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kennoyce
Mar 14, 2012, 8:22 PM
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boadman wrote: jt512 wrote: No, it isn't. A fall factor of 0 is a fall. It occurs when you fall on a rope with no slack and no initial tension—for example, an idealized top rope fall. Hooke's Law predicts that the maximum impact force on the climber will be two times his weight. Jay Sure about that?
I don't know about Jay, but I am, FF = distance of fall (before the rope starts to catch you)/amount of rope (before the rope starts to stretch), so if there is 0 slack, and 0 tension, and a full 200 ft of rope out, the FF will be 0/200 = 0, but the rope will certainly stretch quite a bit before you come to a stop.
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healyje
Mar 14, 2012, 8:31 PM
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jt512 wrote: healyje wrote: [image]http://img209.imageshack.us/img209/5091/1001093b.jpg[/image] I find it a bit hard to believe there isn't decent pro to be had below or to the left of both these bolts. *facepalm*
*facepalm* - exactly.
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boadman
Mar 14, 2012, 8:37 PM
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jt512 wrote: boadman wrote: jt512 wrote: No, it isn't. A fall factor of 0 is a fall. It occurs when you fall on a rope with no slack and no initial tension—for example, an idealized top rope fall. Hooke's Law predicts that the maximum impact force on the climber will be two times his weight. Jay Sure about that? Pretty sure. We're talking plain vanilla Hooke's Law. No dampening. Jay
For the climber at the instant he came to rest, the force would be:
-kx = mg
How do you get 2xthe climbers (I assume you meant mass?) "weight" out of that?
Were you thinking about the load the anchor would see? I could understand that, I think.
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bigo
Mar 14, 2012, 10:01 PM
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boadman wrote: jt512 wrote: boadman wrote: jt512 wrote: No, it isn't. A fall factor of 0 is a fall. It occurs when you fall on a rope with no slack and no initial tension—for example, an idealized top rope fall. Hooke's Law predicts that the maximum impact force on the climber will be two times his weight. Jay Sure about that? Pretty sure. We're talking plain vanilla Hooke's Law. No dampening. Jay For the climber at the instant he came to rest, the force would be: -kx = mg How do you get 2xthe climbers (I assume you meant mass?) "weight" out of that? Were you thinking about the load the anchor would see? I could understand that, I think.
You are thinking statics, but this is a dynamic system. Also, Jay was scrupulous and said as Hooke's law would predict, so it is not what what you see with a actual top-rope that includes friction/dampening.
From
http://en.wikipedia.org/wiki/Fall_factor
which comes from
http://en.wikipedia.org/.../Harmonic_oscillator
(This post was edited by bigo on Mar 14, 2012, 10:04 PM)
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dan2see
Mar 14, 2012, 10:02 PM
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kennoyce wrote: boadman wrote: jt512 wrote: No, it isn't. A fall factor of 0 is a fall. It occurs when you fall on a rope with no slack and no initial tension—for example, an idealized top rope fall. Hooke's Law predicts that the maximum impact force on the climber will be two times his weight. Jay ? , but I am, FF = distance of fall (before the rope starts to catch you)/amount of rope (before the rope starts to stretch), so if there is 0 slack, and 0 tension, and a full 200 ft of rope out, the FF will be 0/200 = 0, but the rope will certainly stretch quite a bit before you come to a stop.
It's a fall. No it's a fall-factor.
It's Hooke's Law. No it's pulley effect.
You're wrong. I'm right.
It's a nitter! No, it's a natter!
Sheesh!
Well, boys, I think the OP's reality check is refreshing and informative.
But there's them that can, and there's them that criticize.
Now shut up, watch, and learn something.
So there! --  --
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ptlong2
Mar 14, 2012, 10:16 PM
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boadman wrote: For the climber at the instant he came to rest, the force would be: -kx = mg
Boadman, the maximum force doesn't occur when the climber comes to rest. In fact, with just Hooke's Law (no damping) the climber never comes to rest.
The point of equilibrium you are thinking of would be if a climber were gently set onto the rope, no fall at all. If he instead falls, even FF=0, he will fall past the point of equilibrium where x=mg/k and stretch the rope further.
It's easy enough to calculate. If the climber falls a distance x the potential energy is mgx. This has to equal the potential energy of the rope which is 1/2*k*x^2. Solve for x and you have x=2mg/k. So, F=kx=2mg.
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boadman
Mar 14, 2012, 10:28 PM
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That makes sense, thanks.
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skurdeycat
Mar 15, 2012, 1:07 AM
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jt512 wrote: skurdeycat wrote: Good lord. There isn't a single post is this thread that is completely accurate, except perhaps the OP, and the one linking to rgold. I don't know where to start. Nor where to stop, unfortunately for you. In reply to: Hanging without bouncing on toprope, the force on the ANCHOR is the sum of the weight of the belayer, the climber, the rope and if you want to be really picky, the masterpoint biner(s). There is no friction, rope elongation, pulley effect, nothing else to consider. No friction? I think you need to think that through a little more. In reply to: aargh, you folks scare me. Is that your best pirate voice? At least it's better than your physics. Jay
Shoot, one missing word and I'm proof of my own statement. Hanging on toprope (the climber and belayer have stopped, the calculus fun is over), the MAXIMUM possible force on the anchor is the weight of everything hanging from it.
I sincerely do appreciate the work you guys do and share here, but sometimes the original question wasn't complicated, and the answer isn't either.
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loginatnine
Mar 15, 2012, 2:16 AM
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ptlong2 wrote: It's easy enough to calculate. If the climber falls a distance x the potential energy is mgx. This has to equal the potential energy of the rope which is 1/2*k*x^2. Solve for x and you have x=2mg/k. So, F=kx=2mg.
Your analysis here is valid for a FF=0 where the rope stretch will be equal to the fall height. That's x in your equation.
In a real fall, it will be something like mg(H+x) = 1/2*k*x^2 where x is the rope stretch and H is the length of the fall before the climber is catch by the rope.
(This post was edited by loginatnine on Mar 15, 2012, 2:18 AM)
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USnavy
Mar 15, 2012, 5:34 AM
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jt512 wrote: The three figure shows the peak at about 40, 22, and 11, respectively. Can you clarify what the units of the horizontal axes are in each of the figures, and explain why peak occurs at different values? Jay
730321 347.9256
730331 350.5929
730341 355.9004
730351 365.2213
730361 374.1467
730371 377.9456
730381 380.442
730391 383.4596
730401 388.9277
730411 394.7827
730421 401.5505
730431 407.5202
730441 414.8244
730451 422.0302
730461 425.8789
730471 428.9844
730481 433.2427
730491 434.0447
730501 438.4572
The number on the left indicate the time stamp, in mS, generated by the strain gauge software. The numbers on the right are the readings, in lbs. The rows are not listed above. What I am trying to get Excel to do is list the time stamps on the X axis and the load values on the Y axis, but I cant figure out how to get Excel to do this. It wont let me change the X axis values, it forces me to use the row numbers.
LostinMaine wrote: In the future, like Jay points out, your figures need a bit of work. Label all axes for each figure - that little bit of time spent helps to strengthen all of the time spent previously generating data to populate the figures. I'm curious about consistency, though. Have you tried taking the same fall with the same style catch multiple times and plotting the results? I wonder what kind of variance you might find.
However, keep in mind, they were not all of the same catch type. Some of the falls were identical, but I was changing the catch types and fall distances, so its not accurate data. I will answer your question in further detail at the end of this post.
MFC wrote: Next time you run some tests, could you measure the peak load from a fall factor=zero fall on a top rope? Thanks. I look forward to further test results.
healyje wrote: I find it a bit hard to believe there isn't decent pro to be had below or to the left of both these bolts.
Anyway, this post marks the start of a detailed and complex study I would like to make regarding the impact forces related on lead falls. Today I went out and took some real lead falls on the load cell and captured some data. I will make another post about that later. But there is a lot I need to do. I want to compare soft and hard catches, low and high fall factors, high and low elongation ropes, heavy and light climbers, and highline falls. Then I want to compare everything to each other, in a thesis paper and on video. However that’s going to takes hundreds of hours of research and testing and countless lead falls, so it will take some time.
It is also going to be rather tricky to isolate all the variables to get accurate results. I have to measure out the rope and fall distance every time to calculate the true fall factor, I have to retie the knot after each fall, I have to let the rope rest, I have to isolate belay device slippage and take into account rope drag, the list goes on and on and on of variables that have to be isolated to be able to truly compare lead falls in a synthesized setting.
But first I need to understand what I am looking at in the graphs.
Specifically I need to figure out how to determine when the fall is fully arrested so I can accurately calculate how long the fall duration is. There is also another issue that I don’t understand, I will explain that problem in the next thread that includes data on lead falls.
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guangzhou
Mar 15, 2012, 9:34 AM
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Very happy people like you enjoy this stuff, it means I will never have to. Instead, I can spend all those hours climbing.
Eman
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skurdeycat
Mar 15, 2012, 11:57 AM
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You are probably using the Line Chart in Excel, if you use the scatter plot (choose the scatter option with lines and no markers) then Excel should plot the x-axis the way you want. While it wouldn't matter in this case, a Line Chart assumes all x-axis readings are equally spaced, a scatter plot will properly space them.
In Office 2007 or later, to add axes titles, click on the chart and go to the Layout tab on the Ribbon.
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LostinMaine
Mar 15, 2012, 4:06 PM
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So you don't waste your time, do you know a statistician that you can talk to... or anyone who has taken a sampling design course? You want replication of each fall type and each catch type to get any useful conclusion. My suspicion is that because of friction in the system and the "real-world" nature of the testing you will get some funky impact force outliers that will need to be balanced by a fair number of replicates (within each fall-catch category). Otherwise, you will do all of this and not get much useful other than some cool figures to consider.
As a side, if you PM me a link to your CSV file, I can make some figures for you.
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cracklover
Mar 15, 2012, 6:12 PM
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jt512 wrote: cracklover wrote: I'm just curious - why would you want to embarrass yourself by putting this in print? I hope that wasn't all you got, because you need something left for skurdeykat. Jay
I don't see why. I assume you're talking about this post:
skurdeycat wrote: Hanging without bouncing on toprope, the force on the ANCHOR is the sum of the weight of the belayer, the climber, the rope and if you want to be really picky, the masterpoint biner(s). There is no friction, rope elongation, pulley effect, nothing else to consider. If the climber or belayer is touching the rock or ground, the force will be less.
... it may be the "wrong" question he's answering, it may even be misleading, but it's not substantively incorrect.
What I found so puzzling in amarius' post was his proud parade of his own ignorance, as if it was a bright shining flag to rally around.
Skurdy, otoh, is just making a factual statement: At rest, climber weight plus belayer weight, assuming both are hanging, equals force on the anchor. True, this is not a particularly enlightening point, but it's nothing to be embarrassed about.
G
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camhead
Mar 15, 2012, 6:21 PM
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guangzhou wrote: Very happy people like you enjoy this stuff, it means I will never have to. Instead, I can spend all those hours climbing. Eman
Yes. I'll never understand the obession with gear, though if you're into it, more power to you. I'm just thinking of this line, however, in an article that's been going around about different climbing types, in which it describes the "Widgeteer":
In reply to: Ironically, though Widgeteers are well-versed in the intricacies of load distribution, impact force, and lobe geometry, they rarely have as keen a grasp of the physiological techniques of climbing itself. |
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jt512
Mar 15, 2012, 7:20 PM
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USnavy wrote: jt512 wrote: The three figure shows the peak at about 40, 22, and 11, respectively. Can you clarify what the units of the horizontal axes are in each of the figures, and explain why peak occurs at different values? Jay Those units were chosen for me by Excel and I cant figure out how to change them. Anyway, they are row numbers. Because the strain gauge conditioner software creates a new row every 10mS, those units also represent time. Specifically, they are the duration multiplied by ten. So if you are looking at the waveform and notice a particular point starts out at say 30 and ends at say 60, that means it covered 300mS total. The reason why the peaks are at different intervals is because Excel uses the starting row number from the dataset I selected as one, not the start of the entire dataset. The entire dataset has over 150,000 points, so scroll through the data to the dataset that’s relevant to the fall I want to view, and I graph just the points I want. So for example, the data I want may land in rows 120,000 – 121,000. So when I chart out the rows in 120k - 121k, the graph shows 1 - 1,000 instead of 120k-121k because I only selected 1,000 rows. This is an example of the dataset that exists in the .csv file generated by my strain gauge conditioner software: 730321 347.9256 730331 350.5929 730341 355.9004 730351 365.2213 730361 374.1467 730371 377.9456 730381 380.442 730391 383.4596 730401 388.9277 730411 394.7827 730421 401.5505 730431 407.5202 730441 414.8244 730451 422.0302 730461 425.8789 730471 428.9844 730481 433.2427 730491 434.0447 730501 438.4572 The number on the left indicate the time stamp, in mS, generated by the strain gauge software. The numbers on the right are the readings, in lbs. The rows are not listed above. What I am trying to get Excel to do is list the time stamps on the X axis and the load values on the Y axis, but I cant figure out how to get Excel to do this. It wont let me change the X axis values, it forces me to use the row numbers.
You should convert the timestamp in Excel to a meaningful 0-based time unit and redo the figures so that the horizontal axis is time. You need to do an X–Y scatterplot, and your data range should be the two columns of data: time and force. It's been a long time since I've used Excel, but if it still (mal)functions the way it used to, if you put the variable names in the first row of each column in your dataset and include that rowin the data range for the figure, Excel will use them as axis labels automatically. (Don't skip a row between the variable names and the data.) If not, you can edit the axis lables manually, as Skurdeykat explained.
Jay
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jt512
Mar 15, 2012, 7:24 PM
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USnavy wrote: healyje wrote: I find it a bit hard to believe there isn't decent pro to be had below or to the left of both these bolts. [T]he FA just wanted to make a playground for the kids to play on, sorta speak.
Does anybody know how to air-pop popcorn?
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njrox
Mar 15, 2012, 7:32 PM
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camhead wrote: Yes. I'll never understand the obession with gear, though if you're into it, more power to you. I'm just thinking of this line, however, in an article that's been going around about different climbing types, in which it describes the "Widgeteer": Ironically, though Widgeteers are well-versed in the intricacies of load distribution, impact force, and lobe geometry, they rarely have as keen a grasp of the physiological techniques of climbing itself.
haha, great link!
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miklaw
Mar 15, 2012, 10:46 PM
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Anyway, good work and congrats for doing something.
Falling from the top rope anchor with 12' of slack, falling 12' (then rope strech adding another 8' to amke 20' of distance all up) is the same as a lead fall from 6' above the belay; they are both factor 0.24 falls.
3 kN is about the ballpark figure bandied around fro a hard sport fall. At that level light belayers will be up at the first draw, did you tie down? (and how come you guys are so light, don't you know that's cheating?)
mikl
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TonyB3
Mar 19, 2012, 7:51 PM
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So USFlotsam has bought an expensive toy to replicate tests that have been conducted by myriad professional institutions with engineers capable of analyzing the data. Woo!
The OP is worthless because of the lack of any sort of comprehensive analysis or controls. All he wants is to fool some n00b into thinking he is some sort of expert, when in fact nothing could be further from the truth.
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blueeyedclimber
Mar 19, 2012, 8:09 PM
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jt512 wrote: USnavy wrote: healyje wrote: I find it a bit hard to believe there isn't decent pro to be had below or to the left of both these bolts. [T]he FA just wanted to make a playground for the kids to play on, sorta speak. Does anybody know how to air-pop popcorn?
I can do that, as long as it doesn't involve math or excel spreadsheets.
Josh
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chadnsc
Mar 19, 2012, 8:38 PM
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blueeyedclimber wrote: jt512 wrote: USnavy wrote: healyje wrote: I find it a bit hard to believe there isn't decent pro to be had below or to the left of both these bolts. [T]he FA just wanted to make a playground for the kids to play on, sorta speak. Does anybody know how to air-pop popcorn? I can do that, as long as it doesn't involve math or excel spreadsheets. Josh
You're screwed Josh, good air-popped popcorn involves both math and an excel spreadsheet.
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guangzhou
Mar 25, 2012, 9:01 AM
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I am amazed this was made a sticky. Really, is RC.com getting desperate. Really should let this get buried before people think the OP actually knows what his doing and talking about.
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majid_sabet
Mar 25, 2012, 3:43 PM
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good work commando,( assuming you really did this yourself and did not fool someone else to do it for you). next time you are in the valley, stop by and I'll get you a free coffee at the lodge.
what brand was the load cell and the receiver ( does it offer serial or USB output )?
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