Finding the spring constant of a rope

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TL;DR Summary
Kinetic Recovery Ropes are commonly used in the off road community to recovered stuck vehicles. But Kinetic ropes are rated by WLL (Working Load Limit) and MBS (Max Breaking Strength) with units in pounds. How can I uses those numbers + mass (weight)*V of my vehicle to guestimate the safe limit?
The 7/8" Kinetic Recovery Rope like this yankum rope is the most common size used for Jeeps, Broncos, and other SUVs. I apologize for English units but that's how ropes are sold and marketed. I've talked to the biggest rope suppliers and they have no idea how to compute the rope's spring constant, how to go size a rope by maximum momentum. All they know is ropes, and ropes are sold by WLL and MBS, which is easy to measure (WLL is just computed as a fraction of MBS).
  • 7/8" Kinetic Recovery Rope "Python" [ WLL 5,700-9,000 lbs] [MBS 28,600 lbs]
  • 1" Kinetic Recovery Rope "Rattler" [WLL 6,700-11,200 lbs] [MBS 33,500 lbs]
  • 1 1/4" Kinetic Recovery Rope "Mamba" [WLL 10,400 - 17,400 lbs] [MBS 52,300 lbs]
I've seen several 7/8" Kinetic ropes broken by 4,500 Jeeps yanking on a vehicle stuck in the mud. My guess is they break from repeated pulls exceeding the elastic limit of the rope.

I'd like to know how fast I can safely go with a given rope size. Assume the stuck rig is an immovable mass. I've yanked out several mud bogged rigs with my 1.25" rope where my 24K lbs winch was worthless. The winch just pulled me to the mud bog.

Related link: Rope wrapped around a rod - belt friction
For winch, the answer is N>9, ie 10 minimum wraps around the capstan (drum).
 
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Answers and Replies

  • #2
jrmichler
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The spring constant is the ratio of the load to the amount of stretch, so some how, some way, you need to measure how much the rope stretches under a known load. And to make it more interesting, you need to do it for a range of loads because the relationship between stretch and load is very likely not linear.

Another thought: you can measure the actual peak loads when yanking a vehicle if you have access to a high speed accelerometer and data acquisition system. Knowing the mass of the pulling vehicle, and measuring the acceleration and velocity vs time, you can calculate the spring constant. Such a test will need an immovable load, such as a dump truck or big tree. The accelerometer will need about 1000 Hz bandwidth and the data acquisition system a sample rate of 1000 samples/sec.
 
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  • #3
B0B
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I can probably get a current vs pull table for my winch link this one provided by Warn

1676775199838.png


I can easily measure current. I'll use single line pulls on the first layer of the capstan up to 8K lbs, then switch to using a snatch ring for up to 16K lbs. I have several snatch rings so I could go even higher using my 12K winch.
 
  • #4
Baluncore
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I've yanked out several mud bogged rigs with my 1.25" rope where my 24K lbs winch was worthless.
That "yank" is a problem with a short kinetic rope or a wire cable. The rope must be long enough to store the energy needed to stop the tow vehicle, or move the towed vehicle.

If you snatch with a short rope, you will tear fittings off one vehicle, which will then be catapulted into the other vehicle. Many people die each year doing that.

A long kinetic tow-rope allows two different vehicles to move somewhat independently through difficult terrain.
 
  • #5
B0B
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You obviously don't understand kinetic ropes if you compare it to a wire rope. Can you provide evidence of Many people die each year doing that?

The famous Arizona fatality was a result of incorrectly attaching to a weak and deadly tow ball. The recovery rig used a tow strap, not a kinetic rope. In that case, a kinetic rope may have helped as it's far more elastic. The tow strap did store enough energy to kill the driver. Even so, attaching to the ball is well known never do in the recovery community.

Properly rigged, if you exceed the elastic limit of the kinetic rope, you break the rope, not the attach point. I mentioned that in my first post.
 
  • #6
Baluncore
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You obviously don't understand kinetic ropes if you compare it to a wire rope.
That "yank" is a problem with a short kinetic rope or a wire cable.
I compare a SHORT kinetic rope, (or a wire cable), with a long kinetic rope.

A kinetic rope stores energy as it is stretched. The total energy stored, without tensile failure of the rope, is proportional to the length of that rope.

I've yanked out several mud bogged rigs with my 1.25" rope where my 24K lbs winch was worthless. The winch just pulled me to the mud bog.
Your winch was NOT worthless, the failure was clearly with your vehicle's anchorage to the ground.

You overcame the poor anchorage by using the unsafe practice of yanking on the rope, which required an extra thick rope, which further encourages the unsafe practice of yanking as a recovery technique. Understand the physics, stop yanking, and use a better anchor.
 
  • #7
B0B
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I compare a SHORT kinetic rope, (or a wire cable), with a long kinetic rope.

A kinetic rope stores energy as it is stretched. The total energy stored, without tensile failure of the rope, is proportional to the length of that rope.
No, it's .5*K*X^2
I'm here to get some help on determining K, not to argue with someone who knows very little about the reality of recovery. Kinetic ropes are very popular and widely used.

Your winch was NOT worthless, the failure was clearly with your vehicle's anchorage to the ground.

You overcame the poor anchorage by using the unsafe practice of yanking on the rope, which required an extra thick rope, which further encourages the unsafe practice of yanking as a recovery technique. Understand the physics, stop yanking, and use a better anchor.
LOL, anchor to what? The grass? How many recoveries have you performed? Have you ever seen https://www.youtube.com/c/MattsOffRoadRecovery ? Not that Matt is exemplary. He yanks for drama, I yank when it's the only option. Matt yanks in reverse, yanks on tow balls, does lots of unsafe yanking. But Matt is good drama and he has millions of viewers. Why don't you scold him?

Yanking is safe if you rig correctly. I was doing recoveries with a kinetic chain 50 years ago. I won't say that was totally safe for the rigs, but I made sure it wasn't life threating. And what's your recovery credentials?
 
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  • #8
Baluncore
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LOL, anchor to what? The grass? How many recoveries have you performed? ... Yanking is safe if you rig correctly. I was doing recoveries with a kinetic chain 50 years ago. I won't say that was totally safe for the rigs, but I made sure it wasn't life threating. And what's your recovery credentials?
Such emotional and ill-informed hostility is rare among professionals.

As a geophysicist, I started out running seismic surveys through rainforest, where there were no roads. I have been solving problems and getting vehicles out of trouble for over 40 years.

Yesterday was a quiet Sunday, I only got one call, and all I did was use a load strap to pull a 2.5 tonne forklift out of a hole, with my old Land Cruiser. I did not need to get the D6 dozer on the float, nor the rotator crane out of the yard.
 
  • #9
B0B
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Such emotional and ill-informed hostility is rare among professionals.

As a geophysicist, I started out running seismic surveys through rainforest, where there were no roads. I have been solving problems and getting vehicles out of trouble for over 40 years.

Yesterday was a quiet Sunday, I only got one call, and all I did was use a load strap to pull a 2.5 tonne forklift out of a hole, with my old Land Cruiser. I did not need to get the D6 dozer on the float, nor the rotator crane out of the yard.
Yet you can't answer the question of when there is no anchor. You spam this thread with nothing to do with determining the spring constant of a kinetic rope. Tow straps store energy, why not use a winch? Sounds like an easy recovery. When's the last time you performed a modified, multi anchor for free SB recovery? You've never been in an area that has zero anchors?
 
  • #10
Baluncore
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You've never been in an area that has zero anchors?
Yes I have. Use more tow vehicles, with a harness to stop the winch vehicle moving, or improvise a simple anchor, in or on the surface.

Get yourself an old agricultural spiked drag harrow, then park on that, but make sure the spikes point down.
That technique will work in the middle of a field to extract tractors and harvesters that sink up to the axles in the soil. It avoids the slippery surface, and takes advantage of any small plant roots or rocks hidden in the soil.
 
  • #11
B0B
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Yes I have. Use more tow vehicles, with a harness to stop the winch vehicle moving, or improvise a simple anchor, in or on the surface.

Get yourself an old agricultural spiked drag harrow, then park on that, but make sure the spikes point down.
That technique will work in the middle of a field to extract tractors and harvesters that sink up to the axles in the soil. It avoids the slippery surface, and takes advantage of any small plant roots or rocks hidden in the soil.

LOL, thanks captain obvious. I use anchor rigs when they're available. Using a snatch ring I get an anchor vehicle connected for free. I use two free anchor vehicles with the modified, multi anchor for free SB recovery. That would be my first choice if possible.

Where would I get more tow vehicles? Guy was stuck for 2 days and I was the only one who'd come to help. Your Land Cruise wouldn't have a prayer of getting to him. Nothing less than 37's & lockers could get to him. No tow truck company would touch it. Given you don't do hard core wheelin with your lame kruser, you've never been in that situation, I understand.

How about the time I had to yank a rolled jeep 20 feet up a hard rock road, cliffs of both sides. Needed to yank him up the hill so we had room to flip him over. My winch pulled me 10 feet and rolled rig 1 mm. 4 jeeps in front of him and it was getting late. Drive 2 hours to town and try to get the required 3 more anchor jeeps? Multiple yanks had him in the perfect position where we could winch from both sides and get him on 4 wheels.

How about recovers that require a two tandem pull to tow a disabled rig for many miles? A kinetic rope is the only way to pull tandem.

I only do kinetic rope recoveries where there's no other choice.
 
  • #12
Baluncore
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This is not a competition.
I am wasting my time.
 
  • #13
B0B
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If you want to help the many thousands of kinetic rope users you could help solve the problem. There are dozens of kinetic rope distributors and what they publish doesn't provide the information needed for momentum bounds. It's easy to measure the mass of a vehicle. It's easy to measure velocity.

THE PYTHON: 7/8 INCH RECOVERY ROPE JEEPS, SMALL TRUCKS, AND SUVS
Best for Jeeps, 1/2 Ton Trucks & SUVs

Breaking Strength: 28,600 lbs.
Stuck? This diameter size is best utilized in anything from Jeeps to 1/2 ton pickup trucks to medium size SUV's. It’s built to stretch and perform. This is the diameter rope used by Matt of Matt's Off Road
For Use on vehicles that weigh 5,700 to 9,000 lbs

I've seen 7/8" rope break and read many more reports using it on 1/2 ton pickup trucks to medium size SUV's. That's why I went double sized when my Jeep weighs 4,300 lbs.

1-1/4” Kinetic Energy Rope for Recovery: Mamba [17,400 lbs MAX static weight]
Best for 1 Ton Utility Bed Trucks, Overland Vans and Earth Roamers
Breaking Strength: 52,300 lbs.

Why did I go with the 1.25" rope two years ago? I knew 7/8" wasn't adequate and didn't think 1" had enough safety factor.

My 1-1/4” Kinetic Rope stretches fine, it's not harsh like my 4" tow strap or my kinetic chain. I suspect the recommendations were written by the first big kinetic rope distributer using cheaper is better. Most buyers rarely use the rope and can typically get buy with the recommended size. All the rest of the distributors copied the recommendations as most buyers differentiate on price. The rope MBS is determined by using D-6268 of the ASTM testing system and WLL 1/3rd of that. Those are useful numbers but don't tell you anything about momentum limits. The MBS is the same for a 1 foot rope and a 30 foot rope.

It's the same with winches. A winch family, say 8K, 10K, 12K , all use the same motor, only the gears differ. The 8K winch is much cheaper and most buy that. But many winch failures are due to exceeding the duty cycle. Even my 3,500 lb 4x4 has a 12K winch. I always use a snatch ring as that halves the load and minimizes the voltage drop. Sometimes I use a Spanish Burton configuration with two snatch rings as that nearly 4x's the pull and I get two anchor vehicles connected for free, although the winch vehicle and anchor vehicle on the first snatch ring side each have 25% of the load and the 2nd pully anchor has 50%. It's fun to hook up. I typically just use one snatch ring and frequently anchor to a jeep next to me.
 
  • #14
erobz
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There is not enough information to determine a ##k## value unless you are doing it experimentally yourself.
 
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  • #15
erobz
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The argument about "yanking" or using an "anchor" to get larger forces is lost on me. The rope (or any portion of the connection) could fail in either mode at practically any load and be potentially hazardous to someone? How do you control that?
 
  • #16
B0B
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There is not enough information to determine a ##k## value unless you are doing it experimentally yourself.
Read the start of the thread. I can measure current to get pull. But you're right, the dozens of venders who sell kinetic ropes publish only WLL and MBS, which doesn't tell you how much momentum you can apply AFAIK. Given the spring constant, rope length and mass of vehicle, you could calculate the max V.

The argument about "yanking" or using an "anchor" to get larger forces is lost on me. The rope (or any portion of the connection) could fail in either mode at practically any load and be potentially hazardous to someone? How do you control that?
Experience, rigging, component selection, and high school physics.
 
  • #17
erobz
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Read the start of the thread. I can measure current to get pull. But you're right, the dozens of venders who sell kinetic ropes publish only WLL and MBS, which doesn't tell you how much momentum you can apply AFAIK. Given the spring constant, rope length and mass of vehicle, you could calculate the max V.


Experience, rigging, component selection, and high school physics.
So lets say you measure the load vs deflection, you find the ##k## value. How do you plan to plan to continue?
 
  • #18
B0B
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So lets say you measure the load vs deflection, you find the ##k## value. How do you plan to plan to continue?
The elastic limit is 30% stretch. From that you can compute the maximum mv
 
  • #19
erobz
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The elastic limit is 30% stretch. From that you can compute the maximum mv
I would use the smaller value of WLL or 30% stretch. They give you that rating, better to try and use it.
 
  • #20
B0B
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I would use the smaller value of WLL or 30% stretch. They give you that rating, better to try and use it.
Exactly, I'll take the min(WLL,25% stretch). WLL is 1/3rd breaking strength, so that should be a safe limit for the rope.
 
  • #21
erobz
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So you find you ##k## value experimentally, lets say hypothetically ##F_{WLL}## is the smaller of the two loads (implying you found ##k## to be larger than the manufacturer would have). How do you calculate the maximal allowable stretch ##x_{max}##?
 
  • #22
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So you find you ##k## value experimentally, lets say hypothetically ##F_{WLL}## is the smaller of the two loads (implying you found ##k## to be larger than the manufacturer would have). How do you calculate the maximal allowable stretch ##x_{max}##?
> ##k## to be larger than the manufacturer would have

They're distributors and they don't even know what ##k## means and hence don't publish it. The rope MBS is determined by using D-6268 of the ASTM testing system and WLL 1/3rd of that. From MBS testing they also noted ##x_{max}## is 30%.

Can you explain:

##F_{WLL}## is the smaller of the two loads (implying you found ##k## to be larger than the manufacturer would have)

How do you go from WLL to ##k##? WLL is invariant to length.
 
  • #23
erobz
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Can you explain:

##F_{WLL}## is the smaller of the two loads (implying you found ##k## to be larger than the manufacturer would have)

How do you go from WLL to ##k##? WLL is invariant to length.
They aren't the same measurement. I'm asking you how you calculate the maximal stretch ##x_{max}## given your experimentally determined ##k## value, and the manufacturers ##F_{WLL}##?
 
  • #24
B0B
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They aren't the same measurement. I'm asking you how you calculate the maximal stretch ##x_{max}## given your experimentally determined ##k## value, and the manufacturers ##F_{WLL}##?
They all publish up to 30% stretch so I'll take their word for it and reduce that to 25% to be safer. I've seen ropes break and read several reports. It's always a rope that had done many yanks previously. My guess is that they're exceeding the elastic bounds. I've never heard of one breaking on the first pull, although I'm sure that's happened. It's more common from multiple breaks. One report I read was from professionals giving a training session, using 7/8" rope. The recoil of the rope did damage to the hood of the simulated stuck vehicle.

A few off road experts (with no physics or engineering knowledge) assert the fibers need 48 hours recovery after yanking. I've had to do 15 pulls over 15 minutes once, but I'm using a double sized rope. Maybe if you yank > 80% elastic limit, then fibers do need time to recover.
 
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  • #25
erobz
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Here is the thing...if you are modeling the rope as a spring with a constant ##k## value it has some arbitrary initial length. If you stretch it some distance ##x## it stores energy ##\frac{1}{2}kx^2##. That amount of stored energy is independent of its initial length.

If a rope is stretching 30% of its initial length (before reaching ##F_{WLL}##), independent of its initial length it doesn't have a ##k## value like a linear spring has a has a ##k## value.

With a single spring that has a ##k## value, if it has initial length ##l## and you stretch it some distance ##x## you will be applying a force ##F##. If it has the same constant ##k## and initial length ##5l##, and you stretch it ##x## you are still applying force ##F##. Do you see the problem? Does that description match your experience with this rope?
 
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  • #26
B0B
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Here is the thing...if you are modeling the rope as a spring with a constant ##k## value it has some arbitrary initial length. If you stretch it some distance ##x## it stores energy ##\frac{1}{2}kx^2##. That amount of stored energy is independent of its initial length.

If a rope is stretching 30% of its initial length (before reaching ##F_{WLL}##), independent of its initial length it doesn't have a ##k## value like a linear spring has a has a ##k## value.

With a single spring that has a ##k## value, if it has initial length ##l## and you stretch it some distance ##x## you will be applying a force ##F##. If it has the same constant ##k## and initial length ##5l##, and you stretch it ##x## you are still applying force ##F##. Do you see the problem? Does that description match your experience with this rope?
I think your saying, just like a spring, ##k## is dependent on length. Put two springs in series and and k prime is ##k/2##

##k## is dependent on length and radius of rope. Rope distributors sell ropes of a given radius but different lengths with the same WLL and MBS. I think MBS is invariant of length.
 
  • #27
erobz
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I think your saying, just like a spring, ##k## is dependent on length. Put two springs in series and and k prime is ##k/2##
I'm saying I have my doubts on whether or not ##k## is independent of initial length for this rope. If you pull a rope of initial length and you do find a ##k## value, I would say that it probably only applies to that particular rope. If it's the same rope material diameter, but different length it could have a different effective ##k## value. So I don't think its wise to make sweeping generalizations until you determine whether ##k## is indeed independent of initial length for this type of rope.
 
  • #28
erobz
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##k## is dependent on length and radius of rope. Rope distributors sell ropes of a given radius but different lengths with the same WLL and MBS. I think MBS is invariant of length.
The WLL and MBS are all independent of length. Thats fine. What is different is how much a rope stretches to reach each. It is that which is going to make comparison across ropes ( length, diameter) difficult. A rope that is 50 ft long will stretch 15 ft to reach WLL, and a rope that is 20 ft will only stretch 6 ft to reach WLL. If this is true, the linear spring model is probably not a good choice (as that is not how a linear spring behaves).
 
  • #29
Frabjous
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A rope that is 50 ft long will stretch 15 ft to reach WLL, and a rope that is 20 ft will only stretch 6 ft to reach WLL. If this is true, the linear spring model is probably not a good choice (as that is not how a linear spring behaves).
I would suggest using strain with a modulus instead of displacement with a spring constant.
 
  • #30
B0B
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The WLL and MBS are all independent of length. Thats fine. What is different is how much a rope stretches to reach each. It is that which is going to make comparison across ropes ( length, diameter) difficult. A rope that is 50 ft long will stretch 15 ft to reach WLL, and a rope that is 20 ft will only stretch 6 ft to reach WLL. If this is true, the linear spring model is probably not a good choice (as that is not how a linear spring behaves).

That's not how ropes work. MBS is invariant of length. But within MBS a rope acts like a spring.

Elastic stretch is invariant of radius and length. It's 30%. But kinetic ropes are only sold in 2 lengths, 30 feet (80% of sales) and 20 feet. K is a function of length. But K30 = 2/3*K20 where K30 is a 30', and K20 a 20'.

Now to the test procedure. For the two immovable objects I'll use a combination of large tree and Jersey barrier. I'll attach a winch dampening blanket to the non-winch end of the line. The back of the Jeep will be anchored with my 4" 30' 40K BS tow strap anchored at two points which yields approximately 80K BS. While tow straps are elastic, they don't stretch much. Hopefully I can ignore the error introduced. Unless it's around a tree, I'll use my 56K soft shackle in the connection.
2023-02-20 12.58.10.jpg


I'll measure the stretch at 2K, 4K, 8K, and 10K using a snatch ring on my 12K winch (yielding 2*cos(alpha/2) pull, where alpha is the angle of two rope segments). I should probably make a 3/4" plywood barrier given I'll need to be near the winch line to measure the current. Perhaps a double winch blanket would be enough.

I would suggest using strain with a modulus instead of displacement with a spring constant.
I think the assertion that a rope within it's elastic limit is not link a spring is wrong, but I could be wrong.
I can measure tensile stress (σ) but how would I measure tensile strain (ε)

See - Spring constant of a rope
 
  • #31
erobz
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It cant both be 30% stretch independent of initial length AND a constant spring rate for a fixed rope diameter.
 
  • #32
B0B
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It cant both be 30% stretch independent of initial length AND a constant spring rate for a fixed rope diameter.
I think I made that clear long ago K depends on length, just like two spring in series make K 1/2. Spring rate is dependent on length as I said.

Maybe I'm not understanding your question or assertion

I think your saying, just like a spring, ##k## is dependent on length. Put two springs in series and and k prime is ##k/2##

##k## is dependent on length and radius of rope. Rope distributors sell ropes of a given radius but different lengths with the same WLL and MBS. I think MBS is invariant of length.
 
  • #33
erobz
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I think I made that clear long ago K depends on length, just like two spring in series have make K 1/2. Spring rate is dependent on length as I said.
What I'm saying is if that is the case, then we shouldn't expect a linear Force vs Deflection graph for a particular rope. The energy stored in the rope will not be ##\frac{1}{2}kx^2##
 
  • #34
B0B
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What I'm saying is if that is the case, then we shouldn't expect a linear Force vs Deflection graph for a particular rope. The energy stored in the rope will not be ##\frac{1}{2}kx^2##
I'm not following. Why won't the energy stored in a rope of length L using the K for a rope of length L not be ##\frac{1}{2}kx^2## ?

And I don't see how it differs for 2 springs in series, 3 springs in series, etc.
Sorry, I took a year of physics 45 years ago. I haven't done any LaTeX in 30 years so I can only copy/paste/edit your markup.
 
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  • #35
erobz
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I'm not following. Why won't the energy stored in a rope of length L using the K for a rope of length L not be ##\frac{1}{2}kx^2## ?

Sorry, I took a year of physics 45 years ago.
Because its spring rate has dependency on the initial length of the rope.

For a linear spring we have that:

$$dF = k dx$$

Where ##k## is a constant.
##x## is the displacement from the initial length.

If we integrate that we get the familiar linear spring force equation (neglecting the minus sign out front of course)

$$ \int_{0}^{F} dF = k \int_{0}^{x} dx \implies F = kx$$

And the work is given by:

$$W = \int_{0}^{x} F dx = \int_{0}^{x} kx dx = \frac{1}{2}kx^2$$

This rope has a ##k## value which is some non-constant function of its length. Its begins with a different differential equation:

$$dF = k(l)dl $$

The work of that type of a spring is given by:

$$W = \int_{l_o}^{l} \left( \int_{l_o}^{l} k(l) dl \right) ~dl $$

It's going to depend on whatever the function ##k(l)## is (which is not a constant if you get 30% stretch up to ##F_{WLL}## for ropes of different initial length)
 

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