My next project.

Problem I see by gearing this down enough to not over rev the winch is the torque you will have multiplied so much that it will be extremely easy to over load the winch and cause some serious damage.
Just my 2c

Cheers Phil
 
Reductions are easy. Here are some examples.

Drive gear has 12 teeth, driven gear has 48 teeth. 48 divided by 12 is 4, meaning that the reduction is 4:1 The driven gear will turn 1/4 the speed of the drive gear.

Another way to arrive at the same conclusion: 12 teeth divided by 48 teeth equals .25 meaning that the driven gear will turn at 1/4 the speed of the drive gear.


Now, if you have multiple reduction ratios, let's say the first reduction is 4:1, the second reduction is 3:1, and there is a third reduction of 2.5:1 you can calculate the total reduction by multiplying all of the ratios as follows:
4 x 3 x 2.5 = 30 meaning the overall reduction is 30:1. The final drive speed of the spool would be 1/30 the speed of the input to the first reduction. If the engine speed was 6000 RPM, and the overall reduction was 30:1, then the spool speed would be 200 RPM.

Inversely, if the engine puts out a torque of 100 in/lbs, the torque after reduction would be 30 times greater, or 3000 in/lbs which equals 250 ft/lbs
 
terrywerm said:
Reductions are easy. Here are some examples.

Drive gear has 12 teeth, driven gear has 48 teeth. 48 divided by 12 is 4, meaning that the reduction is 4:1 The driven gear will turn 1/4 the speed of the drive gear.

Another way to arrive at the same conclusion: 12 teeth divided by 48 teeth equals .25 meaning that the driven gear will turn at 1/4 the speed of the drive gear.


Now, if you have multiple reduction ratios, let's say the first reduction is 4:1, the second reduction is 3:1, and there is a third reduction of 2.5:1 you can calculate the total reduction by multiplying all of the ratios as follows:
4 x 3 x 2.5 = 30 meaning the overall reduction is 30:1. The final drive speed of the spool would be 1/30 the speed of the input to the first reduction. If the engine speed was 6000 RPM, and the overall reduction was 30:1, then the spool speed would be 200 RPM.

Inversely, if the engine puts out a torque of 100 in/lbs, the torque after reduction would be 30 times greater, or 3000 in/lbs which equals 250 ft/lbs
Click to expand...


I'm thinking the saw is going to run around 10500 RPMs (Stihl 009L)


The question is, what gearing do I need in between these two?



The winch is reduced to 8:1 (what speed do I want the spool to turn? (60-80 ft/min cable speed))

http://www.lewiswinch.com/The_Lewis_Winch.html
 
The quoted figures on the winch of 8:1 just refer to its own internal reduction. Crank the handle 8 times, the winch drum turns once. Ignoring frictional losses, basically if you can apply 50 ft/lbs with your arm then you will get 400 ft/lbs out.
If the winch handle spindle is designed to be turned at a maximum of 60 RPM, (thats pretty quick for your arm on a continuous basis) and by the look of it, just bushes for bearings, I would say thats optimistic.
If your saw is running at 10500 RPM and we want to reduce it to 60 RPM, thats 10500/60= 175. Total reduction you need is 175:1 At a input RPM of 10500 your gearbox will need some pretty good bearings.
Now if your chainsaw is 2HP it delivers (approx) 1100 foot lbs force/sec. Now with your winch at the end of 175:1 reduction gears you will have (approx) 192500 foot lbs force/sec!! Enough to tear the winch to pieces.

Cheers Phil
 
Phil, I think there is an error in your calculations there. I don't think there is a chain saw made that will put out 1100 ft/lbs of torque per second. Not even the fabled V8 chainsaw of YouTube fame. To get that kind of torque we are talking about a 10 or 11 liter diesel engine. Are you sure you didn't mean inch pounds?? I ran through the calculations, and come up with 2 HP converting over to 1100 in/lbs. In that case, we're talking 91.66 ft/lbs from the saw engine, and 16041 ft/lbs input at the winch. Still enough to tear the winch apart!

Now, due to mechanical losses and other inefficiencies, I don't think we would see torque values quite that high, but they would still be significant! Additionally, there would be a lot of mechanical loss in power through a gearbox with a 175:1 reduction, not to mention that I don't know where in the world you find such a thing.
 
I think we have to work backwards. The winch is rated at 2200 lb, so as long as the load doesn't exceed that rating then it really doesn't matter what the ultimate torque the engine and gear reduction can deliver. Sure it may be capable of delivering more if you hook 4000 lb on to the winch, but then that is sort of like attaching a long bar to the winch handle and overloading the winch manually. The estimate of 60 rpm at the winch handle is probably not too bad. Now just come up with the 175:1 reduction and clutch.

David
 
OKay, there's a video of this set up working. The one in the video has a belt drive. I wonder if the original designer put that in there for a slip function?

I chose the winch I did so I would have to do less gearing down.

I guess I'll just try some gearing.
 
Dan,

If you can find an old 1/2" or 3/4" hand drill with an output around 500 rpm or less, you may be able to use that gear box as the first stages of your reduction. Hand drill motors run in the 18,000 rpm range so the input would be ok with your chain saw.

David
 
David S said:
Dan,

If you can find an old 1/2" or 3/4" hand drill with an output around 500 rpm or less, you may be able to use that gear box as the first stages of your reduction. Hand drill motors run in the 18,000 rpm range so the input would be ok with your chain saw.

David
Click to expand...

That's a great idea! I have another idea for a something chainsaw powered so that might help there too.
 
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