telemarker Posted December 29, 2004 Share Posted December 29, 2004 I'm sure this has been explained ad nauseum somewhere here, but please someone indulge me the math in a free-falling mass. Tell me if this is calculated wrong or right. If I fall ten feet, and weigh 175 pounds, then is the amount of force loaded on the anchor calculated as such: 175 x 10/s/s? In other words, a fall of 10 feet takes me 2.5 seconds to decelerate, x 175, divided by 2.5 seconds squared? Someone clarify this for me please... Quote Link to comment Share on other sites More sharing options...
Alpinfox Posted December 29, 2004 Share Posted December 29, 2004 Your calculation is not correct. It's pretty tricky to calculate the peak force felt by an anchor in a real climbing scenario because of rope stretch, friction at the top biner, and other elasticity in the system. Keep in mind that you are not stopped instantaneously, but over some amount of time. That said, the value you are looking for is, roughly, the kinetic energy (KE) at the moment of impact which is found by 0.5mv^2. So you have to figure out what your velocity is first. Velocity is the square root of 2Gh where G is ~10m/s^2 and h is the height in meters that you fell. Dave's "Fall Force Machine" will come close to answering your question: www.alpinedave.com Click on fall force machine This one is nice because it allows you to change the variables of the fall such as rope stretch and distances between belayer, top piece of gear, and climber. I don't think he has included some variables (like friction on the biners) in his formula, but it's pretty close. If you are interested in a more ideal physical system and some of the math behind the system, check out this link: http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html Some other equations of interest: To determine how fast you are going after falling for a known time: v = g * t To determine how far you fall in a given time: d = 0.5 * g * t^2 Quote Link to comment Share on other sites More sharing options...
telemarker Posted December 29, 2004 Author Share Posted December 29, 2004 Thanks for the info... My other related question I'll put in the form of a scenario: I weigh 175 lbs. I fall 10 feet total. NOT taking rope stretch, rope slipage through the belay device, or friction through the biner, how much will I theoretically weigh as I come to a stop? Quote Link to comment Share on other sites More sharing options...
Mr._Natural Posted December 30, 2004 Share Posted December 30, 2004 175 unless you shit yourself and become lighter. Quote Link to comment Share on other sites More sharing options...
Alpinfox Posted December 30, 2004 Share Posted December 30, 2004 Telemarker, your question doesn't make any sense. You will still weigh 175lbs when you come to rest. The force applied to your anchor is determined by the distance you fall, the amount of rope out, and the stretchiness of the rope. You have to take into account the stretchiness of the rope because with a rope with no stretch (not physically possible) you would generate an infinite force (obviously not possible). The question I think you are asking would probably be better stated as, "Assume I tie my rope directly to the anchor, I weigh 80kg, and I fall 1.65m above my anchor with no intermediate pro (total fall distance would be ~10ft) and a rope of average stretchiness (say 8%). What force is applied to the anchor?" Well, the answer to that question, according to my calculation is 20.2kN (or roughly 4000lbs). Now here is the part I'm less sure about: If you were being belayed through the anchor with a perfect pulley between you and the belayer, the force felt by the anchor would be TWICE the above value. However, because biners aren't good pulleys, there is a lot of friction there and the anchor doesn't feel quite TWICE the force. I don't know what a reasonable coefficient of friction for a carabiner is, but Dave's fall force machine gives a total force reading of 32kN, so he is apparently assuming a coefficient of friction of 0.8. I'm no physicist, so I hope someone more knowledgable than myself will review this post and correct me if I'm wrong. Quote Link to comment Share on other sites More sharing options...
EWolfe Posted December 30, 2004 Share Posted December 30, 2004 Thanks for the info... My other related question I'll put in the form of a scenario: I weigh 175 lbs. I fall 10 feet total. NOT taking rope stretch, rope slipage through the belay device, or friction through the biner, how much will I theoretically weigh as I come to a stop? I could make the small leap it takes to understand what he meant Quote Link to comment Share on other sites More sharing options...
telemarker Posted December 30, 2004 Author Share Posted December 30, 2004 Yeah, cute! Reading it now, yikes! The thought came to me leading Crackmaster Lambada at Frenchman's. Towards the top of that climb, I was slotting stoppers that have a minimum breaking force of 4kn to 6kn, and I was wishing I could quickly calculate (roughly) how high I would be able to climb above the last stopper, fall, and have it hold (they were bomber placements). I was using a Beal rope, which has something like 8 to 10% elongation, and low impact force, and towards the end of the pitch, with more than 60 feet of rope out. AF, the site you alerted me to is good in answering this question, but is there a "quick and dirty" way to figure out how high one can climb above a small wire, or any piece for that matter, fall and have it hold? Quote Link to comment Share on other sites More sharing options...
Alpinfox Posted December 30, 2004 Share Posted December 30, 2004 Assuming a rope with 9% stretch, as long as the distance between you and your last piece of gear is less than or equal to 1/5 of the total amount of rope out between you and the belayer, you should not generate a force on the anchor higher than 6kN. So, if you have 20m (60ft) of rope out, you can go up to 4m (12ft) above your last piece, fall, and generate a 6kN force on the top piece of gear. How's that for a rough rule? Call it the 1/5th rule. Quote Link to comment Share on other sites More sharing options...
telemarker Posted December 30, 2004 Author Share Posted December 30, 2004 Intrestin'. Thanks Quote Link to comment Share on other sites More sharing options...
slaphappy Posted December 30, 2004 Share Posted December 30, 2004 Think success, not failure... Quote Link to comment Share on other sites More sharing options...
chriss Posted December 30, 2004 Share Posted December 30, 2004 Shouldn't be using the 8-10% number, for the rope stretch, to calculate impact force. The rope manufacturers report those numbers for the static loading test only. You can sometimes get the dynamic elongation from the fall test data. This must be less than 40% to pass the test. Typically they are in the high 30s. But even this number will depend on several other factors, should the fall be different for the test fall. And they always are. chris Quote Link to comment Share on other sites More sharing options...
telemarker Posted December 30, 2004 Author Share Posted December 30, 2004 Think success, not failure... This is a good point. However, I count myself successful when I can push beyond my comfort zone, even if I do lob off. This is what's nice about vertical basalt, it's much safer to fall, thus making hard on-sights or redpoints for me more feasible. Obviously, I won't be so quick to push my limits on ledgy, off vertical terrain. Quote Link to comment Share on other sites More sharing options...
slaphappy Posted December 31, 2004 Share Posted December 31, 2004 I'm not so much talkin about pushin yourself to the falling point, as much as I'm refering to concentrating on "climbing" not "how far can I climb between pro" (falling). Whatever works safely for you . Good to hear you're gettin on it . Quote Link to comment Share on other sites More sharing options...
EWolfe Posted December 31, 2004 Share Posted December 31, 2004 Dude! Like, I be falling MASS! You know? I, like TOTALLY try to make it a safe fall though, right? Clean 'til you know what it mean! Righteous on the falling mass, bro. Word. Mass Falls Mass Balls Quote Link to comment Share on other sites More sharing options...
RuMR Posted December 31, 2004 Share Posted December 31, 2004 175 unless you shit yourself and become lighter. no...the shit will stay in his pants and he will still effectively weigh 175 lbs... Quote Link to comment Share on other sites More sharing options...
drater Posted January 1, 2005 Share Posted January 1, 2005 175 unless you shit yourself and become lighter. no...the shit will stay in his pants and he will still effectively weigh 175 lbs... Unless he doesn't change his pants soon and he's climbing in Vantage in mid-summer. Then the water content will evaporate out of his shitty drawers leaving him somewhat lighter.....and stinkier. Quote Link to comment Share on other sites More sharing options...
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