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I was reading an article about tyrolean traverses,
And it struck me that, say you sling a 120cm sling over a block, its a big block, and the sling only just goes over, sits nicely around the block, and there isn't much slack at all, I know it's incredibly hard to give an answer, but by clipping into that edge of sling that runs at almost 180 degrees, is the weight multiplied increasing wear and cut chance? 8mm slings make me wonder.
Solution, use a bigger sling to reduce the angle or join two together?
And it struck me that, say you sling a 120cm sling over a block, its a big block, and the sling only just goes over, sits nicely around the block, and there isn't much slack at all, I know it's incredibly hard to give an answer, but by clipping into that edge of sling that runs at almost 180 degrees, is the weight multiplied increasing wear and cut chance? 8mm slings make me wonder.
Solution, use a bigger sling to reduce the angle or join two together?
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In reply to zimpara:
It's actually quite straightforward to answer http://www.ropebook.com/information/vector-forces
Longer slings are one answer.
jk
It's actually quite straightforward to answer http://www.ropebook.com/information/vector-forces
Longer slings are one answer.
jk
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In reply to zimpara:
If the slings are the same length, surely you'd need more than two to improve the situation?
If the slings are the same length, surely you'd need more than two to improve the situation?
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In reply to humptydumpty:
Exactly why joining two slings together never makes it long enough!
Jesus christ JK! 5 times the weight if 180degrees... hoo hoo hoo, scary! And then double that if you're being lowered off. :-/
Exactly why joining two slings together never makes it long enough!
Jesus christ JK! 5 times the weight if 180degrees... hoo hoo hoo, scary! And then double that if you're being lowered off. :-/
Post edited at 12:10
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In reply to zimpara:
Not that good with this technical stuff, but could someone explain how a weight of 100kg could become several tonnes on the anchor, just by changing the direction of pull?
Not that good with this technical stuff, but could someone explain how a weight of 100kg could become several tonnes on the anchor, just by changing the direction of pull?
In reply to purplemonkeyelephant:
If the anchors are above the load then the only force on the slings is the weight of the load and they share it. The simple way to think about this is as you move the anchors apart the slings still have to support the load but additionally they're pulling on each other.
To zimpara: That's at 170deg. 180deg isn't practically achievable but if it were then the sling tension would infinite.
jk
If the anchors are above the load then the only force on the slings is the weight of the load and they share it. The simple way to think about this is as you move the anchors apart the slings still have to support the load but additionally they're pulling on each other.
To zimpara: That's at 170deg. 180deg isn't practically achievable but if it were then the sling tension would infinite.
jk
Post edited at 12:38
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In reply to zimpara:
If the angle was 180 degrees the sling tension would be infinity times the weight not 5. However the angle would never be 180 degrees. Mostly a block is narrower at the base so if you can work a sling round it there ends up being some slack. Even if there wasn't any slack the stretch would produce some under load. The final angle when slack was taken up and the sling had stretched would need to be more than 168 degrees to give x5 tension. If you weigh 150 kg (not many climbers would even if laden with gear) the force needed to support you against gravity is about 1.5 kN so with x5 the tension would be about 7.5 kN. The slings are typically rated to 22 kN and I think it would be impractical to get a setup approaching 168 degrees loaded so unless you are very fat or carry enormous loads I wouldn't worry too much about it.
More of a worry would be sharp edges.
> Jesus christ JK! 5 times the weight if 180degrees... hoo hoo hoo, scary!
If the angle was 180 degrees the sling tension would be infinity times the weight not 5. However the angle would never be 180 degrees. Mostly a block is narrower at the base so if you can work a sling round it there ends up being some slack. Even if there wasn't any slack the stretch would produce some under load. The final angle when slack was taken up and the sling had stretched would need to be more than 168 degrees to give x5 tension. If you weigh 150 kg (not many climbers would even if laden with gear) the force needed to support you against gravity is about 1.5 kN so with x5 the tension would be about 7.5 kN. The slings are typically rated to 22 kN and I think it would be impractical to get a setup approaching 168 degrees loaded so unless you are very fat or carry enormous loads I wouldn't worry too much about it.
More of a worry would be sharp edges.
Post edited at 12:54
In reply to zimpara:
The only relatively common climbing situation I can think of where the pulley effect on the anchor results in approx 2x force on it is when bottom roping (i.e. belayer on ground, rope going up to anchor and back down to climber).
So assuming that's what you mean, then it's not just when lowering that you get the 2x effect, it's at all times. The worst case scenario would be the climber falling with some slack in the system and then shock-loading the anchor. On the plus side, you'd always have at least half the climb length of rope out so that would help significantly to soften the shock.
> ... And then double that if you're being lowered off. :-/
The only relatively common climbing situation I can think of where the pulley effect on the anchor results in approx 2x force on it is when bottom roping (i.e. belayer on ground, rope going up to anchor and back down to climber).
So assuming that's what you mean, then it's not just when lowering that you get the 2x effect, it's at all times. The worst case scenario would be the climber falling with some slack in the system and then shock-loading the anchor. On the plus side, you'd always have at least half the climb length of rope out so that would help significantly to soften the shock.
In reply to harold walmsley:
In fairness the question referred to anchors for a Tyrolean traverse so the applied load is already a significant multiple of body weight, it's one area of 'climbing' where relying on over engineered equipment to mitigate poor rigging design probably isn't going to keep you out of trouble indefinitely.
jk
> The slings are typically rated to 22 kN and I think it would be impractical to get a setup approaching 168 degrees loaded so unless you are very fat or carry enormous loads I wouldn't worry too much about it.
In fairness the question referred to anchors for a Tyrolean traverse so the applied load is already a significant multiple of body weight, it's one area of 'climbing' where relying on over engineered equipment to mitigate poor rigging design probably isn't going to keep you out of trouble indefinitely.
jk
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In reply to zimpara:
As an over-simplified rule of thumb, when you have a "V" in a sling or fixed tat (for example, the "V" created by clipping a karabiner into a sling that is wrapped around a boulder and then weighting the karabiner), the angle of the "V" should never exceed the angle between your thumb and index finger. Otherwise, you are amplifying the forces excessively due to an annoying thing called trigonometry.
As an over-simplified rule of thumb, when you have a "V" in a sling or fixed tat (for example, the "V" created by clipping a karabiner into a sling that is wrapped around a boulder and then weighting the karabiner), the angle of the "V" should never exceed the angle between your thumb and index finger. Otherwise, you are amplifying the forces excessively due to an annoying thing called trigonometry.
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In reply to zimpara:
All this talk of infinite force multiplication sounds like witchcraft to me. I bet you all weigh the same as duck.
All this talk of infinite force multiplication sounds like witchcraft to me. I bet you all weigh the same as duck.
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In reply to purplemonkeyelephant:
Its the sort of witchcraft Newton worked out centuries ago. Shame people can't be bothered to understand it. If you do think about this sort of thing rationally it can help you to place better gear and you can be more relaxed about some of the more lurid scaremongering.
Its the sort of witchcraft Newton worked out centuries ago. Shame people can't be bothered to understand it. If you do think about this sort of thing rationally it can help you to place better gear and you can be more relaxed about some of the more lurid scaremongering.
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In reply to purplemonkeyelephant:
Try this simple experiment. Tie a bit of rope / cord / washing line to something solid. Hold the other end with the line just taught. Now get someone to pull the centre of the line whilst you try to keep the line straight. See how much harder you have to pull to keep the line straight than the other person does to pull it out of shape. That's the force multipier.
Try this simple experiment. Tie a bit of rope / cord / washing line to something solid. Hold the other end with the line just taught. Now get someone to pull the centre of the line whilst you try to keep the line straight. See how much harder you have to pull to keep the line straight than the other person does to pull it out of shape. That's the force multipier.
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In reply to zimpara:
it's well worth knowing the mechanics of this type of force multiplier, and IMHO, I would tend to avoid that type of very-tight sling placement.
not got a long sling - use the rope.
it's well worth knowing the mechanics of this type of force multiplier, and IMHO, I would tend to avoid that type of very-tight sling placement.
not got a long sling - use the rope.
In reply to Xharlie:
What a silly rule of thumb
Depending on which hand and how I hold my hand that could be anything up to 95 degrees!
> As an over-simplified rule of thumb, when you have a "V" in a sling or fixed tat (for example, the "V" created by clipping a karabiner into a sling that is wrapped around a boulder and then weighting the karabiner), the angle of the "V" should never exceed the angle between your thumb and index finger. Otherwise, you are amplifying the forces excessively due to an annoying thing called trigonometry.
What a silly rule of thumb
Depending on which hand and how I hold my hand that could be anything up to 95 degrees!
In reply to timjones:
Seems reasonable. I thought anything under 120 was fairly okish, but to aim for 60.
> Depending on which hand and how I hold my hand that could be anything up to 95 degrees!
Seems reasonable. I thought anything under 120 was fairly okish, but to aim for 60.
Post edited at 15:40
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In reply to timjones:
Which is fine. 120deg is the break-even point where you start increasing the load over using a single anchor (while potentially introducing redundancy so still not all bad).
jk
> What a silly rule of thumb
> Depending on which hand and how I hold my hand that could be anything up to 95 degrees!
> Depending on which hand and how I hold my hand that could be anything up to 95 degrees!
Which is fine. 120deg is the break-even point where you start increasing the load over using a single anchor (while potentially introducing redundancy so still not all bad).
jk
Post edited at 15:43
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In reply to jkarran:
I'd generally agree with that, though I try and keep below 90 degrees where possible.
It just seems bizarre to use something that can vary so widely rather than just stating an angle in degrees.
> Which is fine. 120deg is the break-even point where you start increasing the load over using a single anchor (while potentially introducing redundancy so still not all bad).
I'd generally agree with that, though I try and keep below 90 degrees where possible.
It just seems bizarre to use something that can vary so widely rather than just stating an angle in degrees.
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In reply to timjones:
The point is that it's a rule of thumb which is easy to teach, easy to remember and easy to implement when you accidentally forgot to clip your protractor onto your harness...
... because you unclipped your trad. protractor so you wouldn't look like a nob at the gym and forgot it in your gym bag... also, the rule involves thumbs.
The point is that it's a rule of thumb which is easy to teach, easy to remember and easy to implement when you accidentally forgot to clip your protractor onto your harness...
... because you unclipped your trad. protractor so you wouldn't look like a nob at the gym and forgot it in your gym bag... also, the rule involves thumbs.
Post edited at 16:13
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In reply to zimpara:
Yes.
A lot of replies are actually regarding linking two anchors, the OP was about a sling with little slack.
But the same principles of " Triangle of Forces " and "Vector Forces" still apply to the slings,anchors and the tyrolean rope.
> Solution, use a bigger sling to reduce the angle or join two together?
Yes.
A lot of replies are actually regarding linking two anchors, the OP was about a sling with little slack.
But the same principles of " Triangle of Forces " and "Vector Forces" still apply to the slings,anchors and the tyrolean rope.
In reply to jkarran:
its a different case with a sling though, by the time you've got an angle of 120 degrees the tension in the sling will be doubled (over zero angle - not accounting for friction or sharp edges).
its a different case with a sling though, by the time you've got an angle of 120 degrees the tension in the sling will be doubled (over zero angle - not accounting for friction or sharp edges).
Post edited at 17:47
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In reply to zimpara:
if you've got an angle of 160 degrees at the attachment point of the sling around the block (between the strands of the sling) you will be multiplying the tension in the sling by 5.75 ish (not accounting for friction). This will make it more likely to break or cut for sure. Because tyrolean traverses already multiply the force I doubt you've have a lot of lee way (potentially 4 kn in the tyrloean could snap the sling, but accounting for friction but not sharp edges it would be a higher amount in practice)
if you've got an angle of 160 degrees at the attachment point of the sling around the block (between the strands of the sling) you will be multiplying the tension in the sling by 5.75 ish (not accounting for friction). This will make it more likely to break or cut for sure. Because tyrolean traverses already multiply the force I doubt you've have a lot of lee way (potentially 4 kn in the tyrloean could snap the sling, but accounting for friction but not sharp edges it would be a higher amount in practice)
Post edited at 17:49
In reply to CurlyStevo:
If it's on a tyrollean traverse, won't there be a similarly huge angle at the point where the person sags on the line? Thus the horrendous force multiplication happens twice!
If it's on a tyrollean traverse, won't there be a similarly huge angle at the point where the person sags on the line? Thus the horrendous force multiplication happens twice!
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In reply to Landy_Dom:
yeah that's sort of what I was trying to say. Lets say the angle on the tyrolean is 160 degrees with the person in the middle and a weight of 80 kg. Allowing for some bounce I would think 1.5 kn of force from the human could feasibly occur. With a multiplication of 5.75 / 2 on each anchor - 4.3 kn per anchor.
So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn - above the rated strength of the sling. That said friction on the block would decrease the tension in the sling as one side/strand of it would have less affect on the other.
yeah that's sort of what I was trying to say. Lets say the angle on the tyrolean is 160 degrees with the person in the middle and a weight of 80 kg. Allowing for some bounce I would think 1.5 kn of force from the human could feasibly occur. With a multiplication of 5.75 / 2 on each anchor - 4.3 kn per anchor.
So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn - above the rated strength of the sling. That said friction on the block would decrease the tension in the sling as one side/strand of it would have less affect on the other.
Post edited at 18:32
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In reply to CurlyStevo:
I thought the multiplying factor was 1/(2*cos(angle/2)) (matches the 120 break-even point quoted by jkarran). With this expression the multiplier at 160 is only 2.9-ish? Have I got it wrong? How did you get 5.75?
> if you've got an angle of 160 degrees at the attachment point of the sling around the block (between the strands of the sling) you will be multiplying the tension in the sling by 5.75 ish
I thought the multiplying factor was 1/(2*cos(angle/2)) (matches the 120 break-even point quoted by jkarran). With this expression the multiplier at 160 is only 2.9-ish? Have I got it wrong? How did you get 5.75?
Post edited at 18:39
In reply to purplemonkeyelephant:
Think of a setup with two anchor points equalised. On each anchor point you are not only pulling down, you are also pulling the anchors towards each other.
If the angle is low, you are mostly pulling down with a little bit of towards each other
If the angle is very high, you are doing a lot more pulling towards each other than down
The "downwards" bit is your weight / your leader fall / whatever you're loading the anchor with, i.e. always the same, so if you're also doing a lot of pulling-the-anchors-towards-each-other, you have that force on the anchor as well as your downwards pull.
I don't know if that helps.
> Not that good with this technical stuff, but could someone explain how a weight of 100kg could become several tonnes on the anchor, just by changing the direction of pull?
Think of a setup with two anchor points equalised. On each anchor point you are not only pulling down, you are also pulling the anchors towards each other.
If the angle is low, you are mostly pulling down with a little bit of towards each other
If the angle is very high, you are doing a lot more pulling towards each other than down
The "downwards" bit is your weight / your leader fall / whatever you're loading the anchor with, i.e. always the same, so if you're also doing a lot of pulling-the-anchors-towards-each-other, you have that force on the anchor as well as your downwards pull.
I don't know if that helps.
In reply to harold walmsley:
ok consider one side of the sling
the angle is 80 degrees to the vertical and lets use a weight resulting in a mass of 1kn for ease of computation, as the weight is shared 0.5 kn will be on this strand.
The formula for the tension on one side of the sling is: 0.5 (half the weight on each strand) * 1 (mass) / cos 80 = 2.879 kn
Assuming this friction runs the full length around the sling then both sides have this tension. Ofcourse you now need to account for the force on the other strand so multiply by 2 and get 5.75kn.
Going to your break even point of 120 degrees, that doesn't really add up. If you think about the tension in the sling when both strands run perpendicular this is the strongest possible configuration, as soon as you create an angle between the strands at point of load you are increasing the tension in the sling.
ok consider one side of the sling
the angle is 80 degrees to the vertical and lets use a weight resulting in a mass of 1kn for ease of computation, as the weight is shared 0.5 kn will be on this strand.
The formula for the tension on one side of the sling is: 0.5 (half the weight on each strand) * 1 (mass) / cos 80 = 2.879 kn
Assuming this friction runs the full length around the sling then both sides have this tension. Ofcourse you now need to account for the force on the other strand so multiply by 2 and get 5.75kn.
Going to your break even point of 120 degrees, that doesn't really add up. If you think about the tension in the sling when both strands run perpendicular this is the strongest possible configuration, as soon as you create an angle between the strands at point of load you are increasing the tension in the sling.
Post edited at 18:55
In reply to CurlyStevo:
The tension in the sling is the same all the way round without friction. You have calculated it the same as me to get 2.879. Thus the multiplier relating the tension in the sling to the 1 kN force is 2.879. The "upward" force exerted by each side of the sling is 2.879*cos(80) = 0.5 kN thus the net upward force exerted by both sides at the angle is, as it must be, 1 kN. No reason to multiply the tension by 2. The breakeven at 120 adds up perfectly from vector addition of force and trigonometry (just change cos(80) to cos(60)).
The tension in the sling is the same all the way round without friction. You have calculated it the same as me to get 2.879. Thus the multiplier relating the tension in the sling to the 1 kN force is 2.879. The "upward" force exerted by each side of the sling is 2.879*cos(80) = 0.5 kN thus the net upward force exerted by both sides at the angle is, as it must be, 1 kN. No reason to multiply the tension by 2. The breakeven at 120 adds up perfectly from vector addition of force and trigonometry (just change cos(80) to cos(60)).
Post edited at 20:30
In reply to harold walmsley:
Yes I agree without considering friction the tension in the sling is the same all the way around.
Now lets get to the maths.
“Thus the multiplier relating the tension in the sling to the 1 kN force is 2.879.”
The 2.879 kn that I calculated is for half the weight on one side of the sling to where its loaded. Where do you think the other half of the mass goes? The other strand of course. Which is why the total tension in the sling is the force that is being exerted from both sides (assuming no friction) and the multlipler is 5.75.
Think about the simple case here. Take a sling being pulled straight down and rated to 22 kn. If we pull it straight down both strands parallel it can hold 22kn, so the tension in the sling would be 22kn. To get to that we do the same maths before:
For one side we have 0.5 * 22 / cos 0 = 11 kn, as the other strand / side also has 11kn the slings total tension = 22kn.
If we make an angle either side of the sling to the load point of say 20 degrees we get:
For one side we have 0.5 * 22 / cos 10 = 11.17, so total tension in the sling would be 22.33 kn, so the sling may break if its only rated to 22kn.
As soon as we make any angle between the strands we weaken the max force the sling can hold a bit, there is no break even at 120 degrees. The reason with normal anchors there is a break even at 120 is because each anchor is independent. In fact what you have done at 120, is multiplied the force going to each anchor by 2 but as each anchor is only taking half the force, combined they are as strong as a single anchor (in line with the direction of pull). The strongest possible configuration for a belay is the same as for the sling with both lines to the anchor points in the direction of pull. Any angle between the lines multiples the load somewhat and decreases the max load the belay can take.
“The "upward" force exerted by each side of the sling is 2.879*cos(80) = 0.5 kN thus the net upward force exerted by both sides at the angle is, as it must be, 1 kN. “
I can see your mistake here in returning to the value to 1 you are happy to consider both sides of the sling however when calculating the tension you only consider 1 side. Lets take your maths for the simple case and find the tension in the sling when holding 22kn and a zero angle between the strands.
0.5 * 22kn / cos 0 = 11kn tension in the sling, but hang on the sling is rated to 22kn so by your reckoning you could hang a 44kn mass off it! I can assure you that is not correct!
Hope that clears things up.
Yes I agree without considering friction the tension in the sling is the same all the way around.
Now lets get to the maths.
“Thus the multiplier relating the tension in the sling to the 1 kN force is 2.879.”
The 2.879 kn that I calculated is for half the weight on one side of the sling to where its loaded. Where do you think the other half of the mass goes? The other strand of course. Which is why the total tension in the sling is the force that is being exerted from both sides (assuming no friction) and the multlipler is 5.75.
Think about the simple case here. Take a sling being pulled straight down and rated to 22 kn. If we pull it straight down both strands parallel it can hold 22kn, so the tension in the sling would be 22kn. To get to that we do the same maths before:
For one side we have 0.5 * 22 / cos 0 = 11 kn, as the other strand / side also has 11kn the slings total tension = 22kn.
If we make an angle either side of the sling to the load point of say 20 degrees we get:
For one side we have 0.5 * 22 / cos 10 = 11.17, so total tension in the sling would be 22.33 kn, so the sling may break if its only rated to 22kn.
As soon as we make any angle between the strands we weaken the max force the sling can hold a bit, there is no break even at 120 degrees. The reason with normal anchors there is a break even at 120 is because each anchor is independent. In fact what you have done at 120, is multiplied the force going to each anchor by 2 but as each anchor is only taking half the force, combined they are as strong as a single anchor (in line with the direction of pull). The strongest possible configuration for a belay is the same as for the sling with both lines to the anchor points in the direction of pull. Any angle between the lines multiples the load somewhat and decreases the max load the belay can take.
“The "upward" force exerted by each side of the sling is 2.879*cos(80) = 0.5 kN thus the net upward force exerted by both sides at the angle is, as it must be, 1 kN. “
I can see your mistake here in returning to the value to 1 you are happy to consider both sides of the sling however when calculating the tension you only consider 1 side. Lets take your maths for the simple case and find the tension in the sling when holding 22kn and a zero angle between the strands.
0.5 * 22kn / cos 0 = 11kn tension in the sling, but hang on the sling is rated to 22kn so by your reckoning you could hang a 44kn mass off it! I can assure you that is not correct!
Hope that clears things up.
In reply to CurlyStevo:
Impressively clear. It's so clear I'm rubbing my eyes in realisation...
Ps ^ Never seen so many numbers in a post on UKC.
Pss^ I'm as confused as most of the other posters now.
Impressively clear. It's so clear I'm rubbing my eyes in realisation...
Ps ^ Never seen so many numbers in a post on UKC.
Pss^ I'm as confused as most of the other posters now.
Post edited at 02:13
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In reply to GrahamD:
thanks. I'd been puzzled by how this works for years. all makes sense now.
question - how do the forces it work if you have a loop that goes from you to anchor A to anchor B and back to you? (In your example, I hold a loop taught to make two parallel strands, and the other person just pulls on one strand)
> Try this simple experiment. Tie a bit of rope / cord / washing line to something solid. Hold the other end with the line just taught. Now get someone to pull the centre of the line whilst you try to keep the line straight. See how much harder you have to pull to keep the line straight than the other person does to pull it out of shape. That's the force multipier.
thanks. I'd been puzzled by how this works for years. all makes sense now.
question - how do the forces it work if you have a loop that goes from you to anchor A to anchor B and back to you? (In your example, I hold a loop taught to make two parallel strands, and the other person just pulls on one strand)
In reply to jkarran:
Indeed. I find a 16ft sling pretty useful for setting up belays, especially for bouldery gritstone tops like Stanage. And proper full width 1 inch nylon tape too, preferably.
> It's actually quite straightforward to answer http://www.ropebook.com/information/vector-forces
> Longer slings are one answer.
> jk
Indeed. I find a 16ft sling pretty useful for setting up belays, especially for bouldery gritstone tops like Stanage. And proper full width 1 inch nylon tape too, preferably.
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In reply to Donald82:
If it's just a simple triangular loop with equal tension in all 3 strands (ie not knotted to the anchors with a slack top run between anchors) then the load on the anchors increases as compared to the simple case, a strand to each anchor from the load. The tension in the two lower strands remains the same as for the simple case with no top strand but now, additionally you have the taut top strand pulling sideways on the anchors also. The effect is to change the direction the net force acts on each anchor so that it bisects the two strands going to that anchor. The angle made between the two anchor force vectors is now much wider than between the two lower 'load bearing' strands.
Always knot the belay strands to the anchors so the top strand is slack, it reduces the forces and limits extension in the event of a failure. If you're not sure get a decent book or some training.
jk
> question - how do the forces it work if you have a loop that goes from you to anchor A to anchor B and back to you? (In your example, I hold a loop taught to make two parallel strands, and the other person just pulls on one strand)
If it's just a simple triangular loop with equal tension in all 3 strands (ie not knotted to the anchors with a slack top run between anchors) then the load on the anchors increases as compared to the simple case, a strand to each anchor from the load. The tension in the two lower strands remains the same as for the simple case with no top strand but now, additionally you have the taut top strand pulling sideways on the anchors also. The effect is to change the direction the net force acts on each anchor so that it bisects the two strands going to that anchor. The angle made between the two anchor force vectors is now much wider than between the two lower 'load bearing' strands.
Always knot the belay strands to the anchors so the top strand is slack, it reduces the forces and limits extension in the event of a failure. If you're not sure get a decent book or some training.
jk
In reply to harold walmsley:
if you still remain unconvinced then check out this link :
https://books.google.co.uk/books?id=i6H0ESHr3n8C&pg=PA151&lpg=PA151...
"the forces on the sling at 120 degrees are doubled"
if you still remain unconvinced then check out this link :
https://books.google.co.uk/books?id=i6H0ESHr3n8C&pg=PA151&lpg=PA151...
"the forces on the sling at 120 degrees are doubled"
In reply to Xharlie:
I wish to post a correction to my aforementioned rule of thumb. After it was pointed out that the angle in my first post could be over 95 degrees, I went back to my sources and did some reading.
The rule of thumb SHOULD be the angle between index finger and middle finger which is less variable between people and very close to 60 degrees.
For regular climbing, I'd probably consider the first rule of thumb to be acceptable but for a tyrolean, high-line or slack line anchor, closer to 60 would be better because the forces are already hugely exaggerate - see other posts for the maths.
Anyway, I thought it valuable to point out. My apologies.
I wish to post a correction to my aforementioned rule of thumb. After it was pointed out that the angle in my first post could be over 95 degrees, I went back to my sources and did some reading.
The rule of thumb SHOULD be the angle between index finger and middle finger which is less variable between people and very close to 60 degrees.
For regular climbing, I'd probably consider the first rule of thumb to be acceptable but for a tyrolean, high-line or slack line anchor, closer to 60 would be better because the forces are already hugely exaggerate - see other posts for the maths.
Anyway, I thought it valuable to point out. My apologies.
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In reply to Xharlie:
I'm worried about my freakish hands now
I can't get the angle between my index and middle finger to reach anything above about 30 degrees
Quoting rules that involve parts of the human anatomy must surely be hugely ill-advised in safety critical applications?
> I wish to post a correction to my aforementioned rule of thumb. After it was pointed out that the angle in my first post could be over 95 degrees, I went back to my sources and did some reading.
> The rule of thumb SHOULD be the angle between index finger and middle finger which is less variable between people and very close to 60 degrees.
> For regular climbing, I'd probably consider the first rule of thumb to be acceptable but for a tyrolean, high-line or slack line anchor, closer to 60 would be better because the forces are already hugely exaggerate - see other posts for the maths.
I'm worried about my freakish hands now
I can't get the angle between my index and middle finger to reach anything above about 30 degrees
Quoting rules that involve parts of the human anatomy must surely be hugely ill-advised in safety critical applications?
In reply to Xharlie:
Mine's about thirty degrees. Anyway, that's a rule of fingers.
> The rule of thumb SHOULD be the angle between index finger and middle finger which is less variable between people and very close to 60 degrees.
Mine's about thirty degrees. Anyway, that's a rule of fingers.
In reply to Donald82:
Actually, in my example, where you are holding one end of the rope, you are trying to be one of the anchors. Someone in the middle pulling the rope is the belayer. If the rope is straight (180 degrees) it will be impossible to apply enough tension on the rope to eliminate any movement in the middle. Infinite force multiplication
If instead of starting with the rope taught, though, and the belayer starts with the rope already deflected by an angle, the easier it is to resist the pull.
> question - how do the forces it work if you have a loop that goes from you to anchor A to anchor B and back to you? (In your example, I hold a loop taught to make two parallel strands, and the other person just pulls on one strand)
Actually, in my example, where you are holding one end of the rope, you are trying to be one of the anchors. Someone in the middle pulling the rope is the belayer. If the rope is straight (180 degrees) it will be impossible to apply enough tension on the rope to eliminate any movement in the middle. Infinite force multiplication
If instead of starting with the rope taught, though, and the belayer starts with the rope already deflected by an angle, the easier it is to resist the pull.
In reply to CurlyStevo:
No your reply does not clear it up. You have it wrong. If a rope faces an equal pull from both ends the tension is not doubled. Think if you were hanging on single strand from a ceiling. You would be tensioning the rope with the force needed to suspend your weight (mg). At the other end the ceiling must pull up on the rope with the same force (otherwise it would fall). This does not double the tension to 2mg. The situation with a looped sling is analogous. The maths goes as foloows
Without friction the tension in the sling is the same all the way round, say T. If the tensioning rope pulls with a force P and the half angle on the taut tensioned sling loop is "A" each side of the direction of rope pull then the upward pull against the rope from each side of the loop is Tcos(A).
The combined pull from each side of the tensioned loop is must match the rope pull, P. Thus we get:
P=2Tcos(A).
Thus the ratio of tension to Pull , T/P, is given by:
T/P=1/[2cos(A)]
Noting that A is half of the angle made between the parts of the sling itself, if we want to express the ratio in terms of the full angle, F, the expression for the ratio of pull force to sling tension becomes:
T/P=1/[2cos(F/2)]
If you don't now see this I don't know what more i can say to explain so I give up from here on.
PS I think the book is as confused as you are. The doubling effect applies as follows. Two ropes individually connected to 2 anchors but running almost parallel resisting a pull P, tension in each anchor rope
T=P/2,
2 ropes connected to 2 anchors resisting a force P but angle between anchor strands 120 deg, tension in each anchor:
T= P/2 cos (60)
As cos (60 )=1/2 this give T= P
I.e the force is twice what it was with the ideal arrangement. However because there are two strands resisting a single pull the ideal arrangement gives T=P/2 and hence the non ideal arrangement with doubled tension gives T=P. ie the tension with the bad angle is equal to the pull not double the pull i.e the 120 degree angle with 2 ropes is the breakeven angle compared to a single rope, which would also give T=P.
I think the book also confuses the ratio by which the tension is increased and the ratio of the sling tension to the pull force. Don't believe everything you read!
No your reply does not clear it up. You have it wrong. If a rope faces an equal pull from both ends the tension is not doubled. Think if you were hanging on single strand from a ceiling. You would be tensioning the rope with the force needed to suspend your weight (mg). At the other end the ceiling must pull up on the rope with the same force (otherwise it would fall). This does not double the tension to 2mg. The situation with a looped sling is analogous. The maths goes as foloows
Without friction the tension in the sling is the same all the way round, say T. If the tensioning rope pulls with a force P and the half angle on the taut tensioned sling loop is "A" each side of the direction of rope pull then the upward pull against the rope from each side of the loop is Tcos(A).
The combined pull from each side of the tensioned loop is must match the rope pull, P. Thus we get:
P=2Tcos(A).
Thus the ratio of tension to Pull , T/P, is given by:
T/P=1/[2cos(A)]
Noting that A is half of the angle made between the parts of the sling itself, if we want to express the ratio in terms of the full angle, F, the expression for the ratio of pull force to sling tension becomes:
T/P=1/[2cos(F/2)]
If you don't now see this I don't know what more i can say to explain so I give up from here on.
PS I think the book is as confused as you are. The doubling effect applies as follows. Two ropes individually connected to 2 anchors but running almost parallel resisting a pull P, tension in each anchor rope
T=P/2,
2 ropes connected to 2 anchors resisting a force P but angle between anchor strands 120 deg, tension in each anchor:
T= P/2 cos (60)
As cos (60 )=1/2 this give T= P
I.e the force is twice what it was with the ideal arrangement. However because there are two strands resisting a single pull the ideal arrangement gives T=P/2 and hence the non ideal arrangement with doubled tension gives T=P. ie the tension with the bad angle is equal to the pull not double the pull i.e the 120 degree angle with 2 ropes is the breakeven angle compared to a single rope, which would also give T=P.
I think the book also confuses the ratio by which the tension is increased and the ratio of the sling tension to the pull force. Don't believe everything you read!
Post edited at 23:03
In reply to harold walmsley:
Wow. That's pretty scary stuff.
So in english please, what are we looking at?
Wow. That's pretty scary stuff.
So in english please, what are we looking at?
In reply to harold walmsley:
I give up I even quoted a text book that proved my point.
The book is not confused you are.
I give up I even quoted a text book that proved my point.
The book is not confused you are.
Post edited at 01:42
In reply to Robert Durran:
Ok Robert does a sling placed around a rock such that the angle where it is loaded is 120 degrees between the strands approximately double the tension in the sling or not. As compared to the case where a sling is loaded with same force but the strands are approximately parallel.
Ok Robert does a sling placed around a rock such that the angle where it is loaded is 120 degrees between the strands approximately double the tension in the sling or not. As compared to the case where a sling is loaded with same force but the strands are approximately parallel.
In reply to CurlyStevo:
Tension will be equal to the load.
Tension will be half the load.
So yes, but you had double the correct tension in both cases.
> Ok Robert does a sling placed around a rock such that the angle where it is loaded is 120 degrees between the strands approximately double the tension in the sling or not.
Tension will be equal to the load.
> As compared to the case where a sling is loaded with same force but the strands are approximately parallel.
Tension will be half the load.
So yes, but you had double the correct tension in both cases.
Post edited at 08:54
In reply to harold walmsley:
Ok I think I can see here the confusion lays here.
I will explain later, however I still ascertain that:
- As compared to pulling a sling in opposite directions from single points, when loading it so that one of the load points forms an angle between the strands of the sling greater than 0, this does indeed increase the tensional force in the sling. (you refuted this occurred below 120 degrees)
- The greater the angle (up to 180 degrees) the greater the multiplicative factor and that this increases exponentially to infinity at 180 degrees.
- There is no magical value of 120 degrees where less than this value the tensional force in the sling does not increase
- In fact 120 degrees applies double the tensional force to the sling
- 160 degrees applies a multiplication of approx. 5.75 (truncated) as I previously stated but refuted by Harold.
Right now on to where the confusion lays. The UIAA rating for slings ( http://www.theuiaa.org/upload_area/Safety/Standards/Safety-Standards-Pictor... ) is not the tensional force per strand of the sling either side of the pull points. The kn stated are the force it takes to pull the sling apart. This is why I was considering the sling as a single unit when coming to my formulae for calculating the forces on the sling, so we could compare it to the rated strength of the sling in the example given. Much the same as a rope is made of strands but we generally consider the strength of the rope not of the strands.
The tension of the sling considered as a single unit in the setup shown in the UIAA diagram is 22kn. However if we consider the strands either side of the sling ( or indeed the tension within the strand that makes up the sling in a frictionless setup) I would agree the formula changes and we would expect it to be approximately 11kn in the set up shown and agreeing with Harolds formula. However bear in mind a dyneema sling rated to 22kn, as a single unconnected strand has approx. breaking strength of 16 kn. Also that doesn't mean that the tension in the strand that makes up the sling is rated to 22kn, it is not.
Anyway the divide by 2 the Harold added matters not to the calculation of my multiplicative forces as all it does is divide every result by two and the values I calculated remain the same.
I will prove that my original calculations on how forces are multiplied at various angles are correct. I will do this using Harold's Formula and a weight of 1KN.
T/P=1/[2cos(F/2)] (given by Harold)
Ok a pull in opposite directions gives an angle of 0
So Tension in the strand of the sling is
1/2*cos(0) = 0.5 this is our base case
An angle of 90
1/2*cos(45) = 0.707
An angle of 120
1/2*cos(60) = 1
An angle of 160
1/2*cos(80) = 2.879
An angle of 180
1/2*cos(90) = infinitiy
As you can see, as the original tensional force in the strand was 0.5 the multiplicative factors on the tensional forces are 1.414 for 90 degrees 2 for 120 degrees and 5.75 for 160 degrees as I previously stated. Also you can see that as soon as any angle is added between the strands the tension in the sling increases and that this increases exponentially, there is no magic angle (120 degrees for example ) that below this increase does not occur (the only angle with none is zero)
I hope this clears things up
Ok I think I can see here the confusion lays here.
I will explain later, however I still ascertain that:
- As compared to pulling a sling in opposite directions from single points, when loading it so that one of the load points forms an angle between the strands of the sling greater than 0, this does indeed increase the tensional force in the sling. (you refuted this occurred below 120 degrees)
- The greater the angle (up to 180 degrees) the greater the multiplicative factor and that this increases exponentially to infinity at 180 degrees.
- There is no magical value of 120 degrees where less than this value the tensional force in the sling does not increase
- In fact 120 degrees applies double the tensional force to the sling
- 160 degrees applies a multiplication of approx. 5.75 (truncated) as I previously stated but refuted by Harold.
Right now on to where the confusion lays. The UIAA rating for slings ( http://www.theuiaa.org/upload_area/Safety/Standards/Safety-Standards-Pictor... ) is not the tensional force per strand of the sling either side of the pull points. The kn stated are the force it takes to pull the sling apart. This is why I was considering the sling as a single unit when coming to my formulae for calculating the forces on the sling, so we could compare it to the rated strength of the sling in the example given. Much the same as a rope is made of strands but we generally consider the strength of the rope not of the strands.
The tension of the sling considered as a single unit in the setup shown in the UIAA diagram is 22kn. However if we consider the strands either side of the sling ( or indeed the tension within the strand that makes up the sling in a frictionless setup) I would agree the formula changes and we would expect it to be approximately 11kn in the set up shown and agreeing with Harolds formula. However bear in mind a dyneema sling rated to 22kn, as a single unconnected strand has approx. breaking strength of 16 kn. Also that doesn't mean that the tension in the strand that makes up the sling is rated to 22kn, it is not.
Anyway the divide by 2 the Harold added matters not to the calculation of my multiplicative forces as all it does is divide every result by two and the values I calculated remain the same.
I will prove that my original calculations on how forces are multiplied at various angles are correct. I will do this using Harold's Formula and a weight of 1KN.
T/P=1/[2cos(F/2)] (given by Harold)
Ok a pull in opposite directions gives an angle of 0
So Tension in the strand of the sling is
1/2*cos(0) = 0.5 this is our base case
An angle of 90
1/2*cos(45) = 0.707
An angle of 120
1/2*cos(60) = 1
An angle of 160
1/2*cos(80) = 2.879
An angle of 180
1/2*cos(90) = infinitiy
As you can see, as the original tensional force in the strand was 0.5 the multiplicative factors on the tensional forces are 1.414 for 90 degrees 2 for 120 degrees and 5.75 for 160 degrees as I previously stated. Also you can see that as soon as any angle is added between the strands the tension in the sling increases and that this increases exponentially, there is no magic angle (120 degrees for example ) that below this increase does not occur (the only angle with none is zero)
I hope this clears things up
Post edited at 11:23
In reply to Robert Durran:
Not really as I was measuring the tension in the sling as a single unit so I could compare it with the UIAA rated strength in KN. The UIAA rated strength is NOT the tension of a strand of the sling its the strength of the sling as a single unit being pulled apart.
Atleast either way Harold should agree now that any increase in angle of the strands of the sling does indeed increase the tension in the sling and that there is no magic value of 120 degrees where this does not occur (infact this increases the tension of the sling by double), also that the original multiplicative factors that I calculated and he queried were indeed correct.
> So yes, but you had double the correct tension in both cases.
Not really as I was measuring the tension in the sling as a single unit so I could compare it with the UIAA rated strength in KN. The UIAA rated strength is NOT the tension of a strand of the sling its the strength of the sling as a single unit being pulled apart.
Atleast either way Harold should agree now that any increase in angle of the strands of the sling does indeed increase the tension in the sling and that there is no magic value of 120 degrees where this does not occur (infact this increases the tension of the sling by double), also that the original multiplicative factors that I calculated and he queried were indeed correct.
Post edited at 11:26
In reply to zimpara:
Rigging with wide angles causes significant force multiplication (think of a lever if you have trouble with this). Cascading bad rigging, as in the case of a taut tyrolean/slackline attatched to a poorly arranged anchor easily produces forces capable of causing partial or complete failure (think of levering on a lever).
jk
> So in english please, what are we looking at?
Rigging with wide angles causes significant force multiplication (think of a lever if you have trouble with this). Cascading bad rigging, as in the case of a taut tyrolean/slackline attatched to a poorly arranged anchor easily produces forces capable of causing partial or complete failure (think of levering on a lever).
jk
Post edited at 11:19
In reply to CurlyStevo:
So the strength of the actual tape is half the UIAA rating of the sling - fair enough.
I don't think anyone has argued otherwise. The only significance of 120 degrees is that then the tension in the sling is equal to the load (ie at greater angles the tension is greater than the load, so, if we are talking about two separate anchors, the load on each is greater than the load on a single anchor).
> Not really as I was measuring the tension in the sling as a single unit so I could compare it with the UIAA rated strength in KN. The UIAA rated strength is NOT the tension of a strand of the sling its the strength of the sling as a single unit being pulled apart.
So the strength of the actual tape is half the UIAA rating of the sling - fair enough.
> Atleast either way Harold should agree now that any increase in angle of the strands of the sling does indeed increase the tension in the sling and that there is no magic value of 120 degrees where this does not occur.
I don't think anyone has argued otherwise. The only significance of 120 degrees is that then the tension in the sling is equal to the load (ie at greater angles the tension is greater than the load, so, if we are talking about two separate anchors, the load on each is greater than the load on a single anchor).
In reply to CurlyStevo:
T/P=1/[2cos(F/2)] (given by Harold and you appear to be agreed on)
An angle of 160
T/P = 1/2*cos(80) = 2.879
Tension = 2.879 x Pulling force
The 22 kn rating is the Pulling force (at 0 angle), this'll equate to a tension in each strand of the sling of about half that - which is the tension in the sling.
has said by Harold: If a rope faces an equal pull from both ends the tension is not doubled. (if it didn't have an equal pull it would move). Everything he has set out is correct.
To approach understanding the concept from another angle, lets take this setup:
cut a (22kn rated) sling to form a single length of material, fix it at either end and pull. At what force would it break?
T/P=1/[2cos(F/2)] (given by Harold and you appear to be agreed on)
An angle of 160
T/P = 1/2*cos(80) = 2.879
Tension = 2.879 x Pulling force
The 22 kn rating is the Pulling force (at 0 angle), this'll equate to a tension in each strand of the sling of about half that - which is the tension in the sling.
has said by Harold: If a rope faces an equal pull from both ends the tension is not doubled. (if it didn't have an equal pull it would move). Everything he has set out is correct.
To approach understanding the concept from another angle, lets take this setup:
cut a (22kn rated) sling to form a single length of material, fix it at either end and pull. At what force would it break?
In reply to daWalt:
It seems to me we are measuring different things. I was comparing the tension at 0 degrees on the sling with the tension at 120 and 160 degrees on the sling. I was not comparing the tension at x degrees with the load force.
I also calculated a force in KN on the sling at the end of the tyrolean traverse in the same units that the sling is rated not the tension of the continuous strand of a sling (which is not how the sling is rated)
I agree with what you said
"An angle of 160
T/P = 1/2*cos(80) = 2.879
Tension = 2.879 x Pulling force"
However at zero degrees
T/P = 0.5
so
Tension = 0.5 x pulling force
therefore
at 160 degrees there is approx. 5.75 more tension in the sling than at zero degrees (which is its strongest configuration and how it is rated)
( 2.879 / 0.5 = 5.75 approx. )
The reason I calculated the forces in my example like this:
"yeah that's sort of what I was trying to say. Lets say the angle on the tyrolean is 160 degrees with the person in the middle and a weight of 80 kg. Allowing for some bounce I would think 1.5 kn of force from the human could feasibly occur. With a multiplication of 5.75 / 2 on each anchor - 4.3 kn per anchor.
So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn - above the rated strength of the sling. That said friction on the block would decrease the tension in the sling as one side/strand of it would have less affect on the other."
Is because that's how slings are rated they are not rated as the tension in the strand that goes around the sling the sling is in affect a single unit (imagine it as the tension between the biners at either end of the sling that connect it to the pull test equipment in the UIAA diagram I linked, rather than the tension in the strand that forms the sling).
It seems to me we are measuring different things. I was comparing the tension at 0 degrees on the sling with the tension at 120 and 160 degrees on the sling. I was not comparing the tension at x degrees with the load force.
I also calculated a force in KN on the sling at the end of the tyrolean traverse in the same units that the sling is rated not the tension of the continuous strand of a sling (which is not how the sling is rated)
I agree with what you said
"An angle of 160
T/P = 1/2*cos(80) = 2.879
Tension = 2.879 x Pulling force"
However at zero degrees
T/P = 0.5
so
Tension = 0.5 x pulling force
therefore
at 160 degrees there is approx. 5.75 more tension in the sling than at zero degrees (which is its strongest configuration and how it is rated)
( 2.879 / 0.5 = 5.75 approx. )
The reason I calculated the forces in my example like this:
"yeah that's sort of what I was trying to say. Lets say the angle on the tyrolean is 160 degrees with the person in the middle and a weight of 80 kg. Allowing for some bounce I would think 1.5 kn of force from the human could feasibly occur. With a multiplication of 5.75 / 2 on each anchor - 4.3 kn per anchor.
So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn - above the rated strength of the sling. That said friction on the block would decrease the tension in the sling as one side/strand of it would have less affect on the other."
Is because that's how slings are rated they are not rated as the tension in the strand that goes around the sling the sling is in affect a single unit (imagine it as the tension between the biners at either end of the sling that connect it to the pull test equipment in the UIAA diagram I linked, rather than the tension in the strand that forms the sling).
Post edited at 13:08
In reply to Robert Durran:
Well the dyneema tape on the reel that they use for 22kn slings is rated to about 16kn normally . But the tension in the strands either side of the load in the UIAA diagram are 11 kn.
I disagree, Harold was talking about a break even point at 120 degrees and disputed the claim that the book makes that I linked (which I believe was correct). If you compare the forces at zero degrees with 120 degree the forces on the sling are doubled as the book says.
> So the strength of the actual tape is half the UIAA rating of the sling - fair enough.
Well the dyneema tape on the reel that they use for 22kn slings is rated to about 16kn normally . But the tension in the strands either side of the load in the UIAA diagram are 11 kn.
> I don't think anyone has argued otherwise. The only significance of 120 degrees is that then the tension in the sling is equal to the load (ie at greater angles the tension is greater than the load, so, if we are talking about two separate anchors, the load on each is greater than the load on a single anchor).
I disagree, Harold was talking about a break even point at 120 degrees and disputed the claim that the book makes that I linked (which I believe was correct). If you compare the forces at zero degrees with 120 degree the forces on the sling are doubled as the book says.
Post edited at 13:04
In reply to CurlyStevo:
Yes, two strands at 120 degrees will each have the same tension as a single strand and obviously, therefore, double the tension of a double strand (ie two strands at zero degrees)
> I disagree, Harold was talking about a break even point at 120 degrees.
Yes, two strands at 120 degrees will each have the same tension as a single strand and obviously, therefore, double the tension of a double strand (ie two strands at zero degrees)
1
In reply to Robert Durran:
you can't have a single strand sling, and you can't use it as an anchor around a block for a Tyrolean so its irrelevant.
But I'm glad someone else has seen sense and is agreeing that in the context of slings around blocks there is no break even point at 120 degrees. The strongest configuration is near parallel strands, any angle between them weakens the set up. At 120 degrees there is twice as much force on the sling as at zero degrees.
you can't have a single strand sling, and you can't use it as an anchor around a block for a Tyrolean so its irrelevant.
But I'm glad someone else has seen sense and is agreeing that in the context of slings around blocks there is no break even point at 120 degrees. The strongest configuration is near parallel strands, any angle between them weakens the set up. At 120 degrees there is twice as much force on the sling as at zero degrees.
Post edited at 13:18
In reply to all:
This makes me wonder a bit (!!!) about tyroleans that we set up in the 70/80's...
1) Can't remember what the end-point anchors were - probably trees where available but we certainly weren't aware of using longer/multiple slings to reduce the angles.
2) We used to try and tension the tyrolean as much as possible (makes it easier to get across) but were totally unaware that this effectively increases the angle and hence the forces.
3) Totally unaware of the doubling up of forces - 1 on the tyrolean & 2 on the anchors.
We did one across Huncote Quarry in Leics - 5 ropes knotted together (slid down to the middle and then unclipped) and one across Dovedale from Raventor with 3 ropes knotted together (across the whole way to a tree opposite). Funnily enough the double fishermans were complete buggers to get undone.
Makes me wonder whether we were lucky to get away without an anchorage failure or maybe it's just that there's loads of spare capacity in the various rope/sling strengths.
I'd certainly be thinking about the setups a bit more carefully if I did something similar today
This makes me wonder a bit (!!!) about tyroleans that we set up in the 70/80's...
1) Can't remember what the end-point anchors were - probably trees where available but we certainly weren't aware of using longer/multiple slings to reduce the angles.
2) We used to try and tension the tyrolean as much as possible (makes it easier to get across) but were totally unaware that this effectively increases the angle and hence the forces.
3) Totally unaware of the doubling up of forces - 1 on the tyrolean & 2 on the anchors.
We did one across Huncote Quarry in Leics - 5 ropes knotted together (slid down to the middle and then unclipped) and one across Dovedale from Raventor with 3 ropes knotted together (across the whole way to a tree opposite). Funnily enough the double fishermans were complete buggers to get undone.
Makes me wonder whether we were lucky to get away without an anchorage failure or maybe it's just that there's loads of spare capacity in the various rope/sling strengths.
I'd certainly be thinking about the setups a bit more carefully if I did something similar today
Post edited at 13:50
In reply to Michael Hood:
even a tight rope will only magnify the tension in the rope and anchors by about 3. So plenty of lee way left for knots and alike. The main issue would be poor anchor construction IMO.
even a tight rope will only magnify the tension in the rope and anchors by about 3. So plenty of lee way left for knots and alike. The main issue would be poor anchor construction IMO.
Post edited at 14:02
In reply to CurlyStevo:
but to be honest, if you're not arguing about the mathematics that Harold set out, then there isn't much to argue about.
> tyrolean is 160 degrees, 1.5 kn of force . With a multiplication of [5.75 / 2 =2.88] on each anchor - 4.3 kn per anchor.
ok so far.> So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn.
this is where its coming apart - tension in the rope/webbing/sling whatever is 4.3kn........ but to be honest, if you're not arguing about the mathematics that Harold set out, then there isn't much to argue about.
In reply to daWalt:
Oh I am arguing he's wrong about the multiplicative factors of the increase in force with differing angles of sling strands to load point as compared to zero degrees and that the figures I originally quoted that he contested are correct.
And I also think that the forces I estimated for the tyrolean set up I described are correct.
"> So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn"
The tension in the strands of the sling would be half that force, but that is not how the sling is rated. Its not rated for the strand to be able to withstand 22knof tension. If it could the UIAA test ( http://www.theuiaa.org/upload_area/Safety/Standards/Safety-Standards-Pictor... ) would show the sling being able to hold 44 kn of force. The UIAA test gives a force that the whole sling can stand up to when pulled apart not a tensional force that the slingle strand can hold up to. All my workings are in the UIAA frame of reference and hence are twice as big as Harolds but can be directly compared to the UIAA rating.
That's why I calculate the force on the sling as 25kn for the whole unit. The tension in the strand of the sling would be about 12.5kn (note that is above the 11kn that would be present in the UIAA test)
How about you do some calculations to see if the sling would hold up to my example. I think you'll come to the tensional single strand force being 12.5kn and then you'll either then divide the UIAA 22kn by two or multiply your 12.5kn by two as I did to come to 25kn. Either way you'll say the rating of the sling is exceeded.
Oh I am arguing he's wrong about the multiplicative factors of the increase in force with differing angles of sling strands to load point as compared to zero degrees and that the figures I originally quoted that he contested are correct.
And I also think that the forces I estimated for the tyrolean set up I described are correct.
"> So then on the sling with the same angle (and not accounting for friction) you'd get 4.3kn * 5.75 = 25kn"
The tension in the strands of the sling would be half that force, but that is not how the sling is rated. Its not rated for the strand to be able to withstand 22knof tension. If it could the UIAA test ( http://www.theuiaa.org/upload_area/Safety/Standards/Safety-Standards-Pictor... ) would show the sling being able to hold 44 kn of force. The UIAA test gives a force that the whole sling can stand up to when pulled apart not a tensional force that the slingle strand can hold up to. All my workings are in the UIAA frame of reference and hence are twice as big as Harolds but can be directly compared to the UIAA rating.
That's why I calculate the force on the sling as 25kn for the whole unit. The tension in the strand of the sling would be about 12.5kn (note that is above the 11kn that would be present in the UIAA test)
How about you do some calculations to see if the sling would hold up to my example. I think you'll come to the tensional single strand force being 12.5kn and then you'll either then divide the UIAA 22kn by two or multiply your 12.5kn by two as I did to come to 25kn. Either way you'll say the rating of the sling is exceeded.
Post edited at 14:31
In reply to CurlyStevo:
your explanation is piss poor.
and to use the rating of the sling as a compactor is misleading at best and is only likely to lead you up the primrose path.
for anyone else that's struggling to keep up:
tension = force / [2 x Cos (angle/2)]
tension in the left strand is the same as tension in the right strand, is the tension in the sling throughout its length.
your explanation is piss poor.
and to use the rating of the sling as a compactor is misleading at best and is only likely to lead you up the primrose path.
for anyone else that's struggling to keep up:
tension = force / [2 x Cos (angle/2)]
tension in the left strand is the same as tension in the right strand, is the tension in the sling throughout its length.
In reply to daWalt:
At the end of the day the sling around the block with the strands at the load point being 120 degrees apart will only be able to support half its rated force and if the angle was 160 degrees it would be able to withstand only rated force / 5.75 (approx.)
At the end of the day the sling around the block with the strands at the load point being 120 degrees apart will only be able to support half its rated force and if the angle was 160 degrees it would be able to withstand only rated force / 5.75 (approx.)
Post edited at 14:49
In reply to CurlyStevo:
I actually agree with this. The confusion has been partly because (as you pointed out) the sling is rated for a double strand of tape, not a single one - but also because of your somewhat unclear explanation and maths!
> At the end of the day the sling around the block with the strands at the load point being 120 degrees apart will only be able to support half its rated force and if the angle was 160 degrees it would be able to withstand only rated force / 5.75 (approx.)
I actually agree with this. The confusion has been partly because (as you pointed out) the sling is rated for a double strand of tape, not a single one - but also because of your somewhat unclear explanation and maths!
In reply to Robert Durran:
I agree I didn't explain well how I worked it out but the kn force I calculated was in the same units / scale as the rating for the sling and correct.
I agree I didn't explain well how I worked it out but the kn force I calculated was in the same units / scale as the rating for the sling and correct.
Post edited at 15:10
In reply to CurlyStevo:
It is this sort of thing that was confused and confusing.
> ............ the kn force I calculated was in the same units / scale as the rating for the sling...........
It is this sort of thing that was confused and confusing.
In reply to Robert Durran:
I guess but I was only trying to determine if the rating of the sling was exceeded or not. Using the system I outlined you can simply compare the output direcly to the sling rating to see if its been exceeded. In my mind it was clearer as everything was working towards the rating of the sling. The system does function perfectly well for that.
I personally think its also misleading to talk about a break even point of 120 degrees and angles as large as 160 only multiplying the force by 2.87, that's just not the case. The sling is 5.75 times less strong when the angle is 160 degrees. Surely the OP wants to know how much the sling and anchors has weakened by making a poor setup not some proportion of the initial force that is now in some sling when the wouldn't know what proportion of it was in the sling with a better set up. Its next to meaningless unless you do / or know the math.
I guess but I was only trying to determine if the rating of the sling was exceeded or not. Using the system I outlined you can simply compare the output direcly to the sling rating to see if its been exceeded. In my mind it was clearer as everything was working towards the rating of the sling. The system does function perfectly well for that.
I personally think its also misleading to talk about a break even point of 120 degrees and angles as large as 160 only multiplying the force by 2.87, that's just not the case. The sling is 5.75 times less strong when the angle is 160 degrees. Surely the OP wants to know how much the sling and anchors has weakened by making a poor setup not some proportion of the initial force that is now in some sling when the wouldn't know what proportion of it was in the sling with a better set up. Its next to meaningless unless you do / or know the math.
Post edited at 15:43
In reply to CurlyStevo:
It is accurate; the tension in the sling is 2.87 times as big.
No, it has exactly the same strength - it will just have a higher tension. I suspect what you mean is that the load which would generate a tension which would break it is 5.75 times smaller. I suppose you could say that the set up taken as a whole is 5,75 times less strong.
> I personally think its also misleading to talk about...... angles as large as 160 only multiplying the force by 2.87, that's just not the case.
It is accurate; the tension in the sling is 2.87 times as big.
> The sling is 5.75 times less strong when the angle is 160 degrees.
No, it has exactly the same strength - it will just have a higher tension. I suspect what you mean is that the load which would generate a tension which would break it is 5.75 times smaller. I suppose you could say that the set up taken as a whole is 5,75 times less strong.
In reply to Robert Durran:
as big as what? Its 2.87 times bigger than the force coming from the tyrolean, but that's misleading for two reasons. Firstly the sling is not rated as a tension force of the strand the forms the sling, secondly in the strongest possible set up (as per UIAA test) the tension in the sling would be half of the force coming from the tyrolean. That's why I think its misleading, the tension in the sling has actually multiplied by atleast 5.75 as compared to how it is loaded in the UIAA test.
> It is accurate; the tension in the sling is 2.87 times as big.
as big as what? Its 2.87 times bigger than the force coming from the tyrolean, but that's misleading for two reasons. Firstly the sling is not rated as a tension force of the strand the forms the sling, secondly in the strongest possible set up (as per UIAA test) the tension in the sling would be half of the force coming from the tyrolean. That's why I think its misleading, the tension in the sling has actually multiplied by atleast 5.75 as compared to how it is loaded in the UIAA test.
Post edited at 17:39
In reply to CurlyStevo:
As with zero angle (and I should have said 5.75 times a as big)
> As big as what?
As with zero angle (and I should have said 5.75 times a as big)
In reply to CurlyStevo:
To quote the OP "by clipping into that edge of sling that runs at almost 180 degrees, is the weight multiplied". So to paraphrase, the OP asked about the ratio between the tension (wrongly but understandably referred to as "weight") in the sling and the tension ("weight") in a rope pulling on it because it was supporting a person. You calculated something else: the ratio by which the tension in the loop increases when it makes a wide angle compared to the case with an angle of zero. This is a factor of 2 bigger but not what Zimpara asked for.
I confess I assumed the 22 kN rating was the tension and based my assessment of safety factors on this so I was a factor of two out in that part of the assessment.
To quote the OP "by clipping into that edge of sling that runs at almost 180 degrees, is the weight multiplied". So to paraphrase, the OP asked about the ratio between the tension (wrongly but understandably referred to as "weight") in the sling and the tension ("weight") in a rope pulling on it because it was supporting a person. You calculated something else: the ratio by which the tension in the loop increases when it makes a wide angle compared to the case with an angle of zero. This is a factor of 2 bigger but not what Zimpara asked for.
I confess I assumed the 22 kN rating was the tension and based my assessment of safety factors on this so I was a factor of two out in that part of the assessment.
In reply to jkarran:
Thanks. I know to knot the belay strands for an achor - I was just wondering about when you thread a sport route with two bolts at the top that are horizontal to esach other. I guess the low loads as you're not falling and the small angle between strands means it doesn't matter.
Thanks. I know to knot the belay strands for an achor - I was just wondering about when you thread a sport route with two bolts at the top that are horizontal to esach other. I guess the low loads as you're not falling and the small angle between strands means it doesn't matter.
In reply to GrahamD:
Thanks for the reply. I didn't explain myself very well there. i meant if it were like your example but instead of holding a single rope straight you hold a loop of rope straight to form two strands and then someone pulls on just on of the strands. JKarran's explained.
Thanks for the reply. I didn't explain myself very well there. i meant if it were like your example but instead of holding a single rope straight you hold a loop of rope straight to form two strands and then someone pulls on just on of the strands. JKarran's explained.
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