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misanthropic_nihilist


Oct 2, 2005, 6:51 PM
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Does climbing rope obey Hooke's law?
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I did a "real world example" physics homework problem last year, where they said a climbing rope's tension is not simply proportional to extension, as a 'physics spring' is. At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).

Is this true, or did the problem oversimplify/misstate? If anybody wants the actual F(dx) equation that was in the book, I can try to find it.

...It would be nice if engineering reps from the big gear companies monitored this site to answer questions related to unpublished gear specs.


thegreytradster


Oct 2, 2005, 7:00 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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It must.


Or a lot of us would be refered to in ths past tense.


core


Oct 2, 2005, 7:05 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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Think about the kinematics. The forces generated by falls of 4 ft and falls of 40 ft, are quite different. Remember KE=mgh and the tensile force in the rope will be a function of the time required to arrest the falling climber once the ropes slack has been taken out.

I believe it is a good starting point to liken the behavior of a dynamic rope to a linear spring, but I think that your statement of factor 2 fall behavior pushes the linear model too far.


dr.ed
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Re: Does climbing rope obey Hooke's law? [In reply to]
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This url was posted somewhere else around here lately... you might want to take a look at it:
http://www.caimateriali.org/Eventi/Torino/computermodel.html

they model the rope as a spring and a damper.


grk10vq


Oct 2, 2005, 7:17 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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you should ask mr. wizard.


petsfed


Oct 2, 2005, 7:20 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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In reply to:
I did a "real world example" physics homework problem last year, where they said a climbing rope's tension is not simply proportional to extension, as a 'physics spring' is. At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).

Is this true, or did the problem oversimplify/misstate? If anybody wants the actual F(dx) equation that was in the book, I can try to find it.

...It would be nice if engineering reps from the big gear companies monitored this site to answer questions related to unpublished gear specs.

If ropes actually obeyed Hooke's law, then they would not need to "rest" after long falls, they would also act like bungee cords rather than semi-elastic cords.

In addition, were it not for the facts of the matter, posting a rope's "peak force" with its other descriptive info would be pointless. A linear spring doesn't have a "peak force" per se, nor could a general "peak force" be predicted.


dutyje


Oct 2, 2005, 7:29 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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In reply to:
I did a "real world example" physics homework problem last year, where they said a climbing rope's tension is not simply proportional to extension, as a 'physics spring' is. At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).

Is this true, or did the problem oversimplify/misstate? If anybody wants the actual F(dx) equation that was in the book, I can try to find it.

...It would be nice if engineering reps from the big gear companies monitored this site to answer questions related to unpublished gear specs.

If ropes actually obeyed Hooke's law, then they would not need to "rest" after long falls, they would also act like bungee cords rather than semi-elastic cords.

In addition, were it not for the facts of the matter, posting a rope's "peak force" with its other descriptive info would be pointless. A linear spring doesn't have a "peak force" per se, nor could a general "peak force" be predicted.

Wouldn't the peak force of a spring be the point at which the amount of stretch induced in the spring by a given force would require the spring to stretch a distance greater than its fully outstretched length?

Wouldn't the peak force of a rope be that equivalent?


bandycoot


Oct 2, 2005, 8:23 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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No, it does not obey Hooke's law. Otherwise you would spring upward after coming to a stop in a lead fall.


Partner rgold


Oct 2, 2005, 9:36 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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In reply to:
I did a "real world example" physics homework problem last year, where they said a climbing rope's tension is not simply proportional to extension, as a 'physics spring' is. At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).

If your end result was the invariance of maximum force for falls of the same fall factor, then your starting point was indeed Hooke's Law, which says that if you plot the tension in the rope against its percentage elongation, you get a straight line through the origin. The result that you quote is a consequence of Hooke's Law, not some "more realistic" replacement for it.

In reply to:
Is this true, or did the problem oversimplify/misstate?

The actual graph is certainly not a straight line for all possible rope deformations, but is reasonably close for many ropes to a straight line for elongations in the working range of the rope, which is up to about 30%. All climbing ropes are engineered to have lower elongations at low loads (presumably by virtue of the internal friction of the core and sheath), and a number of the newer ropes that qualify for multiple UIAA ratings are engineered to have curves that are not straight lines, since the predictions based on the straight-line model do not agree very well with their published impact numbers.

Certain mathematical features of Hooke's Law are clearly not present in ropes. For example, there is no response to a contracting force. In other words, the tension-extension graph ends at the origin. Moreover, the near absence of a perceptible bounce and the relatively long recovery time all bespeak a damping effect. The Italian model uses the traditional differential equation for damped harmonic motion, which also depends on Hooke's Law but adds a damping term. However, the traditional damping term, which is proportional to velocity, is based on a damping mechanism consisting of a viscous fluid in a cylinder. Whether the mechanisms that damp rope oscillations are analogous to a classical car shock absorber is not addressed in their web presentation and seems to me to require some justification.

One can use models other than the linear one; the graph of a cubic equation T=kx^3, rather than a linear equation T=kx is the starting point for the analysis of what is called a "stiff" rope. But the idea that the model should involve an odd power of the extension x is based on the assumption that compression and extension should produce the same force magnitude, a fact that is certainly not true for ropes.

Perhaps the best way to think about Hooke's Law is that it is, in many cases, a very good linear approximation to the true tension-elongation curve. The fall-factor concept, which is a consequence of Hooke's law, has been confirmed in many tests and remains the fundamental concept in the testing and certification of ropes.

The primary real-world feature that is not accounted for in the analysis you did is the effect of rope friction against rock and through multiple carabiners. The Italian computer model is based on a system of differential equations for damped simple harmonic motion and attempts to account for frictional effects at each carabiner and can be adjusted for a certain amount of rock friction. But at the heart of those equations is....Hooke's Law.


Partner rgold


Oct 2, 2005, 9:49 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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If ropes actually obeyed Hooke's law, then they would not need to "rest" after long falls, they would also act like bungee cords rather than semi-elastic cords.

In reply to:
No, it does not obey Hooke's law. Otherwise you would spring upward after coming to a stop in a lead fall.


A rope could still obey (or very nearly obey) Hooke's Law during the extension phase, and one can, moreover, add to Hooke's Law the assumption of a damping force that accounts for the lack of bounce. Such systems are still based on Hooke's Law.

In reply to:
In addition, were it not for the facts of the matter, posting a rope's "peak force" with its other descriptive info would be pointless. A linear spring doesn't have a "peak force" per se, nor could a general "peak force" be predicted.

A linear spring certainly has a peak force as a response to a specific load, and this is all that is asserted by the peak force data considered for climbing ropes.


petsfed


Oct 3, 2005, 12:53 AM
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A linear spring certainly has a peak force as a response to a specific load, and this is all that is asserted by the peak force data considered for climbing ropes.

I thought the ratings were the peak force the rope will have no matter what the force. Of course that doesn't really make sense, now that I think about it. However, a constant peak force no matter what the input force is could only be possible with a non-simple spring system. That is, more complex than F=kX, even if that complexity is some velocity dependent term. Hooke's Law is a starting point, but it also implies infinite compression or extension, which we know to be impossible. When you stretch things too much, they break. So in that sense, nothing "obeys" Hooke's Law. Its a first order approximation.


antiqued


Oct 3, 2005, 6:57 AM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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See

http://www.marlow-ropes.com/...nager.cfm?page_id=21

for some graphs of stress vs strain. The Superline nylon is a kernmantle rope, the Superline Steelite is high Mw polyethylene, like Spectra fiber. It shows the most elastic behavior.


muskie


Oct 3, 2005, 7:13 AM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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your just workign with a mass spring damper system(you are the mass, and the rope is a spring and a damper) It acts similarly to the shocks on a car in that the spring opposes yoru displacement, the damping opposes your velocity. Therefore it does obey hookes law(ie its a LAW), however this does not account for the damping in the rope so you can do a direct calculation without using differential equations. And for the nerds, this would be considered a second order differential equation.


misanthropic_nihilist


Oct 3, 2005, 7:27 AM
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Thanks for the info.


core


Oct 3, 2005, 7:28 AM
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Without a study, such as the computer model linked above, it's too dificult to say a dynamic rope obeys Hooke's Law. Look, the law is states a linear relationship between between stress and strain; applied force and elongation. This might be true for small displacements, but for large (a relative term) displacements, I suspect that a linear model is not adequate. So, on a 40 ft whipper, the rope probably does not obey Hooke's Law. (Maybe not even for a short ride on a short rope).


daithi


Oct 3, 2005, 7:48 AM
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Thanks for the interesting link. From this data, if it is representative, we'd have to say kernmantle ropes don't obey Hooke's law very well and the linear spring model for the rope, which is used in a lot of simple analysis of fall forces etc., is not a particularly good model! This is a bit of a surprise to me as it seems to give reasonable results.

These load - extension graphs are done at static conditions for each data point (at least for normal materials) so the viscous damping term is not relevant. Therefore, Hooke's law of a linear relationship between load and extension is not a good model of the physics happening in this instance.

I am guessing these non-linearities in the stress-strain relationship arise from the heterogeneities in a cross section of the rope due to its construction, ie. the mantle surrounded by the sheath.


iltripp


Oct 3, 2005, 8:49 AM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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However, the traditional damping term, which is proportional to velocity, is based on a damping mechanism consisting of a viscous fluid in a cylinder. Whether the mechanisms that damp rope oscillations are analogous to a classical car shock absorber is not addressed in their web presentation and seems to me to require some justification.

I don't know what presentation you are talking about, but I think I know where they are coming from with the "viscous fluid" model.

Many polymers such as Nylon exhibit viscoelastic bahaviour. That is to say that they show combined elastic and viscous properties.

A hookean solid (a spring for example) will exhibit perfectly elastic behaviour. It will stretch and recoil according to Hooke's law.

A pure Newtonian fluid exhibits viscous behaviour. Mathmatically, this relationship looks very much like Hooke's law, except that it relates shear stress and shear strain, rather than Force and elongation.

Nylon and other viscoelastics show some combination of these two characteristics. To visualize this, imagine a spring where the bottom is attached to a small plate submerged in a viscous fluid. If you push or pull on the spring, it contracts or extends based on Hooke's law, but there is a dampening effect from the viscous liquid.

Of course, this all for pure nylon, and it fails to take into account the physical construction of the rope. Something else must be done to model the effect of the braiding of the core, the weaving of the sheath, and the interactions between sheath and core.


renohandjams


Oct 3, 2005, 9:09 AM
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At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).
Being a chemical engineer I have found that most things are over simplified to make them easier to understand the basics. Like the ideal gas equation, etc.. there are so many different factors that play into it. If what you said is true, then what if you take the limit of the drop as it approaches zero?

For example:
4 micrometer drop, on a 2 micrometer rope? I doubt the force would be the same, the only way to really know would be to do some lab tests in the physics lab.

Another example would be a 4 mile drop on a 2 mile rope, the object dropping would reach a terminal velocity.


daithi


Oct 3, 2005, 9:09 AM
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A pure Newtonian fluid exhibits viscous behaviour. Mathmatically, this relationship looks very much like Hooke's law, except that it relates shear stress and shear strain, rather than Force and elongation.

In a fluid, shear stress is proprtional to the time rate of strain not shear strain!

For an elastic solid shear stress is proportional to shear strain but in a Newtonian fluid (let's call it a normal fluid) it is proportional to strain rate. This is one of the principal differences betwen a solid and a fluid.


iltripp


Oct 3, 2005, 10:45 AM
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In reply to:
In reply to:
A pure Newtonian fluid exhibits viscous behaviour. Mathmatically, this relationship looks very much like Hooke's law, except that it relates shear stress and shear strain, rather than Force and elongation.

In a fluid, shear stress is proprtional to the time rate of strain not shear strain!

For an elastic solid shear stress is proportional to shear strain but in a Newtonian fluid (let's call it a normal fluid) it is proportional to strain rate. This is one of the principal differences betwen a solid and a fluid.

Oops... :oops:

I believe you are right. I was writing that off the top of my head and it's been a while since I learned and used those equations. Now that I think about it, it was definitely strain rate.


prointraining


Oct 3, 2005, 5:22 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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english please there are 6th graders here thank you

In reply to:
In reply to:
I did a "real world example" physics homework problem last year, where they said a climbing rope's tension is not simply proportional to extension, as a 'physics spring' is. At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).

If your end result was the invariance of maximum force for falls of the same fall factor, then your starting point was indeed Hooke's Law, which says that if you plot the tension in the rope against its percentage elongation, you get a straight line through the origin. The result that you quote is a consequence of Hooke's Law, not some "more realistic" replacement for it.

In reply to:
Is this true, or did the problem oversimplify/misstate?

The actual graph is certainly not a straight line for all possible rope deformations, but is reasonably close for many ropes to a straight line for elongations in the working range of the rope, which is up to about 30%. All climbing ropes are engineered to have lower elongations at low loads (presumably by virtue of the internal friction of the core and sheath), and a number of the newer ropes that qualify for multiple UIAA ratings are engineered to have curves that are not straight lines, since the predictions based on the straight-line model do not agree very well with their published impact numbers.

Certain mathematical features of Hooke's Law are clearly not present in ropes. For example, there is no response to a contracting force. In other words, the tension-extension graph ends at the origin. Moreover, the near absence of a perceptible bounce and the relatively long recovery time all bespeak a damping effect. The Italian model uses the traditional differential equation for damped harmonic motion, which also depends on Hooke's Law but adds a damping term. However, the traditional damping term, which is proportional to velocity, is based on a damping mechanism consisting of a viscous fluid in a cylinder. Whether the mechanisms that damp rope oscillations are analogous to a classical car shock absorber is not addressed in their web presentation and seems to me to require some justification.

One can use models other than the linear one; the graph of a cubic equation T=kx^3, rather than a linear equation T=kx is the starting point for the analysis of what is called a "stiff" rope. But the idea that the model should involve an odd power of the extension x is based on the assumption that compression and extension should produce the same force magnitude, a fact that is certainly not true for ropes.

Perhaps the best way to think about Hooke's Law is that it is, in many cases, a very good linear approximation to the true tension-elongation curve. The fall-factor concept, which is a consequence of Hooke's law, has been confirmed in many tests and remains the fundamental concept in the testing and certification of ropes.

The primary real-world feature that is not accounted for in the analysis you did is the effect of rope friction against rock and through multiple carabiners. The Italian computer model is based on a system of differential equations for damped simple harmonic motion and attempts to account for frictional effects at each carabiner and can be adjusted for a certain amount of rock friction. But at the heart of those equations is....Hooke's Law.


curt


Oct 3, 2005, 6:32 PM
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In reply to:
english please there are 6th graders here thank you

In reply to:
In reply to:
I did a "real world example" physics homework problem last year, where they said a climbing rope's tension is not simply proportional to extension, as a 'physics spring' is. At the end of the problem, we ended up proving that the maximum force on a system is the same for all systems with the same fall factor (i.e. maxForce(4ft drop on 2ft rope) = maxForce(40ft drop on 20ft rope)).

If your end result was the invariance of maximum force for falls of the same fall factor, then your starting point was indeed Hooke's Law, which says that if you plot the tension in the rope against its percentage elongation, you get a straight line through the origin. The result that you quote is a consequence of Hooke's Law, not some "more realistic" replacement for it.

In reply to:
Is this true, or did the problem oversimplify/misstate?

The actual graph is certainly not a straight line for all possible rope deformations, but is reasonably close for many ropes to a straight line for elongations in the working range of the rope, which is up to about 30%. All climbing ropes are engineered to have lower elongations at low loads (presumably by virtue of the internal friction of the core and sheath), and a number of the newer ropes that qualify for multiple UIAA ratings are engineered to have curves that are not straight lines, since the predictions based on the straight-line model do not agree very well with their published impact numbers.

Certain mathematical features of Hooke's Law are clearly not present in ropes. For example, there is no response to a contracting force. In other words, the tension-extension graph ends at the origin. Moreover, the near absence of a perceptible bounce and the relatively long recovery time all bespeak a damping effect. The Italian model uses the traditional differential equation for damped harmonic motion, which also depends on Hooke's Law but adds a damping term. However, the traditional damping term, which is proportional to velocity, is based on a damping mechanism consisting of a viscous fluid in a cylinder. Whether the mechanisms that damp rope oscillations are analogous to a classical car shock absorber is not addressed in their web presentation and seems to me to require some justification.

One can use models other than the linear one; the graph of a cubic equation T=kx^3, rather than a linear equation T=kx is the starting point for the analysis of what is called a "stiff" rope. But the idea that the model should involve an odd power of the extension x is based on the assumption that compression and extension should produce the same force magnitude, a fact that is certainly not true for ropes.

Perhaps the best way to think about Hooke's Law is that it is, in many cases, a very good linear approximation to the true tension-elongation curve. The fall-factor concept, which is a consequence of Hooke's law, has been confirmed in many tests and remains the fundamental concept in the testing and certification of ropes.

The primary real-world feature that is not accounted for in the analysis you did is the effect of rope friction against rock and through multiple carabiners. The Italian computer model is based on a system of differential equations for damped simple harmonic motion and attempts to account for frictional effects at each carabiner and can be adjusted for a certain amount of rock friction. But at the heart of those equations is....Hooke's Law.

He said that ropes do obey Hooke's Law fairly well when they are being stretched--at least for rope elongations that are within the useful working range (about 30% elongation) of the rope. Ropes do not obey Hooke's Law well when they are contracting--without adding in some damping "fudge-factor."

Curt


Partner rgold


Oct 3, 2005, 9:45 PM
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From this data, if it is representative, we'd have to say kernmantle ropes don't obey Hooke's law very well and the linear spring model for the rope, which is used in a lot of simple analysis of fall forces etc., is not a particularly good model! This is a bit of a surprise to me as it seems to give reasonable results.


Don't be too surprised, Hooke's Law ain't dead yet. Instead, be careful how you judge linearity in these graphs. The horizontal scale is much bigger than the vertical scale, which has the effect of emphasizing the apparent curvature. To illustrate, here are two views of the graph from the Marlowe site for the kernmantle-like construction. (The picture won't show up until it is approved. They say 24 hrs, but several days is possible.)

http://www.rockclimbing.com/...p.cgi?Detailed=62139

What I've done here, crudely, is to erase the other curves and trace over the "kernmantle" curve with a fat red line so that it will show up under compression. I've also outlined in blue a rectangle whose sides, on each axis, represent changes of 20%.

In the right-hand version, I've compressed the horizontal scale of the left-hand version so that the two scales agree, i.e. so that the rectangle becomes a square. This gives a properly scaled view of the load-extension curve, which appears to be much better approximated by a straight line once it is viewed in the correct proportions.

But this isn't all. I assume that the 100% number on the load axis represents minimum breaking strength. The load curve is given all the way to the breaking point of the rope. No one would expect the curve over this range to be linear---Hooke's law certainly breaks down as the rope reaches its limit. Moreover, the beginning sections of the curve are also not representative, since under low loads the damping effects of rope construction may well dominate the linear behavior. The middle section of the curve is the portion one would expect to exhibit Hooke's Law behavior, and it is the middle section, in the case of climbing ropes, that is involved in catching falls.

This means that the linearity one expects in the elastic working range is misrepresented by viewing the entire curve here, and a straight line approximation over a smaller range will do even better.

There are, of course, mathematical ways to measure how well the data are approximated by a straight line. Looking at a graph and making a visual determination is mighty subjective, but these subjective eyes don't see any glaring evidence in the middle range of the properly scaled right-hand graph for deciding that Hooke's Law doesn't do a pretty good job.


alpnclmbr1


Oct 3, 2005, 11:54 PM
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Re: Does climbing rope obey Hooke's law? [In reply to]
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[..assumption of a damping force that accounts for the lack of bounce.

That is the second time that you have said that.(along with everyone else...)

In the UIAA rope test. The "bounce" is around 25% of the length of rope. (and it is a steep curve)

also, see the graphon the italian site for a "fixed belay deplacement"
It shows a ~50% bounce.


daithi


Oct 4, 2005, 7:10 AM
Post #25 of 30 (4048 views)
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Registered: Jul 6, 2005
Posts: 397

Re: Does climbing rope obey Hooke's law? [In reply to]
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....which has the effect of emphasizing the apparent curvature.

The curvature, or at least the progressive steepening, is real though not apparent! I do not doubt the fact that large parts of the curve can be estimated by a linear fit. I can divide it into three distinct sections that can each be well estimated by a linear fit all the way out to an extension of 0.21 after which it is clearly non-linear.

The real point is the original question, is the behaviour analogous to a linear spring subject to Hooke's law? This is quite a different question to can parts of load extension graph by plotted by a linear fit and give a good estimation (there is no doubt it can).

In reply to:
No one would expect the curve over this range to be linear---Hooke's law certainly breaks down as the rope reaches its limit.

Absolutely. It is only valid within the elastic working range of the material.

In reply to:
Moreover, the beginning sections of the curve are also not representative, since under low loads the damping effects of rope construction may well dominate the linear behaviour.

It still is a linear relationship at the beginning and passes through the origin. Why are so sure that this is not portion that is analogous to a linear spring and not the middle section? Viscoelastic polymers (at least the ones I have seen) have a linear section at the beginning, although the slope of it depends on how quickly the strain is applied.

In reply to:
The middle section of the curve is the portion one would expect to exhibit Hooke's Law behavior, and it is the middle section, in the case of climbing ropes, that is involved in catching falls.

There is no doubt that it behaves linearly but does the spring constant actually have a physical meaning in this region (can it be related to the product of Young's modulus and cross-sectional area) in the sense it has a physical meaning for a simple linear spring? Also a linear fit in this region comes no where close to crossing the origin, so although it exhibits elastic like behaviour in this region, the model of the linear spring is not a correct one.

I did a linear fit to the data over the entire elongation range and got an R-squared value of around 0.9, which was a bit better than I expected. This means that a simple Hooke's law model will give pretty good results, which explains why all the simple fall force analysis produces pretty good results.

To the original poster I would say that although the linear spring analogy subject to Hooke's law gives reasonable answers (good enough for engineering purposes) this does not mean it is modelling the physics correctly. The real situation is a lot more complex with various different regions exhibiting elastic-like deformation with different rates.
http://www.rockclimbing.com/...p.cgi?Detailed=62159

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