



ensonik
Aug 30, 2010, 6:52 PM
Post #1 of 100
(8002 views)
Shortcut
Registered: Jul 14, 2009
Posts: 134

I am currently reading through the ANAM 2009. On page 32, the analysis of an accident that happened during a pendulum fall is pretty thorough, but mentions the following: "..., if you fall from, say, ten feet to one side of and level with your pivot point, you will have the same speed at the lowest point of your swing as you falling straight down ten feet ..." To say the least, this is surprising to a physics layman like me. To describe this and be sure I really understand the full impact of a pendulum (yes, I over protect them even though I wasn't aware of this small yet non inconsequential detail), I drew up something a small child could have done better but nonetheless illustrates it well: In other words, on both fall types, when I hit point B, I will be going at the same speed? (distances between A and B being equal of course). (Yes, I know, I know: It's in the book ...., but you never know ... errors can slip in. So I want to confirm this on the internets where it's well known that errors are weeded out and only the real information bubbles through to the top)





bill413
Aug 30, 2010, 7:13 PM
Post #2 of 100
(7984 views)
Shortcut
Registered: Oct 19, 2004
Posts: 5674

ensonik wrote: I am currently reading through the ANAM 2009. On page 32, the analysis of an accident that happened during a pendulum fall is pretty thorough, but mentions the following: "..., if you fall from, say, ten feet to one side of and level with your pivot point, you will have the same speed at the lowest point of your swing as you falling straight down ten feet ..." To say the least, this is surprising to a physics layman like me. To describe this and be sure I really understand the full impact of a pendulum (yes, I over protect them even though I wasn't aware of this small yet non inconsequential detail), I drew up something a small child could have done better but nonetheless illustrates it well: In other words, on both fall types, when I hit point B, I will be going at the same speed? (distances between A and B being equal of course). (Yes, I know, I know: It's in the book ...., but you never know ... errors can slip in. So I want to confirm this on the internets where it's well known that errors are weeded out and only the real information bubbles through to the top) As I read that statement, your diagram is in error. "Fall from...level with your pivot point...." So, in your pendulum diaram, point A should be at the same height as your anchor (and as the A in the straight down fall).





nailzz
Aug 30, 2010, 7:33 PM
Post #3 of 100
(7970 views)
Shortcut
Registered: Jun 13, 2001
Posts: 244

What Bill said ... plus: The distance from what you have marked as A>B in your second diagram (the pendulum) would be greater than 10ft. The distance between point A and your anchor would be 10ft. The circumference of the circle would be 2(Pi)r. Your radius is 10ft and you are falling 1/4 of the circle, so your fall distance would be 2(Pi)10ft/4, or 15.7ft. More than 50% longer. I think what the book is saying, though, is that in either case you are moving downward for 10ft. Gravity is constant. It doesn't matter if you fall those 10ft straight downward or at an angle. At the bottom you are going to be travelling at the same speed (not counting things like friction and such). At least that's what I'm taking away from the quoted text.





patto
Aug 30, 2010, 8:19 PM
Post #4 of 100
(7939 views)
Shortcut
Registered: Nov 14, 2005
Posts: 1451

ensonik wrote: To say the least, this is surprising to a physics layman like me. Its conservation of energy. If you fall 10meters and weigh 100kg then your body will have E=mhg worth of kinetic energy (~speed energy). E= 100 * 10 * g(~9.8) = 9800J of energy. In practice a climbing rope stretches so a pendulum fall will be slightly slower as its not a perfect pendulum.





majid_sabet
Aug 30, 2010, 8:30 PM
Post #5 of 100
(7929 views)
Shortcut
Registered: Dec 12, 2002
Posts: 8352

sounds like a case in yosemite . is it ?





TarHeelEMT
Aug 30, 2010, 9:51 PM
Post #6 of 100
(7880 views)
Shortcut
Registered: Jun 20, 2009
Posts: 724

Your chart is wrong. It's 10 vertical feet. The motion of the pendulum just redirects the force that you gather in those ten vertical feet of falling into horizontal motion. At the base of the pendulum, you're traveling sideways with the amount of energy one gathers when falling ten vertical feet  just as if you hadn't been redirected by the pendulum and fell straight down. The pendulum just changes the direction of the force.





cantbuymefriends
Aug 31, 2010, 1:45 AM
Post #7 of 100
(7833 views)
Shortcut
Registered: Aug 28, 2003
Posts: 670

I guesstimate that the speed at the bottom of the the pendulum will be roughly the same as in a straightdown fall before the rope starts to stretch. As mentioned above, the velocity will be in the horizontal direction and not vertical though.





socalclimber
Aug 31, 2010, 3:44 AM
Post #8 of 100
(7806 views)
Shortcut
Registered: Nov 27, 2001
Posts: 2432

Let us not forget that a pendulum low down can take your straight into the deck. I saw it happen once. The person was lucky because it was just dirt. Now if there had been talus, it would have been a whole different story. Also bare in mind that if the swing ends in a corner, the end result can be very bad. You will likely be receiving a "trunk" blow. In other words, slamming the side of your body against the wall. Lot's of vital organs there that can get punctured or ruptured. Good question.





ensonik
Aug 31, 2010, 3:49 AM
Post #9 of 100
(7801 views)
Shortcut
Registered: Jul 14, 2009
Posts: 134

majid_sabet wrote: sounds like a case in yosemite . is it ? It is.





devkrev
Aug 31, 2010, 3:53 AM
Post #10 of 100
(7795 views)
Shortcut
Registered: Sep 28, 2004
Posts: 933

ensonik wrote: I am currently reading through the ANAM 2009. On page 32, the analysis of an accident that happened during a pendulum fall is pretty thorough, but mentions the following: "..., if you fall from, say, ten feet to one side of and level with your pivot point, you will have the same speed at the lowest point of your swing as you falling straight down ten feet ..." To say the least, this is surprising to a physics layman like me. To describe this and be sure I really understand the full impact of a pendulum (yes, I over protect them even though I wasn't aware of this small yet non inconsequential detail), I drew up something a small child could have done better but nonetheless illustrates it well: [image]http://imgur.com/memJH.png[/image] In other words, on both fall types, when I hit point B, I will be going at the same speed? (distances between A and B being equal of course). (Yes, I know, I know: It's in the book ...., but you never know ... errors can slip in. So I want to confirm this on the internets where it's well known that errors are weeded out and only the real information bubbles through to the top) Watch the opening scene in hard grit. Actually watch the whole movie, its good.





Toast_in_the_Machine
Aug 31, 2010, 4:39 AM
Post #11 of 100
(7782 views)
Shortcut
Registered: Sep 11, 2008
Posts: 5184

So this got me thinking about if there was an easy formula for pendulum. Especially looking at it with rope and a drop to calculate the velocity into a vertical wall. It was easy to get to the speed of a full pendulum. (Best web example is here:http://www.worsleyschool.net/...sement/pendulum.html) But my question was more: Knowing the height up and the distance over (and of course the rope length), was there an easy way to calculate (1) the force / tension when the rope "caught" and you began to swing and (2) what would the velocity be when you hit the wall. So I started to put pen to paper, but before I did, I did a quick look on the web and found this gem: What is the proper equation for maximum tension force in a rope for a falling load With the reference to the person's question being, you guessed it RC.com:
answers.yahoo.com wrote: Fall Factor source: The Standard Equation for Impact Force. Goldstone, R. (PDF, 2009) ref: http://www.rockclimbing.com/cgibin/forum/gforum.cgi?do=post_attachment;postatt_id=746 http://www.rockclimbing.com/...hment;postatt_id=746. I thought that was funny. (Now back to my pen and paper....)





majid_sabet
Aug 31, 2010, 6:48 AM
Post #12 of 100
(7747 views)
Shortcut
Registered: Dec 12, 2002
Posts: 8352

ensonik wrote: majid_sabet wrote: sounds like a case in yosemite . is it ? It is. Man, my memories is good but there was a similar case in late 90's on El Cap where a guy took a 15 feet pengi and crushed his side to the wall except he had tons of gear on his shoulder harness ended up with two broken ribs. Rescuers rapped 900 feet and jugged with the injured climber up to the top. pengi falls are extremely dangerous cause you are dealing with down and side forces simultaneously and you can not predict where you may end up.





gunkiemike
Aug 31, 2010, 8:49 AM
Post #13 of 100
(7690 views)
Shortcut
Registered: Oct 1, 2002
Posts: 2263

patto wrote: In practice a climbing rope stretches so a pendulum fall will be slightly slower as its not a perfect pendulum. That's the key. The other statements (ANAM, nailzz, tarheel) that speed would be equal because you've fallen the same vertical interval are only true in the theoretical case of a rigid, frictionless pendulum arm. So now the question becomes: for a typical rope modeled as a Hooke's Law spring, can we (er...rg?) model the final speed of a pendulum fall? The rope stretches (absorbing energy) as it does work against the climber. The tension in the rope increases proportionally (sinusoidally I suspect) as the fall angle increases from vertical. But the calculations are way beyond my pay grade.





fresh
Aug 31, 2010, 10:40 AM
Post #14 of 100
(7632 views)
Shortcut
Registered: Aug 6, 2007
Posts: 1199






ptlong2
Aug 31, 2010, 11:49 AM
Post #15 of 100
(7582 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

Ha ha ha! But the Hooke's Law version is an easy enough problem to solve. Stretch does subtract some of the energy but it also adds to it by increasing the fall distance. The answer is that the difference between the a stretchy rope and a truely static rope is negligable, in theory. Real life, with real ropes and slack added, there's going to be more of a difference but I can't say how much. Either way a big pendo fall into a corner is a bad idea, RG will concur.





darkgift06
Aug 31, 2010, 12:04 PM
Post #16 of 100
(7563 views)
Shortcut
Registered: Mar 16, 2009
Posts: 492

all the calcs & no one added the increase of forces based on the centrifugal force generated by the fall.





gunkiemike
Aug 31, 2010, 12:50 PM
Post #17 of 100
(7533 views)
Shortcut
Registered: Oct 1, 2002
Posts: 2263

"Centrifugal" force is simply the climber's perception of the tension in the rope.





Rudmin
Aug 31, 2010, 12:59 PM
Post #18 of 100
(7524 views)
Shortcut
Registered: Mar 29, 2009
Posts: 606

gunkiemike wrote: "Centrifugal" force is simply the climber's perception of the tension in the rope. Centrifugal force would actually be the climber's perception of a field (similar to gravity) pulling hair cloths, everything away from the axis of the pendulum. It is an artifact of adopting a rotational frame of reference. The tension in the rope is a centripetal force, and is an actual force no matter your frame of reference.





patto
Aug 31, 2010, 2:16 PM
Post #19 of 100
(7483 views)
Shortcut
Registered: Nov 14, 2005
Posts: 1451

I can't believe this is still being discussed. For a perfect pendulum with a perfectly static rope the speed is the same. The tension on the rope simplifies to the extraordinarily simply F=3mg. In other words 3 times the person's weight. For a perfect pendulum with a dynamic rope the speed is less. The tension on the rope will induce stretch the rope and absorb energy. For an imperfect pendulum with a dynamic rope it is all too complicated to work out, too many variables.





jt512
Aug 31, 2010, 3:12 PM
Post #20 of 100
(7461 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21892

patto wrote: I can't believe this is still being discussed. For a perfect pendulum with a perfectly static rope the speed is the same. The tension on the rope simplifies to the extraordinarily simply F=3mg. In other words 3 times the person's weight. For a perfect pendulum with a dynamic rope the speed is less. Maybe you're satisfied with the answer "less." It's obviously still being discussed because others want a little more specific answer.
In reply to: For an imperfect pendulum with a dynamic rope it is all too complicated to work out, too many variables. Too complicated for whom? Are you seriously suggesting that the problem is so complicated that it can't be solved? Jay
(This post was edited by jt512 on Aug 31, 2010, 3:14 PM)





ptlong2
Aug 31, 2010, 5:20 PM
Post #21 of 100
(7429 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

patto wrote: For a perfect pendulum with a dynamic rope the speed is less. The tension on the rope will induce stretch the rope and absorb energy. Yes, less. But the stretch also lengthens the fall which adds energy to the system, nearly enough to make it a wash. For a Hookian rope the difference is maybe a percent or two, depending on the rope modulus you select. If you add damping to the rope it reduces the speed a bit more, but not much.
In reply to: For an imperfect pendulum with a dynamic rope it is all too complicated to work out, too many variables. Got a computer?
(This post was edited by ptlong2 on Aug 31, 2010, 5:29 PM)





jt512
Aug 31, 2010, 6:25 PM
Post #22 of 100
(7400 views)
Shortcut
Registered: Apr 11, 2001
Posts: 21892

ptlong2 wrote: patto wrote: For a perfect pendulum with a dynamic rope the speed is less. The tension on the rope will induce stretch the rope and absorb energy. Yes, less. But the stretch also lengthens the fall which adds energy to the system, nearly enough to make it a wash. For a Hookian rope the difference is maybe a percent or two, depending on the rope modulus you select. If that's true then can you solve for the peak tension as follows? The potential energy (PE) of the fall is converted to kinetic energy (KE) and strain energy (SE) in the rope. That is, (1) PE = KE + SE . (2) PE = mgh = mg(L + s) , where L is the length of the unstretched rope and s is the maximum rope stretch in the fall. (3) KE = (1/2)mv² , (4) SE = (k/2L)s² (from rgold's paper), where k is the rope modulus. Substitute (2), (3), and (4) into (1), to give (5) mg(L + s) = (1/2)mv² + (k/2L)s² . From rgold's paper (6) s = TL/k , where T is the maximum rope tension. Now, substitute (6) into (5), compute v assuming an ideal pendulum, and solve for T. Jay
(This post was edited by jt512 on Aug 31, 2010, 6:26 PM)





ptlong2
Aug 31, 2010, 6:38 PM
Post #23 of 100
(7387 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

jt512 wrote: compute v assuming an ideal pendulum, and solve for T. If you're assuming the speed of an ideal pendulum then you don't need the energy balance equation. T=(k/L)s is also equal to the centrifugal force plus the weight (assuming no vertical component of the velocity at the bottom of the swing). It's easy enough to take that force balance and combine it with the energy balance to get a quadratic in s. Then you have s and v, as well as T.





Rudmin
Sep 1, 2010, 8:25 AM
Post #24 of 100
(7315 views)
Shortcut
Registered: Mar 29, 2009
Posts: 606

s and T are quite linearly related. You solve one and you have the other. The problem is that the centrifugal component of T is a function of v. That is m*v^2/(L+s). Simple enough to solve recursively, but it puts a crimp in your solution. To run it through some simple numbers. Let's say 10 metres of rope with a 100 kg climber and a stiffness of 24 kN per metre. (that's assuming that only the 10 metres of rope is being used). What I get is a velocity of about 50.2 km/hour and a rope tension of about 3kN. In essence, the rope stretch does next to nothing. If we take a quarter of the stiffness, as if you had almost a full rope length out, we get a velocity of 49.8 km/hour. So looks like no amount of climbing rope will help you in a pendulum.
(This post was edited by Rudmin on Sep 1, 2010, 8:26 AM)





ptlong2
Sep 1, 2010, 9:49 AM
Post #25 of 100
(7281 views)
Shortcut
Registered: Aug 10, 2010
Posts: 102

What I posted above is wrong. That is, you can't expect a force balance at the bottom of the pendulum. By coincidence it gives an answer that is very close to that of a numerical approach, at least for the first swing. Rudmin, I'm not sure exactly what you did but when I run your numbers the machine gives slightly lower results. Same basic answer though. Interestingly the tension peak occurs after the bottom of the swing. Now in a real pendulum fall you have a real rope, a belayer and some initial slack. I think it would still hurt.
(This post was edited by ptlong2 on Sep 1, 2010, 9:50 AM)





