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### Recovery calculations, math made simple

Posted: **Wed Jul 11, 2018 12:00 pm**

by **Low_Sky**

The following series of posts is going to go through the steps to calculating a winch recovery. Taken one step at a time, the calculations really aren’t very difficult. I have put together some example scenarios so you can see how it all comes together. I hope some folks find this stuff useful. In real-world application, I don’t expect that any of us whip out the note pad and slide rule to crunch the numbers, but with these calculations you can work out some scenarios at your kitchen table and get a feel for how you need to rig up for them. This is also the math you’ll need to determine for yourself if a piece of recovery equipment is rated high enough for your needs. I’ve added a few notes at the end on that topic.

### Mire Resistance

Posted: **Wed Jul 11, 2018 12:01 pm**

by **Low_Sky**

Mire resistance is the load on a stuck vehicle imposed by the surface it is stuck on. The resistance of a mired vehicle depends primarily on the depth to which it is mired, and the weight of the vehicle and cargo. Not all surfaces are created equal, but the mire factors presented here are a good starting place. If in doubt, adding extra mechanical advantage won’t hurt. For example; soft soupy mud with a fairly hard bottom will offer less resistance than bottomless thick clay. Fresh powder snow offers less resistance than spring snowpack with a hard crust. The clear exception to the mire factor chart below is being high centered or otherwise hung up on rock. Go straight to mire factor 3. Vehicles with smooth skid plates underneath may drag over rocks more easily, but with the mess of crossmembers, bars, small skid plates, etc. underneath a Power Wagon, plan for a hard pull. The best course of action is to jack the vehicle up and build track under the tires so you don’t have to drag it over rocks in the first place.

Multiply the weight of the vehicle (including cargo) by the mire factor to determine the mire resistance.

**Mire Resistance = Vehicle Weight x Mire Factor**

### Grade Resistance

Posted: **Wed Jul 11, 2018 12:03 pm**

by **Low_Sky**

Grade resistance is the load on a vehicle imposed by the angle of the surface it is stuck on, along the direction of travel of the vehicle. An uphill grade adds resistance, and a downhill grade subtracts resistance.

Multiply the weight of the vehicle (including cargo) by the grade factor to determine the grade resistance.

**Grade Resistance = Vehicle Weight x Grade Factor**

### Load Resistance

Posted: **Wed Jul 11, 2018 12:03 pm**

by **Low_Sky**

The load resistance is the combination of mire resistance and grade resistance, and it is the total load required to move the stuck vehicle. The winch only pulls as hard as needed to overcome the load resistance. If the load resistance is 5,000 lbs, a 12,000 lbs winch only pulls with 5,000 lbs to move it (in a single line pull). This is Newton’s third law in action. “For every action there is an equal and opposite reaction”.

**Load Resistance = Mire Resistance + Grade Resistance**

### Available Effort

Posted: **Wed Jul 11, 2018 12:03 pm**

by **Low_Sky**

Available effort is how much the winch can pull (or is rated to pull). Winches only pull their rated capacity with the last wrap on the drum. Additional wraps reduce the available effort because the rope has more leverage against the drum. There are charts available that give the pull capacity for a given number of wraps on the drum. The Power Wagon winch is a bit of special case. Looking at the internals, it’s a Warn M15000 assembled into a configuration to fit in the space behind the Power Wagon bumper. Ram rates the winch at 12,000 lbs. I’ve never heard an authoritative explanation for this, but the theory that makes the most sense to me is that the winch cradle is only rated for 12,000 lbs. Regardless of the reasoning, since Ram rates it for 12,000 lbs, that’s the maximum available effort I use in my calculations. For third and fourth layer pulls, use the M15000 chart.

**M15000 (layer/lbs)**

1/15,000 (PW rated at 12,000)

2/13,890 (PW rated at 12,000)

3/11,520

4/9840

### Mechanical Advantage

Posted: **Wed Jul 11, 2018 12:04 pm**

by **Low_Sky**

Mechanical advantage is the ratio of the force applied by a machine to the force applied to it. To determine the mechanical advantage required for a recovery, divide the load resistance by the available effort and round up to the nearest whole number. This first mechanical advantage calculation is an estimate that will be validated later after accounting for friction losses, which can be substantial. Obviously, a mechanical advantage of one is a single line pull, no blocks and no losses involved; required effort equals load resistance.

Below are examples of common riggings for mechanical advantage. Note that in the last example (compound 4:1), this is a 2:1 system working on a second 2:1 system. The rigging in the second system sees twice the loads seen by the first system and needs to be rated accordingly.

**Mechanical Advantage = Load Resistance / Available Effort, (round up)**

### Required Effort

Posted: **Wed Jul 11, 2018 12:04 pm**

by **Low_Sky**

The required effort is how hard the winch must pull to move the load. In a single line pull, the required effort is the load resistance. If mechanical advantage is needed, the required effort calculation will account for friction in the blocks. To do this, count the number of sheaves in the system, and add 10% to the load resistance for every sheave. Divide this apparent load resistance by the mechanical advantage and you have your required pull. Note that the actual load resistance doesn’t change. The friction in the blocks causes higher tension on the winch side of the block (the fall line) and lower tension on the anchor side of the block (the return line). It all evens out and the actual load resistance stays the same; determined by the load itself, not by the system attached to it.

- RE.PNG (9.57 KiB) Viewed 190 times

This equation looks more complicated than it is, so here’s an example.

A 10,000 lbs load resistance rigged with a 3:1 mechanical advantage (two sheaves).

Required Effort = 10,000 lbs x 1.2 / 3 = 4000 lbs

Now for the validation part. If the required effort just calculated ends up higher than the available effort, more mechanical advantage is required. The initial estimate of mechanical advantage gets us in the ballpark, but if we’re close to the limit, the friction from the blocks can overmatch the winch’s available effort.

### Example Scenarios

Posted: **Wed Jul 11, 2018 12:04 pm**

by **Low_Sky**

Now that we’ve discussed all the steps along the way, let’s work out some examples.

A 9000 lbs Power Wagon is mired to wheel wells on flat ground.

**Mire Resistance = 9000 lbs x 2 = 18,000 lbs**

Grade Resistance = 0 lbs

Load Resistance = 18,000 lbs + 0 lbs = 18,000 lbs

Available Effort = 12,000 lbs

Mechanical Advantage = 18,000 lbs / 12,000 lbs = 1.5, *round up to 2*

Required Effort = 18,000 lbs x 1.1 / 2 = 9900 lbs

9900 lbs (required) < 12,000 lbs (available), *good solution*

A 17,000 lbs Power Wagon and trailer is mired to the hubs climbing a 20 degree slope. In this one the "validation" step is going to do some work.

**Mire Resistance = 17,000 lbs x 1 = 17,000 lbs**

Grade Resistance = 17,000 lbs x 0.34 = 5780 lbs

Load Resistance = 17,000 lbs + 5780 lbs = 22,780 lbs

Available Effort = 12,000 lbs

Mechanical Advantage = 22,780 lbs / 12,000 lbs = 1.9,* round up to 2*

Required Effort = 22,780 lbs x 1.1 / 2 = 12,529 lbs

12,529 lbs (required) > 12,000 lbs (available), *bad solution, use 3:1 MA*

Required Effort = 22,780 lbs x 1.2 / 3 = 9112 lbs

9112 lbs (required) < 12,000 lbs (available), *good solution*

### Static Loads and Shock Loads

Posted: **Wed Jul 11, 2018 12:05 pm**

by **Low_Sky**

All of the calculations above assume static loads, objects are at rest and their weight is what it normally is under Earth gravity. If the rigging is subjected to sudden loads (such as a load shifting, a recovered vehicle gaining and losing traction again on a slope, etc), the loads seen can be more (sometimes multiple times) the weight of the object. The exact magnitude of shock loads can be difficult to calculate because they depend on many variables.

There are some strategies to mitigate the risks of shock loading. Plan your recovery to reduce the potential for loads to shift (secure a load with additional static or winch lines to prevent shifting). Perform uphill recoveries with with winch only, idling or applying very little power to the wheels so you don't outrun the winch if you gain traction. Use the beefiest rigging components you can afford, and rig what you do have for as much advantage as is practical. If a 1:1 pull would do, but there is significant risk for shock loading, rig it as a 2:1 or 3:1. Always plan every pull as if a component might fail, maintain a safe zone around the rigging, use winch liner dampers, and place extra static lines on the load if it may roll over or go downhill if something fails. With wire rope, plan for recoil and winch with the hood up if possible.

### Snatch Blocks

Posted: **Wed Jul 11, 2018 1:26 pm**

by **Low_Sky**

The topic of snatch block ratings comes up fairly regularly. Many blocks bear a Working Load Limit (WLL), and that causes a lot of confusion. The WLL applies to overhead lifting; raising a load off the ground where it may be suspended over people or property you don’t want to drop something on, or the load may be something you don’t particularly want to drop from some height, and working with loads that may be shocked. The WLL is usually derived from a 5:1 safety factor applied to the breaking strength of the block. 5:1 safety factor is really high. Needlessly high for recreational offroad recovery. What you really want to see is a published breaking strength, then you can apply a safety factor that you’re comfortable with and decide if the block is strong enough for your needs.

Because the Power Wagon winch is an M15000 in disguise, I assume it will pull harder than 12,000 lbs as long as it has enough electrical juice and is pulling from deep enough on the drum. If you want to let it do that, that’s your call. If you follow the “double your winch capacity” school of thought for snatch block sizing, you need a block that will hold ~30,000 lbs, apply a 2:1 safety factor and good luck finding (and affording) 60,000 lbs breaking strength blocks. Obviously, blocks that size are out there, but not in the recreational off-road recovery market.

My approach is to use the 12k lbs rating as my maximum available effort. Incorporating 10% friction loss, the most I’ll pull on the first block in a system is 21,800 lbs. If I apply a 2:1 safety factor, my ideal block would have a breaking strength of 43,600 lbs.

If a block you’re thinking about buying doesn’t publish the breaking strength, or the rated capacity and safety factor, skip it. Another hint to look for is the rated rope size. If it’s 3/8”, skip it, a Wagon comes with (and needs) 7/16” rope or larger. If the load limit looks right but the sheave is sized for 7/8” rope, the block is probably way overkill for a Wagon. Using synthetic rope on an oversize sheave isn’t a deal breaker, but oversized sheaves accelerate wear on wire rope by flattening it out during pulls. Plus, blocks for rope that big are inconveniently large and heavy.

I’ve settled on using ARB9000 snatch blocks. Their published breaking strength is 38,500 lbs, which gives me a 1.75 safety factor if I’m pulling to my max. I can live with that. They go for about $110, and at that price I have two of them in the truck. They’re lightweight and they don’t take up much room in the RamBoxes. If I underestimate a pull and stall the winch out at 15,000 lbs, I’m still under the breaking strength of the block.

### Tree strap

Posted: **Wed Jul 11, 2018 2:38 pm**

by **Low_Sky**

The ratings given to tree straps are for a vertical hitch (a straight pull on the sling). Given a long enough sling and a small enough anchor, wrapping the sling around the anchor and connecting to both eyes of the sling leaves you with a basket hitch (the sling makes a U-shape). If the angle between the sling legs is close to zero, the rating of this setup is double the published rating of the sling.

If the anchor is wide enough, or the sling short enough that the sling eyes meet an angle when they come back together, this will derate the "basket hitch" strength of the sling. If the angle between the two legs of the sling is 120-degrees, it will hold the published vertical hitch rating. If the angle between the two legs of the sling is MORE than 120-degrees, it will hold LESS than the vertical hitch rating. Get yourself a longer sling or a skinnier anchor.

Tucking one leg of the sling through the eye of the other leg to form a "noose" around the anchor and get more reach out of the sling is called a choker hitch, and it derates the sling to 70% of the vertical hitch rating.

For example, I found a 3" x 8' tree strap on Amazon, advertised 35,000 lbs breaking strength. If I want a 2:1 safety factor on it, it's a 17,500 lbs sling. If I use it as a winch extension, it's good for 17,500 lbs. If I wrap it around a sturdy 8" diameter tree trunk in a basket hitch, it's good for 35,000 lbs. If I have to choker hitch it around the same tree truck for a little extra reach, it's good for 12,250 lbs. If all I have to work with is a great big 30" diameter tree, and the ends will just barely reach together (160-degree angle between legs), I need to find a longer strap.

### Re: Available Effort

Posted: **Wed Jul 11, 2018 2:40 pm**

by **TankerZak**

Low_Sky wrote: ↑Wed Jul 11, 2018 12:03 pm

Available effort is how much the winch can pull (or is rated to pull). Winches only pull their rated capacity with the last wrap on the drum. Additional wraps reduce the available effort because the rope has more leverage against the drum. There are charts available that give the pull capacity for a given number of wraps on the drum. The Power Wagon winch is a bit of special case. Looking at the internals, it’s a Warn M15000 assembled into a configuration to fit in the space behind the Power Wagon bumper. Ram rates the winch at 12,000 lbs. I’ve never heard an authoritative explanation for this, but the theory that makes the most sense to me is that the winch cradle is only rated for 12,000 lbs. Regardless of the reasoning, since Ram rates it for 12,000 lbs, that’s the maximum available effort I use in my calculations. For third and fourth layer pulls, use the M15000 chart.

**M15000 (layer/lbs)**

1/15,000 (PW rated at 12,000)

2/13,890 (PW rated at 12,000)

3/11,520

4/9840

I've heard an unverified rumor that its an M15000 but it'll only pull 12000 because it's on a 12v system and would run at 15000 on a 24v system. Not sure how true that is. I could verify because i actually know a VP at Warn but i haven't been curious enough yet to pull that lever.

### Re: Recovery calculations, math made simple

Posted: **Wed Jul 11, 2018 4:56 pm**

by **Low_Sky**

https://www.warn.com/m15-12v-heavyweight-winch-47801
Warn literature says the M15000 will do 15k at 12V. I’m not a Warn expert, so I have no clue. I suppose an additional check we could make would be to compare the no-load line speed. The M12000 and M15000 seem to be the same motor geared differently, so have different no-load speeds.

### Re: Recovery calculations, math made simple

Posted: **Thu Jul 12, 2018 2:05 am**

by **TwinStick**

This is the most comprehensive & accurate explanation of recovery calculations I have ever seen posted on this site !!! Kudo's to you sir

Farm boys, & military men agree, off-roading will inevitably lead to some wicked stucks. Sometimes, you just gotta be there to believe it. Ever see a big International Turbo tractor buried ? We ended up chaining 2 railroad ties to each rear tire to get it out. Mud was 4' deep. It was in the late 70's.

Snatch blocks are your friend. I also have 2 & have used them. It really does cut down on the stress to your vehicle. Slows things way down, so you can usually foresee any potential problems.

### Re: Recovery calculations, math made simple

Posted: **Thu Jul 12, 2018 11:06 am**

by **TommyG**

TwinStick wrote: ↑Thu Jul 12, 2018 2:05 am

This is the most comprehensive & accurate explanation of recovery calculations I have ever seen posted on this site !!! Kudo's to you sir

Farm boys, & military men agree, off-roading will inevitably lead to some wicked stucks. Sometimes, you just gotta be there to believe it. Ever see a big International Turbo tractor buried ? We ended up chaining 2 railroad ties to each rear tire to get it out. Mud was 4' deep. It was in the late 70's.

Snatch blocks are your friend. I also have 2 & have used them. It really does cut down on the stress to your vehicle. Slows things way down, so you can usually foresee any potential problems.

Man you called it. When you stick the machine that almost never gets stuck it gets epic. Big heavy stuff sunk to the chassis is a character building experience.

Great thread, thanks for putting it all together LowSky.

### Re: Recovery calculations, math made simple

Posted: **Thu Jul 12, 2018 4:18 pm**

by **olyelr**

WOW !

### Re: Recovery calculations, math made simple

Posted: **Fri Jul 13, 2018 8:53 am**

by **TankerZak**

Low_Sky wrote: ↑Wed Jul 11, 2018 4:56 pm

https://www.warn.com/m15-12v-heavyweight-winch-47801
Warn literature says the M15000 will do 15k at 12V. I’m not a Warn expert, so I have no clue. I suppose an additional check we could make would be to compare the no-load line speed. The M12000 and M15000 seem to be the same motor geared differently, so have different no-load speeds.

Ugh. I used to work in the Warn Factory here building winches many moons ago. I'm fairly sure I've actually built these winches with my own hands. I'm still not an expert, lol. Someday I'll get curious enough to ask my Warn contact to find out... i just don't like her that much so I'd have to be pretty damn curious...