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yokese
Feb 15, 2008, 7:28 AM
Post #26 of 42
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majid_sabet wrote: note; most dynamic rope elongations are between 30 to 38% and yes they act like a bongee cord. GOT IT?
Majid, you're wrong yet one more time (seriously, don't you get tired? because I certainly do)
Those dynamic elongations are measured after a 1.77 factor fall, not in a top-rope. By the way, to pass the test they cannot stretch more than 40%. Therefore, the "guaranteed up to 40% extra stretch" that you mentioned some posts above is, as usual, just another demonstration of your deep ignorance.
Edited to fix the quoting mess...
(This post was edited by yokese on Feb 15, 2008, 6:09 PM)
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theShiba
Feb 15, 2008, 9:41 AM
Post #27 of 42
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1st tree 100% of the load.
2nd tree 100% of the load
3rd tree 52% of the load.
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majid_sabet
Feb 15, 2008, 6:03 PM
Post #28 of 42
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yokese wrote: majid_sabet wrote: yokese wrote: note; most dynamic rope elongations are between 30 to 38% and yes they act like a bongee cord. GOT IT? Majid, you're wrong yet one more time (seriously, don't you get tired? because I certainly do) Those dynamic elongations are measured after a 1.77 factor fall, not in a top-rope. By the way, to pass the test they cannot stretch more than 40%. Therefore, the "guaranteed up to 40% extra stretch" that you mentioned some posts above is, as usual, just another demonstration of your deep ignorance. I did not specify what the FF was or how much FF will creates during a top rope and you started coming with bunch of #a and nonsense. Be a man and accept that you did not know how much the dynamic elongation were till now and They do act like a bongee cord.
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majid_sabet
Feb 15, 2008, 6:12 PM
Post #29 of 42
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theShiba wrote: 1st tree 100% of the load. 2nd tree 100% of the load 3rd tree 52% of the load.
How much does the first anchor (red color) gets ?
100% of the load which is 12 KN ?
is this right ?
(This post was edited by majid_sabet on Feb 15, 2008, 6:13 PM)
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skinnyclimber
Feb 15, 2008, 6:13 PM
Post #30 of 42
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majid_sabet wrote: yokese wrote: If I were a mod I'd consider to move all this bull$hit of "TR tests" from the beginners forums before someone takes them seriously. Unfortunately this self-appointed "rope master" can create quite a bit of confusion. With 90º angle you get 40%+ forces?? Check your basic trigonometry: F_anchor = 2xFxcos(45) = 1.05xF --> an extra 5%, not 40% (and that would be without friction in the biner) Stretch of 40%?? That's a bungee, not a climbing rope. Top ropes generating 18kN??? Please.... most dynamic rope elongations are between 30 to 38% and yes they act like a bongee cord. GOT IT? http://mtntools.com/...e/03DynamicRope.html
Most dynamic cords have an elongation of something like 30% to 40% ON THE FIRST FACTOR 1.78 FALL according to UIAA testing.
Idiot.
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theShiba
Feb 15, 2008, 6:32 PM
Post #31 of 42
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majid_sabet wrote: theShiba wrote: 1st tree 100% of the load. 2nd tree 100% of the load 3rd tree 52% of the load. How much does the first anchor (red color) gets ? 100% of the load which is 12 KN ? is this right ?
The way you drew it, yes. I don't care what the load is. I am giving you the %age of load on the anchor based on the angle of the rope attached to the anchor. For an angle ≤90º, the load on the anchor is 100%. When the angle is >90º there is a reduction factor of the load on the anchor point, as some force is redirected to another point. I approximated the angle you drew to mean that there would be about 52% of the force on that anchor, about 130º.
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onceahardman
Feb 15, 2008, 7:30 PM
Post #32 of 42
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In reply to: I did not specify what the FF was or how much FF will creates during a top rope and you started coming with bunch of #a and nonsense
Actually you did...by bringing up "dynamic elongation"
def:
Dynamic elongation is measured on the first fall of the standard UIAA fall test. Simply put, it is a measure of how much stretch will occur during a 1.78 fall-factor drop — an extreme indicator of how much a rope will stretch during a normal fall.
Odds of a 1.78 ff on a top rope? =0
therefore, 40%=nonsense in this context.
static elongation is a MUCH truer picture of expected rope stretch for a top-rope fall. Even if the belayer stopped taking in rope halfway up the route, while the climber continued climbing...for a 100 foot route, you'd have a 50 foot fall on 50 feet of rope=ff1
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majid_sabet
Feb 15, 2008, 7:38 PM
Post #33 of 42
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theShiba wrote: majid_sabet wrote: theShiba wrote: 1st tree 100% of the load. 2nd tree 100% of the load 3rd tree 52% of the load. How much does the first anchor (red color) gets ? 100% of the load which is 12 KN ? is this right ? The way you drew it, yes. I don't care what the load is. I am giving you the %age of load on the anchor based on the angle of the rope attached to the anchor. For an angle ≤90º, the load on the anchor is 100%. When the angle is >90º there is a reduction factor of the load on the anchor point, as some force is redirected to another point. I approximated the angle you drew to mean that there would be about 52% of the force on that anchor, about 130º.
Well,The anchor with red rope (redirect) does not get 100% but sees 200% or twice as much. This is for the fact that you have load on one end and another load on the opposite side or the other end of the rope.if you build your redirect to handel 12 KN, it will fail cause at 6 KN load, it is already receives 12 KN or twice as much.
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hugepedro
Feb 15, 2008, 7:54 PM
Post #34 of 42
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majid_sabet wrote: Well,The anchor with red rope (redirect) does not get 100% but sees 200% or twice as much. This is for the fact that you have load on one end and another load on the opposite side or the other end of the rope.
Wrong. It will take 1.66 times the force on the climber. In your theoretical example that would be 16.6kn.
majid_sabet wrote: if you build your redirect to handel 12 KN, it will fail cause at 6 KN load, it is already receives 12 KN or twice as much.
Wrong again. If it can handle 12kn it will never fail because there is no way a top rope fall will generate forces that high.
So the point of this thread is . . . what?
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skinnyclimber
Feb 15, 2008, 8:32 PM
Post #35 of 42
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hugepedro wrote: So the point of this thread is . . . what?
To waste our Time... oh yeah, and confuse beginners.
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hugepedro
Feb 15, 2008, 9:24 PM
Post #36 of 42
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skinnyclimber wrote: hugepedro wrote: So the point of this thread is . . . what? To waste our Time... oh yeah, and confuse beginners.
Gotcha. So it's just another normal rc.com thread then, eh?
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drfelatio
Feb 15, 2008, 9:27 PM
Post #37 of 42
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skinnyclimber wrote: hugepedro wrote: So the point of this thread is . . . what? To waste our Time... oh yeah, and confuse beginners.
And to stroke Majid's ego.
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glytch
Feb 15, 2008, 9:55 PM
Post #38 of 42
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yokese wrote: If I were a mod I'd consider to move all this bull$hit of "TR tests" from the beginners forums before someone takes them seriously. Unfortunately this self-appointed "rope master" can create quite a bit of confusion. With 90º angle you get 40%+ forces?? Check your basic trigonometry: F_anchor = 2xFxcos(45) = 1.05xF --> an extra 5%, not 40% (and that would be without friction in the biner) Stretch of 40%?? That's a bungee, not a climbing rope. Top ropes generating 18kN??? Please....
While I agree with your general sentiments re: majid, something's awry with your math on the forces generated by redirects. Using your math, cos(45) = sqrt(2)/2, so 2*F*sqrt(2)/2 = sqrt(2)*F ~= 1.4*F
Other than that, though, right on. This thread borders on silly.
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glytch
Feb 15, 2008, 9:57 PM
Post #39 of 42
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Dave,
Thanks for more clearly explaining why the original question is a) not meaningful without extra data and b) silly even with extra data. I think our resident questioner was trying to get at the issues with a redirect, but his example which includes the edge of a cliff and a (very!) underspecified problem statement is not really helpful towards that end. I'm glad I'm not the only one that noticed the utter lack of meaning in the question as put.
G
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majid_sabet
Feb 15, 2008, 10:02 PM
Post #40 of 42
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(This post was edited by majid_sabet on Feb 15, 2008, 10:34 PM)
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yokese
Feb 15, 2008, 10:11 PM
Post #41 of 42
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glytch wrote: While I agree with your general sentiments re: majid, something's awry with your math on the forces generated by redirects. Using your math, cos(45) = sqrt(2)/2, so 2*F*sqrt(2)/2 = sqrt(2)*F ~= 1.4*F Other than that, though, right on. This thread borders on silly.
Damn.... you're right (damn calculator in radians!).
An apology is due: My apologies, Majid, the extra force is 40% as you correctly stated.
I stand by everything else I said.
Thanks glytch.
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majid_sabet
Feb 15, 2008, 10:18 PM
Post #42 of 42
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yokese wrote: glytch wrote: While I agree with your general sentiments re: majid, something's awry with your math on the forces generated by redirects. Using your math, cos(45) = sqrt(2)/2, so 2*F*sqrt(2)/2 = sqrt(2)*F ~= 1.4*F Other than that, though, right on. This thread borders on silly. Damn.... you're right (damn calculator in radians!). An apology is due: My apologies, Majid, the extra force is 40% as you correctly stated. I stand by everything else I said. Thanks glytch.
I am not here to proof right or wrong
I am here to learn from you guys as well
thank you all and no apologies needed
MS
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