What fall factor will cause me to hit the deck with a typical dynamic rope?
Say I'm leading on a single-pitch climb, or the first pitch of a multi-pitch. The maximum fall factor I could theoretically have without hitting the deck is 1, but that's with a static rope, which of course I wouldn't use in this situation. This raises the question of what is the maximum fall factor I could have without contacting the ground at my belayer's feet. (If the rope stretches, then it's helping me by decelerating me at least somewhat, but I'd really prefer not to touch bottom at all.)
By UIAA standards, static rope stretch for an 80 kg climber is supposed to be no more than 10%. Dynamic rope stretch is going to be greater than this static value.
The UIAA standard fall is a fall factor of 1.77, and rope stretch for that fall is supposed to be no more than 40%. In the first-pitch scenario, you can't actually have a fall that puts this much stress on the rope, since you can't fall past the belayer, but we can be sure than the rope stretch in a first-pitch scenario will be no greater than this. (Again, this is for an 80 kg climber.)
What I can tell from this is that the rope stretch a is in the range 0.1<a<0.4, which is a pretty broad range. Therefore the fall factor f that just barely has me kissing sand is in the range 0.6<f<0.9. What is the actual critical value of the fall factor?
related: What is a typical elongation of a dynamic climbing rope?
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Basing on stretch.
Assume 40% or 1.4 stretch.
Let x be the distance to the anchor from the ground
Let y be the rope out beyond the anchor (free rope)
Total rope out = x + y
Fall factor would be 2y / (x + y)
You touch ground when
1.4(x + y) = 2x
1.4 y = .6 x
y = .43 x
This assumes the anchor point is the ground.
So you cannot even travel 1/2 the distance from the anchor point to the ground.
Need to be aware the rope stretches on both sides of the anchor point.
If you use s for stretch
(1 + s)(x + y) = 2x
(1 + s)y = (1 - s)x
(1 - s) / (1 + s) = y/x
If s = 1.0 then you would hit the ground on any fall. If you fell at the anchor point the rope would not exhibit max stretch but if it did you would touch ground. The rope would double in length and you would touch the ground.
If s = 0 then y/x = 1 which makes sense.
In terms of fall factor
Getting into a fall factor over 1 is reckless climbing
total length
L
fall (to anchor and past)
H
anchor to ground
L - H/2
stretch
S
fall factor = H/L
F
hit the ground when
(1 + S) * L = 2 * (L - H/2)
(1 + S) * L = 2L - H
(1 + S) = 2 - H/L
H/L = F = 1 - S
S = 0
H = L
S = .4
H/L = F = .6
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I don't think the question is really answerable more precisely than you already have, because there are so many variables:
- the original stretchiness of the rope you happen to have - every rope has a different force/elongation curve
- the age of the rope - ropes get less stretchy as they age
- the relative weight of you and your belayer
- the amount of slip the belay device allows
- the amount of slack the belayer allows
Meanwhile outside the realm of theory, routes where there is a significant risk of a grounder even with competent gear placement and belaying are rare, generally well known and require either a leader supremely confident of his/her ability at the grade and climbing style, or special precautions such as a ground level anchor with a sprinting belayer. Or a toprope.
This post was sourced from https://outdoors.stackexchange.com/a/4825. It is licensed under CC BY-SA 3.0.
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