
Lutze
Nov 3, 2013, 12:45 PM
Post #1 of 6
(6447 views)
Shortcut
Registered: Oct 19, 2013
Posts: 4

Hi Does anyone know how to calculate the rope elongation when using two single ropes together? For example when using two Beal Top Gun II: Top Gun II 10.5 mm Dynamic elongation  37 % Elongation under 80KG  9,5 % K = 16000 Notice: I'm well aware of the difference between single, double and twin rope systems.





amarius
Nov 3, 2013, 4:34 PM
Post #2 of 6
(6387 views)
Shortcut
Registered: Feb 23, 2012
Posts: 122

If the two ropes are identical, half. Simplifying this quite a bit  assume both ropes get the same tension, hence each one is experiencing half the weight, hence half elongation.





kennoyce
Nov 4, 2013, 9:47 AM
Post #3 of 6
(6289 views)
Shortcut
Registered: Mar 6, 2001
Posts: 1338

amarius wrote: If the two ropes are identical, half. Simplifying this quite a bit  assume both ropes get the same tension, hence each one is experiencing half the weight, hence half elongation. This is only correct for a static mass hanging from the ropes (static elongation), for dynamic elongation (ie elongation in a fall) this doesn't hold true. The equation for elongation is that elongation is equal to the force on the material (ie rope) times the length of the material to absorb the force divided by the cross sectional area of the material times the material's modulus of elasticity. elongation = (F*L)/(A*E) For two ropes, both the the length and the modulus of elasticity will remain the same and we know that the area is doubled, but just saying that the elongation is half because the area is doubled isn't correct because the force will also increase due to there being a lower elongation. When you do the math (incorrectly assuming that the rope is a perfect spring and that no energy is lost to heat) you find that the force ends up being 1.414 times more than the force on the single rope. This equates to a two rope elongation of .707 times the single rope elongation. Of course we did neglect any energy lost to heat as the ropes stretched, so the dynamic elongation of two identical ropes taking the exact same fall as a single rope would be somewhere between .5 and .707 times the dynamic elongation of the single rope with a rope having less dampening being closer to .707 and a rope with more dampening being closer to .5





amarius
Nov 4, 2013, 11:44 AM
Post #4 of 6
(6268 views)
Shortcut
Registered: Feb 23, 2012
Posts: 122

You are absolutely correct in pointing out that dynamic fall is going to be different This is likely to get long winded :) Let's assume ideal springs, no energy dissipation due to untwisting of rope fibers and internal friction. For anyone curious, here is review of UIAA testing. In short  static elongation is defined on 1m of rope by adding 75kg weight. ( 5kg make sure that the rope is taught, then the weight is changed to 80kg, weight difference is 75kg)  dynamic elongation is defined by 4.8m fall of 80kg weight on 2.5m of rope static test gives as k=75[kg]*9.81[m/s2]/( 1[m]*0.095)=7.7e3 [kg/s2] In the dynamic fall, mass falls 4.8 meters before 2.5 meter rope starts stretching. At that point, rope is starting to accumulate elastic energy, the mass is still falling. I am calling x distance from the moment rope starts stretching. 80[kg]*9.81[kg/m2]=k*x^2/2  80[kg]*9.81[kg/m2]*x This is a quadratic equation, we can solve for x In the original case elongation turns out to be 43%, which shows that there were unaccounted paths for energy dissipation ( Beal says 37%, close enough for these calculations) Of course, the original question was about the effect of combining two ropes. There are two obvious ways to combine two ropes  series, and parallel. It is quite easy to calculate resulting spring constants for both cases Series  k2=k/2; Parallel  k2=2*k; In series it would be stretchier, in parallel  stiffer. Rope elongations, not accounting for energy dissipation, then become  One Rope,  43% Two Ropes in Series  64% Two Ropes in Parallel  30%





Lutze
Nov 5, 2013, 1:35 AM
Post #5 of 6
(6203 views)
Shortcut
Registered: Oct 19, 2013
Posts: 4

Thanks! Very good answers. I've summoned it up into two examples below. To see if I've understood this correctly
Example1, single rope: Dynamic elongation = 37 % m = 10 kg H = 1 m L = 1 m; => fall factor(f) = H/L = 1 K=16000 (Top gun II, 7.5kN => 16000) impact force(F) = mg + mg*sqrt{1+2fK/mg} => F = 1873 N Example2, parallell ropes: F*1.414 = 2648 N (abit less considering energy escaping in the system.) Dynamic elongation*0.5 = 18.5 %





USnavy
Nov 5, 2013, 5:18 PM
Post #6 of 6
(6136 views)
Shortcut
Registered: Nov 5, 2007
Posts: 2667

kennoyce wrote: amarius wrote: If the two ropes are identical, half. Simplifying this quite a bit  assume both ropes get the same tension, hence each one is experiencing half the weight, hence half elongation. This is only correct for a static mass hanging from the ropes (static elongation), for dynamic elongation (ie elongation in a fall) this doesn't hold true. I am not sure it is true period. Ropes do not have a linear modulus of elasticity, and adding another rope changes the modulus of the system all together. Consider the specs of the Beal Joker, which is UIAA certified as a single, twin and half: Number of falls factor 1.77  Béal Guarantee 5 (with 80 kg) 20(with 55 kg/1 strand) >25 (with 80 kg/2 strands) Sheath slippage  0 mm Extension during the first fall  37 % 32 % 29 % Static elongation 80 kg  8% 8%/1 strand 7%/2 strands So adding the second strand only reduces the static elongation from 8% to 7%, and the dynamic elongation from 37% to 29%. The impact force rating for the rope is 8.5kN (single) and 9.2kN (twin). The following is a graph of some testing I did on webbing awhile back. I tested the elongation of 1" webbing, 3/4" webbing, and 1" webbing combined with 3/4" webbing. While webbing is not a dynamic rope, it is still made out of the same material (Nylon 66), and it illustrates the general point I am trying to make.
(This post was edited by USnavy on Nov 5, 2013, 5:34 PM)








