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This is amazing. Finally a video that properly explains why just 'janking' a vehicle out at full speed is never a good idea. Thank you so much Robert. 🙂
Amazingly didactic explanation. It is a great opportunity to understand why is NEVER an option to speed up "fast and furious" for a stuck situation...many times seen, unfortunetly....Best regards from Buenos Aires, Argentina!
In a pinch I have stopped my 2.5 tonne Hilux from 110kmph in well less than your graph suggests, it created smoke and wore my brakes out considerably, but it was a life or death situation. It's a long story, a scary story not one I want to share, but with good brakes you can stop safely in much less than your graph.
Great video - thanks for showing the calculations. How can you estimate the required force to get the casualty vehicle out? A shallow bog in sand would require much less force than a vehicle bottomed out in a mud hole.
G,day Robert great video mate really useful information that I'm planning to remember and implement in safer recovery methods for the future. Found you from your mate Mad Mat 👍
You explained the energy well. But you assumed a fixed average force. Force will be the killer. Can you calculate a maximum impact force for a 30% stretch of the rope or 20% for the strap. I would really like to see the maximum forces that would spike through the rigging if the vehicle is stuck behind a rock or tree stump with zero movement. You can do this with work formula or energy or change of momentum - all newton anyhow.. This instantaneous forces will be very interesting, and then the vehicle doing the recovery will be jerked back also, and it may be interesting to know the hysteresis energy characteristics of the rope. THANKS
@@L2SFBC was about to comment something similar, a winch generates a fairly constant and low power (energy over time), vs a snatch that will take the same amount of energy but apply it over a much lower amount of time producing a power spike thanks for the video, it's not that difficult to figure out the numbers, for the stretching ropes maybe hooks law could help to calculate the energy and force transfers, if you have the time and the energy associated you probably can directly calculate the forces involved (very high probably)
While the amount of energy is proportional to v^2 it's not the case for a maximum force. Let's assume that a bogged vehicle doesn't move and recovering vehicle stretches a snatch strap (or recovery rope), which has a constant coefficient, i.e. F = k * dx. That's what an "ideal" spring would do (recoveries ropes are not ideal springs, but they are not that far). Recovering vehicle will stop at a distance defined by equality of energy: k * dx^2 / 2 = m * v^2 / 2 (k * dx^2 / 2 is a potential energy of a loaded spring). As you see dx = v * sqrt(m / k). Maximum force achieved will be Fmax = k * dx = v * sqrt(m * k), i.e. linear to v . Of course everything depends on k. Tow straps or chains have very high k, so the force will be enormous. Snatch straps and recover ropes, OTOH, have a pretty low k for easy stretching. In any case linearity (approximate) will hold for both, only speed ranges will be dramatically different. The difference in _force_ between 18 km/h and 16 km/h will be ~13%, not ~26%.
Interesting - if the energy is proportional to v^2 as we agree then where does it go when the strap tightens? Are you saying that the maximum force is linear to speed, and the remainder of the energy changes the length of time the force is applied for, in effect? The extra energy of a faster vehicle has to go somewhere. Of course all this is approximate as it ignores the tractive effect of the recovering vehicle which would produce extra force, and makes assumptions about springs etc. I have been thinking of a more complex video around kinetic recoveries but was looking for someone to assist with the physics - if you'd like to help please contact me: l2sfbc.com/contact
@@L2SFBC Force of a stretched strap is F = k * dx, energy is E = k * dx^2 / 2 (where dx is elongation). Force (and its companion - momentum) and energy are two different things, but it's the force that breaks shackles and tears recovery points, not the energy. You can easily see it if you compare chains and snatch straps. Same energy at the same speed, but very different results. Traction does matter, but on asphalt max. traction force of a 3t vehicle will be around 3t (unlikely, even in low range, gas engines don't like low rpms so engine will reach its limit long before tires), probably 1.5 - 1.8t max. on a dirt road (again, it's a tire limit, not engine). If it's that good you may not need a snatch strap at all, towing may be enough. In a mud? I doubt it's even reaching 1t with mud tires, probably ~500kg at max.That's where snatch straps shine as they use accumulated momentum, which you can gain by slowly accelerating on a slippery road.
@@L2SFBC The tow strap will absorb the kinetic energy of the recovering vehicle by stretching. If the recovering vehicle gets off the gas when the strap starts stretching, and assuming no friction, and the casualty vehicle doesn’t move, then all of the kinetic energy is converted to potential energy in the strap. If the strap has a large spring constant (K) then it won’t stretch much and the force exerted by the strap on the two vehicles will be high. If the strap is ‘stretchier’, then it will stretch to a longer length decelerating the recovering vehicle over a greater distance, exerting less force on the two vehicles but storing the same amount of energy.
@@TricksterJ97 thanks John that's a nice clear summary - yes that's why I say there is no multiplication effect with a more elastic strap, as the total energy is the same but how it is applied is different. There are two related concept; same strap/cars, different speeds, and same cars/speed, different elasticity. Victor is talking about the former, and you the latter...I will do more on this when I can and your comments are very useful, thanks.
Good day Robert Very usefull information - Thank you Quick question: the stuck vehicle of 1,000kg - by how much does the weight "increase" because of the forces at play, when stuck. Is the 1,000 kg vehicle the equivalent of 2,000 kg because of the friction or up hill gradient etc. Or does it increase by 20% or 50% or 200%? Cheers
Hi - the force required was 1000kg, which is made up of rolling resistance, gradient, and terrain resistance. The vehicle weighed 3000kg, although its exact weight isn't important for the explanation. Does that clear it up?
thats the fun bit. working it out and thats why you want to clear the wheels to ease the path and so on to make things easier. and it varies on all sorts of things. is it mud or sand etc, how wet, how sticky, is it hung up on debris in the bog all these things affect it. also the vehicle weight is the whole banana as it is on wheels and not a dead weight.
If a light vehicle try to pull out heavy vehicle that stuck in the mud using kinetic energy rope, is there a real danger for the light vehicle to leap backward?
You've just made me realise how dangerous snatch recoveries are. If I'm honest, having seen that chart and its relationship with speed, unless I can do a simple tow I am not even going to entertain further use of a kinetic rope. I just don't feel that it's safe enough.
KILO is the first word in both KILO-gram and KILO-metre. Why is everybody changing the "O" in the word Kilo to AWE, ah, however speled? That first word does not change!!!
Some more computations: let's take ARB snatch strap (8t / 10m / 30%), let's assume that max force (8t) is achieved at dx = 10m * 30% = 3m => k = 8t * 9.81/3m. What vehicle speed is required to achieve that force? Let's take 3t as a pulling vehicle mass. If I didn't mess up with my calculations max speed to break ARB strap is v = sqrt(Fmax * dx / m) = sqrt(8t * 9.81 * 3m / 3t) = 8.8 m/s =31 km/h. Of course it won't be 1t of pulling force, but 8t. Now, for 1t of pulling force we need 1/8 of that, i.e. 1.1 m/s = 4 km/h. Not so hard to control. Medium size winch can easily pull 5t, which is about 20 km/h when using a snatch strap. Another thing is the amount of accumulated energy in the strap and where it goes when (if) it snaps.
@@TricksterJ97 Also how much your recovery points can hold. 8t might be way too much and flying recovery point (along with attached shackle) is not much better than a snapped strap. I'm mostly worried about those and they are not rated (4Runner factory tie-downs). The good part is people use them and there are no stories of them breaking.
Meanwhile max. deceleration in a pulling vehicle at 31 km/h is going to be 8/3 = 2.6g. I would limit max. speed to 1g, i.e. around 12 km/h (3t of pulling force). Long before snatch strap limits. Another consideration is that vehicles are structurally designed to brake to the floor (which is around 1g on asphalt), but not much more.
Do you find this video useful? If so, please tell me why and subscribe...and if you didn't find it useful, subscribe anyway and tell me what you didn't like!
This is amazing. Finally a video that properly explains why just 'janking' a vehicle out at full speed is never a good idea. Thank you so much Robert. 🙂
Glad it was helpful!
Amazingly didactic explanation. It is a great opportunity to understand why is NEVER an option to speed up "fast and furious" for a stuck situation...many times seen, unfortunetly....Best regards from Buenos Aires, Argentina!
Really good explanation. Your presentation of the spread of recovery load over time with snatching vs winching was especially useful.
Great video - thank you for the details.
Glad it was helpful!
In a pinch I have stopped my 2.5 tonne Hilux from 110kmph in well less than your graph suggests, it created smoke and wore my brakes out considerably, but it was a life or death situation. It's a long story, a scary story not one I want to share, but with good brakes you can stop safely in much less than your graph.
Yes those numbers are quite conservative, but didn't need to reflect real-world to make the point.
Great video, very informative.
Thanks please share 👍
Great video - thanks for showing the calculations.
How can you estimate the required force to get the casualty vehicle out? A shallow bog in sand would require much less force than a vehicle bottomed out in a mud hole.
Will cover in future.
G,day Robert great video mate really useful information that I'm planning to remember and implement in safer recovery methods for the future. Found you from your mate Mad Mat 👍
Glad to help
According to many out there, have as much slack as you can and floor it.
You explained the energy well. But you assumed a fixed average force. Force will be the killer. Can you calculate a maximum impact force for a 30% stretch of the rope or 20% for the strap. I would really like to see the maximum forces that would spike through the rigging if the vehicle is stuck behind a rock or tree stump with zero movement. You can do this with work formula or energy or change of momentum - all newton anyhow.. This instantaneous forces will be very interesting, and then the vehicle doing the recovery will be jerked back also, and it may be interesting to know the hysteresis energy characteristics of the rope. THANKS
I do plan to do more on this subject yes
@@L2SFBC was about to comment something similar, a winch generates a fairly constant and low power (energy over time), vs a snatch that will take the same amount of energy but apply it over a much lower amount of time producing a power spike
thanks for the video, it's not that difficult to figure out the numbers, for the stretching ropes maybe hooks law could help to calculate the energy and force transfers, if you have the time and the energy associated you probably can directly calculate the forces involved (very high probably)
While the amount of energy is proportional to v^2 it's not the case for a maximum force. Let's assume that a bogged vehicle doesn't move and recovering vehicle stretches a snatch strap (or recovery rope), which has a constant coefficient, i.e. F = k * dx. That's what an "ideal" spring would do (recoveries ropes are not ideal springs, but they are not that far). Recovering vehicle will stop at a distance defined by equality of energy: k * dx^2 / 2 = m * v^2 / 2 (k * dx^2 / 2 is a potential energy of a loaded spring). As you see dx = v * sqrt(m / k). Maximum force achieved will be Fmax = k * dx = v * sqrt(m * k), i.e. linear to v . Of course everything depends on k. Tow straps or chains have very high k, so the force will be enormous. Snatch straps and recover ropes, OTOH, have a pretty low k for easy stretching. In any case linearity (approximate) will hold for both, only speed ranges will be dramatically different. The difference in _force_ between 18 km/h and 16 km/h will be ~13%, not ~26%.
Interesting - if the energy is proportional to v^2 as we agree then where does it go when the strap tightens? Are you saying that the maximum force is linear to speed, and the remainder of the energy changes the length of time the force is applied for, in effect? The extra energy of a faster vehicle has to go somewhere. Of course all this is approximate as it ignores the tractive effect of the recovering vehicle which would produce extra force, and makes assumptions about springs etc. I have been thinking of a more complex video around kinetic recoveries but was looking for someone to assist with the physics - if you'd like to help please contact me:
l2sfbc.com/contact
@@L2SFBC Force of a stretched strap is F = k * dx, energy is E = k * dx^2 / 2 (where dx is elongation). Force (and its companion - momentum) and energy are two different things, but it's the force that breaks shackles and tears recovery points, not the energy. You can easily see it if you compare chains and snatch straps. Same energy at the same speed, but very different results. Traction does matter, but on asphalt max. traction force of a 3t vehicle will be around 3t (unlikely, even in low range, gas engines don't like low rpms so engine will reach its limit long before tires), probably 1.5 - 1.8t max. on a dirt road (again, it's a tire limit, not engine). If it's that good you may not need a snatch strap at all, towing may be enough. In a mud? I doubt it's even reaching 1t with mud tires, probably ~500kg at max.That's where snatch straps shine as they use accumulated momentum, which you can gain by slowly accelerating on a slippery road.
@@L2SFBC The tow strap will absorb the kinetic energy of the recovering vehicle by stretching. If the recovering vehicle gets off the gas when the strap starts stretching, and assuming no friction, and the casualty vehicle doesn’t move, then all of the kinetic energy is converted to potential energy in the strap. If the strap has a large spring constant (K) then it won’t stretch much and the force exerted by the strap on the two vehicles will be high. If the strap is ‘stretchier’, then it will stretch to a longer length decelerating the recovering vehicle over a greater distance, exerting less force on the two vehicles but storing the same amount of energy.
@@TricksterJ97 thanks John that's a nice clear summary - yes that's why I say there is no multiplication effect with a more elastic strap, as the total energy is the same but how it is applied is different. There are two related concept; same strap/cars, different speeds, and same cars/speed, different elasticity. Victor is talking about the former, and you the latter...I will do more on this when I can and your comments are very useful, thanks.
Good day Robert
Very usefull information - Thank you
Quick question: the stuck vehicle of 1,000kg - by how much does the weight "increase" because of the forces at play, when stuck. Is the 1,000 kg vehicle the equivalent of 2,000 kg because of the friction or up hill gradient etc.
Or does it increase by 20% or 50% or 200%?
Cheers
Hi - the force required was 1000kg, which is made up of rolling resistance, gradient, and terrain resistance. The vehicle weighed 3000kg, although its exact weight isn't important for the explanation. Does that clear it up?
thats the fun bit. working it out and thats why you want to clear the wheels to ease the path and so on to make things easier. and it varies on all sorts of things. is it mud or sand etc, how wet, how sticky, is it hung up on debris in the bog all these things affect it. also the vehicle weight is the whole banana as it is on wheels and not a dead weight.
If a light vehicle try to pull out heavy vehicle that stuck in the mud using kinetic energy rope, is there a real danger for the light vehicle to leap backward?
Small but yes
🍻
You've just made me realise how dangerous snatch recoveries are. If I'm honest, having seen that chart and its relationship with speed, unless I can do a simple tow I am not even going to entertain further use of a kinetic rope. I just don't feel that it's safe enough.
It's safe enough when speeds are kept low..it's when people go all-out first off you find problems, AND also fail to check why the vehicle is stuck.
KILO is the first word in both KILO-gram and KILO-metre. Why is everybody changing the "O" in the word Kilo to AWE, ah, however speled? That first word does not change!!!
Some more computations: let's take ARB snatch strap (8t / 10m / 30%), let's assume that max force (8t) is achieved at dx = 10m * 30% = 3m => k = 8t * 9.81/3m. What vehicle speed is required to achieve that force? Let's take 3t as a pulling vehicle mass. If I didn't mess up with my calculations max speed to break ARB strap is v = sqrt(Fmax * dx / m) = sqrt(8t * 9.81 * 3m / 3t) = 8.8 m/s =31 km/h. Of course it won't be 1t of pulling force, but 8t. Now, for 1t of pulling force we need 1/8 of that, i.e. 1.1 m/s = 4 km/h. Not so hard to control. Medium size winch can easily pull 5t, which is about 20 km/h when using a snatch strap. Another thing is the amount of accumulated energy in the strap and where it goes when (if) it snaps.
@@TricksterJ97 Also how much your recovery points can hold. 8t might be way too much and flying recovery point (along with attached shackle) is not much better than a snapped strap. I'm mostly worried about those and they are not rated (4Runner factory tie-downs). The good part is people use them and there are no stories of them breaking.
Meanwhile max. deceleration in a pulling vehicle at 31 km/h is going to be 8/3 = 2.6g. I would limit max. speed to 1g, i.e. around 12 km/h (3t of pulling force). Long before snatch strap limits. Another consideration is that vehicles are structurally designed to brake to the floor (which is around 1g on asphalt), but not much more.