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billl7
Jan 19, 2006, 11:09 PM
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In reply to: I still question some of the conclusions made for not leaving slack. Only because my basic logic says differently, ....
Bill
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curtis_g
Jan 19, 2006, 11:32 PM
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In reply to: In reply to: - initial slack makes negligable (in theory, if not zero) difference If it makes no difference ito you in theory, you need to find another theory. Jay
Dear Jay, I will restate...with initial slack in the system (your rope), you will only increase your vertical fall velocity and THEN translate it into the SAME AMOUNT of horizontal velocity once the swing starts instead of doing it all over the same radius. It would be like falling in the path of a circle or the path of a 'J' a soft catch would prouce a parabolic fall path. Initial slack does not decrease the amount of total potential energy that will all eventually end up as your horizontal slam. A soft catch is your best bet, not giving out more rope to fall farther. But I wouldn't say keep an uber tight belay, that wouldn't help the climber catch his bearings before he starts to scrape across the rock...but that's headsense, not physics.
again, my conclusion about slack on a traverse...if I was the belay I would give just enough to make sure the first very few feet weren't awkward so that the climber isn't being dragged across the rock before he knows he's falling.
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whoa
Jan 20, 2006, 12:33 AM
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Here's a try. Let me know what you think.
If no slack is let out for a rigid rope L meters long, the potential energy is entirely turned into horizontal kinetic energy at the wall:
e_no-slack = mGL
So, suppose we let out enough slack to generate a free-fall of h meters before finishing with a partial pendulum arc to the wall. When the rope is first taut, the velocity that is in the direction of possible motion (tangent to the arc, perpindicular to the rope) is preserved. But there is no way that the component of motion parallel to the rope can be translated into motion perpindicular to the rope. So maybe the rope absorbs it, maybe the climber absorbs it, maybe the earth gets accelerated ... whatever!
(crude sketch added in edit)
http://groups.msn.com/...=4675556766064520295
The velocity downwards (when the rope is first taut) is:
v = sqrt(2Gh)
The component of v tangent to the pendulum arc would be v * sin(angle), where "angle" is the angle of the rope against the wall.
v_tangent = [sqrt(2Gh)] * [L/(sqrt(h^2 + L^2)]
The energy due to v_tangent is preserved and becomes horizontal energy at the wall:
e_tangent = 1/2 * m * v_tangent^2 = mGh * [L^2 / (h^2 + L^2)] =
= mGL * Lh/(h^2 + L^2) = mGL / (h/L + L/h)
Let R be h/L --- the ratio of the slack-fall to the initial length of the rope. Then,
e_tangent = e_no-slack * (1/(R + 1/R))
but there is also additional horizontal energy due to the partial pendulum, equal to the potential energy of the remaining vertical drop, or:
e_pendulum = mG(sqrt(h^2 + L^2) - h) = mgL * (sqrt(1 +h^2/L^2) - h/L)
= e_no-slack * [sqrt(1 + R^2) - R]
Adding:
e = mGL * [ 1/(R+1/R) + sqrt(1 + (1/R)^2) - R ]
See the graph below. Looks like letting out rope right at the start is a win if you can let out enough rope such that R = 1. That would be sqrt(2) - 1, or about 41% of the (run-out) rope's length.
http://groups.msn.com/...=4675556743304009216
But then I probably goofed somewhere.
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whoa
Jan 20, 2006, 2:57 AM
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Maybe, but I think you can accommodate the only important effect of the rope's springiness (namely, the change in the rope's final length) by including the added length as "slack" in the model. No particular trajectory after tautness was assumed--all we need are conservation of energy assumptions--so it doesn't really matter what the precise trajectory is.
As for the assumption that a tugging on a rope doesn't alter the climber's velocity perpindicular to it, I'm still pretty comfortable with that one. :)
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daithi
Jan 20, 2006, 3:07 AM
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Edit: I have edited the numbers and figures in this post as the original contained a rather large (and embarrassingly basic) error! :oops: Like a certain South Korean human stem cell researcher my reputation is forever tarnished! Thankfully this has nothing to do with my real research! :)
In reply to: v_tangent = [sqrt(2Gh)] * [L/(sqrt(h^2 + L^2)]
I think this is wrong or else I am not visualising your definition of L, h and angle correct (also likely!).
Edit: I had misunderstood your definitions. I see you redid yours using the same definition that I used for R.
I did some numbers too. Here is the setup that I am attempting to model.
http://www.rockclimbing.com/...p.cgi?Detailed=68012
R is the length of slack rope that has been normalised wrt the horizontal distance from the climber to his last runner. The minimum value is unity obviously (no slack - pure pendulum). h is the vertical distance the climber falls before the rope gets taut and the pendulum into the wall begins and theta is the angle between the rope and the vertical fall.
I used a similar method to whoa I think. A combination of the conservation of energy and the equations of motion of a pendulum to calculate the impact velocity at the wall (which was normalised wrt to the impact velocity of a pure pendulum swing 4.429 m/s).
Edit: Similarly, energy has been normalised.
The huge assumption of this is, at the moment the rope gets taut at the start of the pendulum the radial component (in my polar coordinate system) of linear momentum is instantly absorbed by the dynamic rope (this assumes the rope performs a lot better than it does in reality). In reality this will also be a function of time and would need to be incorporated into the model. I have no data on the dynamic mechanical response of climbing ropes so nothing on this is included. This assumption becomes more reasonable as the fall factor goes to zero (lots and lots of rope and small falls). Here is the result.
http://www.rockclimbing.com/...p.cgi?Detailed=68011
Some comments:
1. Initial pay out of slack does not reduce the impact velocity at the wall (compared to a pure pendulum) in this model, it actually increases it. The slack needs to be 1.2 times greater than the horizontal distance before any favourable reduction is achieved. Those who conjectured that the initial increase of kinetic energy due to the longer fall results in a higher impact velocity seem to be on to something. This is not what I expected when I first thought about it.
Edit: The actual increase at the worst possible case, which is a 7% increase of slack, is only 2% in energy or approximately 1% in impact velocity at the wall. This is not significant enough to warrant ever refusing to give slack to a falling climber, since in a real system slack will always be favourable when we take into account rope stretch that takes place over a finite time, as opposed to instantaneously! At 20% slack and above the impact force dramatically reduces with increasing slack.
2. However, as further and further slack is paid out, the impact velocity does get smaller. The rope gets a chance to absorb more of the linear momentum of the falling body.
3. The limits of the model are as R -> 1 (no slack) impact velocity is that of a pure pendulum (4.429 m/s). As R -> OO the impact velocity goes to zero (becomes a vertical fall with the rope absorbing everything). Anyone else doing a model make sure it obeys those physical limits.
4. The highest impact velocity is at a 7% let out of slack, which corresponds to an angle of 60°, with a 1% increase in impact velocity (2% in energy).
5. The rope in this model performs better than it could ever do in reality. It absorbs all the linear momentum in the direction of rope instantaneously before the pendulum trajectory.
6. Although this is a very simple model at least it gives us some physics to discuss as opposed to some pure conjecture and mis-applied physical principles, which the majority of this thread has been thus far!
The final equation plotted is
V = sqrt(2gRcosq)sin q + sqrt(2gR (1-cosq))
Edit:
V = sqrt(2gRcosq)sin^2q + 2gR (1-cosq))
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daithi
Jan 20, 2006, 3:25 AM
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In reply to: But so does the amount of rope out to absorb the additional energy.
The point is only the component of linear momentum that is absorbed by the rope is the component that is parallel to the rope (it would be the dot product of the momentum vector and the elsatic deformation vector of the rope). The rope has no mechanical means (or almost none) to absorb energy in any direction other than along it.
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whoa
Jan 20, 2006, 3:42 AM
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[I see that the delcross post I was replying to has been deleted, so I've deleted my response.]
Daithi, nice pics!!! As for my trig, I've added a pic to make it clearer. (Though I am bowing my head in shame to post it near yours.)
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curtis_g
Jan 20, 2006, 4:09 AM
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First, nicely done daithi. And a convincing explination to the practical conclusion.
In reply to: In reply to: But so does the amount of rope out to absorb the additional energy. The point is only the component of linear momentum that is absorbed by the rope is the component that is parallel to the rope (it would be the dot product of the momentum vector and the vector that describes the rope direction). The rope has no mechanical means (or almost none) to absorb energy in any direction other than along it.
daithi is correct once again. I guess I didn't explain myself fully, or maybe I wasn't read, but (and this is what I believe daithi is getting at), is that the rope does nothing to weaken a force or absorb some force or absorb some velocity. What the dynamics of the rope does is decrease the change in acceleration more spicifically (slang and scientifically) known as jerk. The dynamic rope only elongates the time over which the force is applied and lessens the impulse of the preassure on you from the force.
All the energy is still there! It can't just go away or 'be absorbed' like so many people have claimed. That's day one stuff. Actually its the law of conservation of evergy. IT'S a LAW...you all should go to jail...or at least pay me a hefty fine.
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colkurtz
Jan 20, 2006, 5:21 AM
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In reply to: All the energy is still there! It can't just go away or 'be absorbed' like so many people have claimed. That's day one stuff. Actually its the law of conservation of energy. IT'S a LAW...you all should go to jail...or at least pay me a hefty fine.
you couldn't be more wrong. energy will escape the system.
but
this problem is fascinating. my intuition is that the rope should not absorb the energy instantaneously. and right now i think this is a big factor. as are the properties of the rope.
i would rather see some empirical data if someone is motivated enough.
i am more concerned with the ability to use my feet to control the swing. adding a layer of complexity
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billl7
Jan 20, 2006, 7:16 AM
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In reply to: In reply to: All the energy is still there! It can't just go away or 'be absorbed' like so many people have claimed. That's day one stuff. Actually its the law of conservation of energy. IT'S a LAW...you all should go to jail...or at least pay me a hefty fine. you couldn't be more wrong. energy will escape the system.
Paying out slack to the tune of 250% of the traverse just to break even seems dicey; understanding the model is limited, maybe in reality not so much is needed. However, I do appreciate curtis_g's early remark about having time to get yourself oriented.
Very nice efforts by whoa and daithi. I'm simply outdone here: I degenerated into chasing my tail when trying to find the angle based on dynamic stretch and the dynamic stretch needed the force parallel to the rope which depended on that very same angle; obviously, I'm not looking at it properly.
Hmm, anyone work for one of those car crash test organizations? We could "borrow" a crash dummy with internal accelerometers and probably return it maybe only a little worse for the wear. I'm sure we've got enough of the other needed gear between us. ;-) (Good idea colkurtz.)
Thanks again for sticking with this!
Bill
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curtis_g
Jan 20, 2006, 7:50 AM
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for some reason i find it hard to accept when you say that the rope absorbs energy. it could gain a certain potential engery stored in its stretch and its stretch could slightly warm up the rope as energy will always escape into dissepated heat (negligable), and in a direct vertical fall many things (rope, belay, anything in the system including the two harnesses) are moved and tensioned as well as the climber springing back up as he bounces on the rope.
ok so i read on and you comment that In reply to: as it seems the only way to reintroduce it to the climber's tangential velocity is if the rope significantly springs back well before the climber hits the wall. |
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billl7
Jan 20, 2006, 2:32 PM
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In reply to: for some reason i find it hard to accept when you say that the rope absorbs energy. it could gain a certain potential engery stored in its stretch and its stretch could slightly warm up the rope as energy will always escape into dissepated heat (negligable), and in a direct vertical fall many things (rope, belay, anything in the system including the two harnesses) are moved and tensioned as well as the climber springing back up as he bounces on the rope.
In reply to: ok so i read on and you comment that In reply to: as it seems the only way to reintroduce it to the climber's tangential velocity is if the rope significantly springs back well before the climber hits the wall. I would readily believe that this does happen. as the climber nears a total horizontal vector of velocity, wouldn't the rope regain some of it's pull and toss the climber a bit harder...or possibly the climber would oscillate during his swing that would loot something like the function of y = x sinx (with any number of phase shifts or stretches determined by extraneous factors of the fall) as the climber falls from x = 2 pi in the negative direction towards x=0. just an example/thought.
However, the most efficient way for that shrink of the rope to speed up the arcing leader is if the shrink(s) occur fairly far away from impact where the angles are more favorable for that kind of energy transfer (i.e., if say the first shrink instantly follows the first tension period). As the leader gets closer to impact, more and more of the shrink action only lifts the leader up instead of propelling her towards the wall.
Practically speaking, a shrink just isn't going to happen under conditions as favorable as the last tensioning for transferring energy back to horizontal movement because some of the pendulum arc will have passed and so the angles aren't as favorable.
My guess (don't know) is that the delay in time for the oscillation to reverse plus the energy loss due to friction keep the initial shrink episode relatively weak and relatively ineffective. Throw in a soft catch and there would be even less and weaker oscillations as well (where's Jay! he's probably going to miss another opportunity to crow!! :) ).
Bill
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cracklover
Jan 20, 2006, 3:33 PM
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In reply to: Throw in a soft catch and there would be even less and weaker oscillations as well (where's Jay! he's probably going to miss another opportunity to crow!! :) ). Bill
Actually, I introduced the soft catch way back on page 3. Unfortunately, curtis_g agreed with me, causing me to lose all credibility. :)
Nevertheless, I'm quite convinced that the dampening effect of the soft catch is quite significant. Unfortunately, I haven't studied enough physics to plot out any more than the simple models that have already been done here online (much more gracefully than the scratches I'd worked out on my own, I must say!) But while I *think* I do understand the more complex principles that come into play with the soft catch, I don't know how to plot them out, and of course I wouldn't have the means to test my theories either.
It *is* an interesting problem though. I'd like to see a good physical modelling of it. Anyone know an under-employed physicist climber with some time on his/her hands?
GO
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daithi
Jan 20, 2006, 4:02 PM
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In reply to: But while I *think* I do understand the more complex principles that come into play with the soft catch, I don't know how to plot them out, and of course I wouldn't have the means to test my theories either.
Describe how you think it works in the form of a differential equation, or failing that even the principle and I will do the maths and the numerical modeling part (at some stage!).
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billl7
Jan 20, 2006, 4:27 PM
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In reply to: In reply to: Throw in a soft catch and there would be even less and weaker oscillations as well (where's Jay! he's probably going to miss another opportunity to crow!! :) ). Actually, I introduced the soft catch way back on page 3. Unfortunately, curtis_g agreed with me, causing me to lose all credibility. :)
Which do you think typically causes more pendulum elongation: belay slippage or plain rope stretch? That probably needs some qualification but I'm too lazy right now.
It's tempting to make an effort to take daithi up on the offer to crunch more numbers.
Bill
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cracklover
Jan 20, 2006, 5:39 PM
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In reply to: In reply to: But while I *think* I do understand the more complex principles that come into play with the soft catch, I don't know how to plot them out, and of course I wouldn't have the means to test my theories either. Describe how you think it works in the form of a differential equation, or failing that even the principle and I will do the maths and the numerical modeling part (at some stage!).
I'm just running out to lunch now, and when I get back I need to get some work done.
I'll write back as soon as I can, sorry! :(
GO
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pastprime
Jan 20, 2006, 5:50 PM
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It comes down to this:
In a fall into a corner from a piece horizontal to the climbers position with no slack in the rope, virtually none of the energy is transferred to the rope or absorbed by it; almost all of the energy is absorbed by the climber's body smacking into the wall.
With a small amount of slack out, the climber falls further, generating more energy, and the faller's motion is transfered into swinging motion while still quite a bit sideways from the last piece, so the pull is not parallel to the rope, and the increased energy is still mostly absorbed by the climbers body smacking the wall.
With a whole lot of slack out at the time of the fall, even more energy is generated that is going to need to be absorbed at some time, but by the time the faller hits the end of the rope, the rope is much closer to being lined up with the direction the faller is moving, so much more of the energy is absorbed by the rope before or as translation to horizontal movement occurs.
In the case of little slack, less energy is generated, but it is all absorbed by the climber at the moment of impact. In the case of lots of slack, much more energy is generated, but much more of the total energy is absorbed by the rope. At it's most extreme, in an ideal situation with lots of rope and nothing below to hit, it is like the difference between hitting a solid wall in your car at 30 mph or braking hard from 90 mph.
In real situations, I expect its more like a question of driving into a wall at 30 mph or going into a grove of trees at 60.
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dirtineye
Jan 20, 2006, 6:03 PM
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I just have to know, how many of you posting in this thread lately have taken any big pendulums and done it safely ?
cause from the way you are writing it sounds like you have not done it very much.
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curtis_g
Jan 20, 2006, 6:12 PM
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In reply to: With a small amount of slack out, the climber falls further, generating more energy, and the faller's motion is transfered into swinging motion while still quite a bit sideways from the last piece, so the pull is not parallel to the rope, and the increased energy is still mostly absorbed by the climbers body smacking the wall. With a whole lot of slack out at the time of the fall, even more energy is generated that is going to need to be absorbed at some time, but by the time the faller hits the end of the rope, the rope is much closer to being lined up with the direction the faller is moving, so much more of the energy is absorbed by the rope before or as translation to horizontal movement occurs.
Yes, that was the situation founded by daithi with his explination and I believe the line was drawn at something like a slack payout 250% larger than the horizontal distance of the swing to break even and cause the climber to hit with less speed. That's one hell of a free fall and one I'm not interested in just to 'break even' in terms of final horizontal velocity.
My comment that someone appreciated was that I would enjoy only enough slack so that I can realize that I have fallen before I'm being thrown sideways followed by a soft catch with a gradual slack payout of 5 to 10 ft, and, while more complicated models and introducing a higher level of variables shouldn't be discontinued, I would probably still ask for the scenario I just outlined. A soft catch would do wonders, but the climber will ultimately make the difference.
"I would enjoy only enough slack so that I can realize that I have fallen before I'm being thrown sideways followed by a soft catch with a total gradual slack payout of 5 to 10 ft."
I am starting a physics course on Tuesday with one of my favorite professors who is a big big climber, rock and ice. Denali and a bunch of others. The entire semester is an independant study project presented at the end of the year. Mine will definitely be on tests of climbing gear, dynamics of a fall, friction, something like that, so I'm looking ofrward to making my presentation readable on RC.com and owning my first gold trophy, haha.
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billl7
Jan 20, 2006, 6:23 PM
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In reply to: I just have to know, how many of you posting in this thread lately have taken any big pendulums and done it safely ? cause from the way you are writing it sounds like you have not done it very much.
Since the approach to mathematical modeling generally starts out fairly conservative, because most of us try the easiest things first, the approach could be confused with lack of experience. However, in my case you can accurately chalk it up to inexperience if you like. :)
Bill
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jt512
Jan 20, 2006, 7:05 PM
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In reply to: In reply to: In reply to: - initial slack makes negligable (in theory, if not zero) difference If it makes no difference ito you in theory, you need to find another theory. Jay Dear Jay, I will restate...with initial slack in the system (your rope), you will only increase your vertical fall velocity and THEN translate it into the SAME AMOUNT of horizontal velocity
By looking at the extreme case suggested by Curt, an infinitie amount of slack, it can be seen that you are wrong. The horizontal velocity would be 0 (and, incidentally, the fall factor would only be 1). So, assuming, as we should, until someone proves otherwise (using real equations), that the relationship between slack and horizontal velocity is monotonic, then the more slack in the rope, the less the horizontal velocity.
Edit: Actually, in daithi's model the realtionship is not monotonic.
Jay
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dirtineye
Jan 20, 2006, 7:08 PM
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In reply to: In reply to: I just have to know, how many of you posting in this thread lately have taken any big pendulums and done it safely ? cause from the way you are writing it sounds like you have not done it very much. I don't know how many. Why wait for a response? Tell us what you think from experience. Since the approach to mathematical modeling generally starts out fairly conservative, because most of us try the easiest things first, the approach could be confused with lack of experience. However, in my case you can accurately chalk it up to inexperience if you like. :) Bill
I hate to say this, and I have a degree in math (lots of modeling with PDE) with minor in physics plus grad school math (all behind me now and obviously fading LOL), but I don't give a crap about a mathematical model for this pendulum falling stuff, because one, I know the general principle that as long as the wall is overhanging then you are actually slowing down after you swing past vertical, and two, from actual pendulum falls I have taken, I know that if you do it right you can hit the wall with both feet , with your weight centered between your feet, and absorb the impact with your legs and suffer no ill consequences, period.
I don't blame you for wanting to model, and it woud be interesting, but in the end, when you take the fall two things will matter:
Are you in a good falling position and have you got a plan for meeting the wall?
Keep in mind that you are not a point mass in a theoretical pen and paper model. You can react. You can take action for better or for worse. Pity the poor inanimate dumb as a rock point mass, for he is doomed to smack the wall in a simple collision at whatever velocity the math dictates.
You are not an unyeilding point mass, you are a complicated organism with excellent built in damping devices and intelligence, and if you will only use them correctly, you will come out unscathed.
So if you panic when you fall, you're screwed. If you practice falling and react corectly and quickly, you are not screwed.
IF you want to know more about what I think on this subject, look at the first two pages of this thread. All the examples I gave of how to take these falls were based on actually taking such falls myself. I'm a firm believer in the Arno School of Falling Practice and using your head to judge the falling consequences BEFORE you are in the fall.
BTW, I don't mean to sound like a butthead, I just feel like crap today, LOL
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billl7
Jan 20, 2006, 7:31 PM
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In reply to: I don't blame you for wanting to model, and it woud be interesting, but in the end, when you take the fall two things will matter: Are you in a good falling position and have you got a plan for meeting the wall?
But modeling looks so attractive to someone new to leading, who hasn't yet fallen on lead, who generally climbs on routes that aren't very close to vertical such that a pendulum would probably involve a lot of contact with the rock along the way before the possible impact. I guess I'm just trying to live vicariously through math and experience of others.
Peace, Bill
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jt512
Jan 20, 2006, 7:31 PM
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Edit: the following comments are no longer relevant. They referred to an earlier graph, not the one now pictured above.
I have trouble believing that the slope is so steep. It doesn't seem reasonable that a tiny increase in slack would increase the impact force by nearly 40%.
Note that for pendulum falls originating from a short horizontal distance, you will always be on the "downside" of the curve. If you fall from say 5 feet straight out horizontally from your highest pro, you will almost certainly have at least 7.8 inches of slack in the rope (13% of 5 feet), if there is much total rope in the system at all, and therefore any additional slack in the rope would reduce the horizontal velocity, given the above model.
Jay
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