## / Falls and physics |

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Muppet123 -

*on 09 Dec 2012*Hello just wondering how would you work out a fall facter and from there how do you work out the force exerted on the piece of gear?

Jack B on 09 Dec 2012

In reply to Muppet123:

To approach this problem with physics, and get the absolutely right answer, you need to know a lot of hard to measure things. The position and friction at every runner, the exact dynamics of the belayer and his device etc. Anything else is going to be an approximation.

The first approximation is to ignore friction, and assume the belay is a fixed immobile point with the rope tied to it. The rope stretches evenly, and not very much:

The energy you have to get rid of at the bottom of the fall is your mass (m) times the force of gravity (g) times the height (h) - m*g*h for short.

when you hit the bottom of the rope it will stretch, and absorb the impact. It's if it gets x meters longer, it will pull back with force F=k/l*x where l is the length of rope being stretched, k is the spring constant per unit length. It will absorb energy E=F*x=F*(F*l/k), and the energy it must absorb is that of your fall. Thus m*g*h=F^2/k*l. Re arrange for F and F=sqrt(m*g*h/l*k). That's the force on you, so the force on the gear is F=2*sqrt(m*g*h/l*k) due to the pulley effect.

Notice how h/l appears in that equation, rather than h on it's own. That's the fall factor.

This approximation is probably out by at least 50%. The most important corrections probably are:

1) That the rope stretches a fair bit, so the actual distance you fall is not h but h+x. That makes the algebra a bit harder to follow and increases the force by I guess 10-30%

2) there is friction in the system, particularly at the top bit of gear, so the rope doesn't slide through smoothly. This has two effects, firstly to effectively decrease l, increasing the force on you. Secondly to reduce the pulley effect, reducing the force on the gear. In the limit of infinite friction (rope tied to top runner) the force on both you and the runner becomes F=sqrt(m*g*2*k).

3) Any rope paid out by your belayer. This will increase l so will tend to reduce the force.

4) Any slack in the system.

To approach this problem with physics, and get the absolutely right answer, you need to know a lot of hard to measure things. The position and friction at every runner, the exact dynamics of the belayer and his device etc. Anything else is going to be an approximation.

The first approximation is to ignore friction, and assume the belay is a fixed immobile point with the rope tied to it. The rope stretches evenly, and not very much:

The energy you have to get rid of at the bottom of the fall is your mass (m) times the force of gravity (g) times the height (h) - m*g*h for short.

when you hit the bottom of the rope it will stretch, and absorb the impact. It's if it gets x meters longer, it will pull back with force F=k/l*x where l is the length of rope being stretched, k is the spring constant per unit length. It will absorb energy E=F*x=F*(F*l/k), and the energy it must absorb is that of your fall. Thus m*g*h=F^2/k*l. Re arrange for F and F=sqrt(m*g*h/l*k). That's the force on you, so the force on the gear is F=2*sqrt(m*g*h/l*k) due to the pulley effect.

Notice how h/l appears in that equation, rather than h on it's own. That's the fall factor.

This approximation is probably out by at least 50%. The most important corrections probably are:

1) That the rope stretches a fair bit, so the actual distance you fall is not h but h+x. That makes the algebra a bit harder to follow and increases the force by I guess 10-30%

2) there is friction in the system, particularly at the top bit of gear, so the rope doesn't slide through smoothly. This has two effects, firstly to effectively decrease l, increasing the force on you. Secondly to reduce the pulley effect, reducing the force on the gear. In the limit of infinite friction (rope tied to top runner) the force on both you and the runner becomes F=sqrt(m*g*2*k).

3) Any rope paid out by your belayer. This will increase l so will tend to reduce the force.

4) Any slack in the system.

jimtitt -

*on 09 Dec 2012*In reply to Muppet123:

http://www.ukclimbing.com/articles/page.php?id=647

The force is more complicated, http://jt512.dyndns.org/impact-force-rev1.pdf is a good place to start.

http://www.ukclimbing.com/articles/page.php?id=647

The force is more complicated, http://jt512.dyndns.org/impact-force-rev1.pdf is a good place to start.

jkarran -

*on 09 Dec 2012*In reply to Muppet123:

Calculating a 'fall factor' is easy: Distance fallen / Rope in the system

Going much further than that is considerably more difficult! If it's purely out of curiosity then there are various papers and online calculators that google will turn up. If it actually matters then you probably need to experiment with the specific materials and set-up of interest.

jk

> Hello just wondering how would you work out a fall facter and from there how do you work out the force exerted on the piece of gear?

Calculating a 'fall factor' is easy: Distance fallen / Rope in the system

Going much further than that is considerably more difficult! If it's purely out of curiosity then there are various papers and online calculators that google will turn up. If it actually matters then you probably need to experiment with the specific materials and set-up of interest.

jk

elsewhere on 09 Dec 2012

In reply to Jack B:

Stretchy ropes reduce the force - hence we use dynamic rather than static ropes.

> 1) That the rope stretches a fair bit, so the actual distance you fall is not h but h+x. That makes the algebra a bit harder to follow and increases the force by I guess 10-30%

Stretchy ropes reduce the force - hence we use dynamic rather than static ropes.

Muppet123 -

*on 09 Dec 2012*In reply to Muppet123: Thanks you all :)

Sean Kelly -

*on 09 Dec 2012*In reply to Muppet123: In short, the closer to the belayer and the less runners placed the higher the fall factor, so always place a runner asap above the belayer, especially on multi-pitch. It reduces the fall factor by 50%. If it holds that is!

Jack B on 09 Dec 2012

In reply to elsewhere:

Ah, but the good side of rope stretch is already included in the calculation (otherwise k=infinity and F=infinty). It's not the stretchiness of the rope I left out per se, it's the extra distance you fall as a result of that stretch. I left it out to simplify the maths, otherwise the "solve for F" part means solving a quadratic.

> (In reply to Jack B)

> [...]

>

> Stretchy ropes reduce the force - hence we use dynamic rather than static ropes.

> [...]

>

> Stretchy ropes reduce the force - hence we use dynamic rather than static ropes.

Ah, but the good side of rope stretch is already included in the calculation (otherwise k=infinity and F=infinty). It's not the stretchiness of the rope I left out per se, it's the extra distance you fall as a result of that stretch. I left it out to simplify the maths, otherwise the "solve for F" part means solving a quadratic.

The Ex-Engineer -

*on 09 Dec 2012*In reply to Muppet123:

Getting a rough, first-order approximation of the forces on gear is a bit easier than previous posts possibly make out.

Working out the fall factor (

- The Impact Force rating of the rope in kN from the UIAA single rope test (

- Weight of the falling climber in kg (

You can then use the following simple equation:

Approx. Anchor Force = 5/3 *

> Hello just wondering how would you work out a fall facter and from there how do you work out the force exerted on the piece of gear?

Getting a rough, first-order approximation of the forces on gear is a bit easier than previous posts possibly make out.

Working out the fall factor (

**FF**) is easy. Then to approximate the forces from you only need two additional pieces of info:- The Impact Force rating of the rope in kN from the UIAA single rope test (

**Rope_Rating**)- Weight of the falling climber in kg (

**Leader_Weight**)You can then use the following simple equation:

Approx. Anchor Force = 5/3 *

**Rope_Rating*** Sqrt (**FF**/1.78 ***Leader_Weight**/80 ) mrchewy -

*on 09 Dec 2012*In reply to Muppet123: Doesn't the thickness of the belayer's shoe soles play a part?

Martin W on 09 Dec 2012

In reply to Jack B:

So solve the quadratic! It's taught in secondary schools so it really

The impact force equation

where:

m is the mass of the climber

g is the acceleration due to gravity

f is the fall factor = distance fallen (

K is the rope modulus = Young's modulus of the rope divided by the cross-sectional area of the rope (you can calculate K by re-casting the equation, given the UIAA impact force quoted for the rope, as The Ex-Engineer does in his version).

Multiply F by two to get the force on the top anchor, due to the pulley effect as you say.

As the paper posted by jimtitt shows, trying to account for friction

Basically, trying to calculate

The point about the fall factor is that it is a dimensionless number which is easy to calculate and which, all

> the "solve for F" part means solving a quadratic.

So solve the quadratic! It's taught in secondary schools so it really

__isn't__rocket science.The impact force equation

*ignoring any frictional effects*actually is: F = m.g(1 + SQRT(1+2.f.K/m.g))where:

m is the mass of the climber

g is the acceleration due to gravity

f is the fall factor = distance fallen (

__not__including rope stretch) divided by length of rope between climber and belayerK is the rope modulus = Young's modulus of the rope divided by the cross-sectional area of the rope (you can calculate K by re-casting the equation, given the UIAA impact force quoted for the rope, as The Ex-Engineer does in his version).

Multiply F by two to get the force on the top anchor, due to the pulley effect as you say.

As the paper posted by jimtitt shows, trying to account for friction

__just at the top anchor__makes things a lot more complicated. Trying to do the same thing for friction at the other runners would make things more complicated still, especially if you want to allow for factors such as friction of the rope rubbing against the rock, the friction at some runners changing as the rope comes tight and so forth. And don't forget slippage in the belay device, dynamic belaying, the possibility that the belayer might be pulled off their feet etc etc.Basically, trying to calculate

*exactly*what the impact force is going to be in any particular combination of circumstances quickly becomes an exercise in futility akin to trying to predict exactly where all the balls will end up after breaking off in a game of snooker.The point about the fall factor is that it is a dimensionless number which is easy to calculate and which, all

__other__things being equal, has a very significant effect on the resulting impact force. Simply put: the bigger the fall factor, the more likely it is that the result is going to be painful in some way or another. That's all you really need to know. jimtitt -

*on 09 Dec 2012*In reply to The Ex-Engineer:

Of course if you donīt actually want to work out the forces as an intellectual exercise but really just want to know them then just punch the numbers into Jay Tansmanīs calculator which is the most accurate "simple" model we have at the moment, the derivation is in the link posted above. It ends to overcalculate the loads for a few reasons as would your formula, the climber weight should be adjusted to allow for the rigid body used in the rope test. It is also based on the standard model so doesnīt take into account belay devices, the only model which took these into account no longer being available online.

http://jt512.dyndns.org/impactcalc.html

Of course if you donīt actually want to work out the forces as an intellectual exercise but really just want to know them then just punch the numbers into Jay Tansmanīs calculator which is the most accurate "simple" model we have at the moment, the derivation is in the link posted above. It ends to overcalculate the loads for a few reasons as would your formula, the climber weight should be adjusted to allow for the rigid body used in the rope test. It is also based on the standard model so doesnīt take into account belay devices, the only model which took these into account no longer being available online.

http://jt512.dyndns.org/impactcalc.html

rgold -

*on 10 Dec 2012*In reply to Muppet123:

A long time ago I posted an account of what should be called the Wexler equation, http://www.rockclimbing.com/cgi-bin/forum/gforum.cgi?do=post_attachment;postatt_id=746.

Nothing about this is original with me, or other people, like Attaway, who have given the Wexler equation without proper attribution. See Belaying the Leader, Richard M. Leonard and Arnold Wexler, Sierra Club Bulletin 31 (1946), later reprinted with Leonard as editor in the book Belaying the Leader: An Omnibus on Climbing Safety, Sierra Club (1956).

My account derives the Wexler equation in two ways, the simplest being a conservation of energy argument and a more complicated way being via the differential equation for simple harmonic motion. This equation can be modified to include a viscous damping term, but I didn't write that up. A system of such damped de's, one for each section of rope between protection points, forms the basis of a model promulgated by the CAI---of course, solutions have to be obtained by computer methods.

Jay's work referenced by Jim adds in the frictional effects at the top carabiner but does not include damping effects.

There are a number of other models out there that attempt to model rope behavior more realistically.

A long time ago I posted an account of what should be called the Wexler equation, http://www.rockclimbing.com/cgi-bin/forum/gforum.cgi?do=post_attachment;postatt_id=746.

Nothing about this is original with me, or other people, like Attaway, who have given the Wexler equation without proper attribution. See Belaying the Leader, Richard M. Leonard and Arnold Wexler, Sierra Club Bulletin 31 (1946), later reprinted with Leonard as editor in the book Belaying the Leader: An Omnibus on Climbing Safety, Sierra Club (1956).

My account derives the Wexler equation in two ways, the simplest being a conservation of energy argument and a more complicated way being via the differential equation for simple harmonic motion. This equation can be modified to include a viscous damping term, but I didn't write that up. A system of such damped de's, one for each section of rope between protection points, forms the basis of a model promulgated by the CAI---of course, solutions have to be obtained by computer methods.

Jay's work referenced by Jim adds in the frictional effects at the top carabiner but does not include damping effects.

There are a number of other models out there that attempt to model rope behavior more realistically.

duchessofmalfi -

*on 10 Dec 2012*Does the fall factor include the stretch?

Is it

FF=(distance fallen 'til ropes becomes tight)/(length of rope before stretch)

or

FF=(total distance fallen (climber at rest)) / (length of stretched rope)

Or a combination - possibly this is most useful:

FF=(total fall distance (at rest))/(original length of rope)

Since this gives a good measure ot the energy involved in the fall and what it was held by.

Is it

FF=(distance fallen 'til ropes becomes tight)/(length of rope before stretch)

or

FF=(total distance fallen (climber at rest)) / (length of stretched rope)

Or a combination - possibly this is most useful:

FF=(total fall distance (at rest))/(original length of rope)

Since this gives a good measure ot the energy involved in the fall and what it was held by.

rif on 10 Dec 2012

Those of us who do mainly lower-grade winter/alpine/scrambly climbs can presumably be reassured that we will never have a FF 2 fall, since the energy to be dissipated is proportional to the VERTICAL height descended by the faller. So on a 45 deg slope, max FF is 1.41, etc? Can the engineers and mathematicians out there confirm?

GrahamD -

*on 10 Dec 2012*In reply to rif:

Fall factor is a totally meaningless metric if the faller is being slowed down by other means (such as bouncing off ledges or slithering down snow slopes)

Fall factor is a totally meaningless metric if the faller is being slowed down by other means (such as bouncing off ledges or slithering down snow slopes)

jimtitt -

*on 10 Dec 2012*In reply to duchessofmalfi:

No, your first option is the one.

> Does the fall factor include the stretch?

>

>

No, your first option is the one.

captain paranoia -

*on 10 Dec 2012*In reply to rgold:

My suspicion is that this is simply due to not knowing about the 'Wexler equation'. It seemed to me that Stephen Attaway performed his analysis from first principles. I think it was a pretty good analysis.

For the benefit of others, google xRopes.pdf

> or other people, like Attaway, who have given the Wexler equation without proper attribution.

My suspicion is that this is simply due to not knowing about the 'Wexler equation'. It seemed to me that Stephen Attaway performed his analysis from first principles. I think it was a pretty good analysis.

For the benefit of others, google xRopes.pdf

rgold -

*on 11 Dec 2012*In reply to duchessofmalfi:

As Jim has already said, it is the first option.

The options you gave involving the position with the climber at rest are of no use for computing the peak load, since there is some rebound and any of those options will miss out on the actual maximal extension.

It is an arcane technical point that that you can use a different fall-factor H+/L, where L is the length of rope out as before and H+ is the total fall distance at maximum extension, which is to say the height of the fall H when the rope just becomes tight plus the amount of stretch. This variation has the property that one can still calculate the peak load from this ratio, indeed the equation is marginally simpler, but the ratio itself is unknowable until after the rope has caught the fall and high-speed photographs of the event have been analyzed to determine how much stretch occurred at maximum extension.

In spite of the terminally diminished utility of this approach, you can find an account at www.cwu.edu/~curtiswd/PapersAndPreprints/cmj135-140.pdf, where once again no mention of Wexler's work is to be found.

In reply to Captain Paranoia:

Retracing someone else's steps from first principles does not, in my opinion, absolve one from the responsibility of crediting the person who made the original journey. You wouldn't, for example, announce a first ascent without trying to determine whether or not someone had already been up that way. There are multiple ways authors could have found references to Wexler's work had they tried; the information is out there.

As Jim has already said, it is the first option.

The options you gave involving the position with the climber at rest are of no use for computing the peak load, since there is some rebound and any of those options will miss out on the actual maximal extension.

It is an arcane technical point that that you can use a different fall-factor H+/L, where L is the length of rope out as before and H+ is the total fall distance at maximum extension, which is to say the height of the fall H when the rope just becomes tight plus the amount of stretch. This variation has the property that one can still calculate the peak load from this ratio, indeed the equation is marginally simpler, but the ratio itself is unknowable until after the rope has caught the fall and high-speed photographs of the event have been analyzed to determine how much stretch occurred at maximum extension.

In spite of the terminally diminished utility of this approach, you can find an account at www.cwu.edu/~curtiswd/PapersAndPreprints/cmj135-140.pdf, where once again no mention of Wexler's work is to be found.

In reply to Captain Paranoia:

Retracing someone else's steps from first principles does not, in my opinion, absolve one from the responsibility of crediting the person who made the original journey. You wouldn't, for example, announce a first ascent without trying to determine whether or not someone had already been up that way. There are multiple ways authors could have found references to Wexler's work had they tried; the information is out there.

HardenClimber -

*on 11 Dec 2012*In reply to Muppet123:

http://www.caves.org/section/vertical/nh/52/nh52.html

Article 5, ff page 21 has some interesting graphs, perhaps a bit more applicable to belaying but illustrating some interesting points about what forces are generated when a fall is stopped.

http://www.caves.org/section/vertical/nh/52/nh52.html

Article 5, ff page 21 has some interesting graphs, perhaps a bit more applicable to belaying but illustrating some interesting points about what forces are generated when a fall is stopped.

captain paranoia -

*on 11 Dec 2012*In reply to rgold:

The difference is that, IMHO, Attaway was making no claims for originality, merely presenting an engineering analysis (as anyone with the appropriate skills can do). From the introduction and dedication, the motivation for performing this analysis is very clear; the deaths of two climbers, presumably known to the author, and the desire to increase understanding of rope system safety.

I suspect it never entered Attaway's head that it might be original work, or to make any claims for originality, and he was merely presenting an analysis that anyone competent could have performed. A rather humble approach, rather than one seeking recognition.

> Retracing someone else's steps from first principles does not, in my opinion, absolve one from the responsibility of crediting the person who made the original journey.

*You wouldn't, for example, announce a first ascent*The difference is that, IMHO, Attaway was making no claims for originality, merely presenting an engineering analysis (as anyone with the appropriate skills can do). From the introduction and dedication, the motivation for performing this analysis is very clear; the deaths of two climbers, presumably known to the author, and the desire to increase understanding of rope system safety.

I suspect it never entered Attaway's head that it might be original work, or to make any claims for originality, and he was merely presenting an analysis that anyone competent could have performed. A rather humble approach, rather than one seeking recognition.

*This topic has been archived, and won't accept reply postings.*

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