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Basic climbing math! Kilonewtons explained...
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robmcc


Apr 7, 2004, 8:35 PM
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I don't think ANYONE has EVER considered the Wedgie Factor, either. In the transition from unweighted harness to weighted one, the harness becomes quite intimately well, intimate with the wearer. You're especially familiar with that if you've ever climbed sans underwear. What about that extra 3-6" of give as climber compresses into the harness on a hard fall? That's 6-12" of extra squish when you count both climber and belayer. Even more if we're talking about cows, as long as they're non-rigid cows.

Rob


drunkencabanaboy


Apr 7, 2004, 8:36 PM
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Basic climbing math! Kilonewtons explained... [In reply to]
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In reply to:
I don't think ANYONE has EVER considered the Wedgie Factor, either. In the transition from unweighted harness to weighted one, the harness becomes quite intimately well, intimate with the wearer. You're especially familiar with that if you've ever climbed sans underwear. What about that extra 3-6" of give as climber compresses into the harness on a hard fall? That's 6-12" of extra squish when you count both climber and belayer. Even more if we're talking about cows, as long as they're non-rigid cows.

Rob

EXCELLENT POINT! You are a true scholar of which the likes of us climbers have never seen before. Except maybe for JT - HA - j/k - about JT that is.


drunkencabanaboy


Apr 7, 2004, 8:43 PM
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Drunken texan (W?):

Extra ghetto climbing physics: If a factor 2 fall is like a 2000lb load on the climber, is a factor .3 fall a 2000/(2/.3) = 333 lb load on the climber? Kinda breaks down on real small fall factors though...

I think you are referring to the loads on the anchor - in which case - that number may not be AS FAR off. (think: load to climber ~ 160 lbs + load to belayer ~ 160 lbs)


robmcc


Apr 7, 2004, 8:49 PM
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In reply to:
BTW - it _is_ constant. If you don't think so - you don't know what a spring constant is.

http://www.vishay.com/brands/measurements_group/guide/glossary/spr_con.htm

I know what it is, thanks. :)

I read something yesterday that made the interesting point that ropes typically stretch 7% or so under an 80 kg load. How much should they stretch under an 800kg load, as in a severe fall? 70%? Doesn't happen. Or at least ISTR that max stretch is in the neighborhood of 40%.

Anywho, I'm finding stuff now that contradicts that claim. Some sources say ropes _do_ act like springs, but I've found at least one that says they don't. Until such time as I can do some testing of my own, I'll withdraw the assertion that they don't.

Rob


drunkencabanaboy


Apr 7, 2004, 8:52 PM
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I'll withdraw the assertion that they don't.

Great.

What do you think of my climbing cows photo?


jason1


Apr 7, 2004, 8:54 PM
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now that's some funney $(-)!T....


bigwallgumbie


Apr 7, 2004, 8:56 PM
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Bitchen... are those indusrial strength paper clips load reducing at high forces??

'Ma! I sah besty a rock clamberin!'


drunkencabanaboy


Apr 7, 2004, 8:58 PM
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b----... are those indusrial strength paper clips load reducing at high forces??

'Ma! I sah besty a rock clamberin!'

Yes

Yes they are.

I hear they work as well as keychain biners.

And those a pure-bread texan chocolate milk cows.

YE-HAW.


bigwallgumbie


Apr 7, 2004, 9:15 PM
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From http://www.bealplanet.com/produits/anglais/produit1.html

For the beal 10.2 Flyer II rope:

Imapct force: 7.4 kN
Extension during first fall: 37 %
Static elongation: <= 10% (less than or equal)

So making an ass of u and me:

spring constant for static case:

F = 80 kg * 9.81 m/s^2 = 0.78 kN

k = F/x = 0.78kN/0.1 = 7.8 (kN/%)

for first fall:

k = 7.4kn/.37 = 20 (kn/%)


so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change??

Ghetto physics disclamers: k in kN/%??? a little weird but i think it'll work, also the first fall should be less than 7.4 kN, but that just makes the spring constant bigger, also the elongation for the static case is less than or equal to 10% but whatever, close a fucking nuff. That and bluewaters figures are pretty similar (plus the 2004 BW catalog has a pic of sharma with the reddest eyes I've EVER seen, friggin rad)


robmcc


Apr 7, 2004, 9:15 PM
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In reply to:
What do you think of my climbing cows photo?

Classic. I do have to point out, though, that your cows are not spherical.

Rob


drunkencabanaboy


Apr 7, 2004, 9:33 PM
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In reply to:
(plus the 2004 BW catalog has a pic of sharma with the reddest eyes I've EVER seen, friggin rad)

:lol:


jt512


Apr 7, 2004, 9:35 PM
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In reply to:
From http://www.bealplanet.com/produits/anglais/produit1.html

For the beal 10.2 Flyer II rope:

Imapct force: 7.4 kN
Extension during first fall: 37 %
Static elongation: <= 10% (less than or equal)

So making an ass of u and me:

spring constant for static case:

F = 80 kg * 9.81 m/s^2 = 0.78 kN

k = F/x = 0.78kN/0.1 = 7.8 (kN/%)

for first fall:

k = 7.4kn/.37 = 20 (kn/%)


so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change??

Ghetto physics disclamers: k in kN/%??? a little weird but i think it'll work, also the first fall should be less than 7.4 kN, but that just makes the spring constant bigger, also the elongation for the static case is less than or equal to 10% but whatever, close a f---ing nuff. That and bluewaters figures are pretty similar (plus the 2004 BW catalog has a pic of sharma with the reddest eyes I've EVER seen, friggin rad)

I calculated k = 14.99 for the Beal Booster II. See http://www.rockclimbing.com/...php?t=52735&start=19

-Jay


shank


Apr 7, 2004, 9:38 PM
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In reply to:
Shank, good attempt, but no cigar:

In reply to:
1 kg (kilogram)=2.2046 lbs (pounds)
Correct, but it should be lbs-m (pounds of mass)

I know and you know, but the people that didn't know would still know what I mean I thought.


In reply to:
In reply to:
When you fall you accelerate downward at 9.8m/s/s for 1s before the rope is weighted at all

No, the climber falls for 9.8m. If he was moving at 9.8 m/s then yes, it would take (9.8 m)/(9.8 m/s) = 1 s but the climber starts with a velocity of 0 m/s. So the climbers velocity after falling 9.8 m is:

V = sqrt(2*a*x) = sqrt (2*9.8m/s*9.8m )
V = 13.9 m/s

And the fall takes:

V = a*t
t = V/a = (13.9m/s) / (9.8m/s/s)
t = 1.4 s


And to doulbe check:

x = (1/2) *a*t^2 = (0.5)*(9.8m/s/s)*(1.4s)^2
x = 9.6 m (not 9.8 but i did some rounding so... )


In reply to:
At 7% elongation it will stretch 2m and has to slow you down from 9.8m/s to 0 in that 2m.

Problem: I'm not sure where you got that 7% stretch, but I'm assuming its in relation to either a static loading of the rope or the strech during a UIAA test fall, either way, the amount the rope streches is going to vary on the the imact on the climber which will be related to the fall factor and friction between the rope and biner(s). So, nice idea, but using that 7% is a no go.

So i didn't really help figure out the load, but thought i could help clean up the physics. If i f*cked anything up somebody correct me, its been a good year and a lot of brain cells since i got outa school.

Mike

Out of time, I'll get back for more later.


robmcc


Apr 7, 2004, 9:39 PM
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so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change??

Yeah, that's the kind of stuff I read. Given the static figures, a hard fall should stretch the rope a lot more than it actually does. Like 70-90%, which I think just doesn't happen ever. But whatever. I don't have conclusive data or sufficient credible information, so I'm not drawing a conclusion on that one yet.

Rob


drunkencabanaboy


Apr 7, 2004, 9:45 PM
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Classic. I do have to point out, though, that your cows are not spherical.

Rob

Oh but they are



drunkencabanaboy


Apr 7, 2004, 9:52 PM
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In reply to:
In reply to:
so the spring constant is 7.8 for the static and 20 for the first fall, so the spring constant does change??

Yeah, that's the kind of stuff I read. Given the static figures, a hard fall should stretch the rope a lot more than it actually does. Like 70-90%, which I think just doesn't happen ever. But whatever. I don't have conclusive data or sufficient credible information, so I'm not drawing a conclusion on that one yet.

Rob

Appears you are quite right

http://bstorage.com/speleo/Pubs/rlenergy/Default.htm


shorty


Apr 7, 2004, 10:33 PM
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Let's not be ridiculous people.
I'm offended by the use of heifers with exposed udders in this thread. Please tell me they are using some form of udder shield and that it really was a wardrobe malfunction. The FCC just might try to fine this site. Stay with me while I finish this plate of Rocky Mountain Oysters.

I'm offended by the use of chocolate cows in this thread. Not only should PETA get involved, but using chocolate cows is racist and discriminates against the caramel, vanilla, and strawberry cows. I'm going to call the NAACB ("B" for bovine) as soon as I finish my Big Mac.

I'm offended by the potential injury inflicted on said cows. I would like to see a disclaimer on drunkencabanaboy's post which states "No heifers were injured in the production of this illustration." Next time, please use CG. Save our cattle for burgers! Or steaks, or roasts, or prime rib, or fillets....

I'm offended by the complex calculations which include kilonewtons, foot pounds, accelleration, decelleration, E=mc2, friction, rope stretch, lingerie stretch, wedgie factors, belay butt floss, and the loss of bowel control. We're trying to explain these concepts to cattle, fer chrissake! Cattle have only three things on their minds:
- When do I get to eat again?
- Where do I get to poop again?
- Wow, that new bull really does have big .....
Well, and maybe:
- Isn't that shorty warming up the grill again?


jxl914


Apr 7, 2004, 11:14 PM
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I'm joining this topic a little late and was too lazy to review all 8 pages of stuff. I agree up to the point where you start talking about forces it takes to stop the falling climber. No offense, but I'd love to see how you figured that out (if there is 21 feet of rope and the climber stops after falling 21 feet, there is an infinte force generated). I'd imagine that the effective spring constant of the rope was not considered at all and would likely be non-linear and therefore, really hard to approximate. I also love the "rock psi" whatever that is (I'm guessing your talking about the ultimate tensile or compressive stress the rock can sustain before brittle fracture. I hate to say this but I am actually a "rocket scientist" and I don't feel qualified to takle the subject without alot more data. I think we should all call it quits here guys and gals. thanks for educating us though.


bigwallgumbie


Apr 7, 2004, 11:29 PM
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Hmmm, sounds like jxl914 (WTF??) got aroused by the spherical cows and now feels embaressed.... It's alright man, cows need loving too. You can't fight who you are...


easysteve


Apr 8, 2004, 1:03 AM
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Hey, Red, don't listen to most of these people, you did what you wanted.

Steve


drunkencabanaboy


Apr 8, 2004, 1:32 AM
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Hmmm, sounds like jxl914 (WTF??) got aroused by the spherical cows and now feels embaressed.... It's alright man, cows need loving too. You can't fight who you are...

Maybe (s)he is a said "strawberry cow" who was offended by our complicated physics equations. Notice it was her first post ever!

I was also wondering what happens as our cows approach (but - of course - does not reach) the speed of light on their travels down to earth? Would she perceive the belaying cow's utters shriveling up? While her utters remained pink and supple?

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