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Easy kN

Original Post
William Kramer · · Kemmerer, WY · Joined Jun 2013 · Points: 935

Was wondering if anybody out there had a quick and easy way to figure kN in relation to fall factor?

Example: If a leader fell when he was 50 feet up a route, 5 feet above last placed gear, so fall factor of 0.2 (think that's correct), how would one find the kN load on that last piece of gear?

And leader weighs 200lbs

DanielRich · · Unknown Hometown · Joined Aug 2008 · Points: 5

The biggest factor you need to know is how strechy the rope is.

It would be easier to figure out how much total energy dissipated since that is just height fallen and climber weight but max load on the gear would vary depending on how far the rope streches and how long it takes to stretch.

You could just assume a rope acts like an ideal spring and use hookes law
khanacademy.org/science/phy…

but I am not sure how good that is an approximation I suspect it might be fairly poor.

20 kN · · Unknown Hometown · Joined Feb 2009 · Points: 1,346
William Kramer wrote:Was wondering if anybody out there had a quick and easy way to figure kN in relation to fall factor? Example: If a leader fell when he was 50 feet up a route, 5 feet above last placed gear, so fall factor of 0.2 (think that's correct), how would one find the kN load on that last piece of gear? And leader weighs 200lbs
The only two calculators I have seen that are actually accurate (relative to their limitations) are JT512's from RC.com and Petzl's Fall Simulator. Unfortunately both are offline.

Here is one:

junkfunnel.com/fallforce/

But it is unclear what they are asking for with regard to rope elongation. I am not sure if they are asking for the UIAA dynamic elongation rating, the static elongation rating, or the percentage of elongation at the fall factor entered into the calculator. Without knowing the correct answer, it's just guessing to use that calculator.
Gunkiemike · · Unknown Hometown · Joined Jul 2009 · Points: 3,492

Fall factor is just one factor that goes into peak impact force. Rope stretchiness has already been mentioned, as has climber weight. But the angle of the face and the amount of sliding/bouncing play a huge role. As do belayer movement, rope friction on intermediate gear, and angle of fall (is there a pendulum component?). Bottom line - I wouldn't waste your time trying to get numbers for any particular fall.

Mike Gilbert · · Bend, OR · Joined Jul 2013 · Points: 21

I read somewhere that the average leader fall was somewhere near 7kN if that helps you feel better about some reference? I think of that as 5ish feet above a piece but I don't honestly know anything.

J. Serpico · · Saratoga County, NY · Joined Dec 2009 · Points: 140
Gunkiemike wrote:Fall factor is just one factor that goes into peak impact force. Rope stretchiness has already been mentioned, as has climber weight. But the angle of the face and the amount of sliding/bouncing play a huge role. As do belayer movement, rope friction on intermediate gear, and angle of fall (is there a pendulum component?). Bottom line - I wouldn't waste your time trying to get numbers for any particular fall.
What he ^^^^^ said.

Also, I agree with Robots and Dinos. Chris Harmston of BD determined that top pieces rarely ever saw 10kN and probably saw around 6-7kN worst case. Again though, this assumes dynamic system in a real world environment. Don't remember how he came to this conclusion, but it had something to do with evaluating climbing accidents, among other things.

This data can be found trolling the old rec.climbing archives or reposted on various sites.
teece303 · · Highlands Ranch, CO · Joined Dec 2012 · Points: 596

Finding the force generated by a fall is non-trivial and, honestly, impossible to determine in any real world scenario. That's the whole reason folks were happy to find the fall factor theory existed.

20 kN · · Unknown Hometown · Joined Feb 2009 · Points: 1,346
Robots and Dinosaurs wrote:I read somewhere that the average leader fall was somewhere near 7kN if that helps you feel better about some reference? I think of that as 5ish feet above a piece but I don't honestly know anything.
You probably read correctly, but the author was incorrect. I have seen that claim numerous times, but its often based on the wrong theory. They take that number from an impact force calculator, such as the ones I referenced above, but as others have said the calculator can only calculate the maximum impact force, not the average. The calculator assumes every component in the belay chain is 100% static except for the rope. Unfortunately that is a very incorrect assumption, except under very specific circumstances, which is why the calculators only calculate the maximum impact force possible under the conditions you entered into the calculator.

I have directly measured the impact force on the top piece with a load cell and high-scan-rate conditioner (which is the only accurate way to do it), and I have never hit 7kN. It's possible for sure, but I was lobbing 25' whippers onto the 4th and 5th bolt of a sport route and I never hit over 4kN on the top piece. I even took a whipper on the first bolt and landed below my belayer and I still dident hit 5kN. That said, I am only 155lbs, so someone who was 200 probably could hit 7kN easily, but that's not average.

I would say the average "legit" lead fall where you are at least 3' past the last piece would lob 3-5kN depending on your weight and numerous other conditions. Seven kN is certainly possible, but it's above average.
20 kN · · Unknown Hometown · Joined Feb 2009 · Points: 1,346

Here is the old article I wrote on the subject:

mountainproject.com/v/lead-…

It's quick and informal, but the falls I took in the test were well within the realm of average, and yet I never even broke 4kN on the falls with the soft catches.

Fall 1: - Dynamic catch – 13’ total – 3.09 kN
Fall 2: - Static catch (belayer just stood there) – 10’ total – 3.75 kN
Fall 3: - Running belay (belayer yanked in slack and ran backwards) – 7’ total – 3.91 kN

C Williams · · Sketchy, Blackvanistan · Joined Jul 2013 · Points: 1,556

Thanks 20kn for the great info! In addition I saw a friend take multiple falls onto a #2 rp and it held just fine with no visible damage to the piece. The rated strength is 3 kn...

rgold · · Poughkeepsie, NY · Joined Feb 2008 · Points: 526

You can use T= 0.784 +√[0.615 + (0.562)U(U-1.568)r]; this is the standard equation for fall impact force originally found by Arnold and Wexler in the late 1940's and republished---usually without a hint of attribution---in various other places. This particular form of the equation is from a little article I posted on rc.com some years ago. For a while a .pdf of that article could be downloaded, but the links I've seen---for example in the Wikipedia article---seem to be broken. As of the date of this post, you can still find the pdf at 4sport.ua/_upl/2/1404/Stand….

This equation uses the simplest reasonable rope model and can only be expected to give very rough estimates, which are typically too big, so what you are probably getting in most instances is upper bounds for the actual loads. There are much more sophisticated models, but you won't be running them on your scientific calculator as you can with this one.

The T in the equation is the peak rope tension; this occurs at the instant of maximum elongation. The two parameters are U, the UIAA impact force rating of the rope in kN, a number you can get from any catalog description of the rope, and r, the fall factor. The equation assumes an 80kg falling leader.

There used to be some fall force calculators out there you could trust, but I'm not sure that is presently the case. We can't see the equations used, but sometimes the things said and the information requested indicates the calculator is either unlikely to or completely incapable of yielding even the ballpark standard equation results.

The junkfunnel link posted by 20 kN is highly suspect because it requests the static elongation of the rope with an 80 kg weight. In theory, this would be sufficient to calculate the rope modulus, but not in reality. (The standard equation given above computes the rope modulus from the UIAA impact rating, which means it uses the rope elongation that occurs in a factor 1.7 fall.) Even if the equations used are right, the static elongation data will lead to erroneous answers.

Another calculator that keeps coming up is on the Outdoor Adventures Netwwork at myoan.net/climbart/climbfor…. This one is total garbage; the only reason for posting the link is to warn others not to use it. (I gave up asking them to take it down; I never got so much as an acknowledgement of my email.)

mattm · · TX · Joined Jun 2006 · Points: 1,885
20 kN wrote: I have directly measured the impact force on the top piece... and I have never hit 7kN. It's possible for sure, but I was lobbing 25' whippers onto the 4th and 5th bolt of a sport route and I never hit over 4kN on the top piece. I even took a whipper on the first bolt and landed below my belayer and I still dident hit 5kN. That said, I am only 155lbs, so someone who was 200 probably could hit 7kN easily, but that's not average. I would say the average "legit" lead fall where you are at least 3' past the last piece would lob 3-5kN depending on your weight and numerous other conditions. Seven kN is certainly possible, but it's above average.
Various other sources have seen similar results. Chris Semmel of the DAV reported forces: "in a factor 0.4 fall, forces are in the range of 4.2 to 7 kN, depending on the type of belay and the device used, with the great majority of devices and belay types generating about 5 kN at the runner" - Mark Houston summary of an IFMGA Technical Commission Meeting Report

Petzl reports 6kN in a FF1 scenario using a GriGri (From what I can see in their demo the belay has some movement)

In a high fall factor using a GriGri and Fixed Point Belay (or heinous rope drag etc) you could get above 7kN but the reality is that those types of Falls and Loads are EXTREMELY rare. This is born out by the fact that things like Wired Nuts rated at a MAX of 10kN almost never "fail" due to strength limitations (an idea noted by J. Long in one of his books IIRC)
J. Serpico · · Saratoga County, NY · Joined Dec 2009 · Points: 140
mattm wrote: MAX of 10kN almost never "fail" due to strength limitations (an idea noted by J. Long in one of his books IIRC)
Actually, this data was what I was referring to by Chris Harmston. Perhaps long referrenced him. Or I have it backwards, either way, thanks for confirming my recollection.
Eliot Augusto · · Lafayette, CO · Joined Dec 2013 · Points: 60

It seems to me the median of everyone's range of average falls awfully close to the fall factor without a decimal. FF.5 = ~5kN, FF.3 = ~3kN, FF1 = ~10kN.

Not exactly perfect, but its a decent ballpark.

mattm · · TX · Joined Jun 2006 · Points: 1,885
J. Serpico wrote: Actually, this data was what I was referring to by Chris Harmston. Perhaps long referrenced him. Or I have it backwards, either way, thanks for confirming my recollection.
You're spot on - Looked it up and Long/Gaines reference both CH of BD and Craig Connally from The Mountaineering Handbook (a great book) in their sub 10kN numbers.
runout · · Unknown Hometown · Joined Jun 2013 · Points: 30

I took a almost FF1 onto a C4 .75 in a horizontal and my belayer was tied in because she was so much lighter than me and using a Grigri. The piece held great, one of the lobes moved and looked like it wanted to open up some more but didn't. I didn't hit anything on the way down, but boy, did I feel that in my bones. So did my belayer, who complained about how it hurt her insides too. I wonder how many kNs that was?

patto · · Unknown Hometown · Joined Jul 2012 · Points: 25

Here is a guide along with typical rope stiffnesses.
bealplanet.com/sport/anglai…

My dodgy cut and paste:

IMPACT FORCE EQUATION
F=Mg+Mg*sqrt(1+2fk/Mg)

F = impact force in Newtons
M = falling mass in kg
g = gravitational acceleration
= 9,81 ms-2
K = caracteristics of the rope
(Young’s Modulus X Section of the rope)
f = actual fall factor

value of K as a function of the
maximum impact force of the rope
F = 7,0kN -> K = 13700
F = 7,5kN -> K = 16000
F = 8,0kN -> K = 18500
F = 8,5kN -> K = 21200
F = 9,0kN -> K = 24100
F = 9,5kN -> K = 27100
F = 10,0kN -> K = 30300

runout · · Unknown Hometown · Joined Jun 2013 · Points: 30
Jake Jones wrote: Not the greatest idea (assuming you mean tied in statically with a piece of cord or something similar). If you have somewhere to tie her in, use the end of your climbing rope if you can and allow for a few feet of slack. This will allow her to come off the ground some and also incorporate some dynamic properties of the rope as well. Your top piece of gear (not to mention you and your belayer) will be much better off. Also, consider having her wear a loaded pack to bring your weights closer when she's belaying.
Yeah I have an old rope that I am going to cut up and bring a good section for these situations. But really, I was so close to the ground that a couple of feet would mean hitting the ground.
Kedron Silsbee · · El Paso · Joined Aug 2013 · Points: 0

If I take a section of rope, and elongate it by a factor x, is the tension in the rope a function of x, or is the strain rate of consequence as well (within the bounds of reasonable strain rates expected in a range of falls)? If it is a good approximation that tension in a length of rope is a function just of the strain (elongation) in that segment of rope, is this tabulated somewhere for different models of rope?

rgold · · Poughkeepsie, NY · Joined Feb 2008 · Points: 526

The standard equation uses Hooke's law and so in that model the tension is directly proportional to the strain.

Kedron Silsbee · · El Paso · Joined Aug 2013 · Points: 0

That seems not to be a very good approximation based on these data. One would expect static elongation/.8kn = dynamic elongation/impact force (assuming a test mass of 80 kg for the static elongation). These ratios seem to disagree by a factor of two in some cases. I guess the maximum impact force varies roughly as the square root of the spring constant in the hooke's law model, so perhaps this uncertainty isn't as dire as it first seems, but it seems like no model is going to make much better than factor of 2 level predictions without going beyond the simple spring model.

Guideline #1: Don't be a jerk.

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