Garwood 20K winch mounted in frame

[Markddhumvee]

Could anyone please identify this winch set up I have come across. I would like to convert what appears to be a cog pto drive set up to a hydraulic motor drive. Need to know the correct shaft RPM if possible.
Any help very much appreciated
Mark

 

[Evil Dr. Porkchop]

That’s off an m756a2 pipeline truck.

 

[Markddhumvee]

That’s off an m756a2 pipeline truck.

Thank you - it looks like one of the pictures I posted is not coming out well Ill post it here again to see if we get better results!
Do you have any idea of how I could provide power to the shaft either by installing a hydraulic motor or by obtaining the original drive gear by any chance.
Many thanks for your help!

 

[gringeltaube]

I firmly believe that that winch is basically the same as the 5-tons front winch (#7412382). If so, then the (bare)drum diameter is 6 inches and the reduction ratio, 29:1.
That means, for a cable speed of say 15 ft per minute (1st layer), the drum has to turn at ~9 rpm and the input shaft at about 260 rpm.
(Just apply the rule of three for any other desired speed.)


This thread might be helpful if you wanted to convert to hydraulic drive.

 

[Markddhumvee]

I firmly believe that that winch is basically the same as the 5-tons front winch (#7412382). If so, then the (bare)drum diameter is 6 inches and the reduction ratio, 29:1.
That means, for a cable speed of say 15 ft per minute (1st layer), the drum has to turn at ~9 rpm and the input shaft at about 260 rpm.
(Just apply the rule of three for any other desired speed.)


This thread might be helpful if you wanted to convert to hydraulic drive.

Thats great many thanks. I can get a hydraulic motor that will drive at 240 rpm - you wouldnt happen to know the torque thats needed to drive the shaft by any chance would you?
Again thank you for your help with this.

 

[gringeltaube]

Assuming that the rated pulling capacity (20.000 lbs) was only for the first layer of wire rope on the drum, then the required torque on the drum shaft would be 20,000 / 12 * 3.3 = 5,500 ft-lb

And considering the worm gear type reduction plus all other friction losses combined, the efficiency of our winches is estimated at 65%.

So (if my math isn't totally off) the required torque at the input would be 5,500 /29 * 100/65 = 292 ft-lb

 

[Markddhumvee]

Assuming that the rated pulling capacity (20.000 lbs) was only for the first layer of wire rope on the drum, then the required torque on the drum shaft would be 20,000 / 12 * 3.3 = 5,500 ft-lb

And considering the worm gear type reduction plus all other friction losses combined, the efficiency of our winches is estimated at 65%.

So (if my math isn't totally off) the required torque at the input would be 5,500 /29 * 100/65 = 292 ft-lb

That is great information many thanks. Now to find a hydraulic motor!!!!
Im trying to build a forestry winch is driven from a backhoe hydraulic circuit.
Thank you again!

 

[Markddhumvee]

I just found the spec sheet on the 20000lb winch and it quote RPM 250 and shaft torque at 3427 inch lbs does this seem correct?

 

[gringeltaube]

It seems correct to me...

3427 in-lb /12 = 286 ft-lb.

 

[Markddhumvee]

It seems correct to me...

3427 in-lb /12 = 286 ft-lb.

Thats great then as it should be relatively easy to find a motor providing 300 ft lb at around 250 rpm. Any suggestions to where to source something like this ?
Again thanks for the help.