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daithi
Jan 24, 2006, 3:37 PM
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In reply to: The forces in a two-dimensional elastic pendulum are due to gravity and Hooke's Law:
In reply to: Yeah. Instead, just keep track of the axial rope force and gravity, which are the only forces we're including (ignoring wall-friction, wind resistance, . . .).
What delcross and whoa have neglected in their models of an elastic pendulum is the "centrifugal" force which in addition to gravity acts against the spring. Even the non-mathematical amongst you will be able to appreciate that the faster you swing the rope around the more it will stretch.
If we have a look at the differential equations again.
In the theta direction:
d^2q/dt^2 = -g/r sin q [eqn. 1]
In the radial direction:
(d^2r/dt^2) = -(k/m)*r + (g*cos q) + (r*(dq/dt)^2) [eqn. 2]
Here we have two differential equations in two variables r and theta (q). For the non-mathematical amongst you (who I presume will have left long ago!) I will attempt to explain the terms.
The first equation basically says the angular acceleration is caused by the component of gravity acting in this direction. There is no other force acting in this direction.
The second equation basically says that the acceleration in the radial direction is equal to the component of gravity acting in line with the rope and the centrifugal force (which also acts in line with the rope). These two forces are counteracted (minus sign) by the spring force in the rope (the first term on the right hand side).
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whoa
Jan 24, 2006, 3:51 PM
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Daithi, there is no centrifugal force on the climber. The only forces on the climber are gravity, which pulls down, and the rope, which pulls radially. If you swing a weight around on a rope, there is not some new force on the object pulling it away from you. It is going tangent to you and a force is required to accelerate it towards you to keep it the same distance from you; the rope provides that force.
Probably there is some equivalent way to think of the situation as involving a centrifugal force (despite the common physicists' claim that there's no such thing as centrifugal force). But I don't think there's any force missing from the model. At least I'm not yet convinced.
I look forward to studying delcross's work (thanks!), but I can't until later...
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daithi
Jan 24, 2006, 4:51 PM
Post #128 of 145
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In reply to: If you swing a weight around on a rope, there is not some new force on the object pulling it away from you. It is going tangent to you and a force is required to accelerate it towards you to keep it the same distance from you; the rope provides that force.
What happens if I speed up? The distance increases. Anyway the difference arises because you have chosen to model the motion in catesian coordinates (not that it's incorrect it just makes understanding curved motion a bit less intuitive).
Why have to chosen to model it in cartesian coordinates? In my mind it is a lot easier to understand what is happening physically if you describe it as a polar motion (as both of us did before).
If you derive the equations of motions from first principles in polar coordinates you get the addition of this (dq/dt)^2 term (the centrifugal force term). This is not my assumption or addition, it comes out in the maths. But don't take my word for it. Derive the equations of motion for yourself from first principles and see what you get. I would be interested to see someone else's equations of motion.
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whoa
Jan 24, 2006, 5:41 PM
Post #129 of 145
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I dunno, it seems intuitive enough in x-y coordinates to me. Wouldn't be hard to do something similar on a spreadsheet in polar coordinates. Either way, it's nice to be able to see the trajectories charted to get a sense of what's going on with different assumptions.
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daithi
Jan 25, 2006, 2:39 AM
Post #131 of 145
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In reply to: In reply to: (d^2r/dt^2) = -(k/m)*r + (g*cos q) + (r*(dq/dt)^2) [eqn. 2] Almost, but you left out a term in the first equation for the coupling between r and q. Try this instead: write down the Lagrangian and from that derive the equations of motion.
I agree. See my revised post above.
d^2q/dt^2 = -g/r sin q - ((2/r)*(dr/dt)*(dq/dt))) [eqn. 1]
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antiqued
Jan 26, 2006, 1:34 AM
Post #132 of 145
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I have further refined a SS model with some help from Whoa’s work. My previous model was broader in that it covered arbitrary rope length, lead out distance, lead across distance and slack (frictionless biners). While it imitated a lead fall reasonably well (elongation and impact force) and also gave decent results for a taut pendulum fall, it did not add up the energy balance closely enough - there was a 5-10% surplus of energy stored in the rope.
I took some hints from Whoa’s SS and got to a better energy balance, and then pulled in the damping coefficient approach he used. I could not get reasonable combinations of impact force, elasticity and rope stretch for the standard UIAA fall. Any damping coefficient that produced a reasonable ‘one rebound only’ motion forced unreasonably high elasticity to minimize impact force, and failed to extend the rope sufficiently. I then introduced an additional term, a hysteresis effect, in which the stretched rope does not exert its full elastic tension while contracting during the rebound upwards. This SS does a reasonable casual impression of the UIAA standards for static elongation and impact force. Since the UIAA fall has little rope on the static side of the biner, and the pendulum falls we are looking at don’t stretch too much, the frictionless biner assumption probably does not change these results too much.
The predictions for pendulum energy in the x direction – that is the energy of collision into a wall under a roof, for example when falling off above the roof:
Rope distance belayer to last pro – 8m
Climber height above pro – 1m
Lateral distance of climber from pro – 2m
Rope Slack – Wall impact energy
0 m 1.95kJ (“steel” rope – minimal stretch calculates as 2.29kJ for this fall)
0.1 1.89
0.2 1.83
0.5 1.64
0.75 1.48
1 1.36
1.5 1.16
2 1.02
3 0.82
4 0.68
I still don’t think that this accurately represents best practice – it is all “hard catch” still, and I don’t think that most people are giving a metre of slack or more. I’m pretty damn sure they aren’t jumping that high. But best practice seems to reduce the swing to nearly zero.
(delcross - this stepwise model has the nerve to predict that 85% of the gravitational energy is converted to swing for a "steel" rope)
But it is another step in some direction – unfortunately, work calls, and fortunately, I have a long vacation weekend coming up, so I won’t be able to follow through much for a while.
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healyje
Jan 26, 2006, 9:18 AM
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When you guys get done please post one with the bottom line for us mere mortals. It would seem common sense though that you could take a bunch of steam out of a pendulum with enough slack and air space to play in - but who knows, I could be completely wrong...
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azrockclimber
Jan 26, 2006, 1:55 PM
Post #134 of 145
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haha..okay.
Easy answer, A pendulum fall can be just as bad as decking on the ground... Avoid it, protect yourself, protect your second, and try not to take the hit on the side of your body of you do fall into a wall on a pendulum swing. Vital organs are more exposed there.
And, if you have time to do the calculations in the other posts, bring a graphing calculator when you climb....who doesn't..it just makes it more fun. ya know!!???
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whoa
Feb 1, 2006, 11:39 PM
Post #135 of 145
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Well I do not yet share that conclusion based on my spreadsheet which included modeling the damping effect of the rope. I think we still need to model the actual features of climbing ropes better. It may well turn out that reasonable amounts of slack are good, but I think it's too soon to conclude anything.
Certainly it would be crazy for anyone to rely on our half-assed work on this for actual decisions about safety.
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trenchdigger
Feb 1, 2006, 11:53 PM
Post #136 of 145
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Now, I'm too lazy to actually do the work to look into it, but I would propose that a softer catch via a dynamic belay would be much more effective at reducing horizontal (swinging) speed than simply adding slack to the system. Just another bone to chew on...
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jt512
Feb 2, 2006, 12:12 AM
Post #137 of 145
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In reply to: In reply to: At 20% slack and above the impact force dramatically reduces with increasing slack. In reply to: Okay, these results are much more believable. So, as a rule, we should give the climber more than 20% slack, relative to the their horizontal lead-out. To get a 10% reduction one needs to provide 68% slack. To put this in perspective, a 15 foot traverse results in a speed of about 21 mph at the bottom of the swing.
What is the speed at the bottom of the arc if the traverse is 4 feet and there is no slack in the rope?
Jay
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cintune
Feb 2, 2006, 12:13 AM
Post #138 of 145
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In reply to: Conclusion: Adding slack to a pendulum fall is not a practical means for reducing the speed of the swing.
- This being the case, which it may or may not I suppose, there may still be other quantifiable benefits to the practice. No matter what, every foot-per-second-slower the impact, the less potential damage. Difference between bad sprains or bruises and broken bones. second, giving slack to add even a little more reaction-time to work with can't hurt. Thirdly, it's a shorter lower-off and carry out, if it comes to that.
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rockkid55
Feb 2, 2006, 1:58 AM
Post #139 of 145
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Umm........I'm trying to think of something funny to add to this post, but I doubt that anyone is reading this far into a MATH post. So.
I'll just go away now.
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daithi
Feb 2, 2006, 3:09 AM
Post #140 of 145
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Here is another calculation done modelling the elasticity (very simply)! It was done over a week ago but due to being very busy I didn't have time to post it until now. Sorry.
The physical system this time has replaced the perfectly rigid connection with a spring of stiffness k. The relevance of this physical model to an actual climbing situation is of course very debatable. The dynamic response of a rope is not that of a linear spring! However, it should serve at least some use to understand the effect of elasticity.
http://www.rockclimbing.com/...p.cgi?Detailed=68520
The angle q (theta) is the angle from the vertical, r is the time dependent length of the rope and r0 is the initial length of the rope, which corresponds to the equilibrium position of the spring when there is no force applied.
The differential equations solved were:
d^2q/dt^2 = -g/r sin q - ((2/r)*(dr/dt)*(dq/dt))) [eqn. 1]
(d^2r/dt^2) = -(k/m)*(r-r0) + (g*cos q) + (r*(dq/dt)^2) [eqn. 2]
By defining two new variables u and v as,
dr/dt = u; dq/dt = v,
equations (1) and (2) can be reduced to first order differential equations. Therefore, we have four coupled first order differential equations to solve. I used a 4th order Runge-Kutta method to solve the coupled differential equations. The timestep size was 1E-03 and the equations were marched in time until q = 0 (the point of impact with the wall).
The spring constant is defined as k = EA/r0, where E is Young's modulus and A is the cross sectional area of the rope (I assumed a 10.5 mm rope). Young's modulus was taken to be 2.36E08 N/m^2. The mass was assumed to be 80 kg (UIAA mass). A test was done at initial values of q = 0, r0 = 1 m and a distance dropped of 1.71 m. This corresponds to a vertical fall with a fall factor of 1.71 (the UIAA drop test). The model predicted a peak force at the point the mass stopped of 7.45 kN at an elongation of 40%. The actual experimental figures for the UIAA drop test at a fall factor of 1.71 are 7.4 kN at an elongation of 37% for my Beal rope. Therefore, the value of k is reasonably physical. However, it should be noted that although the peak values are realistic at the point when the mass stops moving, the time dependent motion is not physical as there is no damping.
An example of the trajectory of the elastic pendulum compared to the ideal pendulum is shown below at R = 1 showing the increase falling distance due to the elastic deformation.
http://www.rockclimbing.com/...p.cgi?Detailed=68523
Shown below are the results of the impact velocity at the wall compared to the original model. The impact velocities (r dq/dt) have again been normalised wrt 4.429 m/s, which is the impact velocity of an idealised pendulum at R = 1.
http://www.rockclimbing.com/...p.cgi?Detailed=68521
It should also be noted that in this graph, for the elastic pendulum, R is the initial length of rope at t = 0. Below are some conclusions.
1. Initially at R = 1 there is a slight decrease in the impact velocity even though the mass has fallen a greater distance. The initial velocity and acceleration at this condition are zero, i.e. a pure pendulum.
2. As R increases there is more vertical fall distance and therefore more initial velocity in the pendulum. The rope in the original model performed perfectly in that it absorbed all the linear momentum in the direction of the rope instantaneously at the start of the pendulum. In this revised model it initially absorbs less linear momentum and it is spread out over a finite time interval.
3. The combination of this effect and the increased falling distance due to the elongation results in higher impact velocities at the wall.
4. According to this we need to let out over twice as much slack as the horizontal distance before a decrease is seen in the impact velocity. This is a lot more than the 20% increase in slack with our 'perfectly' performing rope.
Some comments on damping:
In reality the climbing rope is likely to exhibit viscoelastic damping and be over-damped. The damping will probably depend on static pre-load, frequency, dynamic strain rate etc. The Young's modulus will be a complex quantity containing a real part (the elastic behaviour) and an imaginary part (the dissipative term). I have not seen any experimental time dependent data for a kernmantle rope so I really have no idea of the magnitude of the real and imaginary parts of the complex modulus.
As always I cannot guarantee this is devoid of errors!
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curt
Feb 2, 2006, 3:17 AM
Post #141 of 145
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In reply to: In reply to: In reply to: In reply to: - initial slack makes negligable (in theory, if not zero) difference If it makes no difference ito you in theory, you need to find another theory. Jay Dear Jay, I will restate...with initial slack in the system (your rope), you will only increase your vertical fall velocity and THEN translate it into the SAME AMOUNT of horizontal velocity By looking at the extreme case suggested by Curt, an infinitie amount of slack, it can be seen that you are wrong. The horizontal velocity would be 0 (and, incidentally, the fall factor would only be 1). So, assuming, as we should, until someone proves otherwise (using real equations), that the relationship between slack and horizontal velocity is monotonic, then the more slack in the rope, the less the horizontal velocity. Edit: Actually, in daithi's model the realtionship is not monotonic. Jay
I agree with this. The only reason I said that the correct answer was "situational" is because, in real life, feeding out additional slack could cause the climber to hit a ledge or something similar below--perhaps with worse consequences than the initial pendulum fall itself.
Curt
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jt512
Feb 2, 2006, 8:45 PM
Post #142 of 145
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In reply to: 4. According to this we need to let out over twice as much slack as the horizontal distance before a decrease is seen in the impact velocity. This is a lot more than the 20% increase in slack with our 'perfectly' performing rope.
And, based on practical climbing experience, I'm back to not believing the results.
Jay
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daithi
Feb 2, 2006, 9:36 PM
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In reply to: In reply to: 4. According to this we need to let out over twice as much slack as the horizontal distance before a decrease is seen in the impact velocity. This is a lot more than the 20% increase in slack with our 'perfectly' performing rope. And, based on practical climbing experience, I'm back to not believing the results. Jay
This very simplified model is only a tiny part of the real life situation (and I still make no guarantees of it being correct). For example here is a plot of the increase in time to impact with increase in R.
http://www.rockclimbing.com/...p.cgi?Detailed=68803
The worst case scenario in terms of increase in impact velocity is R = 1.36, which corresponds to an increase in time to impact of 22%.
In real world situation the chances of injuring oneself is related to how well one absorbs the impact. Therefore, I think the increase in time is all important in avoiding injuries not necessarily the increase in impact velocity. I personally would prefer to take a higher impact that I was prepared for as opposed to a slightly less severe impact that I hadn't quite got into position to absorb the impact.
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daithi
Feb 2, 2006, 9:47 PM
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In reply to: In reply to: 4. According to this we need to let out over twice as much slack as the horizontal distance before a decrease is seen in the impact velocity. This is a lot more than the 20% increase in slack with our 'perfectly' performing rope. And, based on practical climbing experience, I'm back to not believing the results. Jay
This very simplified model is only a tiny part of the real life situation (and I still make no guarantees of it being correct). For example here is a plot of the increase in time to impact with increase in R.
http://www.rockclimbing.com/...p.cgi?Detailed=68803
The worst case scenario in terms of increase in impact velocity is R = 1.36, which corresponds to an increase in time to impact of 22%.
In real world situation the chances of injuring oneself is related to how well one absorbs the impact. Therefore, I think the increase in time is all important in avoiding injuries not necessarily the increase in impact velocity. I personally would prefer to take a higher impact that I was prepared for as opposed to a slightly less severe impact that I hadn't quite got into position to absorb the impact.
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daithi
Feb 2, 2006, 10:04 PM
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delcross,
Out of curiosity what force and elongation does your values of modulus and damping coefficient produce at the end of a vertical fall with a fall factor of 0.25? Based on simple analysis I would consider values of somewhere around 2.8 kN and elongation of around 14 % to be physical. I would be interested to see what your model predicts. Thanks.
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