
robbiehirsch
Feb 27, 2005, 6:55 PM
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Alright so I'm bored and was trying to figure out how much force is acutally generated in a fall. Now I'm not talking about how to figure out what fall factor something is... I'm talking about how many KN are generated by a fall. Are there any standard equations that take into consideration a rope's stretch, the height of fall, weight of climber, etc?





glyrocks
Feb 27, 2005, 7:05 PM
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Doesn't Petzl's website have some sort of fall simulator? I think it's Petzl, and I think it will tell you what the forces are.





niftydog
Feb 27, 2005, 7:08 PM
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don't know any specifics, but I see it like this; The height is kinda irrelevant. What you need to know is the velocity of the climber just prior to the instant when the rope starts to take weight (could be estimated by knowing that acceleration due to gravity is 9.8m/s/s then you would need to know the time period of the fall). Then, you assume that the rope stretches some distance as it takes the strain, allowing the climber to decelerate to 0 velocity over time  giving you the negative acceleration experienced by the climber. Then, you use good old force = mass x acceleration which, if the units are correct (kilograms and metres), will spit out kNs. am I in the ball park, mathematicians?





ebonezercabbage
Feb 27, 2005, 7:43 PM
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as a sudomathematician (junior math major/physics minor student at RIT) i'd say that you are fairly correct, though there are usually more forces (not actually forces mind you) at hand in these problems than seems. Ask a physics teacher with a climbing background who knows the specifics on rope tollerances. Or pm me and i'll crunch the numbers on what kind of real force is felt after the expansion of the rope. Unless u use a static line, then its simply F=MA.





neurostar
Feb 27, 2005, 8:03 PM
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In reply to: as a sudomathematician (junior math major/physics minor student at RIT) i'd say that you are fairly correct, though there are usually more forces (not actually forces mind you) at hand in these problems than seems. Ask a physics teacher with a climbing background who knows the specifics on rope tollerances. Or pm me and i'll crunch the numbers on what kind of real force is felt after the expansion of the rope. Unless u use a static line, then its simply F=MA. RIT eh? I'm a physics major there. :)





esallen
Feb 27, 2005, 8:55 PM
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nope . . . hate math . . . sux





abraxas
Feb 27, 2005, 9:06 PM
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In reply to: Alright so I'm bored and was trying to figure out how much force is acutally generated in a fall. f=ma you're welcome.





abraxas
Feb 27, 2005, 9:09 PM
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In reply to: as a sudomathematician (junior math major/physics minor student at RIT) i'd say that you are fairly correct, though there are usually more forces (not actually forces mind you) at hand in these problems than seems. Ask a physics teacher with a climbing background who knows the specifics on rope tollerances. Or pm me and i'll crunch the numbers on what kind of real force is felt after the expansion of the rope. Unless u use a static line, then its simply F=MA. Its f=ma in any event, retard. Just make sure you are using the correct value for "a."





verticalcrag
Feb 27, 2005, 9:36 PM
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um..................i cant do math to save my life......................oh well.





paulraphael
Feb 28, 2005, 8:07 AM
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In reply to: as a sudomathematician (junior math major/physics minor student at RIT) i'd say that you are fairly correct, though there are usually more forces (not actually forces mind you) at hand in these problems than seems. Ask a physics teacher with a climbing background who knows the specifics on rope tollerances. Or pm me and i'll crunch the numbers on what kind of real force is felt after the expansion of the rope. Unless u use a static line, then its simply F=MA. there's another complication built into the problem. our simple equations assume that the climber stops accelerating downwards the instant the rope goes tight, but there's no good reason to assume this. so there's a nonlinear element that would be very hard to calculate hypotheticallly. sadly, i don't like math. in fact, when i fall, i'll typically scrape against the rock and bounce and spin a couple of times before i've even finished WRITING the equations. i rarely have them all solved before we get to the emergency room, and then that mean triage nurse always takes my slide rule away.





jw11733
Feb 28, 2005, 9:18 AM
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Man it gets real complicated, but lets start with the simple case where rope stretch is the only play in the system, ie. for a completly static belay. The easy way to do it is with conservation of energy. The amount of "work" done by the rope must equal the energy the climber has just before the rope catches him. This is simply mgh where m is the climbers mass, g is the acceleration due to gravity, and h is the distance the climber has fallen. now use work=mgh = fXd where f is the force generated by the rope, and d is the stopping distance. solving for f, we have f= mgh/d How do we know what d, the stopping distance is? It depends on the rope stretch and the amount of rope out. call this alpha X l where l is the amout of rope out and alpha is the coefficient of stretch, arount 0.06 or so (6% stretch) now we have f=(mgh)/(alphaXl) or write this as (mg/alpha)X(h/l) we see that the force is proportional to h/l, the distance fallen over the rope length, also called fall factor. remember, this does not take into account other factors that increase the stopping distance, ie. dynamic belay, harness stretch, belayer movement, and so on. Plug in some numbers: 80 kg climber, 10m of rope out, 2m above last piece, making for a 4m fall with a 6% stretch we get ~ 5.2 kN if the belayer lets half a meter of rope slip through, we get ~ 2.8 kN. These seem ballpark correct, but now it gets real complicated: The above formula is for the force (really tension) in the rope during a catch, but the force on the top piece depends on friction of the rope through that piece in a very complicated way. For a frictionless biner, the force on the top piece is twice the force on the rope, due to the pully effect. For a top piece with 100% friction, the force is only equal to the force on the rope, BUT this invalidates the above formula, because now only the rope between the climber and the top piece can stretch, making for a factor 2 fall! So fiction on the top piece has two competing effects, on increases the force, and one decreases the force. Where these balance out is beyond me, and probably will only be settled by experiment!





trenchdigger
Feb 28, 2005, 9:33 AM
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In reply to: Man it gets real complicated, but lets start with the simple case where rope stretch is the only play in the system, ie. for a completly static belay. The easy way to do it is with conservation of energy. The amount of "work" done by the rope must equal the energy the climber has just before the rope catches him. This is simply mgh where m is the climbers mass, g is the acceleration due to gravity, and h is the distance the climber has fallen. now use work=mgh = fXd where f is the force generated by the rope, and d is the stopping distance. solving for f, we have f= mgh/d How do we know what d, the stopping distance is? It depends on the rope stretch and the amount of rope out. call this alpha X l where l is the amout of rope out and alpha is the coefficient of stretch, arount 0.06 or so (6% stretch) now we have f=(mgh)/(alphaXl) or write this as (mg/alpha)X(h/l) we see that the force is proportional to h/l, the distance fallen over the rope length, also called fall factor. remember, this does not take into account other factors that increase the stopping distance, ie. dynamic belay, harness stretch, belayer movement, and so on. Plug in some numbers: 80 kg climber, 10m of rope out, 2m above last piece, making for a 4m fall with a 6% stretch we get ~ 5.2 kN if the belayer lets half a meter of rope slip through, we get ~ 2.8 kN. These seem ballpark correct, but now it gets real complicated: The above formula is for the force (really tension) in the rope during a catch, but the force on the top piece depends on friction of the rope through that piece in a very complicated way. For a frictionless biner, the force on the top piece is twice the force on the rope, due to the pully effect. For a top piece with 100% friction, the force is only equal to the force on the rope, BUT this invalidates the above formula, because now only the rope between the climber and the top piece can stretch, making for a factor 2 fall! So fiction on the top piece has two competing effects, on increases the force, and one decreases the force. Where these balance out is beyond me, and probably will only be settled by experiment! Close... kinda. One error is in the fact that the force of the climbing rope is not a constant. It's probably pretty darn close to acting like a spring  tension = constant(k)*displacement. Others, as you noted, involve negleting stretch in the entire rope. Try petzl's website for a better equation. I'll post a link to a previous thread that has a very accurate equation if I can find it. ~Adam~





paulraphael
Feb 28, 2005, 9:45 AM
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"The easy way to do it is with conservation of energy. The amount of "work" done by the rope must equal the energy the climber has just before the rope catches him. " and again, this is assuming that the climber does not continue to gain energy even even after the rope catches. which is a false assumption.





jw11733
Feb 28, 2005, 9:45 AM
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In reply to: Close, but you err in the fact that the force of the climbing rope is not a constant. It's probably pretty darn close to acting like a spring  tension = constant(k)*displacement. Try petzl's website for a better equation. I'll post a link to a previous thread that has a very accurate equation if I can find it. ~Adam~ Very good point! My formula would then only give the average force. This implies that the max force is greater than I calculate. (the max is always greater than or equal to the average) John





jw11733
Feb 28, 2005, 10:05 AM
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In reply to: "and again, this is assuming that the climber does not continue to gain energy even even after the rope catches. which is a false assumption. Also a good point. This is corrected by calling h (in mgh) the total fall distance, including rope stretch, not just the distance to the beginning of the catch, to account for the energy gained during the stopping phase. Another simple way to account for this is to add the weight of the climber (mg) to the final result for the average force. This is because, in my formula work = force X distance, the net force is really the force of the rope minus the weight of the climber, which I neglected. In my example this gives 5.2 kN + 0.8 kN = ~6kN.





paulraphael
Feb 28, 2005, 10:32 AM
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In reply to: This is corrected by calling h (in mgh) the total fall distance, including rope stretch, not just the distance to the beginning of the catch, to account for the energy gained during the stopping phase. Another simple way to account for this is to add the weight of the climber (mg) to the final result for the average force. This is because, in my formula work = force X distance, the net force is really the force of the rope minus the weight of the climber, which I neglected. In my example this gives 5.2 kN + 0.8 kN = ~6kN. well, the total rope stretch is actually an unknown quantity. it depends not only on the wieght of the climber but on the the forces generated as a result of the fall energy , which themselves depend upon the the total rope stretchso you have a twoway interdependence of these values. this is why it's a nonlinear equation and thus a very hard thing to predict accurately with a model. it's not actually the 100level physics problem that it first seems to be.





jw11733
Feb 28, 2005, 11:08 AM
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In reply to: well, the total rope stretch is actually an unknown quantity. it depends not only on the wieght of the climber but on the the forces generated as a result of the fall energy , which themselves depend upon the the total rope stretchso you have a twoway interdependence of these values. this is why it's a nonlinear equation and thus a very hard thing to predict accurately with a model. it's not actually the 100level physics problem that it first seems to be. True, that's why I said it gets real complicated, but its still interesting to do the simple problem. That's the only way fall factor "factors out" easily, which is instructive. You're correct, just as trenchdigger was saying, it depends critically on how the rope stretches. However, if it can be reasonably approximated by a spring f=k X displacement (hey trenchdigger, where is that link?), then it is not nonlinear, and it's possible to get a good estimate.





tradrenn
Mar 2, 2005, 8:13 PM
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In reply to: Doesn't Petzl's website have some sort of fall simulator? I think it's Petzl, and I think it will tell you what the forces are. Yes, last time I check.( 1.5 years ego)





shrubby
Mar 2, 2005, 9:46 PM
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i havent read all the other posts, but figuring out the force actually comes down to the impulse which is equal to the change in momentum which is equal to the force times the time it takes to stop. so basically you would have to know the time that you actually start to put stress on the rope until the time the you stop. this will give you the average force though and really that is the best you can do with equations in this situation without actually having a lot more information about the situation. the time, however, that you begin to stretch the rope and when you actually fall is rather small, but it makes a big difference in the amount of force a piece will feel. im a graduate student in physice by the way.





