# Winch loads and what is happening when you winch

#### mrblaine

##### TJ Guru
Supporting Member
Thanks to firejeep for some diagrams to help me get a clearer picture of what I wanted to rig up to show what happens with loading when you winch. It has taken awhile to get this going due to the visuals not being as clear as I want and some issues with scales for weights. The first were some small digital crane scales. What wasn't clear is there is a hold function that happens when the load quits changing for a few seconds. That became problematic due to an oddity I did not foresee. As I would put a load on the rigging, it would go up 10% above where it would eventually wind up but slowly settle there. The digital scales would lock in a load before the settling finished so it was not clear what you were seeing as far as loads plus hitting the start button and clear button to get them to stay ready to accept load got to be a real pain in the ass. I traded those in for some analog game scales and problem solved.

The first scenario we are going to explore is the most common one that leads to confusion and that is a snatch block pull back to the rig to "double" the power of the winch.

Essentially the way it works is each line at the snatch block will have the load of what the winch is putting out. If it is pulling at 5000 lbs, each leg going through the block is 5000 lbs x 2 for a total pulling power of 10,000 lbs.

Here is a diagram from firejeep and then my redneck rigging to illustrate a rig with a winch on it and the anchor points. We will be exploring these three scenarios.

I needed to show the obvious which is the load the rig has against it. The best way to show that was to rig an anchor line with a scale to show it.

Essentially this scale model shows exactly what was predicted and we won't get too nitpicky over frictional issues adding or taking away from the totals. My version of the second diagram.    • Weeeee, jeepins, ds53652 and 20 others

#### dbbd1

##### go away...
Supporting Member
Thanks. Looking forward to the other scenarios.

#### mrblaine

##### TJ Guru
Supporting Member
Thanks. Looking forward to the other scenarios.
Since no one is saying much, we'll try it a different way. The third diagram which shows a snatch block being used to do a simple change in direction pull is what got me started doing this. The diagram shows that the load at the snatch block anchor will be 50% higher than the other two. I didn't believe that to be true so I set about to come up with a way to prove it one way or another.

So in the diagram below, let's get some predictions from folks if I give one of the loads.
1 has a load of 290 lbs. on it. What do you think the rest of them will be? • Alex01

#### Steel City 06

Assuming that the cart wheels and pulley can be modeled as frictionless, and the angle is 90 degrees, and the angle at 3 is 135 degrees from
the other ropes, the tension in the rope should be the same across all points from 1, 2, and 4 at roughly 290 lbs.

The pulley at 3 sees three loads, all pulling in different directions. Since the tensions from 2 and 4 are the same at 290 lbs, the remaining attachment where 3 is must resist both the loads from 2 and 4. Thus they must be added together to create the load at 3. However, since they are pulling at 90 degrees to each other, or 135 degrees from the anchor rope, we use the cosine of the angle of each rope as a multiplier. Thus each rope from 2 and 4 applies 290 lbs times cos(135), or -290*sqrt(2)/2. Summing the two tensions we get -290*sqrt(2), or -290*1.414, or -410 lbs. The negative sign implies that the force acts in the opposite direction as the other ropes, which is necessary to prevent the acceleration of the pulley.

Thus the load at 3 is 410 lbs, and 1,2, and 4 is 290 lbs. The load at 3 is 41.4% higher than the other points.

If the angle between the two ropes changes, then the load at 3 will also change. A simplified way of calculating it would be L3 = L1*COS(A/2)*2 where L3 is the load at point 3 and A is the angle between the two ends of the rope fed through the pulley. L1 is the load at 1, which in this case is 290 lbs.
The smaller the angle between the two ropes at the pulley, the more tension is applied at 3. At zero degrees, the load at 3 is twice that of the tension applied from 1, which is the same as the snatch block example from before. At 180 degrees, no load is applied at the pulley. Anywhere in between those angles, the load varies with the cosine of half the angle between the primary rope ends.

The only case that 3 has the same load as 1, 2, and 4 is if the angle is set to 120 degrees.

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• Mike_H and ayelix

#### Goatman

Supporting Member
I believe all should have the same load. If the cart is free to move then #1 and #2 are essentially the same. The cart is just a funny shaped piece of the rope. If the pully (#3) is used to change direction only then it too has the same load, just as #4. If none of the pieces do not change the ratio of pull then they are all doing the same amount of work. All should therefore have the same load on them. Assuming no resistance in any of them. (If I remember my school days.)

#### derekmac

To start off, I suck at math and anything related to it, lol.

I would think both 2 and 4 would have the same load (290), but since they want to be pulled together, 3 would see a higher load. How much? Not a clue, lol.

#### mrblaine

##### TJ Guru
Supporting Member
Assuming that the cart wheels and pulley can be modeled as frictionless, and the angle is 90 degrees, and the angle at 3 is 135 degrees from
the other ropes, the tension in the rope should be the same across all points from 1, 2, and 4 at roughly 290 lbs.

The pulley at 3 sees three loads, all pulling in different directions. Since the tensions from 2 and 4 are the same at 290 lbs, the remaining attachment where 3 is must resist both the loads from 2 and 4. Thus they must be added together to create the load at 3. However, since they are pulling at 90 degrees to each other, or 135 degrees from the anchor rope, we use the cosine of the angle of each rope as a multiplier. Thus each rope from 2 and 4 applies 290 lbs times cos(135), or -290*sqrt(2)/2. Summing the two tensions we get -290*sqrt(2), or -290*1.414, or -410 lbs. The negative sign implies that the force acts in the opposite direction as the other ropes, which is necessary to prevent the acceleration of the pulley.

Thus the load at 3 is 410 lbs, and 1,2, and 4 is 290 lbs. The load at 3 is 41.4% higher than the other points.

If the angle between the two ropes changes, then the load at 3 will also change. A simplified way of calculating it would be L3 = L1*COS(A/2), where L3 is the load at point 3 and A is the angle between the two ends of the rope fed through the pulley. L1 is the load at 1, which in this case is 290 lbs.
The smaller the angle between the two ropes at the pulley, the more tension is applied at 3. At zero degrees, the load at 3 is twice that of the tension applied from 1, which is the same as the snatch block example from before. At 180 degrees, no load is applied at the pulley. Anywhere in between those angles, the load varies with the cosine of half the angle between the primary rope ends.
What got me started down this time consuming rabbit hole was the Change of Direction diagram that showed the load at 3 to be about what you predict or 50% higher. It didn't make sense to me so rather than bitch about it, say anyone was wrong, I decided to model it in real life in a smaller scale to find out. I can tell you from the loads on the scales that it is nowhere near the diagram or your prediction.

#### mrblaine

##### TJ Guru
Supporting Member
I believe all should have the same load. If the cart is free to move then #1 and #2 are essentially the same. The cart is just a funny shaped piece of the rope. If the pully (#3) is used to change direction only then it too has the same load, just as #4. If none of the pieces do not change the ratio of pull then they are all doing the same amount of work. All should therefore have the same load on them. Assuming no resistance in any of them. (If I remember my school days.)
The cart works similar to a Jeep in that the rear wheels don't pivot or turn to follow the load. I scoot it around and watch the scales to try and remove as much oddity in the loads as I can to even it out to what "should" be applicable to real world results.

I was going to mount the winch to a set of drawer guides with a scale at the back of it to show the load the winch is imparting to the rig. It was easier and about as accurate to just put one between the cart and anchor point and then try to keep all the loads in the same elevation.

I have to keep the loads kinda small because my anchor point is a jack stand that I have under the axle of a Jeep. Don't want to pull the stand over and drop the rig on the frame.

#### jesseshoots

##### TJ Enthusiast
So in the diagram below, let's get some predictions from folks if I give one of the loads.
1 has a load of 290 lbs. on it. What do you think the rest of them will be?
f1 = f2 = f4 = 290 lbs
f3 = sqrt(f2^2 + f4^2) = 410 lbs

**EDIT: Just realized that SteelCity06 posted the same thing but my browser hadn't refreshed since I opened the window about half an hour ago >.<

• Steel City 06

#### jjvw

##### It's great. No Issues.
Supporting Member
...

I have to keep the loads kinda small because my anchor point is a jack stand that I have under the axle of a Jeep. Don't want to pull the stand over and drop the rig on the frame.

#### mrblaine

##### TJ Guru
Supporting Member
f1 = f2 = f4 = 290 lbs
f3 = sqrt(f2^2 + f4^2) = 410 lbs

**EDIT: Just realized that SteelCity06 posted the same thing but my browser hadn't refreshed since I opened the window about half an hour ago >.< #### jesseshoots

##### TJ Enthusiast
Well I'll be damned. That's not what I was expecting one bit!

#### mrblaine

##### TJ Guru
Supporting Member
Well I'll be damned. That's not what I was expecting one bit!
I posted it.

#### jesseshoots

##### TJ Enthusiast
I posted it.
Yep, I'm full of bad timing today. It must be a Monday.

#### mrblaine

##### TJ Guru
Supporting Member
Well I'll be damned. That's not what I was expecting one bit!
It is exactly what I expected to see which is why I started this mess. I could not get my head around how a COD could run the load up 50% higher.

#### Steel City 06

It is exactly what I expected to see which is why I started this mess. I could not get my head around how a COD could run the load up 50% higher.
What’s the angle of the rope through the pulley? 90 degrees? At about 110 degrees these loads would make more sense

• fuse

#### xxdabroxx

##### TJ Enthusiast
Seems like it works out to be closer to 125% of what load 4 is seeing after all of the losses add up. Would be interesting to see if a 45deg angle would be closer to 175% if 90deg is only 125%.

Clearly an angle of 0deg change is going to impart 0 lbs of force. If 180deg = 2x and 90deg = 1.25x 0deg = 0x

#### pagrey

You can clearly see you have 50lb difference between #1 and #4, that's 20% loss in friction.

20% off of 410 is 328 or pretty much exactly what scale #3 reads.

You've shown pretty well that you have about 20% loss due to friction in your system.

Good science.

Edit: by the way shock loading on #3 will go up to 410 if you give the system a good jolt, dynamic friction is much much smaller than static

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#### xxdabroxx

##### TJ Enthusiast
If you give a good tug between points 2 and 3 toward point 1 do the values equal out at 1, 2, and 4? Just something I thought of to try and combat the friction losses (even though in the real world there will be friction losses)

• jjvw

#### mrblaine

##### TJ Guru
Supporting Member
You can clearly see you have 50lb difference between #1 and #4, that's 20% loss in friction.

20% off of 410 is 328 or pretty much exactly what scale #3 reads.

You've shown pretty well that you have about 20% loss due to friction in your system.

Good science.

Edit: by the way shock loading on #3 will go up to 410 if you give the system a good jolt, dynamic friction is much much smaller than static
I should have mentioned that along with scooting the back of the cart around to get the anchor line as straight as possible and not pulling sideways on the rear wheels, I do bounce all the lines and watch for oddities to even them out so we are just seeing the loads as accurately represented as possible. We will have to ignore some of the frictional issues and just get what we get.

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