How do you calculate how many kN are generated in a fall?

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Messages 61 - 73 of total 73 in this topic << First  |  < Previous  |  Show All  |  Next >  |  Last >>
Steve Grossman

Trad climber
Seattle, WA
May 19, 2012 - 02:52pm PT
Nice Limericks Rich!
bit'er ol' guy

climber
the past
May 19, 2012 - 04:11pm PT
jump
Mike Bolte

Trad climber
Planet Earth
May 19, 2012 - 04:20pm PT
Those are excellent Rich! Very clever.
Chrisw1096

Gym climber
Jan 7, 2013 - 09:12am PT
Well 4 kilonewtons is about 899.23 pounds or 407.8 kg.

So if you weighed about 80kg (average) and fell about 5-6m with a static rope then maybe you could break it
philo

Trad climber
Is that light the end of the tunnel or a train?
Jan 7, 2013 - 09:34am PT
Great post Rgold.
This was turning into rocket science.

The answer is in the question.
I loved this Studly response.
Wish I would have had that reply in my parenting quiver when my kids were little.
kc

Trad climber
the cats
Jan 7, 2013 - 10:44am PT
Well, rated appropriately or not, my feeling is that it's better to place something than nothing at all. After all, you can't be caught by a piece if you don't put one in.
Steve Grossman

Trad climber
Seattle, WA
Jan 7, 2013 - 10:58am PT
Reach for the Aid Screamers for load-limiting peace of mind and torture that little wired nut to the tune of hundreds not thousands of pounds of impact force. Put a couple on your Hula Harness...LOL
Ed Hartouni

Trad climber
Livermore, CA
Jan 7, 2013 - 12:27pm PT
there is an interesting subthread here: the roll of a fundamental understanding of the physical system in explaining our experiences.

From the point-of-view of most people reading this thread, the explanation of even the simplest system is so complicated and arcane that they assume nothing from that explanation can be used to understand a "real" situation.

That's an unfortunate consequence of using a short hand means of explanation, the mathematical description of the system instead of translating it into a simple explanation, which would be quite lengthy.

So the onus is on us to provide the description and the limits of the model used to do the calculation, and how those limits affect the conclusions. However, they have already provided a huge insight.

That insight is simply that very few anchors have ever been tested at the forces generated in a factor 2 fall. There have also been few tests of such situations using modern gear.

We achieve this insight by noting that the calculations we perform for even simple systems give us an idea of what is going on in the dynamics of a fall (where I use the word "dynamics" in its physics sense) including the distribution of forces on the anchor system.

The lack of experience in holding these types of falls is a good thing, people are climbing cautiously enough to avoid such "tests" but if our assumption that our anchor methodologies are able to hold such falls is not based on sound principles, which are also verified by tests, then we have been lucky rather than good.

We should be both. And our calculations tell us we don't actually know. For this particular aspect of climbing, we should know.
Jim Brennan

Trad climber
Vancouver Canada
Jan 7, 2013 - 01:02pm PT
What are the calculations to resolve this problem ?

Michelle

Trad climber
Toshi's Station, picking up power converters.
Jan 7, 2013 - 01:21pm PT
You know, I was seriously considering returning to school for my math degree. I think I'll pass on it. Too much work! But this is an awesome topic.
Ed Hartouni

Trad climber
Livermore, CA
Jan 7, 2013 - 11:48pm PT
part II

the calculation that Richard has repeatedly worked through is simple enough to understand if you have some calculus background. But we can put it into words too.

The climber is out some distance which we take to be the length of rope, and is climbing above the last piece of protection, we can call that the length of runout.

When the climber falls we'll make our first approximation, that the fall is into free space. Obviously there are many other scenarios, but the largest forces are going to be for falls where the energy of the fall is not transferred to, say breaking bones or sliding friction or spinning.

The work done by gravity is just the gravitational force multiplied by the distance the climber falls. This becomes kinetic energy, that is, the climber speeds up in the fall as the length of the fall increases.

We make another approximation by neglecting the aerodynamic drag of the climber falling in air. This drag might be significant in some very long falls, but for the scenarios we're talking about, this is usually a good assumption. Also, the force of the fall without taking into account aerodynamic drag is greater than when it is taken into account.

The climber falls twice the runout length before the rope is a factor. Once this happens, the rope starts to stretch and the force on the climber is the sum of gravity pulling down, and the rope pulling up. The force the rope exerts is proportional to the fraction of its increased length multiplied by a constant which we call the "spring constant" or the "Young's Modulus" of the rope, but the rope model we consider here is that of a spring which is yet another approximation.

You can see that as the climber continues to fall, the fractional length of the rope increases and the therefore the force pulling up increases. The energy of the fall is going into stretching the rope. It does this incrementally, until all of the energy of the fall is used to stretch the rope to the length that "stores" the energy in the stretch.

This is where the calculation stops. We can show what the force is on the rope as a function of time using this method, so the peak load is known.

If the runout length is not equal to the length of rope out to the climber, we assume that whatever the rope runs through does not impede the ability of the rope to stretch under load, we assume a frictionless system.

Certainly this is not an issue for a factor 2 fall, that is, where the runout length is equal to the length of the rope out to the climber. In such a fall, the maximum force is calculated in this manner.

The more length of rope out to the climber, and the less runout the climber is, the smaller the force, the force factor is just the ratio of twice the runout length divided by the length of rope out.

You can see why a low fall factor fall has less force, the fractional stretch of the rope is much less when more rope is out, if the fall is short the energy required to stop it is less. You might have experienced this simul-climbing when your second falls and weights the rope, not much more additional force than the weight of your partner coming onto the rope. This happens when you belay the second, holding a fall with a lot of rope out entails much less force than when the second is at the top of the pitch.

More elaborate models are possible, putting anchors in with realistic friction is possible, and including into the rope model a way of dissipating energy gets rid of the oscillatory behavior we neglected by terminating our calculation with energy conservation. If you read Italian, German or French there are many good articles written by those alpine club safety sections. As far as I know, the American Alpine Club avoids such things entirely, seems somewhat irresponsible.

Knowing what the maximum forces are on the anchors provides an input to the setups you might design to test anchors. It isn't trivial, as jstan related above, to provide a test that is safe to the testers without some initial calculations that give you an idea of what needs to be provided.

Models of the cordelette and the sliding-x are also possible to make that incorporate realistic features and reproduce test results. In fact, I claim that a relatively simple analysis of equalizing systems shows that every such system reduces to combinations of those two... so there isn't much to discover in proposing new ways of torturing webbing, they all reduce to combinations of those two. That's from analysis...

...Where the AAC could come in is how to interpret this for John Q. Climber so that all the "theory" didn't need to be hashed over and good, practical solutions to some of these problems might be created and transmitted for the good of the sport.
JimT

climber
Munich
Jan 8, 2013 - 02:38am PT
One should be clear that the standard model used is ONLY a rope model and not a system model.
One could add to the standard model to give more accurate predictions and Jay and myself have discussed this previously BUT all the additional factors you have to input are variables (in fact even the standard model uses an input value which is variable since the rope factor is not constant over the life of a rope or even from day to day). This would make the model unusable in practice since to account for these we would need to know what the variability is and for most realistic climbing situations we can only make a wild guess. It also makes the maths vastly more complicated since even such an apparently simple thing as friction over a karabiner has six or so variables and then getting to grips with belay systems is another level of difficulty altogether.
For those involved in equipment design or perhaps evolving and teaching techniques a better model could be useful to show directions to go in to reduce the forces but for normal climbers the biggest variable of all can never be included which is gear placement.
Its worth remembering that the standards for equipment are based on experimentation and historic experience, theoretical modelling doesn´t come into the picture.
Understanding strategies to reduce fall forces is useful for the average climber but understanding the importance of good gear placement and general good practice is vastly more important. Badly tied knots and abseiling off the end or the rope are the sort of thing killing people, not failing to accurately estimate fall forces.
Steve Grossman

Trad climber
Seattle, WA
Jan 8, 2013 - 10:37am PT
Between Ed, Rich and Jim T you have all the physics resources you could want for answering climbing related questions.

Thanks guys!
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