Rope Stretch Calculations

I was recently following a 60-meter pitch that begins from the ground. In this case I was climbing on Beal Opera half ropes (35% dynamic elongation). The crux was in the first 20 feet, no communication with belayer above, and given rope stretch I treated it as free soloing. In hindsight, I am trying to figure out how far the ropes would have stretched in the case of a TR fall with a bit of slack. As I understand it a static load would stretch the ropes 10%, so would a TR fall with a bit of slack be somewhere between 10-35% of 60 meters???

Delaney Bray-Stone wrote:

I was recently following a 60-meter pitch that begins from the ground. In this case I was climbing on Beal Opera half ropes (35% dynamic elongation). The crux was in the first 20 feet, no communication with belayer above, and given rope stretch I treated it as free soloing. In hindsight, I am trying to figure out how far the ropes would have stretched in the case of a TR fall with a bit of slack. As I understand it a static load would stretch the ropes 10%, so would a TR fall with a bit of slack be somewhere between 10-35% of 60 meters???

60 X 1.10 = 66m

6m = 19.6 feet.

Had you eaten that burger the night before the rope would have stretched another 6".

Gumby King wrote:

60 X 1.10 = 66m

6m = 19.6 feet.

Had you eaten that burger the night before the rope would have stretched another 6".

Thank god I was on my backcountry rations.

Where do you get the 1.1 from? I thought 10% stretch was for a static load, but in a TR fall with slack I'm assuming there'd be more stretch...

The answer is “it depends.”

If the route is 60 meters long, and your belayer is approx. 200 ft. above you, I would assume that would mean a substantial number of pro placed (either bolted sport route or gear/trad placements).

Depending on how the route meanders and placement extensions, it is conceivable that much of the rope would be compromised and would feel little load/stretch.

An extreme worse case scenario (extreme bends in the rope), the only rope that would feel your fall would be the length of rope from you to the next piece of pro above you (i.e. little stretch, greater force experienced).

Best case, rope runs free the full 200 ft. and your fall would be “held” by the entire rope (i.e. greater stretch, least force experienced).

Also, were you tied in to just one of the doubles, or both strands? Those elongation/stretch numbers you're quoting are for a single strand of the doubles.

Marty C wrote:

The answer is “it depends.”

If the route is 60 meters long, and your belayer is approx. 200 ft. above you, I would assume that would mean a substantial number of pro placed (either bolted sport route or gear/trad placements).

Depending on how the route meanders and placement extensions, it is conceivable that much of the rope would be compromised and would feel little load/stretch.

An extreme worse case scenario (extreme bends in the rope), the only rope that would feel your fall would be the length of rope from you to the next piece of pro above you (i.e. little stretch, greater force experienced).

Best case, rope runs free the full 200 ft. and your fall would be “held” by the entire rope (i.e. greater stretch, least force experienced).

I think those are all bang-on. Rock itself would also contribute friction, unless it' all at least vertical or greater angle.

In addition, the rope itself at that length as a weight that probably isn't neglectible (e.g. it would already be partly stretched). Finally, a ground fall due to extension of the rope isn't directly comparable to a free fall. The rope is stretching after all, and any energy used to stretch it isn't going to be felt upon impact.

Not minimizing the risks due to rope stretch in this context - it's wise to take it into account of course. When I belay from above and cannot see my climbers, I sometimes try to keep the rope a little bit taunt, if the topology of the climb makes sense for it and if my second prefers that approach, mostly for that reason. Depends on the route of course, and what my second preferfs.

aikibujin wrote:

Also, were you tied in to just one of the doubles, or both strands? Those elongation/stretch numbers you're quoting are for a single strand of the doubles.

As half ropes... right, that makes sense.

Franck Vee wrote:

In this particular case the route was essentially straight and slightly overhanging so minimal friction to reduce the elongation. I guess it's a fairly uncommon enough situation to have in combination with a low crux

Keep in mind that the rope behaves like a spring when loaded. As the load increases, the stain (elongation) increases. Under more strain, the total load carried by the rope increases, just like a spring is harder to pull the further you pull it. The result of all of this is a decrease in acceleration as you fall. This is the typical soft fall you feel when taking a normal fall. A simple summary of what I am trying to say is the elongation doesn't happen all at one time, rather it gets applied slowly.

I have taken a 25 foot fall on top rope when almost the whole rope was out. It resulted in touching the ground but very slowly. Unless you have lots of slack, I wouldn't worry about a fall above 10 feet.

I want to point out that obviously as it’s rope stretch and not free falling, the forces and experience is very different. Took a 20-25’ “fall” on TR on a ~68m alpine pitch recently on a single 8.5. Cut up my hand pretty bad and ended up aiding/clipping my pas to gear a lot of the early parts of the pitch to avoid taking another fall. The “fall” felt more like trampoline than anything. It was a pretty graduadual process, as long as the landing is pretty flat and not sharp, I wouldn’t be too worried about falling 15ft up even if you end up touching the ground. Note that this isn’t the same if you’d be nowhere near the stretch limit such as 5ft off the deck.

The big key is to make sure there’s as little slack in the system as possible. Your partner should be aware of the start being hard and pull the rope very tight to try and absorb some of that rope stretch already and make sure there isn’t slack. 70m of rope stretch + 2m dynamic fall*stretch is a lot more stretch than just the static stretch component

Zachary Zwick wrote:

I want to point out that obviously as it’s rope stretch and not free falling, the forces and experience is very different. Took a 20-25’ “fall” on TR on a ~68m alpine pitch recently on a single 8.5. Cut up my hand pretty bad and ended up aiding/clipping my pas to gear a lot of the early parts of the pitch to avoid taking another fall. The “fall” felt more like trampoline than anything. It was a pretty graduadual process, as long as the landing is pretty flat and not sharp, I wouldn’t be too worried about falling 15ft up even if you end up touching the ground. Note that this isn’t the same if you’d be nowhere near the stretch limit such as 5ft off the deck.

The big key is to make sure there’s as little slack in the system as possible. Your partner should be aware of the start being hard and pull the rope very tight to try and absorb some of that rope stretch already and make sure there isn’t slack. 70m of rope stretch + 2m dynamic fall*stretch is a lot more stretch than just the static stretch component

In this situation my concern stemmed from not knowing when the rapid falling would become decelerated falling.

def. agree with your comment regarding slack.

Delaney Bray-Stone wrote:

In this situation my concern stemmed from not knowing when the rapid falling would become decelerated falling.

Well, if you treat dynamic rope as a simple ideal spring following Hook's law you can make some estimations. 

Hooks law says  F = k *dx, here F is returning spring force, k is spring constant, dx is displacement. Static test is done with 80kg weight, it is capped to 10% elongation, this would be, roughly, 6m for 60m rope ( we disregarding weight of the rope and reality). Then k =80kg/6m. Edit - the units are not correct, but that is on purpose, just simplification to make understanding easier

So, at this point you can tabulate returning force, this is what would climber feel pulling up/slowing down-

1m "fall" - 15kg
2m "fall" - 26kg

and so on, till the returning force equals climber's weight. For example, max displacement for 50kg climber (~110lb) would be 3.75m ( around 12ft ). Again, this is static approximation - it is likely to be a bit longer. But, assumptions are for free hanging rope, in reality there, usually, is extra friction, it would make this displacement shorter.

I know that means absolutely nothing - but the feeling of someone experiencing TR on extremely long rope - first 1-2m of fall on rope without extra tension will feel like no rope at all. It is imperative to have belayer understand this, especially if there is no extra friction in the system. Heck, I sometimes TR routes in gym that can only be TR after leading something easier, and even with ~30m rope falling on starting moves feels like no help from  rope if I want to climb without any tension. 

There have been a number of injuries in Indian Creek where a toprope fall low on a 30-40m pitch (belayed from the ground) lead to broken or sprained ankles. I can't say if these were poor belaying or rope stretch alone, but certainly cause for concern. 

2 notes on the ideal spring model:

1) There may be a case for using dynamic elongation (8000 N / 35%) rather than static (800 N / 10%) for calculating the k.  Ropes under constant tension take a while to attain their final length, and I suspect static elongation is measured after at least a minute.  I.e. static elongation includes longer-term rope state changes that don't apply in a fall.  Obviously a 3x larger k means a much quicker and less dangerous fall arrest for TR.

2) Once you figure out the k, simply divide mg by k to get the fall length at the apex (where the body attains its maximum speed).  And the body's speed at the apex is equivalent to a free fall from half that height.

E.g. if L = 60m, impact force = 8kN, and the elongations are 10% / 35% then:
- static k = 800N / (60m * 10%) = 133 N/m, so the apex occurs after 800/133 = 6m (duh) and the speed at the apex is equivalent to a 3m free-fall
- dynamic k = 8000N / (60m * 35%) = 380 N/M, so the apex occurs after 800/380 = 2.1m and the speed is equivalent to a ~1m free-fall