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davemacd76
Aug 5, 2009, 5:45 AM
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Hey Hey,
This came up the other day...
My friend was rappelling from a bolted anchor. It had to rap rings about a foot apart. He was concerned about rapping through both because he thought he would be creating the death triangle, and magnifying forces.
I argued that although it does increase forces on anchors, since he was only rappelling it would be fine, and justified because then he would not have to rappel off a single anchor. I really think I was right, I was just curious to get some input on the subject.
Thanks again.
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bennydh
Aug 5, 2009, 6:29 AM
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How about a really long sling drooped like an 'M' shape through both bolts to reduce the death triangular forces, and a little texas rope trick. No death triangle, leaves no trace.
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jt512
Aug 5, 2009, 6:31 AM
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If you rap from only one of the bolts, the load on that bolt will be 100% of your weight. If you rap through both bolts, then the load on each bolt will be about 70% of your weight. Thus by rapping through both bolts you reduce the load per bolt and gain redundancy.
The force multiplication of the American Triangle is relative to slinging each of the two bolts independently, in which case, if you ran your rope through the two slings, each bolt would bear 50% of your weight.
Jay
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majid_sabet
Aug 5, 2009, 7:02 AM
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davemacd76 wrote: Hey Hey, This came up the other day... My friend was rappelling from a bolted anchor. It had to rap rings about a foot apart. He was concerned about rapping through both because he thought he would be creating the death triangle, and magnifying forces. I argued that although it does increase forces on anchors, since he was only rappelling it would be fine, and justified because then he would not have to rappel off a single anchor. I really think I was right, I was just curious to get some input on the subject. Thanks again.
when you are rappelling, the load on the anchor is pretty much static meaning the peak force on the anchor is about 1KN assuming you are 100 kg. If you do not monkey around like a combat commando while rapping, the peak force should stay @ 1KN.
if you run the rope thru few bolts then peak force will be divided to each anchor.
Now , if you were belay a leader on the same anchor then you are ( belayer) under a static condition which could turn in to dynamic situation and that could turn the same peak force on the anchor from 1KN to 20KN
once you or the leader takes fall on it.
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bennydh
Aug 5, 2009, 7:15 AM
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majid_sabet wrote: davemacd76 wrote: Hey Hey, This came up the other day... My friend was rappelling from a bolted anchor. It had to rap rings about a foot apart. He was concerned about rapping through both because he thought he would be creating the death triangle, and magnifying forces. I argued that although it does increase forces on anchors, since he was only rappelling it would be fine, and justified because then he would not have to rappel off a single anchor. I really think I was right, I was just curious to get some input on the subject. Thanks again. when you are rappelling, the load on the anchor is pretty much static meaning the peak force on the anchor is about 1KN assuming you are 100 kg. If you do not monkey around like a combat commando while rapping, the peak force should stay @ 1KN. if you run the rope thru few bolts then peak force will be divided to each anchor. Now , if you were belay a leader on the same anchor then you are ( belayer) under a static condition which could turn in to dynamic situation and that could turn the same peak force on the anchor from 1KN to 20KN once you or the leader takes fall on it.
Majid, you are a master of clarity and the engrish language. Could you please add a diagram with multi-colored arrows so that I may better understand your post?
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majid_sabet
Aug 5, 2009, 7:44 AM
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bennydh wrote: majid_sabet wrote: davemacd76 wrote: Hey Hey, This came up the other day... My friend was rappelling from a bolted anchor. It had to rap rings about a foot apart. He was concerned about rapping through both because he thought he would be creating the death triangle, and magnifying forces. I argued that although it does increase forces on anchors, since he was only rappelling it would be fine, and justified because then he would not have to rappel off a single anchor. I really think I was right, I was just curious to get some input on the subject. Thanks again. when you are rappelling, the load on the anchor is pretty much static meaning the peak force on the anchor is about 1KN assuming you are 100 kg. If you do not monkey around like a combat commando while rapping, the peak force should stay @ 1KN. if you run the rope thru few bolts then peak force will be divided to each anchor. Now , if you were belay a leader on the same anchor then you are ( belayer) under a static condition which could turn in to dynamic situation and that could turn the same peak force on the anchor from 1KN to 20KN once you or the leader takes fall on it. Majid, you are a master of clarity and the engrish language. Could you please add a diagram with multi-colored arrows so that I may better understand your post?
sorry, I assume you were not retard after readying jay's but here, this image may help. also I speak 4 different languages and lived and traveled thru 30+ countries and changed 6+ passports.How about you ?
working on your half ass Spanish pinche loco ?
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bill413
Aug 5, 2009, 12:38 PM
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If you actually rig an ADK, you have a fully complete triangle - one sling looped through both bolts, one leg going straight across, the other two legs coming down to your anchor point. Bad.
If you have enough slingage to rig that, you should instead rig the "magic-X" on the same bolts. Good.
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j_ung
Aug 5, 2009, 1:48 PM
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majid_sabet wrote: bennydh wrote: Majid, you are a master of clarity and the engrish language. Could you please add a diagram with multi-colored arrows so that I may better understand your post? sorry, I assume you were not retard after readying jay's but here, this image may help. also I speak 4 different languages and lived and traveled thru 30+ countries and changed 6+ passports.How about you ? working on your half ass Spanish pinche loco ?
Good stuff, Majid.
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markpaul
Aug 5, 2009, 2:11 PM
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It is a ADK but what you're forgetting is that a ADK is only dangerous when the bottom angle is really obtuse. The more acute the bottom angle the less force each anchor takes. As you descend the the first few feet the angle will already be within a reasonable range that you will not be placing such a huge load on the anchors.
So if you read below, 100 lbf on your 'ADK' will only place 82 lbf per anchor at 30 degrees bottom angle. As you approach 0 degrees on the bottom, you will approach 50 lbf per anchor on top.
Load per anchor based on a central, perpendicular application of 100 lbf of force
Bottom angle Load per anchor (V arrangement) Load per anchor (triangle arrangement)
0° 50 lbf 50 lbf
30° 52 lbf 82 lbf
60° 60 lbf 100 lbf
90° 70 lbf 130 lbf
120° 100 lbf 190 lbf
150° 190 lbf 380 lbf
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gimmeslack
Aug 5, 2009, 2:13 PM
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Color me stoopit - I'm REALLY confused...
OP is asking about a bolted rap station and we're discussing leaving slings, texas rope trick, using only one of the bolts, and other shennanigans.
WTF????
PUT YOUR ROPE THROUGH BOTH RINGS AND RAP!!
(This post was edited by gimmeslack on Aug 5, 2009, 2:15 PM)
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jt512
Aug 5, 2009, 3:32 PM
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markpaul wrote: It is a ADK but what you're forgetting is that a ADK is only dangerous when the bottom angle is really obtuse. The more acute the bottom angle the less force each anchor takes. As you descend the the first few feet the angle will already be within a reasonable range that you will not be placing such a huge load on the anchors. So if you read below, 100 lbf on your 'ADK' will only place 82 lbf per anchor at 30 degrees bottom angle. As you approach 0 degrees on the bottom, you will approach 50 lbf per anchor on top. Load per anchor based on a central, perpendicular application of 100 lbf of force Bottom angle Load per anchor (V arrangement) Load per anchor (triangle arrangement) 0° 50 lbf 50 lbf 30° 52 lbf 82 lbf 60° 60 lbf 100 lbf 90° 70 lbf 130 lbf 120° 100 lbf 190 lbf 150° 190 lbf 380 lbf
The figure for 0 degrees for the AT is wrong. It should be 70 lbf, not 50.
Jay
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markpaul
Aug 5, 2009, 3:35 PM
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70? How? 0 degrees would be both anchors on top of each other or you so far away that you're approaching a 0 degree angle. Please explain how you came up with 70 lbf.
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Carnage
Aug 5, 2009, 4:32 PM
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it has to do with the fact that in an ADK setup, the anchor is not just experiencing a downward force from the climber, but also a pull towards the other bolt. The net force on the bolt is somewhere in between the two@ around 70 lbf
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markpaul
Aug 5, 2009, 5:17 PM
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Carnage wrote: it has to do with the fact that in an ADK setup, the anchor is not just experiencing a downward force from the climber, but also a pull towards the other bolt. The net force on the bolt is somewhere in between the two@ around 70 lbf
But at 0 degrees you do not have a horizontal vector, you are now dealing only with a downward vector so there are no other forces in play (hence 50 lbf per anchor point). The resultant force comes from the addition of your two vectors, once they are in line (at 0 degrees) you don't get any extra load from anywhere. This of course is impossible as you will never reach 0 degrees, this is just a limit that will never be reached but approached the further you move down the rope or the closer your anchor points. So at 0 degrees you are at 50 lbf per anchor point as all your force is directly downward.
Your 70 lbf will be correct but at some angle greater then 0 and less than 30...
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Carnage
Aug 5, 2009, 5:28 PM
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markpaul wrote: Carnage wrote: it has to do with the fact that in an ADK setup, the anchor is not just experiencing a downward force from the climber, but also a pull towards the other bolt. The net force on the bolt is somewhere in between the two@ around 70 lbf But at 0 degrees you do not have a horizontal vector, you are now dealing only with a downward vector so there are no other forces in play (hence 50 lbf per anchor point). The resultant force comes from the addition of your two vectors, once they are in line (at 0 degrees) you don't get any extra load from anywhere. This of course is impossible as you will never reach 0 degrees, this is just a limit that will never be reached but approached the further you move down the rope or the closer your anchor points. So at 0 degrees you are at 50 lbf per anchor point as all your force is directly downward. Your 70 lbf will be correct but at some angle greater then 0 and less than 30...
i did the math, i see the 0 horizontal force. you're right i guess.
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ptlong
Aug 5, 2009, 5:43 PM
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Carnage wrote: markpaul wrote: Carnage wrote: it has to do with the fact that in an ADK setup, the anchor is not just experiencing a downward force from the climber, but also a pull towards the other bolt. The net force on the bolt is somewhere in between the two@ around 70 lbf But at 0 degrees you do not have a horizontal vector, you are now dealing only with a downward vector so there are no other forces in play (hence 50 lbf per anchor point). The resultant force comes from the addition of your two vectors, once they are in line (at 0 degrees) you don't get any extra load from anywhere. This of course is impossible as you will never reach 0 degrees, this is just a limit that will never be reached but approached the further you move down the rope or the closer your anchor points. So at 0 degrees you are at 50 lbf per anchor point as all your force is directly downward. Your 70 lbf will be correct but at some angle greater then 0 and less than 30... i did the math, i see the 0 horizontal force. you're right i guess.
Try doing it again.
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jt512
Aug 5, 2009, 5:53 PM
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markpaul wrote: 70? How? 0 degrees would be both anchors on top of each other or you so far away that you're approaching a 0 degree angle. Please explain how you came up with 70 lbf.
A triangle cannot contain an angle of zero degrees, so we can either view the force at zero degrees as either a theoretical value, the limit of the force as the angle approaches zero; or a practical value, the force when the angle is very close to zero. Either way, the answer is 70 lbf. The practical interpretation is applicable to the OP's question, pertaining to treating a rappel with the rope running through two rap rings as an American triangle. Except at the very start of the rappel, the strands of the rope will be nearly parallel, so the force on each bolt, ignoring friction and lack of smoothness in the rappel, would be 70 lbf.
Jay
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rgold
Aug 5, 2009, 5:57 PM
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Markpaul's single bolt load value for a zero degree angle at the bottom of the "v" is at least misleading, in the sense that it assumes the anchor configuration changes as the sling angle approaches zero.
Of course, there can never have a zero degree angle there if there are two bolts side-by-side, so the question ought to be, what does the individual bolt load approach as the sling angle approaches zero. If L is the load applied at the bottom of the "v," then the bolt load does not approach L/2, as markpaul suggests, but rather (L/2)*sqrt(2) assuming no frictional effects. In other words, with no friction at the rings and a 100 lbf load at the "v," the load on an individual bolt can approach roughly 70 lbf, but cannot get any lower (markpaul's table suggests 50 lbf for this case). This isn't merely hypothetical, since it describes what happens if you thread the rope through the rings for rappelling---the rope forms an ADT whose sling angle approaches zero as the rappel progresses.
Calculations like these, which are widely promulgated, do not take into account the friction of the slings around the anchor points. When the anchor points are pitons or bolt hangers, the friction might be quite significant, but I've never seen any measurements. It has become "common knowledge" that the friction of climbing rope around a biner reduces tension to about 2/3 its value on the load side, but in an ADT the rope does not make a 180 degree angle and so this friction coefficient would presumably not apply.
For anyone interested in generating some numbers, here is a formula: L is the load applied to the bottom of the "v," theta is the "v"-angle, r, a number between 0 and 1 inclusive, is the friction effect factor, with r=1 corresponding to no friction and r=0 corresponding to so much friction there is no tension in the horizontal strand (so essentially the v-configuration rather than the triangle configuration). The formula calculates B, the load on a single bolt.
I've checked this result but would be grateful for corrections.
By the way, for calculational purposes, rappelling loads will always be at least double body weight and usually more than that, even if you are careful.
Edit: Sorry for the duplication: Jay got done saying much the same thing while I was typing this.
(This post was edited by rgold on Aug 5, 2009, 6:00 PM)
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ptlong
Aug 5, 2009, 6:05 PM
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Carnage wrote: markpaul wrote: Carnage wrote: it has to do with the fact that in an ADK setup, the anchor is not just experiencing a downward force from the climber, but also a pull towards the other bolt. The net force on the bolt is somewhere in between the two@ around 70 lbf But at 0 degrees you do not have a horizontal vector, you are now dealing only with a downward vector so there are no other forces in play (hence 50 lbf per anchor point). The resultant force comes from the addition of your two vectors, once they are in line (at 0 degrees) you don't get any extra load from anywhere. This of course is impossible as you will never reach 0 degrees, this is just a limit that will never be reached but approached the further you move down the rope or the closer your anchor points. So at 0 degrees you are at 50 lbf per anchor point as all your force is directly downward. Your 70 lbf will be correct but at some angle greater then 0 and less than 30... i did the math, i see the 0 horizontal force. you're right i guess.
If it makes you feel any better, whoever did the wikipedia page for the ADT also got the math wrong.
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Carnage
Aug 5, 2009, 6:26 PM
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ptlong wrote: Carnage wrote: markpaul wrote: Carnage wrote: it has to do with the fact that in an ADK setup, the anchor is not just experiencing a downward force from the climber, but also a pull towards the other bolt. The net force on the bolt is somewhere in between the two@ around 70 lbf But at 0 degrees you do not have a horizontal vector, you are now dealing only with a downward vector so there are no other forces in play (hence 50 lbf per anchor point). The resultant force comes from the addition of your two vectors, once they are in line (at 0 degrees) you don't get any extra load from anywhere. This of course is impossible as you will never reach 0 degrees, this is just a limit that will never be reached but approached the further you move down the rope or the closer your anchor points. So at 0 degrees you are at 50 lbf per anchor point as all your force is directly downward. Your 70 lbf will be correct but at some angle greater then 0 and less than 30... i did the math, i see the 0 horizontal force. you're right i guess. Try doing it again.
what did i mess up? i got tan(0)=x/50 (x being the horizontal force. since tan(0)=0 x=0
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ptlong
Aug 5, 2009, 6:41 PM
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Carnage wrote: what did i mess up? i got tan(0)=x/50 (x being the horizontal force. since tan(0)=0 x=0
You're forgetting that the rope between the bolts is under the same tension.
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markpaul
Aug 5, 2009, 6:45 PM
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ptlong wrote: Carnage wrote: what did i mess up? i got tan(0)=x/50 (x being the horizontal force. since tan(0)=0 x=0 You're forgetting that the rope between the bolts is under the same tension.
Thats the whole point, there is no rope between the points. 0 degrees means anchors in same location. This is not realistic but a limit.
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rgold
Aug 5, 2009, 6:45 PM
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Carnage, perhaps you aren't representing the physical situation correctly? If there is no friction at the anchor rings, then the horizontal force is equal to the tension in the sling strand because of the pulley effect.
Markpaul, you seem to misunderstand what a limit means. In all the configurations leading to the limit, there is rope between the two bolts, and this fact determines the limit. You want the limiting value to be the same as the value associated with the limiting configuration, and this is not true.
Moreover, as Jay and I both mentioned, it is the limiting value that has practical significance for the ADT configuration.
(This post was edited by rgold on Aug 5, 2009, 6:53 PM)
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ptlong
Aug 5, 2009, 6:56 PM
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markpaul wrote: ptlong wrote: Carnage wrote: what did i mess up? i got tan(0)=x/50 (x being the horizontal force. since tan(0)=0 x=0 You're forgetting that the rope between the bolts is under the same tension. Thats the whole point, there is no rope between the points. 0 degrees means anchors in same location. This is not realistic but a limit.
Rgold may correct me on this one, but I don't think that what you're talking about is a limit. In your interpretation of the zero angle case the triangle ceases to exist.
If you instead look at successively smaller angles you will find that the force converges on the result L/2 * sqrt(2). So even with your interpretation it would still be true that for 0.000000000001 degrees the force would be very close to 70 lbs.
edit: oops! Rgold beat me to it.
(This post was edited by ptlong on Aug 5, 2009, 6:57 PM)
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ptlong
Aug 5, 2009, 7:07 PM
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rgold wrote: For anyone interested in generating some numbers, here is a formula: L is the load applied to the bottom of the "v," theta is the "v"-angle, r, a number between 0 and 1 inclusive, is the friction effect factor, with r=1 corresponding to no friction and r=0 corresponding to so much friction there is no tension in the horizontal strand (so essentially the v-configuration rather than the triangle configuration). The formula calculates B, the load on a single bolt.
How about r = exp[-µ*(pi-theta)/2], where µ = coefficient of friction ?
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