drum, say 300 rpm. So you need to work out the reduction between the motor and the axle.
This is also a ratio. In this case, the motor does 1100 RPM, and you need 300 RPM.
1100/300 = 3.66 :1, so this is the reduction required.
Now let's assume a 2" dia. pulley on the motor. The one for the axle would need to be 3.66 x 2", or 7.33" diameter.
The actual formula for calculating pulleys is as follows:
Revs of motor x diameter of motor pulley = revs of tumbler drive axle x diameter of pulley required.
1100 x 2" = 2200.
2200/300 = 7.33".
Same answer.
You can use any combination of pulley diameters, and axle and drum diameters, so long as the overall reduction is the same.

Heng Long Panther with 3:1 gear reduction