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Klimmer
Mountain climber
San Diego
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(4) Klimmer's account begins with the incorrect statement that momentum=impulse. The correct version is that impulse is equal to the change in momentum. The next error is the formula suggesting that impulse = Ft, which is only true if the force involved is constant over the time interval in question, which is certainly not true of the tension in a climbing rope. The product Ft needs to be replaced by the integral of F(t) dt.
rgold,
The basic algebraic formula of mv = Ft is correct, without doing calculas, and taking the derivative of time.
It's a little hard to throw in the delta symbol in ST.
The change in Momentum = The change in Impulse, is correct, in this case.
(If the example was a golf ball at rest first, being teed-off then you could say The change in Impulse = The change in Momentum, Ft = mv )
In this case the falling climber has Momentum = mv, first.
The climbing rope is applying an Impulse = Ft, second, to stop the falling climber, which is equal to the momentum of the climber. The longer the impulse time the less the impulse force.
The t of the impulse would have to be accurately timed. Time for impulse would begin the moment the dynamic climbing rope started to take any load of the falling climber, until the moment the climber came to a complete and full stop. So F in this case would be the sum of all forces, during the entire impulse time.
That is why we have dynamic climbing ropes to extend impulse time. If we climbed on steel cables, then the impulse time would be unbelievably short and the impulse Force would be unbelievably high. So high in fact our bodies would be dismembered, and this is a very good reason we don't climb on static ropes.
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Karl Baba
Trad climber
Yosemite, Ca
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Use Yates screamers if you think you're going to fall on a piece that may or may not break
Or even how to account for factors would affect that?
the Devil is in the details. I doubt the result of most people's calculations actually matches the actual force on the gear because of so many factors like friction, the give in the body and whatnot. Has anybody tested how theory matches up with reality?
Theoretically a fall on your daisy chain should mess you up and/or rip your gear but many people have got away with them.
Peace
Karl
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rgold
Trad climber
Poughkeepsie, NY
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Karl, at this point, there are quite a few papers comparing theory to reality, and a number of approaches, more complex than the standard equation mentioned here, that have very good correlation with measurements made with real climbers and real belayers. Unfortunately, much of these accounts are in Italian and German and are not well-known in this country.
For an idea of some of this work, look at
http://user.xmission.com/~tmoyer/testing/Simulation_of_Climbing_and_Rescue_Belays.pdf
(scroll past initial frames on rescue configurations).
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Klimmer
Mountain climber
San Diego
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klimmer, if you need a Δ look for it here:
http://www.supertopo.com/climbing/thread.php?topic_id=351467&msg=352411#msg352411
you calculation isn't a productive approach to calculating the force...
the better way of looking at the problem is through the work done...
Ed,
Yes, I agree it could be done that way too. I suppose distance would be easier to measure accurately than time. d = the point at which a load is on the rope to the full dynamic stretch length of the rope.
ΔKE = ΔWork.
ΔKE = ΔW
1/2mv^2 = Fd
Solve for F:
F = 0.5mv^2/d
v = gt
Thanks for the character tip!
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JimT
climber
Munich
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It certainly is the case that jt´s calculator will give you the best value you are will get at the moment.
The debate at that time was to some extent at cross purposes as I was discussing a true model and others a standard model.
My objections were of a more technical nature in that the calculator (and the standard models) fail to reflect the complexities which occur in a real-life fall, belaying etc. This is not something which would really concern a normal climber but not very helpful for the work I was doing at that time where an accurate (and horribly complex) model would have allowed the identification of the forces at all points in the system at any point in time.
The subtlety is that I wrote "To produce a theoretical force calculator at this time is impossible...." not "a useful force calculator...)" since at the moment we still have to put measured values (rope characteristic and karabiner friction) into any calculator and some of these measured values are very variable. Currently for research we still have to do drop tests since the standard model is not detailed enough.
Not falling off is the key to survival, model or not!
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Ed Hartouni
Trad climber
Livermore, CA
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klimmer,
you want the total energy to be mgh where h is the total length of the fall
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Klimmer
Mountain climber
San Diego
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OK, got it.
ΔKE = ΔGPE
ΔKE = ΔW
So therefore, . . .
ΔGPE = ΔW
mgh = Fd
Solve for F:
F = mgh/d
h= length of fall in meters (m)
d = dynamic stretch of the climbing rope in meters (m)
Ok, easier still.
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rgold
Trad climber
Poughkeepsie, NY
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Klimmer (and everyone else), I promise this is my last word on the subject of these calculations.
Both your original approach and the latest version have the same mistake, which is that you are treating the maximum tension in the rope as if it was constant during the entire time the rope stretches to arrest the fall.
In the original treatment, you had mv=Ft, but what is needed is
and of course with this you can't divide both sides by t any more.
In the latest version, you have mgh=Fd. Here you treat F as if it was constant throughout the entire stretching process, but in fact F is a function of the stretch x, and so rather than Fd you need
and so you can't divide both sides by d.
There is another error in the second approach, which is that mgh isn't correct, at least if, as I imagine Ed intended, h denotes the distance the climber falls. The reason mgh isn't the correct change in potential energy is that the climber goes an additional distance, I think symbolized by d according to you, due to the stretch in the rope, making the net change in potential energy mg(h+d).
The link I provided has a derivation based on the conservation of energy approach you attempted; I believe this is the way Leonard and Wexlar originally derived the "standard model" formula
where F is the maximum rope tension, w is the weight of the climber, K is the rope modulus (which has to be calculated from the UIAA impact rating of the rope), and R=h/l is the fall factor.
If you just want an answer for the UIAA standard 80 kg climber, the following expression can be obtained from the formula above:
where as before R is the fall factor, U is the rope's UIAA impact rating, and F the maximum rope tension in kilonewtons
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Ed Hartouni
Trad climber
Livermore, CA
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actually I intended h to be the total distance... which included stretching the rope...
but I gave up with the derivation since you had done such a good job on your rockclimbing.com link deriving the expression in two ways...
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Ksolem
Trad climber
Monrovia, California
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I’m no mathematician, but I’ve got enough common sense to understand that every fall will generate a unique amount of force. Even the same fall taken twice will be unique (your knott will be tighter or looser, your belayer will be situated differently, the rope will be less dynamic from the previous fall...)
I’m with RGold on the use of double ropes. If the deal is really serious I like two belayers as well.
And for a climb which involves thin delicate pro I like having belayers who have the skill and presence of mind to perform a good soft catch.
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Hawkeye
climber
State of Mine
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If the deal is really serious I like two belayers as well.
if the deal is really serious the best outcome is dont fall. the next best outcome is recognize that if falling is possible and you are not willing to take that risk, then you had best be a good downclimber.
the most important part of all of this when serious injury is a possibility? keeping your wits about you. placing very small pieces just might give you the extra mental confidence that you need to pull it out. all situations are different and only experience will get you out.
frankly, your ability to keep your wits about you is far more useful than calculating how much force you will exert on a questionable piece. by wits i mean the ability to exercise the judgement that will tell you how much danger you are in and whether you have the ability to climb through it, or bail before you get into a drastically dangerous situation.
by the way, this is not taught in gyms.
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Seamstress
Trad climber
Yacolt, WA
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I generate roughly half the force that my husband does. Nonetheless, I have been known to place two blue camelots, cry to reduce my weight, and invoke the name of "christ" as in Chapin's tune about naming the only person who can save me now.
I find it very helpful to attempt to understand the math. It is a good realituy check on just what the consequences might be. Love to do some of the match and compare answets from the competing sources....
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pell
Trad climber
Sunnyvale
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Will the microstopper hold or pop?
It depends on many-many factors- rock quality, and surface of "brass-rock" contact, and an angle between direction of pull and a microstopper's wire, and a history of microstopper's usage (metal fatigue is accumulated every time it is loaded) - and a whole lotta other I don't know and can't imagine.
The only way to know - fall and see what'll happen.
It's like a car driving - the only way to know whether you will survive on your way to home or will be killed in a road accident is to jump in your car, start it and drive home.
We all gonna die.
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donini
Trad climber
Ouray, Colorado
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Climbing safety is an experiential thing, testing in the lab is of little consequence.
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donini
Trad climber
Ouray, Colorado
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The testing goes on in the development stage of climbing related products, how you use them is the most critical factor.
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mission
Social climber
boulder,co
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If you use a safety bend, the force generated in a fall will never exceed twice your body weight. The rope goes up from your harness to the carabiner, bends around the lower part of the biner, and goes down your belayer.
As Hugh Banner once said, "If it's below the waist, it's not a runner."
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Ksolem
Trad climber
Monrovia, California
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May 10, 2012 - 12:15am PT
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If you use a safety bend, the force generated in a fall will never exceed twice your body weight. The rope goes up from your harness to the carabiner, bends around the lower part of the biner, and goes down your belayer.
Maybe I'm just thick, but that post is harder to understand than all the equations...
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WBraun
climber
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May 10, 2012 - 12:19am PT
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Ksolem
That's just yokel speak for ....
"Don't worry everything will be all right"
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Dr.Sprock
Boulder climber
I'm James Brown, Bi-atch!
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May 10, 2012 - 12:29am PT
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just put the rope in a bucket of water right after you fall and measure the temp change of the water, deduct .098 for each 1000 feet of elevation,
it takes longer to cook muffins in denver than SF, right?
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