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Partner tisar


Jul 2, 2005, 3:40 AM
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Shockload theory / shockload on the second placement
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This is a spin-off of the last Anchor analysis thread
. Thought it would be better to give it an own thread.

In reply to:
I said that the force on the next piece should depend in part on how much the rope stretched before the first piece failed. If the first piece were to fail quickly then the next piece would not receive much of a shockload. On the other hand, if the first piece were to hold longer, the rope would stretch, and then the next piece would be loaded with, essentially, a less dynamic rope (in the sense that spring force increases with stretch).

-Jay

As this is a very common theory (at least on rockclimbing.com) but I've never found anything about it in the literature, I would be very glad if someone could point out a link/thread/discussion on that. I did a forum search but with no satisfying results.

I tried to talk about that with the instructor of the trad class I took recently. He completely disavowed the existence of said phenomenon. His argument: The energy lost in deformation/friction of the failing piece plus the energy absorbed by the rope just reduces the load on the second piece. The given (and now missing) rope stretch would be just the representation of energy already absorbed and therefor cannot increase the load on the second piece.

I'm no physician and both approaches seem reasonable to me... maybe the answer is out here?

Thanks!

- Daniel


samanddusti


Jul 2, 2005, 6:12 AM
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I think you mean 'physicist'


Partner tisar


Jul 2, 2005, 6:40 AM
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In reply to:
I think you mean 'physicist'

:oops: of course...

- Daniel


petsfed


Jul 2, 2005, 9:04 AM
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I can make a strong argument, but I can't come up with anything in the literature right off hand.

What it comes down to is this: If you have 2 systems with identical amounts of energy, but one with the rope already stretched, the peak force on the piece will be less in the one with the rope unstretched. That's a simple fact.

Now lets look at the two piece case. After the first piece fails, we can claim (due to conservation laws) that some of the energy will already be absorbed before any load hits that second piece. However, the work of a rope is twofold: to take energy out of a system and to make the impact take more time. So while the work done (in this case force inegrated over time) to the piece may stay constant, its peak force will be higher, perhaps even higher than what hit the initial piece. We are worried about peak force. The other thing to remember is that 30 ft/sec/sec is pretty fast and unless you put your second piece really close to your first piece, you'll be back up to the same speed that you hit your first piece with by the time you hit the second piece. There is simply not enough time to unstretch the rope between your topmost piece and the next piece. Rope recovery time is measued in minutes, not milliseconds.

And while there is a limited amount of energy in the system, its a high limit, your mass times your distance from the ground times the acceleration of gravity. In other words, until you hit the ground (or the rope stops you) there's always energy to spare.


alpnclmbr1


Jul 2, 2005, 9:44 AM
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In reply to:
What it comes down to is this: If you have 2 systems with identical amounts of energy, but one with the rope already stretched, the peak force on the piece will be less in the one with the rope unstretched. That's a simple fact.

No, it isn't.

"rope already stretched" = energy

Basically, you are saying A+B > A

Really? Thats good to know.


Going off you first statement of "identical amounts of energy"

Then it would be A = B
with a time differential


As far as the rest of your ramblings.
The best guess seems to be that the "relaxation wave" in a rope travels around the speed of sound. So that would be mili-seconds.


dirtineye


Jul 2, 2005, 10:36 AM
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In reply to:
In reply to:
What it comes down to is this: If you have 2 systems with identical amounts of energy, but one with the rope already stretched, the peak force on the piece will be less in the one with the rope unstretched. That's a simple fact.

No, it isn't.

"rope already stretched" = energy

Basically, you are saying A+B > A

Really? Thats good to know.


Going off you first statement of "identical amounts of energy"

Then it would be A = B
with a time differential


As far as the rest of your ramblings.
The best guess seems to be that the "relaxation wave" in a rope travels around the speed of sound. So that would be mili-seconds.

how much physics have you had alpnclmbr1?


dirtineye


Jul 2, 2005, 10:44 AM
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petsfed, take a look at this. It's posted by a, "genuine physicist". Thanks to Rgold for sending me the link a few years ago when this topic was making one of it's orbits around rc.com.

from rec.climbing

mhu...@bu.edu (maohai huang) writes:
> Ken.Cl...@cs.cmu.edu wrote:
> : When the
> : next piece is significantly lower than the one that failed, the rope
> : has time to relax a bit, dissipating energy that would otherwise
> : contribute to peak fall force (in the form of spring energy.


> so, how quickly does the rope relax? how fast does waves propagate in a
> rope when the first piece pops ? the rope must be acclerating really fast
> since there is a force of many kN solely dedicated to accelerating a couple
> of pounds of rope, until the next piece starts to hold.


To answer this, you have to say what you mean by relax. Some
relaxation will take place as fast as the shock wave can travel the
rope, l/v (where v is the speed of sound an l the rope length). On
the other hand, we all know that ropes need to relax for minutes
(hours? days?) to regain their full energy absorbing potential. The
ultimate answer is somewhere between quite fast and forever.


First, let's consider the fast side of the coin: As any physicist
knows, the speed of longitudinal waves in a string is given by


v = sqrt(F/mu),


where
F, the rope tension, might be in the 2-10 kN range, and
mu, its mass per unit length, is around .07 kg/m


So v is around 150-400 m/s. Approximately.


Even at the slower speed, the relaxation wave will travel, say 15
meters of rope 15 meters of rope in 0.1 second. That's pretty fast -
certainly faster than you will fall (terminal velocity is on the order
of 50m/s). I think it is safe to assume a good deal of relaxation
occurs in practice, even if include factors like friction which I left
out for simplicity.


>


> Maybe there are too many factors for my feeble mind to grasp, how about an
> example: there are two pieces X feet apart, a climber of 150lb is falling
> on a single rope from 5 feet above the first piece, and 30 feet above a
> static belay. The rope stretches at 8% with 80kg load. The rope density is
> 65g/m. The first piece pops after a half of the climber's kinetic engergy
> (after falling 10ft) is absorbed by the rope. To simplify the matter,
> suppose the popping was instant and the piece absorbs no energy. The
> qustion now is: what is the maximum force the 2nd piece has to take in
> order to hold as a fuction of X? what kind of distances X has to be
> so that the peak force on piece #2 to be less 50% of the force that the
> #1 piece had when it popped?


>



To get a rough approximation, assume the pieces are far enough apart
for the rope to relax significantly and subtract the work done on the
rope (that's half the potential energy of falling 10 feet - i.e. 5
feet worth of free-fall) from the fall energy. Better still, subtract
the 5 feet from the distance fallen to determine a new fall factor.


OK, force goes up with the square root of fall factor:


f = k*sqrt(ff)


In the scenario you presented, we have a fall distance of 2X+10, so
without to RP, the force is


f = k*sqrt( (2X+10)/30 )


with the RP, the force is


f_rp = k*sqrt( (2X+5)/30 )


A reasonable value for k is k = 1348 lbf, so we get the following table:


X (ft) f f_rp (forces in lbf)


1 852 651
2 921 738
4 1044 887
7 1205 1073
10 1348 1230
15 1556 1456
20 1740 1651
25 1906 1825


Thus, the failed RP will in theory reduce peak loads by about 100-200
lbf. In no case do you get a reduction by 50% of the force at the
point of RP failure, and that's making some generous assumptions in
the calculations.


This should come as no surprise, since fall force increases with the
square root of fall energy. Take away half the energy and you've
reduced the force by less then 1/3.


Hope this helps!


Ken


Edited to paste the text, thanks stymingerfink.


stymingersfink


Jul 2, 2005, 11:01 AM
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check your link, i'm curious where it points to but end up misdirected


dirtineye


Jul 2, 2005, 11:08 AM
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In reply to:
check your link, i'm curious where it points to but end up misdirected

Stupid links! I just pasted the text in there. Thanks.


stymingersfink


Jul 2, 2005, 11:25 AM
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so by his conclusion:

"This should come as no surprise, since fall force increases with the
square root of fall energy. Take away half the energy and you've
reduced the force by less then 1/3. "

may we surmise that even if the pieces were placed relatively short distances apart, the actual reduction of impact force on the second piece would be minimally less than the initial impact force on the top piece? (and that the rope will still stretch to absorb energy and increase the time that force is applied to the second piece)




jt512


Jul 2, 2005, 11:44 AM
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In reply to:
This is a spin-off of the last Anchor analysis thread
. Thought it would be better to give it an own thread.

In reply to:
I said that the force on the next piece should depend in part on how much the rope stretched before the first piece failed. If the first piece were to fail quickly then the next piece would not receive much of a shockload. On the other hand, if the first piece were to hold longer, the rope would stretch, and then the next piece would be loaded with, essentially, a less dynamic rope (in the sense that spring force increases with stretch).

-Jay

I tried to talk about that with the instructor of the trad class I took recently. He completely disavowed the existence of said phenomenon. His argument: The energy lost in deformation/friction of the failing piece plus the energy absorbed by the rope just reduces the load on the second piece. The given (and now missing) rope stretch would be just the representation of energy already absorbed and therefor cannot increase the load on the second piece.

I think he's confusing energy and force. The rope stretch and friction do indeed reduce the kinetic energy of the falling climber. If the next piece were impacted with as elastic a rope as the first piece, then the maximum force on the second piece would be reduced, assuming that the climber hasn't had much time to re-accelerate. But if the rope is less elastic when the second piece is impacted, the force will be greater than in the above scenario. Consider, in theory, a rope whose entire stretch was used up just as the first piece popped, and assume that the rope hasn't recovered at all when it impacts the second piece. Then the maximum force on the second piece would be high, in spite of the reduced energy of the falling climber.

-Jay


Partner tisar


Jul 2, 2005, 12:26 PM
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In reply to:
I think he's confusing energy and force. The rope stretch and friction do indeed reduce the kinetic energy of the falling climber. If the next piece were impacted with as elastic a rope as the first piece, then the maximum force on the second piece would be reduced, assuming that the climber hasn't had much time to re-accelerate. But if the rope is less elastic when the second piece is impacted, the force will be greater than in the above scenario. Consider, in theory, a rope whose entire stretch was used up just as the first piece popped, and assume that the rope hasn't recovered at all when it impacts the second piece. Then the maximum force on the second piece would be high, in spite of the reduced energy of the falling climber.

-Jay

Maybe you're right that I (don't know if he did) mixed up energy and force in the OP.

Nonetheless: Consider the fall would be on a static rope. Then the impact on the second piece would be the same as on the first one (as no rope stretch took place) minus the reduction of falling speed provided by the the first piece.

Now to the dynamic rope:

Extreme a): 1st piece pops immediatly. There's no reduction of falling speed, but also no stretch in the rope - you fall on the 2nd piece as if there was no 1st one.

Extreme b): 1st piece pops when rope is stretched to the max. We now have (idealized) a static rope, the force on the second piece would be exactly the same as on the first one the very moment it failed.

As I see, this also refers to every case between the two extremes: The force on the second piece (no regain of stretching ability given) would be at max the same as on the first one in the moment it failed.*
Thus in the (theoretically) worst case, the second piece is loaded as if the first one didn't exist - but there's no way the impact on the second piece could be higher than on the first one.

Maybe I'm missing something. Correct me, if I'm wrong...

- Daniel

* edited to clarify


Partner tisar


Jul 2, 2005, 1:09 PM
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Addition, as this might be the most irritating point in the whole discussion:

For sure the maximum force on the second piece might be reached much faster, (worst case again) maybe even immediatly. But this wouldn't affect the placement as the resulting force on the piece is the same.

- Daniel

(still wishing his english would be precise enough to discuss such topics the easy way :? )


thegreytradster


Jul 2, 2005, 8:53 PM
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The practical answer to this question is; when things get out of hand, pile in the gear, use screamers if you have them and revert to "the leader must not fall"


blondgecko
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In reply to:
In reply to:
I think he's confusing energy and force. The rope stretch and friction do indeed reduce the kinetic energy of the falling climber. If the next piece were impacted with as elastic a rope as the first piece, then the maximum force on the second piece would be reduced, assuming that the climber hasn't had much time to re-accelerate. But if the rope is less elastic when the second piece is impacted, the force will be greater than in the above scenario. Consider, in theory, a rope whose entire stretch was used up just as the first piece popped, and assume that the rope hasn't recovered at all when it impacts the second piece. Then the maximum force on the second piece would be high, in spite of the reduced energy of the falling climber.

-Jay

Maybe you're right that I (don't know if he did) mixed up energy and force in the OP.

Nonetheless: Consider the fall would be on a static rope. Then the impact on the second piece would be the same as on the first one (as no rope stretch took place) minus the reduction of falling speed provided by the the first piece.

Now to the dynamic rope:

Extreme a): 1st piece pops immediatly. There's no reduction of falling speed, but also no stretch in the rope - you fall on the 2nd piece as if there was no 1st one.

Extreme b): 1st piece pops when rope is stretched to the max. We now have (idealized) a static rope, the force on the second piece would be exactly the same as on the first one the very moment it failed.

As I see, this also refers to every case between the two extremes: The force on the second piece (no regain of stretching ability given) would be at max the same as on the first one in the moment it failed.*
Thus in the (theoretically) worst case, the second piece is loaded as if the first one didn't exist - but there's no way the impact on the second piece could be higher than on the first one.

Maybe I'm missing something. Correct me, if I'm wrong...

- Daniel

* edited to clarify

I'm afraid I'm going to have to correct you here.

Think of it this way...

Look at the force on the piece as a function of time distance as the rope rope stretches - it starts at zero, increases to a peak as the rope dissipates the energy, and decays away to a constant as you slow to a stop. Now, the energy dissipated in that fall is directly related to the area underneath the force-stretch curve (it's been much too long for me to remember the exact relationship) - that is, for a given fall length, the area under the curve is constant.
The role of the stretch in the rope is to "stretch" this curve on the distance axis. Remembering that the area under the curve is constant, that means that the peak force must come down.
On the other hand, in a truly, absolutely static situation, the distance axis is compressed to zero, and thus the peak force becomes infinite. Of course, this never happens in reality, but the argument here is how far towards this extreme "pre-stretching" the rope takes you.

Edit: just a small comment... in the club I started climbing in, we used to top-rope using static. I guess there's static and then there's static, but this stuff still had about 2% stretch. That meant that, on a 25m climb, if the climber fell off within the first 2 m, they would find themselves standing back on the ground. Even a small amount of stretch makes a big, big difference.

Edit #2: As pointed out by rgold, the work done is actually the equal to the area under the force-distance curve, not the force-time curve. I knew there was something not quite right, but it was too late in the day to work out what. :oops:


maxclimber1w


Jul 2, 2005, 10:09 PM
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So, correct me if I'm wrong, but it (the first piece failing) could reduce the force on the 2nd piece if it is near enough (very) to the 1st;

if the 2st is really far down, the force will be about equal or slightly greater because the rope has rebounded considerably (but the fall factor has increased);

and if the 2nd piece is at a medium/close distance, the force will be maximum because the rope will not have rebounded and the climber will have regained his momentum


alpnclmbr1


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Hang a thousand pound rock on a full length 60 meter rope.

Cut the rope with a razor and how fast do you think the end of the rope is going to shoot upwards. Faster than a bullet at 300 feet per second? Easily.


redzit


Jul 2, 2005, 11:03 PM
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Wow, very impressive physics going on here. Here's my two cents. By the way, I have some experiance writing papers on rock climbing falls, (one is to be published, I'll try and find out where) on falls from the upper vertical face of an over hang. so this is kinda familiar teritory. BUT ANYWAYS

SKIP TO HERE TO AVOID RANT
in order to safely make any conclusions about "shockloading" some constants have to be set with all the possible variable in a fall scenario, and they have to represent the worst case scenario, jsut to be safe.

So, here are My constants for a rope that stretches first and then pulls/destroys the first anchor.

- The rope stretches its full capasity and length, ergo it cannot stretch anymore, and is essencially static.

- The first piece of pro did not fail until the force on it is equal to the greatest force a human body can sustain, (about 12 kN)
(btw, SI units just for consistancy with UIAA standards)

These first two, I believe to represent the worst case for there variables. It also jsut so happens that the UIAA has a standard that for a factor 1.7 fall, a 4.8m fall on 2.8, given a specified length of rope and an 80kg mass, the force must never exceed 12 kN, and must have no more than a 40% elongation

- Factor 1.7 fall on the first piece

To further the worst case scenaro:

- The belay point IS the second anchor, (any distance away will padd the fall more, hance more worst case) and the belay is fixed by webbing to the fattest bolts imaginable, and the belay is grigri, and you belayer is superman starting to go cold turkey on smoking so hes kinda jump and grabs hold of the rope super fast.

Assume the mass of the rope and the first piece of pro that fails to be negligible.

At this point the math becomes simple, relatively speaking.

The distance between the two pieces of pro must be in a ratio of 1:16

this can then tell you the total energy of the climber the moment jsut before the rope takes load. you can also determine the elogation of the rope, (max 40%) and there for determine the distance and time over which the absorbtion of the 12 kN took place, allowing you to figure out how much Kinetic energy the climber still has.

The climber then falls to his next piece on a rope that is, for all intent and purposes, now a static line.

But let me save you some math.
a fall greater than one meter in length on a static rope will generally create enough force to break a climber's back, due to the sudden stop at the end. so if your pro is more than a meter apart, there is a chance, you could be screwed, and superman will have to take your dangling, broken, and by now bleeding carcas to the ER.

The flip side of this worst case scenario is that the first piece blows off because of the wind jsut as your starting to fall, in which case its a normal,although long, fall.

Everything else is in between people.


AND JUST BECAUSE i know someone is going to be all smart and point out that the worst case scenario doesn't represent reality well at all, that most falls arn't going to be like that...well yes i realize that, but this is all theoretical reasoning, and there for can only prove any given theory or idea if it can be proven at an infinate point/most extreme case.


All standards on rope testing by the UIAA are from the UIAA testing guide lines Dynamic Mountaineering Ropes of both EN, Section: 892, and UIAA, section: 101. Up to date as of June 2004.


Partner tisar


Jul 3, 2005, 5:21 AM
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This seems to get out of hand...

Just to clarify: The question is not weather or not the second piece is loaded to failure if the rope is completely streched out.

The question was and is still if the existence of an additional piece placed closely to the first one leads to a higher force on the second.

If this would be true one would try to avoid placements close to another to prevent shockloading the second one in case the first one blows. If not, there's reasonable cause to place another piece close to a manky placement to raise the chance that at least one of them holds the fall.

To avoid another complicated explanation, I tried to get some graphics done, which hopefully display my argument:

http://home.versanet.de/...all-on-one-piece.jpg

http://home.versanet.de/.../one-piece-fails.jpg

As I stated before I don't see that the second piece could ever be exposed to a peak force which is higher than the max impact the first one would have had to withstand. A 'shockload' in the sense that there's an additional force appearing out of... whatever, I just can't see.

In contrary I'm pretty sure that a) the energy the blow of the first piece takes out of the system and b) the regain of stretching ability of the rope (both not counted in when I did the graphics) reduce the force on the second piece enormously.

- Daniel


kel_e


Jul 3, 2005, 9:10 AM
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Hi Tisar,

If the two pieces are close together, so close that there is not enough time for the rope to recover it's elasticity before loading the second piece, wouldn't it be better to equalise the two pieces?


sspssp


Jul 3, 2005, 9:28 AM
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In reply to:
- The rope stretches its full capasity and length, ergo it cannot stretch anymore, and is essencially static.

Well then, you had better be d*mn glad the first piece failed. Otherwise, you would have one very broken body after your rope turned into a piece of chain... :wink:


karlbaba


Jul 3, 2005, 9:47 AM
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Re: Shockload theory / shockload on the second placement [In reply to]
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I know folks love to analyze fall scenarios with geeky physics that fail to consider all the factors and forces that bear on the situation (which would be unbearably complex)

That's good fun, but, just as sort of a disclaimer for any newbies that might think that they need to think about these arguments too hard in order to be safe, I feel like I should say:

The real safety issue in these scenario is whether hit simething on the way down until your rope stops you, and if you managed to avoid the rope flipping you over (cause maybe it was between your legs.)

In a real situation, if you don't hit anything, you're going to be fine with impact forces on your body in any scenario where God is not out to get you, and a good piece is going to hold you. (assuming modern rope)

Doubt it? Here's some back up. I took a whipper once which was hard enough to break the cable of a #4 stopper. I fell about 12 feet further and a wired hex with the same strength cable held my fall. Total fall, maybe 40 feet, distance from the ground after I fell, pretty darn close! Stress on my body, very little, it felt like a bungy jump.

My guess is the vast majority of pieces either pull right away before the rope stretches much, or they hold, but hey, that's a guess.

Screamers are a good call for sketchy pieces.

Peace

karl


dirtineye


Jul 3, 2005, 1:00 PM
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Re: Shockload theory / shockload on the second placement [In reply to]
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http://www.impact-force.info/anglais/impact3.html



I like geeky physics stuff, but I still like what karlbaba wrote.


geezergecko


Jul 3, 2005, 4:31 PM
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Re: Shockload theory / shockload on the second placement [In reply to]
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One would have thought that the UIAA or some alpine organization would have tested this. So far in this thread there have been logical arguments or anecdotal experience but no actual measurements. As Douglas Adams once said "Assumptions are the things you don't know you are making". Goran Kropp pulled an undercammed #3 Camalot and then proceeded to break the biner on the second piece (#2 Camalot), rip out the third piece (#1 Camalot), and finally rip out the fourth piece (#3 TCU). That suggests shock loading to me. The Beal website reference above that showed the actual fall factor to be higher than the theoretical fall factor due to rope drag through the carabiners would indicate that gear placed farther apart would result in lower actual fall factors. Sounds counter-intuitive despite being correct. It all makes double ropes look like a good investment - wider spaced gear per rope (as compared to a single) to lower the actual fall factor and if one rope shock loads then the other is bungie fresh. Maybe the British have this thing figured out as they almost always trad with doubles.


brutusofwyde


Jul 3, 2005, 5:38 PM
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Re: Shockload theory / shockload on the second placement [In reply to]
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In reply to:
That's good fun, but, just as sort of a disclaimer for any newbies that might think that they need to think about these arguments too hard in order to be safe, I feel like I should say:

The real safety issue in these scenario is whether hit simething on the way down until your rope stops you, and if you managed to avoid the rope flipping you over (cause maybe it was between your legs.)

In a real situation, if you don't hit anything, you're going to be fine with impact forces on your body in any scenario where God is not out to get you, and a good piece is going to hold you. (assuming modern rope)

Doubt it? Here's some back up. I took a whipper once which was hard enough to break the cable of a #4 stopper. I fell about 12 feet further and a wired hex with the same strength cable held my fall. Total fall, maybe 40 feet, distance from the ground after I fell, pretty darn close! Stress on my body, very little, it felt like a bungy jump.

My guess is the vast majority of pieces either pull right away before the rope stretches much, or they hold, but hey, that's a guess.

Screamers are a good call for sketchy pieces.

Karlee, you're my hero. As a lowly physics nOOb this discussion was making my head swim and my eyes cross, seeing purple cows and asterisks, and thinking "But what if I hit something on the way down?" and "Shouldn't I be using screamers on these pieces?"

Ratchet and I hope to get out your way some time before this year ends.

Brutus

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