winch PTO gear ratio ?

I have look at all of the TM's that looked appropriate.. but I can not seem to find the gear ratio of the winch PTO for the M35 front mounted winch.. Anyone know the answer or where I can find it.. Thanks

You should have started by doing a simple search for "PTO ratio"....
2nd page: PTO speed, here....


G.

Nice link, as always...great info G! I was surprised at 27hp...not that a worm drive needs a crazy amount. I remember it being in that area when I was setting up the oil motor for the deuce winch project for the front of the Mog.

spicergear said:

...not that a worm drive needs a crazy amount......

Click to expand...

Following may help to calculate, or better say estimate HP needed to operate our winches at rated power (and over), using the popular formula:
HP = torque x rpm / 5252


Winch facts:

• PTO low range for the Spicer 3053 is 650rpm @ 1000 engine rpm; equals 520rpm at 800rpm (engine idle)
• 520rpm, propeller shaft => 22.6rpm, drum (23:1 ratio)
• Bare drum diameter is 5.0”, so 1 drum revolution takes 1.5ft rope (first layer)

Winch is rated 10,000lbs @ 15ft/min, bare drum), so:

• 10,000lbs on bare drum => 2333 ft.lb torque needed on drum shaft
• 22.6 rpm => 34ft/min, so 15 ft/min => 10 drum rpm

2333 x 10 / 5252 = 4.5HP, as rated.


Now, all losses included, the estimated efficiency for a typical worm gear reduction is not more than 60%. So we actually need more like 7.5HP at rated speed, but prob. up to 17 HP at actual speed, in LOW.....!?

Interestingly, with the original shear pin resisting around 300ft.lb, a winch operating at the real “limit” and only 15ft/min would then draw 13 HP from the PTO, which is +/- what the book shows... (What they didn't tell is how to achieve that low pull-in speed).....


G.

Here's yet another permutation of winch application...Gringletaube, please weigh in here. I'd been trusting the formula you cited but not sure what it's based on, so maybe that's the problem.

I'm looking at repurposing a winch taken from a scrapped 5-ton, by using it in a stationary configuration to haul stuff up a steep slope. In my case, the winch would be adapted to be driven by a variable-speed AC motor (really a 3-phase motor driven by a VFD, for those who want to know). So I'm looking to determine my line-pull with one possible motor candidate, which happens to be a 3HP/1750RPM model.

Reorganizing your formula to solve for torque:

Torque = (HP x 5252)/RPM

The HP is just 3 in this case, from my motor ID plate.

RPM (at drum, as per your example) would, at design RPM of the motor, be 1750/23 = 76RPM. (but note that RPM can go down to as low as zero, using the VFD, as well as maybe double the 1750 ID-plate rating, though I'd not do that for real).

Thus, at my AC motor's design RPM, I'd get a torque of (3 x 5252)/76 = 207 ft-lbs. So at the full (12") drum condition, I'd get only 207lbs of line pull? Seems intuitively wrong from a 3HP motor.

But playing with that version of the formula a little more, say I reduced the motor's RPM using the VFD (which will still give something close to rated HP because of the VFD electronics) - say I dropped the RPM by half. Per the formula, that'd mean I'd get TWICE the line-pull? Nah...can't be right. And of course, as I drop the RPM closer to zero, the line-pull would blow up to something astronomical.

I must be mis-using the formula - let me know what's up with that. Seems like it's only appropriate to use it to calculate the input HP vs any other variable, but that's not how formulas are supposed to work...

Thanks - Dave

Your plugging in deuce winch specs for a 5ton winch, I bet your numbers are off a bit.

True, though the formula is generic.

I just manually checked the winch ratio; it's 30:1, which means my drum would turn at 58RPM, not 76RPM, with my AC motor running at ID-plate RPM of 1750 RPM. I only had time to eyeball the drum ID but it looks like it's prolly 5", too. Not much impact on the outcome of the calculation, though.

--Dave

I seem to remember something about with the truck idling in 1st gear, 6 wheel drive, winch pto engaged and on 1st layer the truck would not overdrive the winch. In other words, on dry land under normal traction, the winch would reel in about the same speed as the truck would move forward? Am I having a senior moment?

Bump for ratios and other good questions