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Kn calculators

Original Post
Jeff Luton · · It's complicated · Joined Aug 2016 · Points: 5

Yesterday my wonderful girlfriend took her first whip, 20 feet, it was rad!
On a black totem which is rated 6Kn.
After we got home I got curious as to how close she came to that 6Kn. So we used el google and found a Kn calculator which kinda tripped me out.
Using my statistics as a 150 pound human with 40 feet of rope out and taking a 20 foot fall on dynamic rope generates roughly 5.74Kn.
Now if I would take me again, 150 pounds, 200 feet of rope out, and a 100 foot fall, the calcutor would still spit 5.74Kn. Which to me seems a bit crazy, but just crazy enough to believe.

Curious what you mountain project scientists have to say!

Mike · · Unknown Hometown · Joined Dec 2013 · Points: 30

That number seems high... Maybe doesn't account for any motion of the belayer. Also it seems unlikely that the two results should be identical. Does it ask for weight of the belayer?

amarius · · Nowhere, OK · Joined Feb 2012 · Points: 20
Jeff Luton wrote: Yesterday my wonderful girlfriend took her first whip, 20 feet, it was rad!
On a black totem which is rated 6Kn.
After we got home I got curious as to how close she came to that 6Kn. So we used el google and found a Kn calculator which kinda tripped me out.
Using my statistics as a 150 pound human with 40 feet of rope out and taking a 20 foot fall on dynamic rope generates roughly 5.74Kn.
Now if I would take me again, 150 pounds, 200 feet of rope out, and a 100 foot fall, the calcutor would still spit 5.74Kn. Which to me seems a bit crazy, but just crazy enough to believe.

Curious what you mountain project scientists have to say!

Calculators are based on math, according to the math, fall forces depend only on "Fall Factor" - the ratio of fall distance to the length of rope in the system. Based on quoted numbers, FF is the same in both cases 20/40=0.5, 100/200=0.5, leading to the same impact force.

Hopefully, rgold sees this thread and adds something  relating this to real world events.
Drew L · · Boulder · Joined Jan 2018 · Points: 0

I think it's hard to have a calculator spit out a real number. Different ropes stretch at different rates which spreads the load over different periods of time. Then there's all the environmental factors that can affect the forces (soft vs hard catch, belayer pulled up vs not, phase of the moon). 

NegativeK · · Nevada · Joined Jul 2016 · Points: 40
Jeff Luton wrote: Yesterday my wonderful girlfriend took her first whip, 20 feet, it was rad!
On a black totem which is rated 6Kn.
After we got home I got curious as to how close she came to that 6Kn. So we used el google and found a Kn calculator which kinda tripped me out.
Using my statistics as a 150 pound human with 40 feet of rope out and taking a 20 foot fall on dynamic rope generates roughly 5.74Kn.
Now if I would take me again, 150 pounds, 200 feet of rope out, and a 100 foot fall, the calcutor would still spit 5.74Kn. Which to me seems a bit crazy, but just crazy enough to believe.

Curious what you mountain project scientists have to say!

A 100' fall obviously has more energy to dissipate than a 20' fall, so of course it feels crazy that the peak force is the same.

The idea, though, is that stretchy ropes spread the higher energy of the 100' fall over a longer distance than the 20' fall. In other words, you're going to feel a tug on your harness for much longer if you go for a giant screamer -- but the peak tug will be the same if the fall factor is the same.

A joule is a measure of energy. Conveniently, a joule is a newton*meter. That means you integrate the tug over the distance the rope is catching you and you get the total energy of the fall being dissipated.

(all climbers are spherical cows in a vacuum)
curt86iroc · · Lakewood, CO · Joined Dec 2014 · Points: 274

doesnt seem unreasonable to me. a 150 lb person just hanging statically on a single piece in a normal belayer-lead climber system loads that single piece with 1.3 kn... (150 lb x 2) / (225 lbs/kn). rough approximation...

tradryan · · Unknown Hometown · Joined Feb 2008 · Points: 63

What's the calculator you're using? (link please)

Good point about the fall factor being equal in both scenarios (longer fall but more rope = more stretch). Hence the important notion that short static falls (e.g. onto a dyneema PAS or daisy can generate phenomenal forces). 

5.74 kN seems a bit high to me also for these types of whipper scenarios. The calculator must have some built in assumptions (rope stretch varies widely between different makes and models of rope, as one example). Probably also assumes the belayer is a static ground anchor. The "flex" of the belayer moving upward at the moment of the fall would reduce the peak load in a real world scenario. 

Jeff Luton · · It's complicated · Joined Aug 2016 · Points: 5
http://ferforge.tripod.com/Srt002.htm
There’s the link to the calculator I used Tradryan.
In the fall I did give her a bit of slack in hopes to get the rope from behind her leg (fucking liebacks). I was fairly static in a hanging belay until locking off which sucked me about a foot and a half into the wall.
Rope being used was that tommy caldwell bipattern that was on sale on backcountry not long ago, I want to say it’s a 9.4mm
I’d say she was about 8 feet above the totem, a foot of slack from me, me getting sucked into the wall, and her total fall length

Edit to add: would the distance from past anchor be at the end of the fall? Or how far above the piece? I assumed at the end
Kyle Tarry · · Portland, OR · Joined Mar 2015 · Points: 448
Jeff Luton wrote: http://ferforge.tripod.com/Srt002.htm
There’s the link to the calculator I used Tradryan.

Be very careful with this calculator.  It is giving incorrect results for calculated fall factor, which means that the rest of the results are total garbage.  You can't even get the FF below 1.0 no matter what inputs you put in, so it's definitely got some bad math.

Example 1: 40 feet of rope out, 10 feet above the last piece (20 foot fall) = FF0.5.  This calculator calls it FF1.25 (!!)
Example 2: 100 feet of rope out, 5 feet above last piece (10 foot fall) = FF0.1.  This calculator call is FF1.0.

Who knows if the other math in this thing is right, but at the very least the FF in the scenario you experienced (40 feet of rope, 10 feet above gear) is FF0.5, not FF1.25 like they say, so the force will be WAY lower than they predict.
NegativeK · · Nevada · Joined Jul 2016 · Points: 40

You also can't calculate force without info from the rope.

I'm also kind of amused that the calculator thinks that static ropes exert exactly twice as much force on the faller as dynamics.

Jeff Luton · · It's complicated · Joined Aug 2016 · Points: 5

Sweet that calculator is trash.
Can somebody recommend a better one? Appreciate the responses guys!

Serge S · · Seattle, WA · Joined Oct 2015 · Points: 688

In some old threads here people mentioned calculators from 10-15 years ago that at least got the math right under some simplifying assumptions.  But, as the community's understanding of the factors that go into it improved, it seems the people who somewhat knew what they were doing gave up and took their calculators down.  So now only the completely clueless ones remain ;)

The impression I got from those threads:
 - a tiny fall is 3kN on the top piece
 - a pretty big fall is 5kN on the top piece
 - the most people realistically encounter is 7kN on the top piece

NegativeK · · Nevada · Joined Jul 2016 · Points: 40
Malcolm Daly · · Hailey, ID · Joined Jan 2001 · Points: 380

I think Petzl has a fall calculator as well. It would be interesting to compare. 

Kyle Tarry · · Portland, OR · Joined Mar 2015 · Points: 448
Jeff Luton wrote: Sweet that calculator is trash.
Can somebody recommend a better one? Appreciate the responses guys!

You can roughly ballpark it by assuming that the max force is proportional to the square root of the fall factor, using the known datapoint from the UIAA test as a starting point.

Most ropes might be around 8 kN for ~FF2.0 (rounding a bunch here), FF0.5 from your scenario is 1/4, square root of that is 1/2, so maybe 4 kN max force in your scenario?  Keep in mind that the UIAA test has a completely static belay so real world results are likely to be lower.  You also get lower results with a human than a steel mass (Petzl results here), so that further lowers it.  Maybe 2-4 kN is a reasonable range for force on the piece.

Note that in the Petzl results referenced above were 5 kN on the piece with FF0.7 and an 80 kg (176 lb) climber, and a fixed point Grigri.  So your scenario is probably less than that, given lower fall factor, lighter climber, moving belayer.  I'd throw my dart at 3 kN, with a +/- 2 kN uncertainty.  
Martin le Roux · · Superior, CO · Joined Jul 2003 · Points: 416
Kyle Tarry wrote:

You can roughly ballpark it by assuming that the max force is proportional to the square root of the fall factor, using the known datapoint from the UIAA test as a starting point.

Most ropes might be around 8 kN for ~FF2.0 (rounding a bunch here), FF0.5 from your scenario is 1/4, square root of that is 1/2, so maybe 4 kN max force in your scenario? 

That's a pretty good approximation. The equation for the simple model is actually

T = w + sqrt(w^2 + 2*k*r*w)

where

T = maximum rope tension (impact force) in kN
w = climber's weight in kN = 0.00445 * weight in pounds
k = rope modulus, which can be estimated as U * (U - 1.568) / 2.791 where U is the published UIAA impact force in kN
r = fall factor

In this case w = 0.67 kN and r = 0.50. If one assumes that U = 8.0 kN then T = 4.25 kN.

(You can easily verify that if you set w = 0.78 kN and r = 1.78 then T = U, the UIAA impact force).

Here's rgold's note on this topic:  4sport.ua/_upl/2/1404/Stand…. He mentions that this equation was first published back in 1946 (possibly earlier). As he and many others have cautioned, it's based on highly simplified assumptions that are routinely violated in real-world conditions.

Marty C · · Herndon, VA · Joined Aug 2008 · Points: 70

Martin - nice summary.

Serge S · · Seattle, WA · Joined Oct 2015 · Points: 688

Here is a calculator that tries to model rope non-linearity, internal rope friction, top carabiner friction, belayer lifting, and belay device slippage:  

http://fallsimulator.000webhostapp.com/FallSim.html

Certainly wrong but maybe less wrong than anything else publicly available ?

The belay device slippage model is linear with a force threshold - below the threshold there is no slippage, above the threshold it slips at a rate proportional to the difference between actual force and threshold force.  If somebody knows of a better model, please let me know.  A large threshold (e.g. 10kN) effectively means GriGri.  The defaults I picked - 1kN threshold and 0.5 kN/(m/s) slope - are just made up numbers that seem to give believable slippage totals for some scenarios I tried; I don't claim that the defaults represent anything real.

Don't put in very small length numbers (rope length < 0.1m or fall length < 0.001m) - it currently has problems with those, starting with (but not limited to) the 1-millisecond simulation step granularity.

Guideline #1: Don't be a jerk.

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