I'm trying to reduce the rpm of a combustion motor. It operates at 2000 rpm with 3HP, and I need it to work at around 12 rpm (no matter the power). I've made the math with sprocket and chains, but the biggest sprocket is of 472 mm of diameter, which is too big. Is there any cheap and easier way to achieve this? Thanks!
2000/12 is about 167, so you need to gear it down a lot.
To a sensible 1st-order approximation, you don't want to gear down by more than 10:1 in any one stage (I'm not a mechanical engineer; if one of them speaks up feel free to believe them).
If I were throwing this together from surplus parts, or even building a one-off machine or a short production run, I think I'd run belt drive to a gearbox with my desired total reduction. So maybe 4:1 belt reduction to a 40:1 box, or 8:1 belt reduction to a 20:1 box.
One stage of reduction on a belt drive to one more stage of reduction with a gear drive wouldn't be horribly impractical. Because of the large torque multiplication, you'd need an enormous belt drive on the final stage which is why I'm suggesting gears or chains.
If you're building 10,000 a year, then you can afford to get an integrated motor/gearbox -- I suspect that's another story.
Or I'd go with 200:1 reduction because it's a nice round number, and run the motor at 1200 RPM.
Keep in mind that if the motor can do 3HP then the potential torque output at 12 RPM is around 1300 foot-pounds*, which is ginormous. Size your shafts accordingly or put in torque-limiting clutches, and don't get in its way.
* For those who noted my earlier math error -- I did the torque calculation from the RPM and power, without regard to gear ratio (and I used a calculator). So it should be correct.