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USnavy


Mar 14, 2012, 1:16 AM
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Top rope fall analysis w/ strain gauge analyzer
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I recently purchased a 500Hz. strain gauge analyzer and 5000 lbs load cell that I can use in the field with my laptop, so I decided to do some testing on TR falls today. This is just some quick non-scientific testing to familiarize myself with the equipment, so don’t take anything in here too seriously. I took six falls of varying length and catch method (i.e. hard catch, soft catch, ect.). The graphs below represent the most serious of the six falls. Basically I wanted to determine what would be the highest impact force I could produce on the anchor while top roping without doing something extremely dumb, like taking a huge TR whipper on static rope.

Here are the specs:

-160 lb climber
-175 lb belayer
-Mechanically locking belay device
-9.9mm Maxim Glider, 9.5 kN impact force rating, 29% dynamic elongation, 5% static elongation
-Hard catch (the belayer just stood there and locked off)
-50 feet of rope out, 12 foot free fall, 20 foot total fall distance (approximate, not measured)

To start with, some really crappy pictures of the set up:


Next, the fall graphed out:



This next graph represents the loading that was in the upper 50% of the fall range. The peak load was just shy of 700 lbs. so this graph shows the loading above 350 lbs. The numbers on the X axis represent the duration in milliseconds multiplied by ten.



This last graph captures what I am going to call the “high” or “mean peak” meaning it shows how long the force stayed in the upper 10% range of the fall range, or above 600 lbs. This graph would be useful to compare different falls of the same fall factor, but different fall lengths, because it shows how long the anchor experiences the peak force of the fall. As you can see the loading on the anchor was only in the top 10% for 160mS.



So the maximum impact force on the anchor with a 12 foot free fall on TR was 3.09 kN. Keep in mind, I pulled out a fair amount of slack before I let go and my belayer gave me an intentionally hard catch. So your average TR fall will likely produce less of an impact force on the anchor, unless you weight a lot more than me.


(This post was edited by USnavy on Mar 14, 2012, 3:21 AM)


moose_droppings


Mar 14, 2012, 8:03 AM
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USnavy wrote:
I recently purchased a 500Hz. strain gauge analyzer and 5000 lbs load cell that I can use in the field with my laptop, so I decided to do some testing on TR falls today. This is just some quick non-scientific testing to familiarize myself with the equipment, so don’t take anything in here too seriously. I took six falls of varying length and catch method (i.e. hard catch, soft catch, ect.). The graphs below represent the most serious of the six falls. Basically I wanted to determine what would be the highest impact force I could produce on the anchor while top roping without doing something extremely dumb, like taking a huge TR whipper on static rope.

Here are the specs:

-160 lb climber
-175 lb belayer
-Mechanically locking belay device
-9.9mm Maxim Glider, 9.5 kN impact force rating, 29% dynamic elongation, 5% static elongation
-Hard catch (the belayer just stood there and locked off)
-50 feet of rope out, 12 foot free fall, 20 foot total fall distance (approximate, not measured)

To start with, some really crappy pictures of the set up:
[img]http://img209.imageshack.us/img209/5091/1001093b.jpg[/img]
[img]http://img696.imageshack.us/img696/5688/1001096h.jpg[/img]
Next, the fall graphed out:

[img]http://img338.imageshack.us/img338/8853/totalg.jpg[/img]

This next graph represents the loading that was in the upper 50% of the fall range. The peak load was just shy of 700 lbs. so this graph shows the loading above 350 lbs. The numbers on the X axis represent the duration in milliseconds multiplied by ten.

[img]http://img3.imageshack.us/img3/5476/53675516.jpg[/img]

This last graph captures what I am going to call the “high” or “mean peak” meaning it shows how long the force stayed in the upper 10% range of the fall range, or above 600 lbs. This graph would be useful to compare different falls of the same fall factor, but different fall lengths, because it shows how long the anchor experiences the peak force of the fall. As you can see the loading on the anchor was only in the top 10% for 160mS.

[img]http://img52.imageshack.us/img52/2038/peek10.jpg[/img]

So the maximum impact force on the anchor with a 12 foot free fall on TR was 3.09 kN. Keep in mind, I pulled out a fair amount of slack before I let go and my belayer gave me an intentionally hard catch. So your average TR fall will likely produce less of an impact force on the anchor, unless you weight a lot more than me.

Thanks for your time and work.

Now, who the hell one starred this effort. Stand up and out yourself and why.

I don't use the star system but thought I'd counter there one star with a five star.


jt512


Mar 14, 2012, 8:44 AM
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Re: [USnavy] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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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


skurdeycat


Mar 14, 2012, 9:20 AM
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Re: [USnavy] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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Can you measure the force exerted by the climbers fist on the belayers face after a 20ft toprope fall?

Skurdey


LostinMaine


Mar 14, 2012, 9:21 AM
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Good stuff. Thanks for posting.

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.


MFC


Mar 14, 2012, 9:22 AM
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Re: [USnavy] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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Thanks for posting this up - very interesting.

If I may, I have a question/suggestion.

According the the standard equation for impact force, a fall on a top rope with zero fall factor (basically just weighting the rope) will produce a peak force on the rope (and felt by the climber) of twice his body weight before returning to just one body weight.

Most climbers I have talked to find this to be unbelievable/untrue.

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.


LostinMaine


Mar 14, 2012, 9:22 AM
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Re: [skurdeycat] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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skurdeycat wrote:
Can you measure the force exerted by the climbers fist on the belayers face after a 20ft toprope fall?

Skurdey

Sure. What's the mass of the fist and what is its acceleration?


amarius


Mar 14, 2012, 9:49 AM
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Re: [MFC] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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Not the OP, but here is quick answer.
If just hanging off the top anchor with full mass of M, the top anchor attachment will experience 2*M*g force.
This is, most likely, a bit hard to believe :)
So, to simplify this a bit - hanging off the top anchor could be reconfigured as a fixed pulley


MFC


Mar 14, 2012, 10:29 AM
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I am not referring to the "pulley effect" of the top piece of pro.

If the standard equation of impact force is true/accurate then weighting the top rope (FF=0)produces an instantaneous impact force of 2x body weight, and the pulley effect would effectively double that (neglecting friction at the carabiner).

My original question was to verify that the impact force equation is valid or not for zero fall factor on a top rope.


amarius


Mar 14, 2012, 10:50 AM
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FF=0 is equal to static loading.
It is a bit easier to understand this if there is no friction on the system.

Climber rope pulls down with M*g
Belayer rope pulls down with M*g - remember, no frictional losses. Why are these two forces equal? If they were not equal, according to Newton's Laws, something would have to be moving, and we are talking about a static load.

Climber rope pulls down with M*g, belayer rope pulls down with M*g, anchor pulls up with 2*M*g. Why 2*M*g - nothing is moving, forces have to be cancel out, two M*g down, the one going up must be equal to the sum.

As a matter of fact, if one examines the Force vs time over longer period, the one with force oscillation, observation that the loading settles around 300lb is quite obvious, with swings to above ~325 and below 250. Why 300lb, and not 320lb? - possible explanations - additional friction somewhere, i.e. climber touching something, rope rubbing against the rock, instrument not calibrated/erroneous. BTW, I am not implying calibration or erroneous reading issues, the possibility is just listed

Adding rope over the couple of carabiners friction doesn't change the final static loading on the anchor, but the forces get too complicated for the verbal analysis

Hope this helps


MFC


Mar 14, 2012, 11:15 AM
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I believe Richard Goldstone's article " Standard Equation for Impact Force" is still on the rockclimbing.com forum archives.

I hope this address is still accurate:
http://www.rockclimbing.com/cgi-bin/forum/gforum.cgi?do=post_attachment;postatt_id=2957;

It is a good read!


jt512


Mar 14, 2012, 11:22 AM
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Re: [amarius] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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amarius wrote:
FF=0 is equal to static loading.

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.

In reply to:
As a matter of fact, if one examines the Force vs time over longer period, the one with force oscillation, observation that the loading settles around 300lb is quite obvious, with swings to above ~325 and below 250. Why 300lb, and not 320lb?

The simplest explanation would be because of friction between the rope and the top carabiners.

In reply to:
Adding rope over the couple of carabiners friction doesn't change the final static loading on the anchor . . .

Yes it does. It reduces the tension in the rope on the belayer's side of the anchor.

In reply to:
Hope this helps

Hope springs eternal.

Jay


(This post was edited by jt512 on Mar 14, 2012, 11:24 AM)


Partner cracklover


Mar 14, 2012, 11:36 AM
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Re: [amarius] Top rope fall analysis w/ strain gauge analyzer [In reply to]
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amarius wrote:
FF=0 is equal to static loading.
It is a bit easier to understand this if there is no friction on the system.

Climber rope pulls down with M*g
Belayer rope pulls down with M*g - remember, no frictional losses. Why are these two forces equal? If they were not equal, according to Newton's Laws, something would have to be moving, and we are talking about a static load.

Climber rope pulls down with M*g, belayer rope pulls down with M*g, anchor pulls up with 2*M*g. Why 2*M*g - nothing is moving, forces have to be cancel out, two M*g down, the one going up must be equal to the sum.

As a matter of fact, if one examines the Force vs time over longer period, the one with force oscillation, observation that the loading settles around 300lb is quite obvious, with swings to above ~325 and below 250. Why 300lb, and not 320lb? - possible explanations - additional friction somewhere, i.e. climber touching something, rope rubbing against the rock, instrument not calibrated/erroneous. BTW, I am not implying calibration or erroneous reading issues, the possibility is just listed

Adding rope over the couple of carabiners friction doesn't change the final static loading on the anchor, but the forces get too complicated for the verbal analysis

Hope this helps

I'm just curious - why would you want to embarrass yourself by putting this in print?

GO


amarius


Mar 14, 2012, 11:51 AM
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Thanks for you corrections.
I have to agree, I made erroneous statements.



Tension on the belayer side is defined by the Capstan equation
Tbelayer=Tclimber*exp( - miu * angle)

miu - is the friction coefficient between the surfaces, angle in radians is how much arc coverage surface provide. In the most general case miu is different for dynamic and static friction. Dynamic friction values are, typically, lower.

Force on the anchor would then be Tbelayer+Tclimber
Rough estimations are just that since rope friction over carabiners should depend on the construction of the rope and carabiners - that is very specific to set up.

Two carabiner setup will result in angle > pi, exact value depending on the angle that carabiners form. miu is a bit more difficult - published values range from .15 to .25

So, for angle of pi ( 3.14), miu of ( .15/.2/.25 )
Tanchor = (259 245 232) lb of force, for the 160lb climber
Not quite the same as the measurement. As a general observation - lower friction will result in higher anchor loading.
Two extreme cases
- no friction - Tanchor == 320lb, the pulley
- super friction - belayer ends is unloaded ( aka rope tied directly to the anchor ) Tanchor=160lb.

Just to clarify - that is the static loading value. As jt512 pointed out, FF=0 will include the rope stretching out.


bigo


Mar 14, 2012, 12:16 PM
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jt512 wrote:
...Hooke's Law predicts that the maximum impact force on the climber will be two times his weight.

I don't disagree, Hooke's law is specific to stiffness only, but I believe the above only stands true for an un-damped response which seems like a pretty rough assumption for a FF=0 fall.

I would definitely agree that friction at the top biners causes the anchor load to not equal 2x the climber weight.


dan2see


Mar 14, 2012, 12:27 PM
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Great stuff, USN!

Real data, in a real situation, simply presented.

Your experiment surpasses the thousands of myths and millions of words about anchor dynamics.

Daniel the scientist wrote:
Why do climbing engineers have hunched shoulders, and sloping foreheads?

When you ask one to describe the real forces on a TR anchor, he'll hunch his shoulders and say "I don't know".

When you show him USN's experiment, he'll slap his forehead and say "I knew it!"


skurdeycat


Mar 14, 2012, 12:28 PM
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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.

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.

aargh, you folks scare me.


bigo


Mar 14, 2012, 12:48 PM
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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.

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.

aargh, you folks scare me.

Go belay someone half your weight and try to lift your legs off the ground while they are hanging free on the rope.


healyje


Mar 14, 2012, 12:54 PM
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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.


jt512


Mar 14, 2012, 1:01 PM
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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


(This post was edited by jt512 on Mar 14, 2012, 1:12 PM)


jt512


Mar 14, 2012, 1:02 PM
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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.

*facepalm*


jt512


Mar 14, 2012, 1:05 PM
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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


kennoyce


Mar 14, 2012, 1:06 PM
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skurdeycat wrote:
Good lord. There isn't a single post is this thread that is completely accurate,

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).

Quoted for posterity.


boadman


Mar 14, 2012, 1:16 PM
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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?


jt512


Mar 14, 2012, 1:20 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?

Pretty sure. We're talking plain vanilla Hooke's Law. No dampening.

Jay

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