



brunoschull
Mar 5, 2011, 2:02 AM
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Hi, First, I have posted lately about force calculation stuff. I have received overwhelmingly helpful feedback, and detailed explanatory comments. Thanks! I really appreciate that. There seems to be a great depth of experience and knowledge in this area on the site. Cool. I made a Excel spreadsheet to make these calculations easy to play with. Of course, they are approximations, but it's still interesting to learn about the possible forces in climbing situations. If would be great if somebody felt like checking the equations, or seeing of the numbers generated are reasonable. It's easy to make a mistake. Some notes: I attached two versions, new (.xlsx) and old (.xls) Excel. Don't know if they will both work. The main fields you can change are the mass, the distance fallen and length of rope deployed, and the K value. The K value table and equations are there for reference. The different forces for climber, protection, and belayer were from my best understanding of how these forces are mathematically related. Comments welcome, and please feel free to modify/use/change/discard as wanted. Safe climbing, Bruno

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Force calculator.xlsx
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Force calculator.xls
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notapplicable
Mar 5, 2011, 7:45 AM
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Interesting. Do your calculations assume a static belay? According to that spreadsheet, the only pieces of pro that could reliably hold a factor 2 fall would be Camalots larger than 0.75 and even then, they are only rated to 1 KN more than the 13.04 KN that the pro will be subjected to. C3's, all Metolius cams and pretty much all nuts only have a max rating of 10 KN. Edited to add: Of course your anchor should be equalized enough to prevent any one piece from taking all the force. Perhaps a more interesting consideration would be the 12.5 KN (or so) that the individual piece would be subjected to if you redirect off a single piece of the anchor, as I often see people do.
(This post was edited by notapplicable on Mar 5, 2011, 7:50 AM)





brunoschull
Mar 5, 2011, 8:53 AM
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Thanks for checking out the spreadsheet. The equation I used (or intended to use!) was the equation generated by rgold. The equation appears to be widely accepted. Of course, it's an approximation, but I think it's probably the best approximation we have. Of course, I could easily have made a mistake, so perhaps the forces are too highthis is why I am asking for help. Assuming for the moment that the calculator is working as intended, then your observation is interesting, and that's the kind of thing that got me into looking at the forces. All the more reason to do what you can to avoid a factor 2 fall in the mountains! Bruno





brunoschull
Mar 5, 2011, 9:01 AM
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Sorry to post againmissed you inquiry about the belay. The type of belay would seem to influence the forcesthere are some nice examples (don't know how accurate but they seem OK) on the Beal site comparing forces in different situations, including different belays. Devices like a Gri Gri appear to generate higher forces, while belay plates, old style figure eights, and so on, generate lower forces. I assume that the force equation takes into account some kind of dynamic belay approximation, although I am not sure what assumptions are incorporated (I will have to go back and read more, although perhaps others can directly answer that question). What I am sure of is that the equation takes into account the elasticity of the rope. Here's hoping this conversation might generate some more informative commentary from the community! Thanks again, Bruno





JimTitt
Mar 5, 2011, 9:22 AM
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The formula assumes a fixed belay. The type of belay device (and the amount of slip it allows) is the biggest single influence on impact force, more than the fall factor. To produce a theoretical force calculator at this time is impossible, the number of variables is bigger than you would think and most are not completely understood. At the very simplist you have to rely on a tested value for the rope since if anyone has ever worked this out thoeretically they are keeping it very close to their chests (and rightly so, a valuable bit of information it would be). So you might as well just do a lot of drop tests varying the parameters since you will have to anyway to get the complete rope characteristics and produce a mathematical model of the results which is the basis of the Beal calculator. Jim








jt512
Mar 5, 2011, 12:03 PM
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brunoschull wrote: Hi, First, I have posted lately about force calculation stuff. I have received overwhelmingly helpful feedback, and detailed explanatory comments. Thanks! I really appreciate that. There seems to be a great depth of experience and knowledge in this area on the site. Cool. I made a Excel spreadsheet to make these calculations easy to play with. Of course, they are approximations, but it's still interesting to learn about the possible forces in climbing situations. If would be great if somebody felt like checking the equations, or seeing of the numbers generated are reasonable. It's easy to make a mistake. Some notes: I attached two versions, new (.xlsx) and old (.xls) Excel. Don't know if they will both work. The main fields you can change are the mass, the distance fallen and length of rope deployed, and the K value. The K value table and equations are there for reference. The different forces for climber, protection, and belayer were from my best understanding of how these forces are mathematically related. Comments welcome, and please feel free to modify/use/change/discard as wanted. Safe climbing, Bruno You have successfully programmed the standard equation for impact force.* Your results agree with mine to within rounding error. If you want to test your results further, you can check them using my online impact force calculator. In my calculator, the first column of results uses rgold's equations, so your results should agree with those. To make the results comparable, set the "friction factor" in my calculator to 0.4, and make sure you enter the other numbers in the units indicated. *That said, you could use a little help with Excel programming. Instead of typing that whole megillah in cell B10, you could have just typed B8*.6, and in B9 you could have just entered B7+B10. For your table of "K values," try this: Type rgold's equation for converting rope impact force rating to rope modulus in cell E3. Don't type in 7.00 for the rope impact force; instead type D3. Then copy and paste that equation, without changing anything, to cell range E4 to E9, and watch the magic! Jay
(This post was edited by jt512 on Mar 5, 2011, 12:07 PM)





JimTitt
Mar 5, 2011, 12:33 PM
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Can´t you change K as well please! K is the bulk modulus which definately isn´t what you are using, the normal symbol is E. jim





brunoschull
Mar 5, 2011, 12:49 PM
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Thanks, thanks, and thanks again. Yes, it's true, I'm an Excel novice. I imagine my programing must be like the sound of nails on a chalk board to experienced Excel programmers. I will make the changes you suggest, but I probably won't post a new version. I'm just happy it generates relatively accurate numbers, accepting all of the approximations and assumptions. As I said, I encourage others who may be interested to take the rough model and modify as they want/need. Bruno





Rudmin
Mar 5, 2011, 2:59 PM
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Off the top of my head, a more robust force calculator could be split into 3 different systems, rope, climber belayer. You have the rope. They are definitely not perfect springs. I am sure that a little bit of testing could create a much better rope model, they are probably not linear, and they certainly have a lot of damping. You also have changing rope tension along the length as some of the load gets taken by friction of the holding piece of pro and other pieces that redirect the ropes path. For the climber, you have the initial velocity that they fall at, the relative position of their fall from the last piece (penduluming). Then you also have the spring between the climbers centre of gravity and the rope, that is their body, their harness, and the knot stretching. This probably reduces peak forces too. End then you have the belayer, how much rope they let slip, whether they are belaying off of themselves or the anchor. If off themselves, then they limit the force at their end to their body weight usually. At least for some distance of travel. I think that if you correctly modeled all of these factors you could get a pretty realistic peak force.





notapplicable
Mar 5, 2011, 7:20 PM
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The math behind these calculators is way over my head but I have a, possibly stupid, question for you guys. Are there any adjustments made for when the FF = 2 and you do not have a top piece of gear and the resulting pulley effect? Is that something that can, or even needs to be adjusted for? It seems like the impact force should actually go down when you go from a FF of 1.9 to a FF of 2.0





jt512
Mar 5, 2011, 11:26 PM
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notapplicable wrote: The math behind these calculators is way over my head but I have a, possibly stupid, question for you guys. Are there any adjustments made for when the FF = 2 and you do not have a top piece of gear and the resulting pulley effect? Is that something that can, or even needs to be adjusted for? It seems like the impact force should actually go down when you go from a FF of 1.9 to a FF of 2.0 That's not even a remotely stupid question. The rgold equation for the impact force on the belayer assumes that the rope is passing through a top anchor. Therefore, if you use that equation with a fall factor of two the result you get for the impact force on the belayer will be the limit of the force as the fall factor approaches 2, assuming that the rope passes through a top anchor. However, a true factor2 fall can only happen if the rope does not pass through a top anchor. Therefore, the result will be correct (given the model) for a fall factor of 1.999, but not an actual factor2 fall. In the case of a real factor2 fall the model will give the correct force (given the model assumptions) for the climber, but the forces on the belayer and the anchor will essentially be equal to the force on the climber. Jay





notapplicable
Mar 6, 2011, 11:12 AM
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Well, thats what I was thinking but with a climber weight of 165 lbs. and a rope rating of 8.5 KN, you get nonfriction adjusted numbers of... On climber 8.64 On belayer 5.76 8.64? On anchor 14.40 8.64? If that is correct, or even close to correct, that is a profound reduction in force exerted on the gear placement(s) holding the fall. The obvious trade off being that the belayer is going to have to overcome 1.5 times the force they would in a redirected fall. Which is not an inconsequential increase if you are talking about a tube style device. Exactly what influence those numbers should have on the decisions I make about belaying off a multipitch anchor is anything but clear though, at least to me, because they assume a static belay, which would be near impossible in either scenario. Let the game of highly variable and still mostly theoretical tradeoffs begin...





notapplicable
Mar 6, 2011, 11:25 AM
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I will say though, discussions about multipitch anchor and belay arrangements are pretty interesting to me and I have followed most that have gone on during my time on this site. I've also read several from the archive and this is the first time I remember seeing this (IMO significant) distinction made between the theoretical peak force on the gear seen in a FF1.9 scenario with a redirect and an actual FF2 scenario where there is no redirect. This may old news to everyone else but I'm inclined to think this supports my preference for belaying directly off the harness/anchor. And more specifically, for NOT redirecting off a single piece of the anchor.





hafilax
Mar 6, 2011, 11:57 AM
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I made some plots at some point of the force on the belayer and force on the anchor for various high FF scenarios. I'm not sure where that would be at this point. My take home was that if you're going to redirect in a fall onto the anchor or a low first piece situation then that anchor has to be really strong. If you're going to catch the fall directly be prepared to be hit hard and hold on for dear life but the force on the anchor will be less. My question at that time was, does the rope recover at all between a piece pulling and it catching again on the next? If it doesn't recover at all then there is no benefit to placing gear that pulls. If it does recover then some energy is removed from the fall when a piece pulls. The next approximation for a rope would be to add some damping which would essentially be the same as the damped harmonic oscillator which has pretty simple solutions especially for the critically damped situation. This type of damping depends on the velocity of the mass. The data I've seen made it look like a rope is slightly underdamped since the falling object can bounce a bit. Friction like damping is another option but the solutions are a little messier since the equations become nonlinear with the friction force switching direction at the extremes of the motion. At one point I started working through the math of the damped spring but I can't remember how far I made it. It should be pretty easy to calculate the forces involved.





notapplicable
Mar 6, 2011, 5:03 PM
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hafilax wrote: My question at that time was, does the rope recover at all between a piece pulling and it catching again on the next? If it doesn't recover at all then there is no benefit to placing gear that pulls. If it does recover then some energy is removed from the fall when a piece pulls. The reading I have done makes it sound like the rope does recover a significant amount of it's elasticity almost instantly. I think Jay has commented on that more than once. Which is nice, at least when you are talking about a series of marginal placements where the second may have a better chance of holding after the first piece "slows you down" before pulling. That said, I feel there are some rather severe trade offs to the practice of placing a "ripper piece" when it is the first and only piece. Even if the reductions in peak impact forces are large. The most important being that you will begin to catch the fall locked off in the completely wrong direction relative to the eventual direction of pull and you will likely be lifted towards that first piece for the fraction of a second that it holds. At which point you will be slammed down on to an essentially static anchor, whipped around as the direction of pull changes and you will be both braking in the wrong direction and your hand will be in the wrong orientation to provide the greatest mechanical advantage for locking off at your chest, instead of behind the hip. I think the belayer stands a much better chance of being able to control the rope, even with the increased impact, if you just take it straight on to the harness/anchor and have prepared for that from the beginning. In that interest, I typically will belay palms up until a second piece has been placed off of the anchor, that way if the first piece blows I will have that small advantage when I reverse braking directions.
(This post was edited by notapplicable on Mar 6, 2011, 6:28 PM)





cmceown
Apr 27, 2011, 7:34 PM
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There's a really good discussion on the Beal site on the impact of rope drag on fall factor: http://www.bealplanet.com/sport/anglais/facteurdechute.php The basic thesis is that rope drag across the carabiners used for protection will increase the fall factor significantly by effectively shortening the amount of rope available to absorb impact forces. There's also a great article on the paci website regarding the maximum braking force of various belay devices: http://www.paci.com.au/downloads_public/equipment/19_Belay_Device_Theory.pdf While the latter requires a bit of reading between the lines, it suggests that the maximum braking force that can be applied by a normal human being on an ATC style device is around 2kN (before the rope starts to slip through the device). This is based on the brake hand being able to exert around 300N of force onto the device (about 30kg/66lbs). If you assume there is no rope drag in the system (except friction on the top anchor), and that the ratio of force felt by the climber to force felt by the belayer is about 60:40, then with an ATC belay, the maximum force that could be exerted on the climber is about 3kN  therefore the peak force seen by the anchor is a mere 5kN. Forces felt by the belayer in excess of this would cause rope slip in the device. An autolocking device, on the other hand, which might have a peak braking force of 5 or 6kN before seeing rope slip would put a theoretical peak force of up to 12.515kN (again, based on the maximum braking force exerted, not on the force of the fall). In the autolocking case, this analysis depends on the belayer not lifting off the ground  I'd be interested to know what sort of force needs to be applied to the belayer to make this happen  I'm guessing it's on the order of 1kN for most people, and maybe half of this if you're giving a dynamic belay. I'd suspect that the dampening effect provided by the lifting belayer is also quite significant in reducing peak forces  would love to know how to calculate the force reduction for a 1 metre lift, say ... So on this basis a nonlocking belay device combined with a lifting belayer / dynamic belaying technique would seem to keep peak forces on the anchor well below 1012kN and probably explains why most of us don't have ruptured kidneys.





cmceown
Apr 27, 2011, 9:28 PM
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Oops ... should have credited Jim Titt with the article on the PACI website ... hope I paraphrased correctly!





