By Avery Nelson
From Boulder, CO
Feb 12, 2008
| Kind of related but not totally...
Petzl has a fun little fall *force* simulator applet on their site, if you search for 'Fall Simulator' |  |
By Hbar
Feb 15, 2008
| brenta wrote: Allow me to be a bit pedantic here. If s were proportional to l for fixed w, the impact force, which is just k*s/l, would be independent of the fall factor. I believe I understand your main idea: The higher the fall, the more energy is at play. The more rope is involved, the easier it will be to absorb that energy. An ideal spring stores energy by stretching. Of course, the more it stretches, the higher the forces on the climber and the anchors. The way ropes are built nowadays, forces that hurt a climber or damage an anchor are usually well below the rope's tensile strength. So, what matters is the maximum force that a climber or an anchor may withstand. Let this force be F and let k be the force required to double the length of a piece of rope. Then A = F^2/2k. Let f be the fall factor. If we omit s from the left hand-side of the equation, we get f < A/w. If we include it, we get f < A/w - F/k. So, the effect of accounting for s is additive rather than multiplicative, and A does not depend on w, but otherwise I concur with your analysis.
As far as I can tell, our only disagreement here is that we are defining variables differently. I'm going to try and use your variables to address at least one of your comments, though I admit to the real possibility that I misunderstand your choice of variables.
Denoting the maximum force the climber can take without injury or anchor failure as F we have the relation that F=ks/l exactly, and F is no longer a variable, but rather a parameter due to our choice of the extreme value (remember I am only trying to look at inequalities, which means that the change occurs at the extreme case). Then we arrange, and get s=Fl/k. Note that s is then proportional to l (proportionality constant F/k) for investigations of the critical point where the fall force is fixed at the point where any more force, no matter how small, would lead to failure of the climbing system. Investigation of this point is all that is required to establish my inequalities.
A(w) indicates a functional dependence on w, rather than multiplication by w. I agree that the differences are additive.
To see exactly how our results are related, we write w(d+s)=Al, taking the energy absorption capacities of the rope as proportional to its length as discussed above. Then we have wd = Al-ws, then using the result above relating s and l, we get wd = Al-wFl/k. Moving to the inequality, and manipulating, we get d/l = f < A/w-F/k, the equation you wrote above. The A I initially used is related to this A as A' = A-wF/k, which is certainly dependent on w. Doing the problem this way we can start directly from my original inequality (written slightly more explicitly), wd < A'(w)l, and rearrange in one more step to get f < A'/w. Given a fixed climber weight, these approaches are exactly equivalent physically and mathematically as demonstrated by my derivation of their equivalence. The difference is simply different cosmetic choices of parameters for the theory as I demonstrated above.
Finally, thanks for the great discussion. I see you are from Boulder. I am likely going to be in Boulder during July at a summer school, and will be needing some locals to partner up with if I am going to get any climbing in . . . you interested? | |
By brenta
From Boulder, CO
Feb 16, 2008
| I agree that we are just defining A differently. I was going by your original definition:
Hbar wrote: E(absorbed by rope)=A*l, where A is a constant dependant on the rope, and l is the amount of rope out
but, of course, you can use the "adjusted" A to reach the same fundamental conclusions.
As for July, send me an email when the time comes. If I'm in town (I have a couple of trips lined up) and you are happy to climb easy stuff... | |
By rgold
From Poughkeepsie, NY
Feb 16, 2008
| If anyone is interested in what I believe to be clear derivations of the standard equation for fall forces, I posted two such derivations on rc.com a a downloadable .pdf file. One derivation is via the conservation of energy approach and the other derivation is via the more or less standard differential equation for simple harmonic motion. Clicking the following link will download the file.
One fact that is absolutely clear from either derivation is that the "L" in the fall factor "H/L" ratio refers to the amount of rope that is actually involved in absorbing fall energy.
As for fall factors in climbing situations never exceeding two, there are two comments to make.
First, it is true that if the fall factor starts off bigger than one, then if the belayer pulls in slack during the fall, the fall factor will be increased. The comment was made that the belayer couldn't pull in enough slack during a fall for this to make any difference. This doesn't seem to me to be entirely true, because the effect of pulling in slack is relative, it depends on how much rope is out, and small amounts are "bad." For example, a leader who falls from two feet above the belayer without any pro in has a factor-2 fall. If the belayer sees that the leader is about to pitch and manages to pull in a foot of slack just as the leader goes, then the leader falls 3 feet and is stopped by a foot of rope; fall factor is now 3.
Another situation in which the fall factor is greater than 2 occurs if a climber falls after directly connecting to an anchor with a quickdraw. Let's say it is a short quickdraw with the webbing part approximately the length of a carabiner. If the climber moves above the anchor as far as possible and then slips, he falls six carabiner lengths, but the energy-absorbing sling material is only one carabiner length, and the tension in it is determined by the fact that the fall factor is 6. This will, in theory, break some gear, and there is at least one known case when such a fall broke a carabiner. | |
By brenta
From Boulder, CO
Feb 16, 2008
| rgold wrote: (Sorry, as far as I can tell the Mountain Project system will not let me give an alias for this URL, the system insists on truncating the full URL and printing out part of it.)
Enclose the alias in double square brackets, followed by the URL without any intervening space. Also, replace semicolons with ampersands. Like this. (The easiest way to examine the source of this post is to hit the "quote" button.) | |
By rgold
From Poughkeepsie, NY
Feb 16, 2008
| Brenta, thanks for the alias tips, I've edited my post. Replacing the semicolon from rc.com with an ampersand was what I didn't know to do. | |
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