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tweeek


Jan 13, 2009, 10:48 PM
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majid_sabet


Jan 14, 2009, 1:37 AM
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Re: [tweeek] Need help calculating forces / fall factor! [In reply to]
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tweeek wrote:
Hello physics junkies!

I need some help calculating forces placed on my climbing gear in a slightly different than normal situation. And by slightly, I mean alot :)

I constructed a large scale rope swing under a nearby trestle. (see diagram below). Anchor is set in the middle of trestle (actually 2 anchors, one for each rope) they hang flush with the bottom of the truss system @ dead center. two ropes are attached to anchors, the other ends to your sit & chest harness. Walk back along the truss system until the ropes are nearly tight... and jump.

Now my question is this. What would the fall factor be in this type of system. I don't know how to calculate it, as you are not falling straight down onto the rope (In which case it would be 100ft fall : 100 ft rope = Factor 1) I don't think it would be as severe as a factor 1 as there is no shock loading and the process of weighting the rope is smooth and constant.

That being said, it is obviously much more force than body weight alone, as the centrifical force when you are at the bottom of the pendulum is quite significant.

All of this force is then distributed between: 2 Ropes, Two Anchors M.P., 4 anchor points (1" tubular webbing slings)

Any help??

Thanks!

Video: http://www.youtube.com/watch?v=Hp1S3g_11UA

[image]http://members.shaw.ca/emailtech/Niagara265FeetHigh.jpg[/image]

You got any pictures of you anchor, slings, knots and everything else between the bridge and you ?

I see some scary stuff and I think you should worry about those more than the falling forces.


adatesman


Jan 14, 2009, 1:59 AM
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hafilax


Jan 14, 2009, 2:20 AM
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Re: [adatesman] Need help calculating forces / fall factor! [In reply to]
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I don't understand why this got kicked out of the lab. Is there no room for theory in the lab?

I don't have time for a complex theory so here's the spherical cow guess. Treat it like a pendulum and ignore air drag and the stretching of the rope.

Energy is conserved so E=PE+KE (potential plus kinetic energies).

The initial PE=mgr where r is the length of the rope.

PE=0 at the bottom so KE=1/2mv^2=mgr
v^2=2gr

The force on the rope will be the centripetal force plus gravity
F=mg+mv^2/r=mg+2mgr/r=3mg

So to first approximation the force on the rope will be 3 times the force of gravity.

If I'm bored I might try to incorporate stretch with a Hook's law approximation.


patto


Jan 14, 2009, 3:07 AM
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Re: [hafilax] Need help calculating forces / fall factor! [In reply to]
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Nice work there Halifax.

3 times body weight tells us what should already be obvious. The nominal forces involved are nothing to be concerned about.

The big concern in any of these things is ensuring the rope swings freely and doesn't rub against anything.


shoo


Jan 14, 2009, 3:10 AM
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Re: [tweeek] Need help calculating forces / fall factor! [In reply to]
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Fall factor is not a relevant measure here. Arguably, the fall factor is 0, since you are free falling for exactly 0 meters with any given length of rope out. You can measure force, as shown above, but not fall factor.


tweeek


Jan 14, 2009, 4:41 AM
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brenta


Jan 14, 2009, 5:16 AM
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Re: [hafilax] Need help calculating forces / fall factor! [In reply to]
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hafilax wrote:
If I'm bored I might try to incorporate stretch with a Hook's law approximation.

I guess I was bored a few months ago, and I wrote a simple model with Hooke's law taken into account, but no friction. Here's a trajectory computed by Octave. I didn't focus on force, because I was interested in the speed at which one would slam into a dihedral if falling during a traverse. I'm a bit too tired tonight to change the model, but the rope modulus is 20 kN, so it's not difficult to estimate it at roughly 2 kN from rope elongation.



majid_sabet


Jan 14, 2009, 5:54 AM
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tweeek wrote:
In reply to:
You got any pictures of you anchor, slings, knots and everything else between the bridge and you ?

I I do not have any pictures, but I will do my best to describe the system.

There are 4 shoulder legnth 1" tubular webbing slings tiedd with a water knot. Each of these slings are girth hitched around a different railway tie on the top of the trestle. Each sling has 1 locking biner on it.

I have 2 60' legnths of 8mm static cord. Each legnth is tied into a loop with triple fishermans. These loops are looped through 2 biners each and back down (Like a 2 piece cordalette) and tied together with a figure 8, makeing each loop 15' long before the knot is tied.

One the master point of anchor 1, 2 locking beaners are placed, gates opposed and rope one is clipped in at the half way point with a figure 8. 1" tubular webbing acts as a second sheath for this rope and extends another 12 feet incase of rubbing on the truss system.

On the master point of Ancher 2, 1 locking biner attaches a GriGri with the rope at halfway. Below the grigri, on the breakhand side is a hand ascender with a pulley on the back and has the break hand rope threaded through. This forms the 3:1 ratcheting pulley system used to haul back up. The hand ascender system sits at the top below the grigri until the swing is finished, then lowered by the weight of the hand ascender. Haul and reset 2 or 3 times until jumper is at top.

Rubbing is possible at the extreme far end of the swing right before the 2nd pendulum begins. This is why the extra sheath is added to Rope 1. An extra sheath on rope 2 is not possible due to the ratcheting pulley system. Inspection of the ropes / sheath after the first 3 jumps showed no signs of wear at all, other than a little dirt transfer from the steel.



Now, I am trying to understand dthe forces put onto the rope... If I read correctly, I have been told 2 things... the force is 3x that of gravity, or the force is 3x that of body weight? Sorry, Im not sure if I read that correctly.

The reason I ask, is I am wondering if the forces are signifigant enough to warrant me from retiring a rope used in this swing after X amount of uses, as I would with a rope that has been exposed to hard falls while I am climbing.

Thanks so much for all your help!!

are you jumping on these two static 8 mm ropes?


JimTitt


Jan 14, 2009, 8:10 AM
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Re: [tweeek] Calculating Fall Factor / Forces on a pendulum swing [In reply to]
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It´s like hafilax says.
A more elegant formula which derives from his is
F= M g (3-2 cosX) where X is the starting angle from vertical.

And the answer, as he says is 3 x your weight, as long as you start with the rope tight!

Plenty of baffing stuff on pendulums in The Pendulum Lab. http://monet.physik.unibas.ch/...r/pendulum/lroom.htm


cantbuymefriends


Jan 14, 2009, 10:13 AM
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Re: [tweeek] Need help calculating forces / fall factor! [In reply to]
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tweeek wrote:
Now, I am trying to understand dthe forces put onto the rope... If I read correctly, I have been told 2 things... the force is 3x that of gravity, or the force is 3x that of body weight? Sorry, Im not sure if I read that correctly.
The reason I ask, is I am wondering if the forces are signifigant enough to warrant me from retiring a rope used in this swing after X amount of uses, as I would with a rope that has been exposed to hard falls while I am climbing.

Thanks so much for all your help!!
Are you absolutely sure that you know what you are doing/talking about, cause this question indicates that you don't have a clue?

Neither gravity nor body weight is a force, per se. They just "form" a force when combined. You know, F=m*a (where "a" in this case is gravity)

So, "three times bodyweight" means that you take the force that Gravity normally exerts on Your Bodyweight and multiply that by 3.
If your weight is 100 kg (220 lbm) the gravitational force on your body is 980 N (almost 1 kN)
So 3 times bodyweight means in this case a load or force of about 3 kN.
Ok?


brenta


Jan 14, 2009, 3:24 PM
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Re: [cantbuymefriends] Need help calculating forces / fall factor! [In reply to]
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Weight is a force. If you go to the moon, your mass does not change much. Your weight does.


cantbuymefriends


Jan 14, 2009, 4:18 PM
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My 2kg dumbell is a weight. How much does that change if I bring it to the moon? WinkWink

I could be right http://en.wikipedia.org/wiki/Body_weight
" the term body weight is overwhelmingly used in daily English speech and in biological and medical science contexts to describe the mass of an organism's body."

but since even NASA states that weight is a force, I guess I'm proven wrong.SmileSmile
http://www.grc.nasa.gov/...irplane/weight2.html

sorry 'bout that...


tweeek


Jan 14, 2009, 8:30 PM
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fxgranite


Jan 14, 2009, 8:58 PM
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Re: [tweeek] Calculating Fall Factor / Forces on a pendulum swing [In reply to]
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I think an interesting question to ask would be what happens to that 3x force if the rope isn't completely tight when you jump. Would it go up, down, or remain unchanged? My intuition says up but that could be completely wrong (as it often is in physics). I can't play around with it right now but I may when I get home this evening.

Also, if I'm visualizing this correctly, fall factor would actually start becoming a valid measurement the closer to the anchor point you get (ie, more slack on jump). I don't think it would ever get very high though, unless you're falling straight down of course.

on a side note: that swing looks awesome

(This post was edited by fxgranite on Jan 14, 2009, 8:59 PM)


Tree_wrangler


Jan 14, 2009, 9:27 PM
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In reply to:
Is there anyone with any constructive critisism of the anchor system?

Yeah.

Dump the girth hitches. Even if they were acceptable tie-ins, you'll have to keep in mind that your system does not evenly distribute the force among them, excepting the brief instant in which you are directly below the master point.

Instead, tie your water knot around the railroad ties. You can loop the webbing around an extra time to create a cinch if that was why you were working with the girth.


Tree_wrangler


Jan 14, 2009, 11:01 PM
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In reply to:
I think an interesting question to ask would be what happens to that 3x force if the rope isn't completely tight when you jump. Would it go up, down, or remain unchanged?

Up.


fxgranite


Jan 14, 2009, 11:39 PM
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Re: [Tree_wrangler] Calculating Fall Factor / Forces on a pendulum swing [In reply to]
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Tree_wrangler wrote:
In reply to:
I think an interesting question to ask would be what happens to that 3x force if the rope isn't completely tight when you jump. Would it go up, down, or remain unchanged?

Up.

One would expect then that the maximum force you could generate on the anchor was if you dropped straight down. I guess that makes sense.


austin.timm


Jan 14, 2009, 11:59 PM
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tweeek wrote:
There are 4 shoulder legnth 1" tubular webbing slings tiedd with a water knot. Each of these slings are girth hitched around a different railway tie on the top of the trestle. Each sling has 1 locking biner on it.

I highly suggest going to a wrap 3 pull 2, or wrap 4 pull 3 type system. Girth hitches aren't very good to start with not to mention that we don't know how creosote stained 1'' tube webbing does under load.


Tree_wrangler


Jan 15, 2009, 12:07 AM
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In reply to:
I highly suggest going to a wrap 3 pull 2, or wrap 4 pull 3 type system.

Yes. Although he does already have 4 points.

In reply to:
we don't know how creosote stained 1'' tube webbing does under load.

Good point. Normal "treated" wood literally eats metal, etc. You could probably swing on one set all day, or two days running, but if you exposed the slings to the treated wood, and then packed them away for a month (or whatever, I don't know), they might be compromised.

I'd change them regularly, if not for every setup.


gunkiemike


Jan 15, 2009, 11:57 PM
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Re: [Tree_wrangler] Need help calculating forces / fall factor! [In reply to]
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Tree_wrangler wrote:
Good point. Normal "treated" wood literally eats metal, etc. You could probably swing on one set all day, or two days running, but if you exposed the slings to the treated wood, and then packed them away for a month (or whatever, I don't know), they might be compromised.

I'm not aware of any data to indicate that short term exposure of nylon or Dyneema to treated wood, esp. wood that has seen years of weather exposure (hence no solvents left), presents any problem.

The compatibility of wood fasteners with treated wood is irrelevant to the OP's question, as I read it.


(This post was edited by gunkiemike on Jan 16, 2009, 12:26 AM)


Tree_wrangler


Jan 16, 2009, 12:10 AM
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I said "might". I have no idea what wood treatment eats, aside from iron and aluminum. The fact that those metals (and others) are quickly destroyed by wood treatment suggests rather strongly that the chemical is highly reactive....to what? Go look it up if you're that interested. I don't know the answer, but it's worth consideration

In reply to:
And you'd better tell the world's carpenters that metal nails, screws, and brackets are incompatible with threated wood. Too bad for all those decks and outdoor structures that will fall apart in a month.

They use screws specifically designed for treated wood. Regular nails and screws are rapidly destroyed. Even the hot-dipped treatment-friendly nails go downhill pretty quick. Think I'm wrong? Put a galvanized screw into treated wood and take it back out in a year and look at it.


(This post was edited by Tree_wrangler on Jan 16, 2009, 12:16 AM)

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