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sungam


Oct 5, 2008, 12:42 PM
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hafilax wrote:
When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.
Simplest view. Cunning. The force on the anchors is higher, but has the tension increased?
Perhaps yes...


jt512


Oct 5, 2008, 12:48 PM
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colatownkid wrote:
jt512 wrote:
angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight that is, a little more than triple the climber's weight.

Jay

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

In his paper, Richard refers to it as a "convention" to account for the reduction in force on the belayer's side of the rope due to friction between the rope and the carabiner. I presume that it is based on empirical data.

In reply to:
also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

The peak force of twice the climber's weight for a fall-factor-0 fall is a consequence of using Hooke's Law to describe rope stretch. As you'll see, Richard provides two separate derivations of this result.

Jay


colatownkid


Oct 5, 2008, 12:54 PM
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jt512 wrote:
colatownkid wrote:
jt512 wrote:
angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight that is, a little more than triple the climber's weight.

Jay

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

In his paper, Richard refers to it as a "convention" to account for the reduction in force on the belayer's side of the rope due to friction between the rope and the carabiner. I presume that it is based on empirical data.

In reply to:
also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

The peak force of twice the climber's weight for a fall-factor-0 fall is a consequence of using Hooke's Law to describe rope stretch. As you'll see, Richard provides two separate derivations of this result.

Jay

just finished reading it; thanks.


Partner rgold


Oct 5, 2008, 12:55 PM
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Colatown, the 2/3 value for carabiner efficiency is an empirically derived value that has been stated over and over in various places, and now has the status of "folklore." I don't know the original reference. I have seen 1/2 stated, but much less often. The fact is that the number will be different for different rope-carabiner combinations and probably, like other friction coefficients, has a sliding and a static value.

The fact that the peak tension in a rope with applied weight W is 2W is a simple conservation of energy consequence of Hooke's Law, viewed as a reasonable approximation to the behavior of a climbing rope. (Yes, there is some acceleration involved, since a climber who weights a rope will drop as the rope stretches.) The result assumes that the climber's weight is instantaneously applied to the climbing rope with no slack in the rope. Using various strategies to ease the weight onto the rope could, in theory, keep the peak rope tension no greater than the climber's weight, but this would not apply to catching a top-rope fall.

Whether the real result is more or less than 2W would be a matter for experiment. I'd guess a little less than 2W because of the internal damping effects of rope construction, but given the manifold uncertainties and variations in real-life practice, 2W is a very reasonable working estimate.

Apparently, I derived this result in the .pdf jt512 refers to, though I'd have to check back to be sure.

If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers, or an estimation that knowing the real figures will not contribute in any meaningul way to practical behavior in the field.


colatownkid


Oct 5, 2008, 1:00 PM
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Re: [rgold] tension [In reply to]
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rgold wrote:
Colatown, the 2/3 value for carabiner efficiency is an empirically derived value that has been stated over and over in various places, and now has the status of "folklore." I don't know the original reference. I have seen 1/2 stated, but much less often. The fact is that the number will be different for different rope-carabiner combinations and probably, like other friction coefficients, has a sliding and a static value.

The fact that the peak tension in a rope with applied weight W is 2W is a simple conservation of energy consequence of Hooke's Law, viewed as a reasonable approximation to the behavior of a climbing rope. (Yes, there is some acceleration involved, since a climber who weights a rope will drop as the rope stretches.) The result assumes that the climber's weight is instantaneously applied to the climbing rope with no slack in the rope. Using various strategies to ease the weight onto the rope could, in theory, keep the peak rope tension no greater than the climber's weight, but this would not apply to catching a top-rope fall.

Whether the real result is more or less than 2W would be a matter for experiment. I'd guess a little less than 2W because of the internal damping effects of rope construction, but given the manifold uncertainties and variations in real-life practice, 2W is a very reasonable working estimate.

Apparently, I derived this result in the .pdf jt512 refers to, though I'd have to check back to be sure.

If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers, or an estimation that knowing the real figures will not contribute in any meaningul way to practical behavior in the field.

indeed--i find many "bibles" to be quite inadequate for my tastes. on the other hand, i suppose one need not necessarily understand the underlying science in a numeric sense to appreciate the lessons to be learned. personally, i appreciate a mathematical understanding of climbing situations.

as you suspect, you did indeed justify T=2W in the linked paper.

as for the empirical 2/3, i'll guess i'll just take that one on faith. Wink


jt512


Oct 5, 2008, 1:25 PM
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rgold wrote:
If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

Stop it. You're kill me.

Jay


joeforte


Oct 5, 2008, 5:26 PM
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Re: [angry] tension [In reply to]
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angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

You are right, the anchor would see about 200 lbs... :Engineers cringe:


milesenoell


Oct 6, 2008, 10:11 AM
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lemon_boy


Oct 6, 2008, 10:24 AM
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jt512 and rgold are on the money, per usual. angry gave a pretty decent blue-collar interpretation (minus the initial peak load).

by far the scariest thing in this entire thread is the quality (or lack of) that the US Navy is providing students these days.


mtnrock


Oct 6, 2008, 5:14 PM
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Re: [hafilax] tension [In reply to]
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When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.

This is what i was trying to say and i was just checking thanks for the help


mtnrock


Oct 6, 2008, 5:21 PM
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haha this post got so off topic i just had a simple question on tention and it went into all this stuff i can't really say it was simple because i can't word anything well but how can i put a diagram so i can show what im talking about more clearly. also does any one know the spring constant of a new dynamic rope?


jdefazio


Oct 6, 2008, 7:21 PM
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Re: [mtnrock] tension [In reply to]
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mtnrock wrote:
haha this post got so off topic i just had a simple question on tention and it went into all this stuff i can't really say it was simple because i can't word anything well but how can i put a diagram so i can show what im talking about more clearly.

Run, Forrest. Run.


onceahardman


Oct 7, 2008, 1:00 PM
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mtnrock, at great risk of furthering misunderstanding, I'm going to try.

The situation is static. Nothing is moving. Climber of weight X is hanging from a normal slingshot-type belay.

Since he is not moving, the tension in the rope between the climber and the 'biners on the toprope anchor is also X. X=X, no motion.

Between the toprope anchor 'biner and the belayer, you also have tension = X, since the rope is not moving.

You now have TWO downward tension vectors, each = X. 2X going downward. The anchor is not moving, so there must be an UPWARD tension vector on the anchor of 2X, or else the anchor would have to be accelerating.

Does that help?


Partner cracklover


Oct 7, 2008, 1:50 PM
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jt512 wrote:
I claim that a toprope anchor should be built to withstand a force of at least 3000 lb.

I agree with both your reasoning and your result.

Furthermore, to demand this does not put an undue burden on the anchor builder. It's a pretty easy task to achieve.

Two strands of military spec webbing equalized in a power-point: 18 kN each. Conservatively lowered by 1/3 for the knots = 24 kN, or 5,400 lbs.

If you substitute in 7mm cord (with half the breaking strength), you should still be fine, so long as you use loops.

As for using gear for anchor points, again, not an undue burden. Assuming your weakest point of a three-point anchor system has a failure point of 6kN, and the anchor is reasonably well equalized, it should hold the 13 kN we're aiming for.

GO


shockabuku


Oct 7, 2008, 2:14 PM
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jt512 wrote:
rgold wrote:
If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

Stop it. You're kill me.

Jay

Damn Jay, that's right up there with "all your base are belong to us."


jt512


Oct 7, 2008, 2:18 PM
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shockabuku wrote:
jt512 wrote:
rgold wrote:
If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

Stop it. You're kill me.

Jay

Damn Jay, that's right up there with "all your base are belong to us."

I blame all my typos on lack of caffeine.

Jay


hafilax


Oct 7, 2008, 2:23 PM
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Minee arre usuallly due too tooo muucch cafffeine.


mtnrock


Oct 7, 2008, 6:08 PM
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yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system


jt512


Oct 7, 2008, 6:32 PM
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mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay


sungam


Oct 7, 2008, 6:35 PM
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jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay
Shhhhh, you'll confuzle them.


mtnrock


Oct 8, 2008, 1:53 PM
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thats not what im saying in both situations there is no acceleration so T doesn't increase


colatownkid


Oct 8, 2008, 2:48 PM
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jt512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

mtnrock wrote:
thats not what im saying in both situations there is no acceleration so T doesn't increase

um...i'm pretty sure that is what you're saying, if i'm not mistaken.


Partner robdotcalm


Oct 8, 2008, 3:38 PM
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jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

rob.calm


jt512


Oct 8, 2008, 3:53 PM
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robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay


Partner robdotcalm


Oct 8, 2008, 4:09 PM
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jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Yes, especially if I (first) stood in front of the parked car and then in front of the car going 30 mi/hr. I'm not sure, though, that I'm willing to do that experiment. Would you do it for me?

r.c

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