Winch re-design questions/sanity check.

[dpsmith85]

I'm currently in the process of redesigning a winch system and wanted to go over some numbers and make sure I'm thinking about it correctly. The current design has been in place for about 10 years but probably wasn't used but once or twice a month for about 10 to 15 minutes each time. I've attached a picture of the current setup with gear ratios etc.

Here is the process I am following through the system:

1. 3/4 hp motor at 1725 rpm produces 27in-lb (36in-lb x .75)

2. 27in-lb x 5 x .9 (Assuming 10% friction loss) = 121.5in-lb
entering the gear box. (This is a 5:1 belt and pulley)

3. 121.5in-lb x 40 x .75 (25% friction loss for gear box) = 3,645in-lbs exiting gear box

4. 3,645in-lbs x 2.3 x .9 (Assume 10% loss) = 7545 in-lbs total torque for system (the last stage is a chain drive)

5. Compare 2000lbs x 2.75 (winch drum radius) = 5,500 to the 7,545 theoretical max and we should be good correct?

That being said here are some problems I am seeing with this design. The gearbox involved is a Boston Gear 721 40:1 with the following specs output torque 737in-lbs and input of 0.68hp. So am I correct in saying that the teeth on the gears should never feel more than 737in-lbs of torque? If so this system is seeing 3,645in-lbs which is 5 times larger than the manufacturers spec! It looks like the mistake was made by putting a pulley in front of the gearbox thus increasing the input torque beyond the rated spec. I'm throwing this all out there to do a bit of a sanity check and make sure my head is in the right place on this. Thanks in advance for any comments/feedback on this situation.

 

[BrianE22]

Your numbers make sense to me -

 

[dpsmith85]

Do you agree that the manufacturers spec is exceeded? I'm making sure that I'm understanding everything correctly. It's one of those situations where in my mind they are exceeding it so bad that it is making me second guess my thinking. If the manufacturer specifies an input h.p. of 0.68 then they are basically saying "Don't put more than 0.68 h.p. (or 24.5 in-lbs) of torque in to this gearbox or you're going to have a problem". Is that correct?

 

[geesamand]

I agree you are probably over spec on the reducer. Your calculation seems sound. Remember that for drives like this with intermittent use, the torque is the real spec for gearbox performance. Our company makes our own reducers and the sales staff (who select and size reducers per application) think in terms of how much torque they need to transmit at the output shaft. This torque limit is driven by the gearing design. So the BG reducer is a 737 in-lb box and you need a 3500 in-lb box, give or take service factor.

I have not sized a BG unit before, but perhaps the HP limit is a thermal rating. It would need to go well above 2-pole motor speed to have .68 hp and 737 in-lb output in the same application. Just a conjecture on that though.

David

 

[dvd]

I would not use a friction loss factor on the torque multiplication in a reducer. The loss is to the power where heat is generated and power dissipated; the torque is multiplied by the gearing and doesn't really disappear.

Plus your motor can develop even higher torque than the full-load torque, at the pull-out load it might be closer to 200% of full-load. The reducers usually allow for brief excursions with higher torques.

Do you need all of the torque that was being developed, or was the motor-to-reducer-input reduction done in order to slow down the winch? If it was to slow down the winch, you could possibly run on a VFD at 1/5 of the synchronous hertz.