# tractor to small engine with gear reduction

I want to run an implement requiring a minimum of 35 hp at 540 rpm with a small engine running 3600 rpm with a 6:1 gear reduction. What horsepower engine would be required to get the min hp to run the implement?

• Does the implement need a constant speed? Or will it vary? With a big enough fly wheel, once the implement reaches the required speed then horse power becomes less important. – Sir Swears-a-lot Dec 7 '14 at 7:54

You'll need about 40 HP @ 3600 rpm at a minimum in order to run it. Here is my math (please correct me if I'm wrong):

• 35 HP @ 540 rpm = 340.41 ft lbs (convert HP to Torque)
• 340.41 / 6 = 56.735 ft lbs (6:1 gear reduction)
• 56.735 @ 3600 rpm = 38.89 HP (minimum size of engine if actually running at 3600rpm)

You'll need to look at the spec of any engine to ensure the speed and hp actually mate up. Some small engines may product 40 HP, but will do it at say 5000 rpm, which would mean @ 3600 rpm the HP output may be much lower. 40 HP is going to be the minimum number to shoot for, I'd think.

• Your math looks good to me, assuming that what the implement actually needs is the same amount of torque that it would get at 35 HP @ 540 rpm, i.e. 340.41 ft lbs. Of course, it will end up with 340.41 ft lbs of torque at 600 rpm, which again is 38.89 HP (as it must be; energy is conserved). – Nate Eldredge Dec 5 '14 at 23:19