UKC

I can understand how to calculate Fall Factors but Newtons and Kilo-newtons are way beyond me that I would appreciate some advice on what impact force on Kns I put on the system?

Outside I have only had two falls. The first when I was 16 meters up a route with only one piece of protection at around 13 meters. I fell a total of 6 meters giving a fall factor of 0.35.

I can understand that bit of maths from the UKC calculator but what would all this conver to in Newtons?

I've put on a bit of weight since the fall but at the time I think I was only 60kg.

The second fall was about an inch so I aint worried about that one. I don't like falling.

I'm 73kg today, and seriously worried if my old kit, such as slings and extenders connected to my quickdraws are way past their shelf-life.

In reply to The Lemming:

Comprehensible information here:
http://www.thebmc.co.uk/Feature.aspx?id=1477

In reply to The Lemming:

You dont need the calculation: don't use slings or webbing that are very worn or that have had long term UV exposure.

PS In your scenario you should have fallen over 7m (slack in system and rope stretch).

In reply to The Lemming:

2 things

73Kgs is still quite light (60Kgs is very light)
soft goods have a much shorter life than the metal work
how old are the slings etc?

In reply to rallymania:

You are exceptionally kind.

In reply to The Lemming:


Something not right, there, 16m up, 3m above your gear and you fall, I reckon you would go at least 10m unless the rope was bow-taught and the 2nd pulled you off (so to speak!!!)

Chris

In reply to The Lemming: simply put F=MA or the mass of the falling object mulitplied by the Accelleration of the object (in this case 9.8ms^-2) is equal to the force in newtons (1000N = 1kn and that roughly equates to 980kg). There are more complex equations to take into account the initial velocity and final velocity and distace fallen etc (google kinematic equations).

simply put if you fall 9.8m onto the ground you will hit it with a force of approximately 73*9.8 = 715n (0.715kN). this is as simple as it gets and when you take into account you fall past a piece of gear before your weight comes on to it and the elastic properties of rope, bodies etc, it quickly gets more complex and confusing.

In reply to The Lemming: I would pay far more attention to my judgement and a common sense approach rather than worrying about the maths and the science.

Al

In reply to Gaston Rubberpants: I disagree, I think a rudimentary understanding of the maths and physics of falling objects helps you understand the limitations, and indeed the capabilities of your gear.

In reply to higherclimbingwales: Yes when you are sat at your desk considering which gear to buy. I have visions of climbers hanging from one hand about to fall off and trying to work out the maths and the physics as the strength slowly drains from their fingers. I stand by my argument. That doesn't mean that I don't have a rudimentary appreciation of the science though.

Al

In reply to higherclimbingwales:

Ah,
Highclimbers physics is wrong i'm afraid.

What he/she has worked out is the force generated by sitting on the end of a rope (gravity). When you fall off a climb you'll accelerate towards the ground at 9.8m/s, and quickly pick up a decent speed to write home about.

The deceleration, and hence force, caused by the rope stopping you will be much higher as it happens faster.

Dynamic ropes stretch when you land on them slowing the rate of deceleration and then the system gets complicated when you try and work out forces on individual components.

In reply to Surge:

my physics is not wrong, what I said is factually correct. I was merely highlighting the acceleration due to gravity which is actually 9.8 m/sec/sec.

Also, The rope elongates the deceleration and reduces the force.

In reply to The Lemming:

You can work out the peak deceleration, generally the point in time at which the rope stretch becomes >0 (assuming locked belay about to slip and good ropework). At that point in time is the peak force F=ma produced during the fall.

After that point assuming a good system, rope slippage and elasticity in the rope the deceleration tends to 9.8m/s^2 ie the force due to gravity and you are static dangling on the rope.

In reply to higherclimbingwales:

you gotta watch for that rope induced elongating deceleration!

In reply to The Lemming:

Head hurts. Of those that have done the sums so far, have you come to a general consensus of how many Kns I put into the system when I fell?

I'm only after a ruff estimate, so that I can put my fall into context of all the safety numbers stamped onto my kit.

Cheers

In reply to The Lemming:

It's so hugely dependent, that's the point. If you stop instantly the force is infinite.

In reply to The Lemming:

As I said there is no need. If its worn or bleached replace it, if not its probably OK.

In reply to Surge: The physics is right, as an average acceleration due to gravity is 9.81 ms-2 so in theory no matter how high I were to jump from my FORCE will always be about 981N (I weigh about 100kg).

The real problem (as you rightly point out) is the speed (velocity if we're splitting hairs) you'll build up on the way down. The greater you're speed the more energy you'll have when you hit the deck, that's when problems start...

In reply to The Lemming: I am assuming you are still alive and not writing these posts from beyond the grave. in which case, I wouldn't worry about it as it's obviously well within the SWL of your gear!

In reply to tmather: And *your before the punctuation and grammar police arrive.

In reply to tmather:

Nope, it's the deceleration that produces the force. Not the build up of velocity. I could stop from the speed of light to 0m/s over 1000 years and produce less force on the rope than stopping from 10m/s over 0.00000000000001s

In reply to The Lemming:

I'd guess way less peak force than the numbers stamped on the kit. Way too many variables to be more accurate than that.

In reply to John Roberts (JR): Not suggesting velocity produces force, I'm stating that it produces energy (i.e. Joules as opposed to Newtons)

In reply to The Lemming:

This is why fall factor is used, as it allows you to work out at least a comprehension of the scale of force involved.

See if this makes sense

http://www.camp4.com/rock/index.php?newsid=231

Like it says, you'll probably break your back before a 12kn rated piece of gear.

In reply to tmather:

I know but it's mainly irrelevant unless you look at it in the way that you're dissipating the kinetic energy 1/2mv^2 over time which is basically deceleration. You must know the rate at which you slow down on the rope which has very little to do with gravity and nothing at all to do with your velocity.

In reply to The Lemming:

Hang on your rope statically (all or part of it) and find the percentage by which it stretches. Call this P.

If your mass in Kg is M then you can work out (to a pretty good approximation) the maximun tension in your rope during a fall with fall factor F by using the formula:

140 x M x the square root of (F/P)

eg if your mass is 73kg and say your rope stretches by 5% (not sure if this is realistic) when you hang on it, you will generate in your 0.35 factor fall:
140 x 73 x sqrt(0.35/5)
= 2700 newtons

Your bodyweight in newtons is about 10 times your mass (ie 730 newtons), so this maximum tension is about 4 times your own bodyweight (ie it's like 4 of you hanging on the rope at the same time - which is nothing for aclimning rope.

A factor two fall gives anout 9 times bodyweight with 5% stretch. Again, I can't imagine this would be a problem. Remember, of course, that the runner which holds the fall will be loaded with twice this.

In reply to tmather:
You and highclimber are stating facts that are correct, but they are not the answer to the question. You are calculating the force exerted on you by gravity. The OP wants to know the force exerted on the gear. They are not the same.
If you are hanging from a runner, the force on it is double the force on you - since the second has to be exerting the same force to keep you there.
When arresting a fall, the force on the runner is the force exerted by your weight, plus the force required to bring you to a halt from your downward plummet. This second force is much higher, and very difficult to calculate. It depends on how fast you are falling (which you can work out), the rate at which the rope stretches (which may not be linear, so is hard to calculate), and the friction in the system ( no chance of calculating it).
In short, it's so hard to come up with a figure that's accurate enough to be useful that no one bothers. If you want to know, you need to set up a test and measure the force.

In reply to higherclimbingwales:

This is, I am afraid, bollocks.
The force due to gravity at any time is 715N. this has nothing at all to do with hitting the ground.

In reply to higherclimbingwales:

If the force you are referring to is your 730N, then I am afraid this is also bollocks. As you decelerate, the force on the rope will always be greater than 730N, however stretchy the rope. It will only be 730N once you have stopped.

In reply to John Roberts (JR):

More bollocks


And more bollocks

In reply to Hooo: Fair point sir

In reply to tmather:

Yet more bollocks.
Unless you are jumping from a spacecraft, you won't need to worry about average acceleration due to gravity. What do you mean by MY FORCE? What you are talking about is your weight (ie the force on you due to gravity). You have this all the time, falling or not. Stating it here gets you no nearer working out whether the rope might break.

In reply to Robert Durran:


I was assuming an elastic rope (ie a climber of twice the weight will stretch the rope twice as much) - probably fairly realistic.

Any slack in the system will (contrary to the popular myth) increase the maximum tension in the rope.

Friction through runners will increase the maximum tension - a bit harder to deal with.

In reply to The Lemming: There are many variables involved in a fall when climbing, as said, force, mass, acceleration, also inertia, resistance, weather conditions etc. We used to do a lot of similar calculations at uni, albeit, different scenarios. You can make it as complicated or as easy as you want, but, I would just read the information provided by the BMC, rather than ask people on here because most people really don't understand what they are attempting to explain, some parts are correct, most isn't.

Just read the BMC's advice, they are usually pretty good.

In reply to Robert Durran:

I would say bollocks, but I'm not rude. My first assumption is that you're a Maths or Physics teacher, I hope that's not your normal feedback... anyway.

Climbing ropes do not obey hooke's law. If they did, you wouldn't need to let them "rest" after a fall. Therefore your assumptions break down and P is inaccurate. It's a good approximation but it's very simple, it isn't a truly elastic system.

There is a peak force, that's obvious. The peak force, matches the point at which the deceleration is greatest (it isn't constant for the time you slow down to stationary), that's obvious. Can you draw me a graph of force (y) against time (x)? When is F(max) with respect to rope elongation (where the rope doesn't obey Hooke's law.)

What is clear is that the F(max) is a factor of many different parts of the system. If it wasn't a grigri would be a perfectly sane belay device for trad climbing.

I wasn't particularly clear in my second point but the force on the rope does tend to your mass x 9.8 m/s^2 as that's the force on the rope when you are hanging on it. Simple.

What is clear is that the forces created shouldn't ever really be a problem for the rope.

I'm 73kg today, and seriously worried if my old kit, such as slings and extenders connected to my quickdraws are way past their shelf-life.

What's clear to me is that he should discard the old kit if, for no other reason, than he will be able to sleep at night!

In reply to The Lemming:


It's not trivial to calculate and any estimate will be just that, an estimate. Your weight has a lot less bearing on the forces exerted than the type/condition/run of the rope and your belayer's behavior.


Well you know that the force exerted on the kit was less than it's rated for (I'm guessing 12-14kN for the weakest link) since you walked away from the experience. That's not actually very useful information since modern gear tends to be very conservatively rated and the fall you took wouldn't have come close to troubling it. In reality it's more likely to have applied somewhere in the region of 2 to 5kN (give or take a bit) at the gear and a little more than half that at your harness (rope tension).

As for your kit getting old: Chances are it's not as bad as you fear but if you've lost confidence in it then you should replace it. Simple.

As to falling from 16m over a single runner at 10m: I'd expect you to fall more like 10m than 6. Your fall won't have come close to overloading your kit. That said, falling from 16m onto a single runner is far from what I'd consider safe!

jk

In reply to John Roberts (JR):


If it was a little more versatile and a little less porky it would be.
jk

In reply to John Roberts (JR):

Some dicussion here with more equations that you can through a stick at:
http://www.physicsforums.com/showthread.php?t=234907

Once upon a time as a PhD student I wrote a presentation on the "physics of falling" that put all this down and had a pretty accrutate description of the rope behaviour. Goodness knows what happened to it!

As you say, ropes do not obey hooke's law, otherwise we'd bungy after the fall! They stretch to aborb the energy of the fall; then relax afterwards.

In reply to Davy Virdee:


Sorry, I'll qualify that better, certainly any expression of elongation/springyness will use Hooke's law as a starting point, but expressing a climbing rope as a "spring" is a simplification, but a good way to get your head around it.

In reply to John Roberts (JR):
[...]
[...]
>[...]

F =ma the "a" when stopping is your speed when the slack is taken up divided by the time taken to stop.

"a" will be a lot bigger than 9.8, so the peak force will be a lot bigger than "your mass x 9.8"

I wish it were possible to draw on forums...

In reply to The Lemming:

For those of a more mathematical bent, this might be of interest:

http://www.cwu.edu/~curtiswd/PapersAndPreprints/cmj135-140.pdf

In reply to The Lemming:

999th andy how will a be bigger than 9.81ms/s

In reply to 999thAndy:

No I know, i really do, read my posts! When you are stopped the force will be your mass x 9.8 (for the 3rd time!)

In reply to fizzpup:

because your slowing down due to the rope pulling you back up has nothing to do with gravity. The force on the rope on has something to do with gravity once you're hanging on it still.

Roughly,

If I crash a car and go stop 50m/s to 0m/s in 1 second a is 50m/s^2

If the car weighs 1000kg then the force is 50x1000 = 50kN

In reply to John Roberts (JR):

I read this

"There is a peak force, that's obvious. The peak force, matches the point at which the deceleration is greatest (it isn't constant for the time you slow down to stationary), that's obvious. Can you draw me a graph of force (y) against time (x)? When is F(max) with respect to rope elongation (where the rope doesn't obey Hooke's law.)

...

I wasn't particularly clear in my second point but the force on the rope does tend to your mass x 9.8 m/s^2 as that's the force on the rope when you are hanging on it. Simple."

That reads to me as if you're stating the peak force will tend to 9.8m. If that's not what you meant, you should have said what you think peak force will tend to.

In reply to 999thAndy:

You make even less sense, why would the force tend to a distance unit? The force on the rope over time will tend to

F = ma
F = your mass x 9.8ms^-2

Because when the rope stops you and you become stationary, that is the force on the rope!?

This is a very rough graph of the force on the rope over time....

http://www.jr-climbing-files.co.uk/Force_on_rope.jpeg

In reply to John Roberts (JR):

Here's some experimental data to show it:

http://biomech.me.unr.edu/wang/abstracts/rope_drop.htm

In reply to The Lemming:

Your rope and its ability to stretch are the most important things that stop you breaking your protection. If you had a static, non-stretching rope and fell 6 m at a weight of around 75 kg, you would create anough force to break your biners. However, because your rope does stretch and absorb energy your slings and your biners don't break and you are saved.

Your slings are probably fine, esp. if they are Dyneema, as it is a tough, long lasting material (see DMM 'How to break slings'). Plus, there isn't slack in the system when they are acting as quickdraws. Slings are not designed for factor 2 falls but are fine at full extention.

Of all the things you want to check on - I would focus on your rope being in tip top condition.

In reply to Lumbering Oaf:

The formula for maximum force he arrives at is precisely equivalent to the one I worked out and put in my post at 1753 yesterday. We both assumed Hooke's Law (I might think harder about this assumption if it rains when I am away climbing next week.....). I have to say that his maths was rather long-winded though! I put g =10 and my percentage stretch relaced k in the formula (by hanging on the rope you can effectively measure k)

Most of the other stuff in this thread is bollocks.

In reply to John Roberts (JR):


9.8m is 9.8 times mass, not 9.8 metres!

In reply to fizzpup:
a=dv/dt is acceleration when you hit the deck it is instantaneous
ie Dv/0 =>infinity. This is why we use dynamic ropes so the deceleration is more gentle on our soft little bodies.

In reply to The Lemming:

Ignoring friction and such like this is my wooly thinking.

The force of a falling person is F=Ma for constant acceleration, or more generally F=M*dv/dt. This is just their weight.

If you remove the pulley and apply an equal force upwards of F=Ma, then the total force on the falling human is 0, but they are still travelling at constant velocity downwards. You have to apply a decelerating force.

The peak decelerating force is when dv/dt is greatest. I have no idea when this occurs but it's the reason rope stretch reduces the peak force, because dt is maximised and the force is 'spread out.'

However, the total decelerating force over time in the rope must be exactly equal to the force required to bring the person to rest, the falling human never goes up. (We aren't bungee jumping.) Obviously the instanteous force has to be greater than their falling force in order to stop them, but the total cannot be greater.

The force on the anchor starts at 0N. It then rockets almost instantaneously to 20M + a portion of the decelerating force for a time t, then settles at 20M.

So what force is needed to decelerate a falling body?

It's Fd = M * dv/dt. Integrating for the total force gives Fd = M*(v at end of decelerating period - v at start of period). We know the climber ends at rest so Fd = 0 - M*velocity of falling climber.

v^2 = 2as so v=sqrt(20s) (assuming a = g = 10 and s is the distance of the fall)

So Fd = 20 * mass of climber * distance of fall

So total force on anchor = 20 * mass of climber + Fd

Total theoretical max force on anchor is climber stops instaneously is

F = 20M(1+distance of fall)

In your case, distance is 6m, so the total force was 140 * mass which is 10.2kN

Could be total balls though.

In reply to Eric the Red:

I forgot a square root. Told you it was balls.

A mammut galaxy rope has an impact force of 9.2KN, assuming an 80kg load. The further you fall, the more rope there is to stretch so this should be fairly constant.

I reckon this is a sensible and conservative estimate of the load on your gear, given you were lighter.

In reply to Robert Durran:

You would do, you were right assuming Hooke's law, I guess it's fine for engineering purposes to simplify it to Hooke's law, (my best guess is it gives you an overestimate anyway which is good in this situation), but it's not the true physics of the situation.

A quick google of "viscoelasticity climbing rope" found this:

http://www.sigmadewe.com/fileadmin/user_upload/pdf-Dateien/Physics_of_climb...

I haven't got time to read it but it seems to have a go.

Was I right? Have a good half term.

In reply to John Roberts (JR):

Jeez, that's a bit much for most people - keep it simple! I think that just goes to show why using Hooke's Law as an approximation is valid!!

In reply to Lumbering Oaf:

Ha,

Then again, it can't be right as a HO model because there's hardly any oscillation at all.

In reply to The Lemming:

All this physics is a waste of time anyway. Watch E11 and you'll know that your biggest concerns are really gear popping for one reason or another, and then hitting the deck because of it.

If you are "seriously worried" about your old kit, just replace it. Really, how much is you life worth???

As an aside this thread is amazing! A couple of people have got onto the right lines with the maths but there are some embarassing theories flying around. I think some people need to go back to school!

In reply to John Roberts (JR):
, I guess it's fine for engineering purposes to simplify it to Hooke's > http://www.sigmadewe.com/fileadmin/user_upload/pdf-Dateien/Physics_of_climb...



The author says the simple harmonic oscillator model - as we're discussing based on simple Hooke's law - doesn't agree with experiment, hence his expansion etc

In reply to Calder:

Yes and no, if the gear's marginal you're better off using a thinner rope (might not rip then), that's the physics of it.

In reply to John Roberts (JR):

So you're suggesting the OP buy thinner ropes rather than replacing his quickdraws?

Fact is, E11 shows it takes a hell of a lot to make a (relatively) very weak piece of gear fail. A typical climber is not often going to test his gear to that sort of extreme, so getting hung up about relatively poor estimations of the loads they are subjected to is a waste of everyone's time.

Any sensible engineer would see the practical solution straight away and propose the OP replace his draws/slings.

In reply to John Roberts (JR):
Hmmm, I would say a rope with a lower imapct force rating, not necessarily a thinner rope...

Single 11mm Beal Apollo Beal Guarantee 7.7Kn
Single 9.1 Beal Joker Beal Guarantee 8.2Kn

In reply to creag:

Alright alright, at least you know what I meant. As this thread has gnerally been about assumptions then generally thinner ropes - lower impact force.

In reply to Calder:
To which I would add: you were 16m up with only one piece of gear in!? Unless that was the only possible placement, I think you should worry a lot more about your protection regime than the physics!

In reply to Calder:

You said the physics was a waste of time. It isn't really, most of the time it underlies the engineering. I'm not an engineer, I *was* a physicist...

You're right, if the OP doesn't feel comfortable on his gear then replace it. End of.

In reply to John Roberts (JR):


This is interesting, as it shows that the "Maximum force depends only on fall factor and not on rope length" theoretical result does not hold very well for short rope lengths: it increases by about 50% as the rope length increases from 0.7m to 3.5m. The article suggests this is the effect of knot tightening, rather than deviation from Hooke's Law.

It also shows that bouncing does occur, at least to some extent.

In reply to Mark Bull: Absolutely don't pretend to understand the maths, but I have attended several courses at the Torre di Padova, as well as the materials testing department at Padova University to destruction test climbing equipment. Great fun!
http://www.youtube.com/watch?v=9h2byiZ7btw&feature=related
The example shows a static (no belay, but dynamic rope) onto a snapgate Krab, gate open.
We did numerous "hands on" tests of impact forces in different scenarios, using a vast range of belay devices simulating a 2m fall length: Quickdraw placed 1m above the belay device. 80kg weight attached to rope 1m above quickdraw. Thus total of 2m fall with 2m of rope out. (Fall factor 1).
Impact forces measured at the quickdraw were, for a range of devices and belayers, between 550 and 700 daN. If you're like me, then it helps to convert that to Kn for your Krab rating (so, between 5.5 and 7kn), and kilograms of force for my brain to compute (so approx between 550 and 700kg of force.
They also said (if my Italian is correct), that for a given fall factor, the impact forces are a constant, as the longer the fall, the greater the rope available to absorb the energy, but I could be wrong on this.
They summarised by saying that in general, the impact forces for a factor 1 fall, irrespective of distance were as follows: Impact on lead climber: 500 daN, impact on last quickdraw: 800 daN. Impact on belay device: 300 daN ("300kg"), force needed by belayer's hand to arrest fall: 30daN ("30kg").
This assumes only one piece of gear, with the belayer fully anchored (so no "energy absorption" from belayer leaving ground in that sense), no rope drag etc. In other words, a worst case scenario for the example given.
A shocking discovery for me at least was that if an abseiler "bounces" while abseiling, they can quite easily increase the force on the anchor to 200-300 daN ("2-300kg"). A sobering thought...

If you're interested in seeing any other videos of what they test on the tower, do a search on "Torre di Padova" on youtube. All in Italian, but still quite interesting.

In reply to Mark Bull:

http://www.beal-planet.com/sport/anglais/forcedechoc.php

This is quite nice and simple. Uses hooke's law but as you would expect also says that the impact force changes over the life span of the rope - thus again it isn't simply to do with hooke's. I wonder genuinely how close the approximation is... and actually how the rope's material reacts during the fall (no knots etc)

Anyway this has gone on long enough!

In reply to John Roberts (JR):


The physics aren't a waste of time, but in the case of falling off, the maths are so complex and have so many variable factors that unless you are able to accurately measure friction coefficients and stretch factors of the rope, you stand almost no chance of getting any meaningful numbers out at the end of it.

Even with accurate measurements, you still have factors such as slack rope in the system and responsiveness of the belayer to take into account, both of which can make a huge difference.

In reply to The Lemming:

As mentioned by 'kean' above, I would suspect the single biggest thing in taking the 'sting' out of any fall, is the 2nd being lifted off the ground, almost a pulley effect. The very opposite shock-loading a fixed length of rope.

Chris

In reply to EeeByGum:

I know, I'm only really talking about the rope bit of the system but of course you're right about the complexity in reality. Reality is different from the theory, doesn't mean you can't try and get close with theory!

In reply to Davy Virdee:


I spent 13 years working for Troll, the webbing and harness specialists.

I am also the PhD student Davy refers to. The thesis, "The Development of Fall Arrest Technology using Textiles" was published at the University of Leeds in 1987 (or 1988? It was a long time ago).

To save you the trouble of wading through the details of over 300 pages, it's all condensed down to a few paragraphs in the excellent "Rock Climbing - Essential Skills and Techniques" by Libby Peter. This is the MLTUK's training handbook.

Essentially, peak impact forces on a runner are typically between 3kN and 7kN depending on the fall factor, the rope, the belay device and the person operating it. This agrees with the data shown in http://www.thebmc.co.uk/Feature.aspx?id=1477 quoted in the first reply to this post.

If you are concerned about the strength of your quick draws, there are 3 main things to consider:
UV degradation - the more the sling has faded from its original colour, the more strength it will have lost
Edge damage - even the smallest cut in the edge of the webbing has a disproportionate effect on strength
Stitching damage - if the stitching is starting to unravel, this will start to have an effect on the strength of the sling. Although visually obvious, this has less effect than the other two but usually indicates that the sling is reaching the end of its useful life.

Hope that helps.

In reply to The Lemming:

Thank you everybody for your comments and advice. I won't pretend to understand most of it, because I don't. Maths and physics were not on my list of options and I really was crap at them.

I've tried to wade through this discussion this morning and I think that the consensus is that I did not put more that 8 or 9 Kns into the whole system during my fall?

After reading a few discussions on this subject and about ropes I shall be replacing my slings and extenders between the carabeners on my quick-draws over the winter months.

The whole concept of Kns, is beyond me.

In reply to The Lemming:

You don't really need to do all the maths as the rope manufacturer has already (effectively) done this for you by giving you the impact force for the UIAA standard drop test. That is 80kg doing a factor 1.73 (I think) fall and typically gives 7-9kN. This is lower than most pieces of gear so failure by overloading is very unlikely (unless taking a whipper onto a single micro cam or wire). Of course, that applies to new gear. Worrying about whether old gear is still strong enough is another thing entirely. If in doubt, replace it. You don't want to be leading above it and wondering whether it's ok to fall!

In reply to Jimbo C:
Hmm, 7 to 9 kN for a new rope on the climber, that translates under normal circumstances into 11 to 15 ish on the gear.

In reply to The Lemming: Maths was (and probably still is) compulsory, not optional!

In reply to Andy Perkins:


Hmm, that's not the conclusion I would draw from the sling testing I have done. My results suggested that cuts in the dge of webbing had an entirely proportionate effect - ie you can bascially measure the percentage width reduction and translate it directly to a percentage strength reduction.

The one thing we found did that had an disproportionately large effect on strength was abrasion of the surface of the sling. That's the one to watch for I'd say - it doesn't look so obvious as 'damage' as an nick in the edge but can often have a more severe effect.

In reply to The Lemming: Are the newtons youre referring to olivia and Andy?

In reply to Submit to Gravity:

Ho yea, I must have missed the complex explanation about calculating newtons with complex equations module at school level.

In reply to jimtitt:

In the case of a factor 2 fall I thought that the force on the climber would be the same as the force on the anchor since there is no rope going back down to a belay and generating additional force on the top piece via the pulley effect. Please correct me if I'm wrong (meant sincerely, this interests me).

Granted, in a normal fall, the pulley effect comes in but the fall factor would be less, tho I'm not sure if impact force and fall factor have a linear relationship. What is the highest reported load on the top piece that you're aware of?

In reply to Jimbo C:

In a FF2 the force on both ends of the rope are the same, as of course they are in any situation where the rope doesn´t run through any intermediate gear such as bringing up your second.
In most lead falls (and the one the OP describes and is being discussed) the force on the gear is considerably higher than on the climber, 1.6:1 being a commonly used factor which is a bit conservative since there is plenty of evidence that it can get up to at least 2:1. This is one of the arguments against using part of the belay as the first protection piece since you have increased the possibility of it becoming overloaded and the belay failing.
Not suprisingly, since no-one is out there taking mighty whippers on routes equipped with load cells, there isn´t much information on the maximum loads on gear in normal climbing but testing gives a top runner force of 12kN for dynamic belaying and 20kN for static belaying.

In reply to jimtitt:

Cheers.

In reply to Andy Perkins: It was 1987, Andy. Leeds has still got a copy in the library. (For what it's worth I thought it was very good by the way!)

In reply to John Roberts (JR):
I can't be bothered reading the whole thread but from the little bits I've seen so far John Roberts knows what he's talking about. High Climber doesn't.

In reply to The Lemming: Climbing ios all in your head anyway . If you haveany worries about your gear replace it and boost your confidence. If you are not worried it wont break (there is a huge safety factor built in).

Besides Christmas is coming make everyones prese buying easy for them and treat yourself!

In reply to The Lemming: In reply to The Lemming: Your question is (in effect) actually a standard A level physics question. Here's an overview.
When you fall there are two things that determine how hard you will hit the rope when you stop; how far you fall and how quickly the rope stops you. In general terms, the further you fall, the more force on the rope (pretty obvious) and the faster the rope stops you, the greater the force on the rope will be (often missed).
Now the physics...
The force on the rope when it stops you will be equal to your rate of change of momentum. So what does that mean..?
Momentum is your mass (in Kg) x your speed (in metres/second). You can work this out if you know how far you've fallen using...
Final speed squared = initial speed squared + (2 x gravity x distance fallen)
...but as your initial speed will have been 0 this simplifies to...
Speed squared = (2 x gravity x distance fallen) and, as we know gravity on Earth is basically 10 (actually 9.81) then we can further simplify this to..
Speed squared = 20 x distance fallen.
Of course, we actually want to know the speed, not speed squared, so the equation we need for final speed is...
Speed = Square root of (20 x distance fallen)
This speed will be the speed that you are going when you hit the rope. It ignores air resistance, which is broadly irrelevant anyway and certainly is irrelevant when you consider that the fall length will always be an estimate anyway.
A 1m fall will make this speed about 4.5m/s, 2 m fall about 6.4m/s, 5m about 10m/s, 8m about 12.7m/s, 10m about 14.1m/s, 15m about 17.3m/s and 20m about 20m/s.
You work out your momentum using...
Momentum = speed x mass, so just multiply your fall speed by your mass. The 6m fall would be made of a fall to the point where you hit the rope plus a further flight while the rope was catching you. The 'free-flight' component of the fall would therefore be less than 6m and can only be worked out if we know how far past your last runner you were (in your case, 3m). From your fall description you were 3m above that last runner with a total of 32m of rope out..?A 3m free fall would give a final speed of about 7.8m/s so your fall momentum would be...
Momentum = mass x speed ; in this case Momentum = 60 x 7.8 = 468Kgm/S (don't worry about the kgm/s bit, it's just there so I know I've been good!).
All the above was knowable. The problem in calculating force on the rope is that you need to know how long you took to stop and personally I don't employ someone with a stopwatch as a seconder. Your rope is designed to stretch to slow you to a halt. This is for two reasons; one, if you stopped all at once the force on the rope would be enormous and the the rope would break, two, even if your outside stopped all at once your internal organs wouldn't! This is referred to as massive internal injury.
The Edelrid Boa (as an example 9.8mm rope) is designed to stretch by a maximum of 38% in a fall situation. This means that it is designed to bring your speed to 0m/s while the rope stretches by a maximum of 38% of the total rope between the leader and the belay plate at the time of the fall. This is why falls are less serious for the rope if you fall when you have a lot of the rope in use. Not all of this stretch will be taken up in every fall though, the greater the impact force the more of the stretch will be employed. This now becomes a bit complicated...
We need to know the time it takes to stop. This is determined by the stopping distance (how far you travel down while the rope is catching you) and your fall speed and is given by...
Time = (2 x stopping distance) / max fall speed
The stopping distance will be given by...
Stopping Distance = Total fall length - height you were above the last runner.
From your fall description you would fall a further 3m while the rope was slowing you to 0m/s. THIS IS A GUESS BASED ON YOUR DESCRIPTION - I do not claim that it is correct. The numbers below may therefore be rubbish.
Using the equation above; Time - (2 x stopping distance) / max fall speed
We can substitute in the numbers to give...
Time = ( 2 x 3m) / 11.8m/s
to give your slowing down time as 6/11.8 = 0.51 seconds.
This now allows us to calculate the force on the rope as we have taken the momentum to be 468Kgm/s and the time to 'dissipate' it as 0.51 seconds we have...
Force = Momentum / Time taken to stop
so in this case...
Force = 468/0.51
which gives
Force on rope = 918N or 0.92kN
This is well within the spec for the Edelrid Boa rope which has a max impact force quoted at 8.8kN (and, I would assume, yours).

In terms of the maths - it is, as you say, complicated. Fortunately nice people invented the idea of fall factors which makes the process easier to think about. A friend of mine works on the basis that he adds up the fall factors from all the falls on a rope, when they add up to 2, he discards the rope.
Time for a cup of tea (or equivalent!).
Dave

In reply to higherclimbingwales: Sorry - this is a standard misconception (and rather dangerous). The fall impact is not determined by the fall itself but by the arresting of the fall. Please see my reply to the original article.
Dave

In reply to Eric the Red: Sorry - your logic contains an error. You suggest that the upward force arresting the fall can never be greater than the downward force. In fact it has to be or a deceleration would not occur. Please see my reply to the original article.
Dave

In reply to David Pye:
>[...]
>[...]

Stopping distance = the distance the rope stretches by in arresting your fall (roughly calculated as a %age of the total rope paid out). This was clearly put in one of the linked articles above.

In reply to The Lemming: Mate, I too have no idea about the maths behind fall factors, infact I dont even care, I just give the gear the once over and if it looks ok its good to go, If you are worried, im sure you can send all the "soft goods" to Troll etc and tell you if they are ok? and they will, re-sling the draws if the connectors are knackered.

In reply to David Pye:
"which gives Force on rope = 918N or 0.92kN"

Whilst you admit your calculations may be off it's quite easy to see this is the case as just the tension in the rope with the climber sitting on it would be 0.6kn (ish) never mind with a reasonable fall. I would expect the max tension in the rope due to the fall to be closer to 3kn (also it would differ either side of the top biner).

Can you also explain your logic here:
Time = (2 x stopping distance) / max fall speed

I can't logically see how you can know the time that the fall will be arrested over without knowing the springness of the rope but maybe I'm missing something. You are not assuming constant de-acceleration or something are you?

In reply to John Roberts (JR):

"http://www.beal-planet.com/sport/anglais/forcedechoc.php"

Nice link - using that equation Mr Lemmings fall generated an impact force of around 3.5kn (same thing as the tension in the rope on that side up to the top runner). The beal equation doesn't account for standard belay devices which will slip or deformation of the body or belayer movement so in practice the forces are likely to be lower.

In reply to The Lemming: Listening to manufacturers is better than sifting through bullshit and sense from the ukc mob. Here is what Mammut have to say:

"The impact force is the maximum force which affects the load in a standard fall, when the rope absorbs the fall energy by its elongation. It is the measurement for the «hardness» of the fall. Ropes with higher impact force, when holding the fall, produce a stronger «jolt» in the falling body and on the safety system. In standard tests the impact force for single and twin ropes may not exceed 1200 daN and for half ropes ‹ 800 daN (approx. 800 kp).
The practical relevance of the impact force is relatively small because it is measured with the standard static fall test, i.e.: the fall rope is completely fixed. In practice, however, a fall is almost always caught dynamically. Belay devices (Munter, figure eight, ATC, etc.) have a certain rope path, and their attachment to a central point, or on the harness, brings a dynamic effect. Through dynamic belaying a large part of the fall’s energy is dissipated and so the impact force is reduced. Measurements by Mammut of typical sport climbing falls show, that with dynamic belaying the difference in impact force between two different ropes is barely discernable. It’s therefore important to provide a truly dynamic belay."

The standard test (mentioned above) used to calculate the 'maximum impact force' value for your rope will be a factor 1.77 fall, with a static belay. Most half ropes tested this way have an impact force value of around the 6kn mark.

In reality, there are other conflicting factors - complications as discussed above by mammut, and earlier in the thread.

Fortunately these conflicting factors and complications aren't going to be all that important for most of us to bother calculating, because the sort of falls you are worried about have a considerably lower fall factor than in the standardised test. The thing to take away from all this nonsense is this:

Even if you take a factor 1.77 fall, you probably wont generate enough force to damage a 12kn rated piece of gear. For a fall factor lower than one, you almost certainly wont.
So you dont need to fork out loads of cash to replace things every time you fall. Gear works, really well, if used right. So long as you have the skill to use all of your gear well, you will be fine. Spend the money on coaching if you still have doubts - skill is worth so much more than a shiny new rack!

Hopefully that is a bit more helpful to you than much of this maths/ego discussion.

In reply to jacobjlloyd:
"The standard test (mentioned above) used to calculate the 'maximum impact force' value for your rope will be a factor 1.77 fall, with a static belay. Most half ropes tested this way have an impact force value of around the 6kn mark."

Yeah but that test is done with a weight of only 50kg, in pratice if you fall on single half rope the actual impact force is much the same as a single of similar quality and for the 80kg test weight would be more in the region of 8kn impact force

Also the force on the gear can be (nearly!) as much as twice the impact force on the climber.

I agree with your post in sentiment though. If I factor twoed on to a single nut I may bin it especially if it was a small nut, but I wouldn't bin the biners or the rope unless they looked damaged mainly because the belay device would slip at around 2-3 kn so you'd probably never get all that close to generating enough force to breaking the gear (as long as it was rated to 12kn)

The force is determined by the mass (of the climber) and the acceleration (F=MA) but the acceleration is very hard to determine as it is governed by the elasticity of the rope.

A crude and simple method would be to determine (estimate) the overall elongation of the rope and assume a constant acceleration while the rope was stretching. BUT the problem is how do you work out the elongation of the rope - there is some data from the rope manufacturers for extreme falls (ie elongation for an 80Kg fall factor 1.77 fall). But this isn't directly applicable to smaller falls. You could take the model a bit further by estimating the stretch from this published data (say linearly related to the FF and the mass) but by the time you've done this you may be way off.

I'd imagine the elasticity of the ropes is reasonably non linear in practice.

I recall typical big leader falls generate up 6kn (determined by the rope) on the climber (hence up to double this on the gear (friction losses at the top piece drop the forces to < double). So unless you are using small gear and taking huge falls it is all well within limit.

However, the placements are generally the weak point, even for small gear (where the pressures generated on a tiny nut can crumble the rock before the wire snaps).

In reply to David Pye:

That the solution is a standard A level physics problem is going to be a blow to the number of professors and PHD´s that have battled with the problem for decades, mostly to no great sucess!
There are a few modifications of the standards model from Magdefrau which get nearer the correct solution but near is fairly relative, 20% here or there is pretty good. The best approach is from Manin who made a mathematical model from a large number of test results and used this to modify the standard model, this was the basis of the old Beal calculator which is no longer available.

A cursory glance at the load curves from a test drop show exactly why the problem is so difficult as changing one parameter means another one obeys another rule. A complex subject which can anyway be ignored by using the `it didn´t break principle´which is the basis for most gear standards.

In reply to jimtitt:

Do you have any links to papers on the subject.

Just interested, I'd been thinking that you could take two approaches, either applying Newtons second law and deriving a differential equation similar to that of a spring/mass/damper to look at forces or simply look at energy generated in the fall. I'm suspecting that failure of the rope or anchor would more closely correlate to energy than force (and a lot easier to calculate).

In reply to Removed User:

This one is the starting point for most modern attempts, best of luck!

I can dig out some more tomorrow sometime.

Jim

In reply to jimtitt:

Forgot the link!
http://personal.strath.ac.uk/andrew.mclaren/Pavier.pdf

Jim

In reply to CurlyStevo: The equation drops out of two of the standard applied maths motion equations - granted you have to substitute one into the other to get to it. Whilst you are quite right about the static force of the climber hanging on the rope, the rather small force I calculated won't be a mile out, we're dealing with a very small fall (in mechanics terms) here. Lobs of almost this scale are pretty common.
On the other hand - unless we're going to get the rope manufacturers to give us he specifics of the rope extension for certain initial impact forces we'll never get to an exact answer. My calculations were partly done to allow me to explore the subject and, I am now content, the tolerances built into the gear are pretty high.
Dave

In reply to jimtitt:

Thanks,

I'll have a look in more detail tomorrow but a quick skim confirms he's applying Newton's second law. I'd like to see what sort of correlation you get between energy absorbed and falls to failure. I'll maybe have a go at that.

Also interesting to note that small radii runners have a considerable effect on falls to failure. I wonder if the upper tested size of 6mm is really the point at which radius no longer weakens rope or whether it's larger. Something to think about when buying krabs...

In reply to jimtitt: Predicting the force in a fall situation without observations (measurements) of the actual fall is extremely hard. I am totally happy to acknowledge that with no fixed parameters except a faller, a total length and a rope elasticiy the problem is horrendous. What we need is someone from a car manufacturer safety department - the situation is the same one that they have when designing materials for the front of dashboards etc, except at 90 degrees to ours. In his case the fall is well described and Newton's laws can be applied by breaking the fall into the two sections - though it did take me a while to think out how to get round the elasticity problem.
Dave

In reply to Removed User: The point you make here is something that I wanted to take into consideration, but don't know how to. I assume that the failure force for the rope is given for an impact 'in line' with the rope when in reality there will be a considerable component across it. Having said this, I rather suspect that you and Eric have a greater knowledge of the subject than my 'nearly 30 years ago degree' is allowing me. I shall watch your further posts and hope to learn.
Dave

In reply to Removed User:

Normally you expect to get up to radii of 30 to 50 times the rope diameter before there is no weakening at least with other kinds of cordage. Tape for example is tested over a 50mm drum (if I remember correctly) and needs to be de-rated for use with karabiners.
Since the biggest normal karabiner is 6mm and rope failure at the karabiner is unheard of it is academic anyway, you have more problems with the radius inside the knots.

In reply to David Pye:

The standard model unfortunately ignores friction on the runner(s) which means it only applies to direct falls which are very rare, Jay Tansman has however done this one based on Richard Goldstones model at least for the top piece which is usually the one with the greatest effect. http://jt512.dyndns.org/impact-force-rev1.pdf
This paper explains pretty well the other factors as well and includes the climbers weight which is of considerable importance. He has an online calculator as well which is probably the most accurate one available though he and I have some disagreement about a few things, you get a better result if you correct the weight input by using a factor of 80% http://jt512.dyndns.org/impactcalc.html
You will notice your calculations seem very wide of the mark!

However it must be remembered that the biggest single influence on impact force is the belay method, not the rope and that in tests it is clear that normally the elasticity of the rope plays a minor role, in fact a lot of the time a static rope would be satisfactory.

In reply to jimtitt: Many thanks for the link to Jay Tanzman's paper. Very illuminating. Are you in contact with him? If so it would be great if he would allow a versio of the graph at the end of the paper to be published where many climbers would see it. If nothing else, given his numbers, it should make folk very aware of the dangers of cross loading krabs given that relatively low fall factors are giving forces around their quoted sideways breaking strains - or am I missing something?
Dave

In reply to David Pye:

As I pointed out your calculations were out by over a factor of 4 (based on the beal estimation linked in my last post) which I think is a quite a lot as your (mis)caluclation was actually far closer to the force of the climber just sitting on the rope than the factor 0.375 fall Lemming described.

Lemming would have fallen around 6 meters before the rope started to tighten and would have then gone a few meters further after that deaccerating. This IMO is quite a large fall in the grand scheme of things placing a reasonably large force on the gear, hence why I could see your maths was well out.

As I point out in my last post you can't use the standard equations of motion to calculate the time the fall would have occured over (based on the distance the climbers fall was arrested over) as they are based on CONSTANT ACCELERATION, which is fine for the part of the fall until the rope starts to arrest the climber but is completely inaccurate for working out how long the fall will take once the rope comes tight hence why your calculations were so far out

Often in these type of estimates people use hookes law and approximate the rope to a spring, which although is inaccurate is still at least making a reasonable estimate.

You'd do well to listen to Jim he is pretty much the resident expert in these matters.

Stevo

In reply to David Pye:

I do have contact with Jay, sometimes even on polite and friendly terms! (He´s a bit abrasive occasionally but certainly smart as they come). The other guy he worked on this with, Richard, is professor of mathematics at one of Americas top uni´s and certainly knows his stuff.
Where he could publish it where more climbers would see it I don´t know, it´s commonly known on rc.com where there is a platform for rather esoteric stuff but in the UK there is nowhere suitable.

And it is worth noting that his work is for the static belay condition which is worst case, it is well known that most moderate falls are in the 7kN region but the problem is ensuring that one is in the moderate fall area and haven´t accidentally drifted into a scenario where the higher numbers start such as a tangle in the rope preventing dynamic belaying.

Jim

In reply to CurlyStevo: Thanks - I can see now that I made a simple mistake. I worked the numbers out based on a 3m fall followed by a 3m stop - which is clearly rubbish and suggests I had temporarily sent my brain for a walk (fall?). This error, unfortunately, makes the route I was trying to take a dead end as the 'collision with the rope' can only be modelled if we know the elasticity of the rope, the amount of slippage etc (which is what I was trying to find a way round). You are also quite right in the comments about the acceleration, the best that my method could give, even if we knew the length of the two fall components accurately, would be the average force - which is clearly of no real use.
If I may further take of your time..?
Is dF/dt controlled by k, or does the polymer behave differently at different force levels during the pre-catastrophic stretch? I appreciate this is beyond Lemming's query.
Dave

In reply to jimtitt: The 7kn figure is that the impact force or the force on the top runner just for the beal estimation which is based on the uiaa impact force result, you'd be talking a fair bit greater than factor 1 falls for a 80kg climber to generate 7kn impact force (on a rope with uiaa 8kn impact force) which adds up if you think that a 1.77 fall factor fall is generating 8kn.

One thing I've always wondered is are fall factors really a good estimation - is a factor 1 fall with 1 meter of rope paid out really the same impact force as one with 50 m of rope paid out?

In reply to David Pye: The beal estimation assumes static belay with no slip and rope already tight to leader when they fell, much the same as you were attempting to do, all meaning that any info we get from Lemming to reverse engineer in to yr equation would be invalid even if constant acceleration is an ok approximation ( which I doubt). In any case why bother when beal have done it for us. I agree the result from beal calcs is not applicable to real world falls but it does give us a worst case upper bound.

As for yr later questions Jim would better answer those, my maths is starting to get ropey in the areas I don't use now adays. I do still use a lot of matrix / vector maths and as you maybe guessed the equations of motion a fair but still

Stevo

In reply to CurlyStevo:

That´s 7kN on the top piece.

The problem is we can get fairly near all the rope-side values with a bit of fudging for a few things by looking at fall data and correcting the theory BUT the great unknown is always the belayer. From a weak-handed wimp with a worn out ATC up to a horribly abrupt device like a Cinch they are all possible. Without actually measuring at the time how well the belayer stopped the faller a very accurate calculation will give you a completely wrong answer! In effect one can only say a)worst case and b) normal and be prepared that b turns into a.

As to fall factors, they are useful for the theory but have no real relevance otherwise, a 3cm FF2 doesn´t frighten many people but a 30m FF1 does for good reason. Hitting the ground is the worst case!