tree grower
Silver Member
- Joined
- Dec 29, 2007
- Messages
- 206
- Location
- Cuttingsville, VT
- Tractor
- Ford 1210, Bobcat 742B, John Deere 1050
I am fabbing a M6 Braden winch into a psuedo-Farmi for firewood skidding. If I Google Braden winch, there is an excellent 40-page manual. On page 39 is a table which includes max input RPMs and line speed in Ft per minute for many Braden models, but not the M6. I emailed Pasccar, the current manufacturer of Braden, and they sent me an old M6 brochure with some specs, but not all. Soooo, we resort to extrapolation.
With a worm gear reduction of 34:1, I assume 34 revs of the worm = one rev of the gear = one rev of the winch drum. With a 5" winch drum diameter, one rev = 16". At 300 RPM input speed, that is 4800" or 400 ft per minute. I think 100-200 ft/ min is plenty fast enough for a log to be travelling behind a winch, so I would need to cut my input speed in half--easy to do with a jack shaft and a small sprocket. (PTO speed 300 rpm to jack shaft, 12 tooth sprocket #60 chain to 24 tooth sprocket on worm)
Request critique of my assumptions and calculations. Thanks in advance.
With a worm gear reduction of 34:1, I assume 34 revs of the worm = one rev of the gear = one rev of the winch drum. With a 5" winch drum diameter, one rev = 16". At 300 RPM input speed, that is 4800" or 400 ft per minute. I think 100-200 ft/ min is plenty fast enough for a log to be travelling behind a winch, so I would need to cut my input speed in half--easy to do with a jack shaft and a small sprocket. (PTO speed 300 rpm to jack shaft, 12 tooth sprocket #60 chain to 24 tooth sprocket on worm)
Request critique of my assumptions and calculations. Thanks in advance.