How much Force/Load doing a Snatch recovery...?




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  1. #1
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    Default How much Force/Load doing a Snatch recovery...?

    On request from ChrisStoffel from CT here it goes:
    Many 4WD'er, seasoned or new to the "game" are either too scared to do a Kinetic recovery or to relaxed when doing it...

    Snatch recoveries were discussed for a few years now and there are still many opposing views, so best to start fresh.

    Doing a Snatch Recovery with a snatch strap/rope (kinetic strap/rope, usually nylon) what is the expected load in kg or ton for the below scenario?

    Recovery 4x4 Weight: 2700kg (close approximate)
    Recovered/Stuck 4x4: Unmovable (dead weight)
    Recovery 4x4 speed: 35km/h before the load slowed it down. (approximate as per GPS and speedo) (yes I know this is too fast. "regulation" speed is about 20km/h)
    Recovery 4x4's traction surface: Damp compacted black sand. (= good traction)
    Recovery 4x4's tyres: 285/70R17 MT's, deflated to 0.6bar (just to get even more traction)
    Centre transfer locked for 4x4 (as it's a part time)
    Manual hubs locked.
    Low range 2nd.
    Front and Rear diffs not locked.
    Auto box, so it can keep pulling without stalling.

    In the above scenario, where a vehicle is so deeply stuck, a snatch recovery is not an option. But this is a scenario to determine what is the expected maximum load during such an "unsafe recovery"

    Recovery force calculations can be made, but we will always wonder what would it be in "real-life"...??
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Well using a basic kinetic energy calculator you are bringing 127604 joule into the recovery.
    it depends on what force you want to calculate? but you are transferring all that joule into the strap and whatever it is attached to. I'm too uneducated in physics to tell you what that translates to kg on the attachment.

    https://www.calculatorsoup.com/calcu...cs/kinetic.php

    joule converts directly to newton meter.

    https://www.translatorscafe.com/unit...e-watt-second/

    To work that out you need to know how far the strap stretched so you are missing some of the data to complete the calculation. If you know that you can plug it into this formula. All the other stuff you supplied means nothing. You need the speed, weight and how far the car went before stopping. IE the strap stretch distance.
    (the impact from a horizontal moving object one)

    https://sciencing.com/calculate-forc...t-7617983.html
    Last edited by Stephan van Tonder; 2018/04/16 at 06:41 AM.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    I've seen a flow chart somewhere (think it's even been on this forum) where you can calculate the total effort (Kg force) required to recover a vehicle. Factors include surface, slope, depth, mechanical damage, snatch block losses etc. I'll see if I can find it.

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    Default Re: How much Force/Load doing a Snatch recovery...?


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    Default Re: How much Force/Load doing a Snatch recovery...?

    Each snatch rope has a certain value in term of energy it can absorbs. You need to stay within that limits when snatching, thus the force (energy) apply with the pulling vehicle should not more than the rope can take. Thus there will be a max speed for your vehicle you can snatch with, else the rope will snap. And obviously the recovery points should also be strong enough.

    if that speed cant recover it, you need a stronger rope for a higher speed.
    Last edited by JLK; 2018/04/16 at 03:11 PM.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Quote Originally Posted by Kiri View Post

    Recovery force calculations can be made, but we will always wonder what would it be in "real-life"...??
    Don't trust those calculations that people so freely throw around. To calculate the resultant force on the rope is easy. To calculate the force it will take to recover, much less so.

    In a snatch recovery it's not as simple as one object of a specified mass, moving at a specified speed, moving another with known properties.

    That vehicle is stuck, with wheels pushed up against objects of unknown angles and unknown rigidity.

    There is no way in hell to accurately guess those.

    Throw into that the effect of the recovered vehicle running its wheels to assist, and you'll quickly realise that any attempt to get accurate figures are futile.

    Maybe somebody can write an app where you input your wheel outside diameter, the depth it's penetrated into the ground, from that it can then calculate the magnitude of the vectors for each wheel, based on vehicle mass and angles. To test the ground density it can use the depth that your cellphone penetrate the ground when dropped from 1m at perfect right angles to the ground, using it's short side. Then you need to input tyre friction coefficient, based on compound type and thread pattern, and also the surface speed of the tyre, based on wheel circumference, diff ratio, Tcase ratio, gear ratio and engine RPM.

    ...or use common sense and experience?
    Last edited by Sakkie; 2018/04/16 at 03:33 PM.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Yes I agree 100% that calculations can be with some degree of accuracy, but with so many variables with a stuck vehicle, you can be out by some margin.

    When you by a kinetic strap or rope, the only rating you will find is the breaking strength of the rope (if you are lucky). That unfortunately give you little to idea of the application. In kinetic recovery, more is not necessary better, actually the opposite. Using a 12t rope on a Jimny will be almost the same as using a chain.

    ARB propose to choosing a strap with breaking strength of 2 to 3 times of the recovered vehicle's weight. So at 2.5 ton you're looking at 5t to 7.5t rated gear. Some suppliers in SA will tell you minimum 10 ton, some will go as high as 12 ton.

    There are differences between one rope and another and so with straps. There are even bigger differences between strap and rope.
    Ideally you want the most elastic rope/strap with the highest breaking strength. That is not so easy, as gaining one end, you lose another. Rope in general outperform strap for that balance. Most "name brand" straps claim (even though they don't state it) around 20% stretch. Most rope splicers claim 30 to 40% stretch.

    Now these claims of stretch are at WHAT load?
    If we do not know what load exist during a snatch, how do we know what kinetic gear to pick?

    Therefore the reason for this thread. Are there any members that have done tests or can say from experience what is the recommended rating for a particular scenario? Some time ago a few of us did a few "tests", but 1st some brainstorming...
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Unless you want to carry multiple ropes, one for each weight class, just take the highest load spliced rope for your vehicle.

    With that you can recover anything from a Jimny right up to, and depending on circumstances, even above your own vehicle weight.

    All you do is adapt your snatch speed according to the mass of the vehicle being recovered.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    I would think you can calculate the exact force in the tow strap given you have enough info on the tow strap.

    How deep the vehicle is stuck doesn't matter, what you'll do is to calculate the worst case scenario force. The worst case being when the vehicle stuck can't move an inch. You will be effectively pulling against a dead anchor. and with this in mind the recovery gear will need to stop the recovery vehicle completely, traction and wheel size also doesn't matter, only speed.

    But rather than calculating it I would also as suggested buy the biggest applicable kinetic rope and operate within the manufacturers guidelines. And if you're going to do full on kinetic recoveries, then make very sure about your recovery points.

    I have VERY limited experience of snatch recoveries on 4x4 vehicles. The only things I have recovered with snatch ropes is combines and articulated tractors with heavy grain carts, so the recovery weights ranged between 16 and 20 tons. The ropes we used were strong enough to not be worried about.
    One thing I learned there by doing sometimes 5 or more recoveries (all kinetic) per day after rains when the fields were wet in different situations etc, is to properly asses the situation and do the right prep work before even thinking about recovering the piece of equipment. We once left a combine in the field for 4 days simply because the risk of trying to recover it and possible damage was not worth it.

    I think the importance of prepping the stuck vehicle, is often overlooked. Sometimes a vehicle on the limits of being able to be recovered can be very safely recovered just by unpacking it first and secondly taking a shovel and removing excess soil where needed.
    Too many times I see people get stuck and then they just want to hook up another vehicles and pull it out, but they don't want to get dirty and properly clear paths for the stuck vehicle to get out...
    In all the recoveries we did we never even tried to recover something that has some weight on that we could remove. Yes it was on a bigger scale and removing 2 - 20 tons from the stuck piece of equipment made all the difference but still the same principle counts when recovering 4x4s.
    Last edited by WillemT; 2018/04/17 at 08:30 AM.

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    Default Re: How much Force/Load doing a Snatch recovery...?

    The force that the rope can handle is specified. You don't have to calculate that, just stay within it.

    As for the recovery attempt force, that too is easy to calculate. Mass of recovering vehicle multiplied by the speed of the recovering vehicle.


    Calculating the force it will take to recover a stuck vehicle......? I can give you several unknowns that each will have a massive effect on the force required.

    This is where experience and common sense kicks in, and totally overshadows analysis paralysis.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Quote Originally Posted by Kiri View Post
    Most "name brand" straps claim (even though they don't state it) around 20% stretch. Most rope splicers claim 30 to 40% stretch.
    Quote Originally Posted by Kiri View Post
    Recovery 4x4 Weight: 2700kg (close approximate)
    Recovered/Stuck 4x4: Unmovable (dead weight)
    Recovery 4x4 speed: 35km/h before the load slowed it down.
    Recovery force calculations can be made, but we will always wonder what would it be in "real-life"...??
    Quote Originally Posted by Sakkie View Post
    As for the recovery attempt force, that too is easy to calculate. Mass of recovering vehicle multiplied by the speed of the recovering vehicle.
    It is easy to get confused with different measures, e.g. momentum = mv, Energy = Work = 1/2mv2 ; Force = ma; Force = Work x Distance etc. Sakkie calculated the momentum which is relevant, but it is not Force.

    Let me attempt a calculation:
    The speed must be converted to m/s: 36 kmu = 10 m/s
    Lets assume that the tow rope is 10 meters long and stretches 30%, so recovery vehicle stops in 30% x 10 = 3 meters.

    Work = Energy = Force x Distance
    So Force = Energy / Distance = 1/2mv2 x 1/d = [2700 x (10)2] / [2x3] = 45000 Newton = 45 kN

    This is equivalent to 45000/10 = 4500 kg = 4.5 tonne.

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    Default How much Force/Load doing a Snatch recovery...?

    Firstly even if a vehicle is very stuck and an attempt at digging/ clearing a path has not helped, no snatch attempt should be done at a balls to the wall “i have to get it out at the first attempt” pace.

    Take it easy, get a “feel” for the load and if needed, add a little bit more on the next attempt.

    Very few snatch recoveries need the max force that can be applied via the rope.
    As said before, common sense should prevail.
    My Jimny has recovered much bigger vehicles stuck in terrible places, but it took 5 to 7 attempts, each with a bit of progress.
    And the prescribed safety measures kept everything and everyone healthy.
    Last edited by Flipside; 2018/04/17 at 12:21 PM.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    So if the recovered vehicle didn't move, d=0, meaning that the force calculation cannot be done, as dividing by zero is not allowed?

    If I ever had to use formulae for calculating these, I'd rather use momentum change and Impulse. Basically the stuff used for calculating collisions. After all, a snatch recovery can be seen as a controlled collision.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Quote Originally Posted by Sakkie View Post
    So if the recovered vehicle didn't move, d=0, meaning that the force calculation cannot be done, as dividing by zero is not allowed?

    If I ever had to use formulae for calculating these, I'd rather use momentum change and Impulse. Basically the stuff used for calculating collisions. After all, a snatch recovery can be seen as a controlled collision.
    D is the distance over which the force was applied, ie the 3m elongation of the strap.


    I would use kinetic energy and spring energy...

    the kinetic energy should be E = 0.5 m v^2 = 0.5 x 2700 x 100 = 135 000 J (J, kg and m/s)
    Then if you take the spring energy formula E = 0.5 k x^2 (J, N/m and m)
    135 000 = 0.5 x k x 9
    k = 30 000 N per meter
    for the spring f = k x (N , N/m and m)
    so f would be f = 30 000 x 3 = 90 000N = 9000kg = 9 tons
    Last edited by WillemT; 2018/04/17 at 12:57 PM.

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    Default Re: How much Force/Load doing a Snatch recovery...?

    Should the force not be calculated using the momentum force of the vehicle doing the recovery.
    With maximum force applicable just at the moment before the movement of the recovered vehicle..
    Last edited by Hedgehog; 2018/04/17 at 12:54 PM. Reason: I
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Quote Originally Posted by LouisF View Post
    It is easy to get confused with different measures, e.g. momentum = mv, Energy = Work = 1/2mv2 ; Force = ma; Force = Work x Distance etc. Sakkie calculated the momentum which is relevant, but it is not Force.

    Let me attempt a calculation:
    The speed must be converted to m/s: 36 kmu = 10 m/s
    Lets assume that the tow rope is 10 meters long and stretches 30%, so recovery vehicle stops in 30% x 10 = 3 meters.

    Work = Energy = Force x Distance
    So Force = Energy / Distance = 1/2mv2 x 1/d = [2700 x (10)2] / [2x3] = 45000 Newton = 45 kN

    This is equivalent to 45000/10 = 4500 kg = 4.5 tonne.
    The force is not constant over the distance so the simple energy = force x distance doesn't apply..

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    Default Re: How much Force/Load doing a Snatch recovery...?

    Quote Originally Posted by Hedgehog View Post
    Should the force not be calculated using the momentum force of the vehicle doing the recovery.
    With maximum force applicable just at the moment before the movement of the recovered vehicle..
    I would look at the worst case where all the energy need to be dissipated in the strap. So the stuck vehicle can't move and the recovery vehicle needs to be brought to a complete stop (no energy left in the vehicle) by the recovery gear. Everything else is less force is the rope/strap.

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    Default Re: How much Force/Load doing a Snatch recovery...?

    I hope all of the above is purely for academic purposes.........

    When it comes to a snatch recovery, "as soft as possible" applies. As Flip said, just work your way up until the vehicle is free, or you've reached the safe limit.

    Now you can go calculate the "safe limit".....it being the maximum speed that the recovery vehicle (taking it's mass into account) should not exceed in order to not break the strap, or cause damage to either of the vehicles.

    Again, the strap is easy.

    Anybody want to have a go at guessing the forces required to cause damage to the vehicles?

    Those are the important forces. Nothing else.
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    Default Re: How much Force/Load doing a Snatch recovery...?

    Quote Originally Posted by Sakkie View Post
    I hope all of the above is purely for academic purposes.........

    When it comes to a snatch recovery, "as soft as possible" applies. As Flip said, just work your way up until the vehicle is free, or you've reached the safe limit.

    Now you can go calculate the "safe limit".....it being the maximum speed that the recovery vehicle (taking it's mass into account) should not exceed in order to not break the strap, or cause damage to either of the vehicles.

    Again, the strap is easy.

    Anybody want to have a go at guessing the forces required to cause damage to the vehicles?

    Those are the important forces. Nothing else.
    Agree 100%
    The only thing I would use a calculation like this for is to know what a specific strap can handle in terms of max permissible speed for a vehicle strap combination... for instance, you sit with a small strap and a heavy vehicle... You can't just increase your speed everytime or you're looking for trouble. Somewhere you should know that I'm at the maximum that the strap or rope can handle... and then not to even talk about straps vs ropes where some straps can only be used a few times in short succession before needing time to recover their memory...

    And then as you rightfully stated, the strap or rope is just one small part in a snatch recovery!

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    Default Re: How much Force/Load doing a Snatch recovery...?

    More often than not, the recovery point on one of the vehicles is the weakest link, and then it gets ugly very fast. Regardless of any calculations. A Pofadder for example, has a lot of safety margin built into its rating...
    That is why safe procedures have been developed and if possible, they should be applied as necessary.
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