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Question about climbing and fall forces generated

  1. Sep 10, 2010 #1
    I'm using the formulas on Wikipedia and the cited Goldstone article to try and figure out how much force I'll put on climbing hardware when taking a fall. The numbers in Excel all look the same despite rope stretch?

    It's relevant because with aid climbing, fall distances are very short (and with lots of rope out there's alot absorbed in stretch) but the gear can be margional. Some aid climbing anchor pieces take less than 2kN of force.

    How can I change this so rope stretch properly shows me how impact force on the climber (or anchor) changes??

    Attached Files:

  2. jcsd
  3. Sep 11, 2010 #2
    Sorry, the Fall Factor and impact force link is http://en.wikipedia.org/wiki/Fall_factor

    The formula I'm using in Excel, which changes very little despite a big change in rope stretch is: =$C3*$C4 + SQRT(POWER($C3*$C4, 2) + 2*$C6*$C3*$C4*C10)

    =mg + [(mg)^2) + 2mgkR]^.5

    Where m=mass (80k), g=gravity (9.8), k=rope stretch (~9.5), and R=fall factor (changes from .1 to 2)

    Thanks for any help!
  4. Sep 12, 2010 #3
    For one thing, you have a typo in your formula for the modulus of elasticity. When you fix that, though, there still isn't a huge change.

    From my climbing days, I seem to remember that you should get a pretty huge difference in force if you go from a fall factor of 0.1 to 2. Unfortunately I don't really have time to look more closely, but hopefully someone will (or perhaps you can find another typo. I would guess that the Wikipedia article should have the correct math.)
  5. Sep 14, 2010 #4
    Thanks for your reply. I fixed the typo you caught and also tried a couple of alternatives but I don't see the impact force increasing with fall factor.... There should be a huge increase as charted here: http://www.rockclimbing.com/cgi-bin/forum/gforum.cgi?post=2375113#2375113

    One thing that would make a big difference is the "modulus of elasticity" applied to fall factor. There were a couple different derivations of the formula for it but I got the same results (no big change in impact force). I'm just copying over final equations in the paper so I don't see why this shouldn't work. :(

    Maybe fall distance should make a difference with impact force? In a simple system, shouldn't this look like Mass X Gravity X Distance??

    Thanks for any help!
  6. Sep 14, 2010 #5
    I´d check the k thing. In Wikipedia, k is rope modulus, not stretch.
    I´d expect RM values to be much higher than 9.5N.
    (k must have the dimension of a force, otherwise the dimensions in the formula don´t work out). Maybe there is a problem with the units in UIAA rating/RM conversion?
    Last edited: Sep 14, 2010
  7. Sep 14, 2010 #6
    I wrote my calculus professor and attached the most recent spreadsheet here, thanks.

    Attached Files:

  8. Sep 15, 2010 #7
    I still think you might be more careful with your units. Just look at the weight of your climber in the spreadsheet. And some units are not specified.
  9. Oct 29, 2010 #8
    I posted a question that is identical in nature, earlier in the week. Unfortunately I did'nt get any responces. The spread sheet here posted in the thread is definitely flawed, as some of the forces are too much and there is hardly a change in forces as the fall factor increases.

    Somebody out there... please figure this out. What is important to me is the maximum impact force on the anchor. Just like the person that started this thread, I am using gear rated in 2Kn to 4Kn. I need to know how far I can fall without gear failure. This could mean as small as a foot. Once again, here is the wikipedia link. I am also including a calculator link that appears weak on values, as well. The calculator is the only one of its kind online out there for the climbing community and it's wrong. Let's bang this out.


    http://shockforce.com/ [Broken]

    Incase you don't know:

    Last edited by a moderator: May 5, 2017
  10. Oct 30, 2010 #9
    In C6 you have a 5 instead of C5

    The number you have for the UIAA impact force rating is wrong. The UIAA specifies
    a maximum of 2540 pounds for this number.

    As a result of these, your value of k will be much too low. The rope will appear to
    be incredibly elastic, so the deceleration of the fall will ad little to the maximum
  11. Oct 30, 2010 #10
    Hi SnowLurk, thanks for following up on the thread. It's been a while since I looked at the spreadsheet and i don't remember specific steps taken to correct what's above. Units did get fixed and the formulas here (use this post as the xls starting point) are a copy from Goldstone's article. ...None of my guess work.

    I'm not seeing what WW wrote about $c5 values. The numbers seem to match a couple other calculators online but 'may be looking at different file. Two things I saw that were interesting in the numbers (please point out other problems) are:

    1) a climber at rest puts a little less than twice his weight on the top anchor

    2) doubling the length of rope does NOT change the forces, for the same fall-factor fall. I don't get that.

    Attached Files:

  12. Oct 30, 2010 #11
  13. Oct 30, 2010 #12
    The number you should end up with in c6 should be the force needed to stretch the rope to twice its length (in newtons). (or 100 times the force to stretch it 1%). What you have is obviously much too small.
  14. Oct 30, 2010 #13
    What you're writing about c6 makes sense, the less force it takes to stretch the rope, the more impact would be absorbed by the rope, imparting less force on the climber and anchors. The unit is kilo-newton though and 9 is a typical median number.... UIAA lists a maximum number of 12Kn and I took 9 by comparing some popular 9.8-10.2mm single ropes at REI.

    Rope Label > Impact Force

    UIAA Specs
    http://www.theuiaa.org/safety_standards.php [Broken]

    REI compare link
    Last edited by a moderator: May 5, 2017
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