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


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Trad climber
Topic Author's Original Post - May 8, 2012 - 07:11pm PT
Does anyone know how many kNs are generated when you take a vertical fall with your gear at your feet? Or even how to account for factors would affect that? Like the dynamic nature of a rope, the quality of your belay partner, or the integrity of a gear placement?

How big of a fall would be required to blow a well placed micro stopper that was rated at 4kN?

Social climber
So Cal
May 8, 2012 - 07:14pm PT
How fat are you?

1/2 MVsq

That's the biggest variable.

Yer gonna die.

A long way from where I started
May 8, 2012 - 07:17pm PT
I think it depends on whether or not you have a blue Camalot on your rack.

Trad climber
May 8, 2012 - 07:18pm PT
The answer is in the question.

Social climber
Joshua Tree
May 8, 2012 - 07:19pm PT

Trad climber
Topic Author's Reply - May 8, 2012 - 07:37pm PT
I'm 190lbs(ish)

Feel safe falling on a well placed blue camalot, but question how much of a fall a micro stopper is good for.

Trad climber
Placerville, California
May 8, 2012 - 07:38pm PT
dont kill o' newton, dude.
he's boss.

philadelphia, pa
May 8, 2012 - 08:11pm PT
Dunno about the one linked above, but the one by probably has the best mathematical model of what's going on during a fall. IIRC he improved upon the "Standard Model" put forth by RGold (I think that was his work) and put it online 'round about when Petzl pulled their online calculator.

Trad climber
Oaksterdam, CA
May 8, 2012 - 08:12pm PT
F=sqrt(2mgA f)

Where f is the fall factor.

To calculate the force produced by a fall you need the springiness (or spring constant per unit length A). This can be calculated from the stretch percentage of the rope (%x) under a weight. This is quoted for a weight of 80 kg on most ropes.


For a Mammut Gym Rope ((world wide web) this is 9%.

A = 80 * 9.81 / 0.09 = 8720

In our example our 76.2 kg climber could produce a maximum force when the fall factor is 2.

Fmax=sqrt(2 * 76.2 * 9.81 * 8720 * 2) = 5.1 kN

This is the maximum force on the climber. The maximum force on the protection holding him would be more than this as it has to hold the force of the climber + the force that is transmitted down to the belayer. If no force was lost then the protection would get twice the force on the climber, but only 66% of this force gets transmitted to the belayer the rest gets lost as heat due to friction on the carabiner - no wonder it gets hot!

So the force on the belayer (Fbelay) equals 66% of the force on the climber (Fclimb).

Fbelay=0.66 * Fclimb

And the force on the protection (Fprotect) equals these added together.

Fprotect = Fbelay + Fclimb = 1.66 * Fclimb

So the maximum force our climber could produce on his protection during a fall would be

Fprotect = 5.1 kN * 1.66 = 8.5 kN

Enough force to break nuts # 1-5!

If our example climber were to place protection a 3 meters then take a 2 meter fall he would produce a force of

Fprotect=( sqrt(2 * 76.2 * 9.81 * 8720 * 2/4) * 1.66) = 4.2 kN

But if he were to place protection as high as 7 meters before taking a 2 meter fall

Fprotect=( sqrt(2 * 76.2 * 9.81 * 8720 * 2/8) * 1.66) = 3.0 kN

Now if his friend who is twice as heavy (24 stone or 152.4 kg) was to try the same climb and take the same fall onto the 7 meter protection he would produce a force of

Fprotect=( sqrt(2 * 152.4 * 9.81 * 8720 * 2/8) * 1.66) = 4.2 kN

Mountain climber
San Diego
May 8, 2012 - 08:18pm PT
Momentum = Impulse

mv = Ft

Solve for F:

F = mv/t

F = N (1000N = 1kN)
m = kg
v = gt
t = seconds (to solve t see below)
g = 9.8m/s^2

d = 1/2gt^2

Solve for t:

t = square root (2d/g)

Trad climber
May 8, 2012 - 08:33pm PT
I just place my pro as good as I can and try hardly not to fall.

There're too many factors should be involved in calculations. It's nearly impossible to predict a pull force even in lab environment without thorough equipment testing. E.g., when DMM engineers tested slings they get a noticeable different results for the same setups in different tries, ref. (e.g., line 2, 16mm Nylon, min 10.5kN, max 11.4kN, assuming an average 10.95kN we get more than 4% deviation.
Spider Savage

Mountain climber
The shaggy fringe of Los Angeles
May 8, 2012 - 08:44pm PT
Fun thread.

Old School: The leader must not fall.

I saw a chart once that said a 150 lb weight falling 10 ft will generate 2000 lbs at impact.

That is how much I weighed and how far I fell onto my left ankle one time. There was damage.

Realistically, there are so many variables to the forces, if you have a specific placement on a specific route you could go to a lot of trouble to work it all out. Or you could do what it takes to climb it and not overload the piece.

I have used micro stoppers and the little tiny brass things. The small aluminum stoppers hold pretty well in short falls. I did a big tension traverse off a #1 stopper once. There was a point when It would have been a long bouncing fall but the going was only 5.5 or so.

Experience will take you there. Unless you die on the way.

Sport climber
May 8, 2012 - 09:21pm PT
190lbs? Holy f*ck you're gonna die.
Chickenhead Climbing Gear

Social climber
Philadelphia, PA
May 8, 2012 - 10:04pm PT
Fwiw, JT credits both Attaway and Goldstone. In any event, from reading the paper that accompanies his calculator I suspect it is better than most, and many out there are worse than horrible.

Trad climber
Poughkeepsie, NY
May 8, 2012 - 10:40pm PT
A few comments.

(1) Jay Tanzman's fall calculator,, is now the best thing available on the web that I know of. It uses Jay's improved version of the "standard model." Chances are that standard model type calculations overestimate peak loads, but they are still quite useful as worst-case scenario estimates.

(2) The account I wrote of the "standard model," mentioned by Adatesman,;postatt_id=2957; has been been presented over and over again by different authors, including Attaway, often, unfortunately, without a hint of attribution to the real originators. The credit belongs to Richard Leonard and Arnold Wexlar, who first published an account (plus quite a bit more) in the Sierra Club Bulletin in 1946. In this matter, those guys were way ahead of their times.

(3) The equation given by snowhazed is incorrect. For example, it says that if you just hang on the rope (fall factor=0), the maximum tension in the rope will be 0 kN (good news for manky rap anchors…). See Equation~(9) of the above-linked account, or Equation (13) for a version more adapted to calculation. But if you just want some numbers, use Jay's calculator.

(4) Klimmer's account begins with the incorrect statement that momentum=impulse. The correct version is that impulse is equal to the change in momentum. The next error is the formula suggesting that impulse = Ft, which is only true if the force involved is constant over the time interval in question, which is certainly not true of the tension in a climbing rope. The product Ft needs to be replaced by the integral of F(t) dt.

(5) I'm not sure I agree with Jim's claim, quoted by Del Cross, that a useful mathematical model is impossible because of the number of variables. You can't include all variables in a model, otherwise it wouldn't be a model, it would be the real thing, so the challenge is to include the "right" number of variables. Counterbalancing the problems with a mathematical model omitting variables is the fact that variables in real-world tests sometimes confound each other and lead to so much variation that one cannot use the test to decide between different approaches. This situation is particularly relevant to falls caught by human belayers, whose performance varies enormously from trial to trail.

A long way from where I started
May 8, 2012 - 11:31pm PT
Ha! I knew R Gold would show up and blow all of the bullshit away. But even with a perfect understanding of the physics available on the internet, 99.9999% of us neanderthals will simply wiggle in the best pro we can, stroke our blue Camalot a few times, and go for it.
Captain...or Skully

May 8, 2012 - 11:34pm PT
Given the options available, that's just what I'd do.
Except the Blue camalot part. Those things are junk.

Trad climber
Topic Author's Reply - May 9, 2012 - 12:31am PT
Thanks for the great replies.

I guess what it comes down to is trying to figure out if there is a rating in kNs that denotes that a piece of protection is aid gear as opposed to something that is appropriate for a trad rack.

Sounds like those really small brass microstoppers have very limited applications in moderate trad climbing (<5.10), and that I should stick with a set of regular BD stoppers or DMM walnuts or something similar?

lawrence kansas
May 9, 2012 - 12:57am PT
Some of the micro stoppers have higher kn ratings than some smaller cams. For instance in the Black Diamond set only the two smallest sizes are for direct aid. The largest size 6 is rated 8kn
while the smaller master cams are rated below that
Micro stoppers are great because they fit in places cams do not go (DMM brass offsets) and they do not add much weight to your rack. Sometimes something is better than nothing.

Trad climber
Poughkeepsie, NY
May 9, 2012 - 01:09am PT
The physics stuff is sometimes an interesting exercise after the fact, to try to estimate what kinds of forces might have been involved. It also can help to debunk outlandish claims, such as one made on another site that top-rope anchors need to be able to withstand 20 kN loads, when even the overestimating standard model indicates you'll never manage to get near 10 Kn, even with almost inconceivable worst-case scenarios.

As for micro gear, I think most of us it for free climbing. Success depends on many things. For example, the falls have to be short, which means after a few feet you ought to consider the piece worthless, the placements have to be made with great care and attention to the the local geometry, and the rock has to be excellent so that the trinkets don't break out of their placements.

Sometimes, with enough endurance and discipline, you can get in a string of pieces not far apart that will probably work because potential falls are kept very short, and other times you can place two or three pieces in one spot that may hold a longer fall.

Double-rope technique is especially useful in combination with small gear. You can try for a series of overhead placements, with new ones made before the previous placement is below knee level. With a bit of luck, your falls will be upper-belayed. The reason for double ropes is that if an overhead placement blows with a single rope, the large loop of slack created contributes to a long fall and so a big impact on the next piece down. With double ropes (properly managed by the belayer), there is no penalty for a blown overhead piece, because the other rope through the lower piece does not acquire slack when the overhead piece pulls.

For example, here's a shot of Dave Birkett using micro gear on The Walk of Life, perhaps most famous for being initially graded E12 7a, the hardest grade ever given to a route in the UK.

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