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hosh
Apr 7, 2004, 4:17 PM
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... :?:
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shank
Apr 7, 2004, 4:46 PM
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Force=Mass x Acceleration
But in figuring out the force on gear that is stopping you,
Force=Mass x DEcceleration
So you have to determine how fast you are falling just when the slack come out of the rope, then figure in rope stretch and all that good stuff to determine how much time it will take for you to come to a complete stop. Figure your DEcceleration from that. I'll work out an example and post it later if anyone is interested.
Steve
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drunkencabanaboy
Apr 7, 2004, 4:49 PM
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In reply to: Force=Mass x Acceleration But in figuring out the force on gear that is stopping you, Force=Mass x DEcceleration So you have to determine how fast you are falling just when the slack come out of the rope, then figure in rope stretch and all that good stuff to determine how much time it will take for you to come to a complete stop. Figure your DEcceleration from that. I'll work out an example and post it later if anyone is interested. Steve
LOL - no one is interested :D HAHA
sorry - I'm just being a jerk because this whole thread is a laugh and i need to "stop watching" it.
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jt512
Apr 7, 2004, 4:50 PM
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In reply to: As stated in a previous post.... 1Kn = 225 Lbf (although you incorrectly stated pounds...not pounds of force!)
The pound is a unit of force. "Pounds of force" is, therefore, redundant.
-Jay
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andypro
Apr 7, 2004, 4:51 PM
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Edit: I posted too slow. This was meant for shank.
Eesh. That will require calculus that I've voluntarily forgot! But hey...if you up to it, lets see force over time graphs :twisted: That'll fix this mess...
P.S.-it wasn't the puff of smoke from the grassy noll, it was friggin Jackie O!!!!
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shank
Apr 7, 2004, 7:03 PM
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Ok. Someone correct me if I'm wrong here:
1 kg (kilogram)=2.2046 lbs (pounds)
1 m (meter)=3.28083 ft (feet)
All conversions are rounded from here on.
29.4 m (96 ft) of rope out
9.8 m (32 ft) lead fall
7% elongation (rope stretch)
I'm gonna use sport climbing as my example here.
You have a mass of 80 kg (176 lbs.), you are 29.4 m (96 ft) off the deck, and 4.9 m (16 ft) above the last bolt with an anchored belayer and no slack or friction in the system at all. You fall. This is a factor .3 fall (fall factor= (2 x distance above the bolt)/length of rope between you and belayer=(2x4.9)/29.4=9.8/29.4=.3)
When you fall you accelerate downward at 9.8m/s/s for 1s before the rope is weighted at all. with a static rope this is where your spine breaks, but with a dynamic rope it starts to strech and slow you down (decellerate). At 7% elongation it will stretch 2m and has to slow you down from 9.8m/s to 0 in that 2m.
So at 9.8m/s it will take you around .2s to go that 2m which is a decceleration of 49m/s/s roughly.
So the force on you is
49m/s/s x 80kg=3920N (newton) or 3.92 KN (kilonewton)
Also the belay has to exert this much force to stop you so the force on him is the same, but the force on the top gear is twice this amount, which is 7.84 KN.
Any corrections or questions?
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dirtineye
Apr 7, 2004, 7:17 PM
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YOU left out the friction over the biner. I believe Rgold and someone else covered the fact taht friction over the top biner reduces the force felt by the belayer by about 2/3.
Notice that if this friction over the top biner affects the force felt by the belayer, it must also affect the fall factor, and so using the theoretical fall factor is incorrect.
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robmcc
Apr 7, 2004, 7:22 PM
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Isn't this about the point where we start with some simplifying assumptions? I propose the following:
1) We replace the biner with a frictionless pulley. Clearly incorrect, but it puffs up our numbers, and as everyone knows, Size Matters!
2) We replace the climber with a spherical cow of uniform density. Just because. No real preference for what kind of cow, but in deference to the fact that this thread has been milked for all its worth, I'd stick to dairy cows.
Rob
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drunkencabanaboy
Apr 7, 2004, 7:24 PM
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In reply to: YOU left out the friction over the biner. I believe Rgold and someone else covered the fact taht friction over the top biner reduces the force felt by the belayer by about 2/3. Notice that if this friction over the top biner affects the force felt by the belayer, it must also affect the fall factor, and so using the theoretical fall factor is incorrect.
And don't forget about air resistance! ROFL - and the pointless thread starts all over again just as it began.
the person who commented about fall factors was right.
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drunkencabanaboy
Apr 7, 2004, 7:26 PM
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In reply to: 2) We replace the climber with a spherical cow of uniform density. Just because. No real preference for what kind of cow, but in deference to the fact that this thread has been milked for all its worth, I'd stick to dairy cows.
this is - in my opinion - the best idea yet.
Maybe the cows we get chocolate milk from would be even better?
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drunkencabanaboy
Apr 7, 2004, 7:31 PM
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In reply to: So the force on you is 49m/s/s x 80kg=3920N (newton) or 3.92 KN (kilonewton) Also the belay has to exert this much force to stop you so the force on him is the same, but the force on the top gear is twice this amount, which is 7.84 KN. Any corrections or questions?
actually - i do have a real comment - this is incorrect i believe. For this simple reason:
the force from the dynamic rope is not constant.
There is a peak force that is higher than the # you quote.
Blah - this thread will go on forever. Next we will be talking about the quantum possibility that all the molecules in the rope will simultaneously phase through the biner.
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robmcc
Apr 7, 2004, 7:34 PM
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Yep, I think you're right, DCB. That's probably the average force, not peak.
I think I'm to the point where I'd rather do something more fun than play in this thread. Like get a big mouth full of lemon juice. Right after sticking my tongue in a blender.
Rob
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drunkencabanaboy
Apr 7, 2004, 7:36 PM
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In reply to: Yep, I think you're right, DCB. That's probably the average force, not peak.
you DO agree with my chocolate milk cow statement too right? That it central to my whole thesis
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robmcc
Apr 7, 2004, 7:39 PM
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In reply to: In reply to: Yep, I think you're right, DCB. That's probably the average force, not peak. you DO agree with my chocolate milk cow statement too right? That it central to my whole thesis
Of course. That idea was so clearly an improvement over my suggestion that I was too embarassed to address it directly. Chocolate cow it is. Just make sure it's a cow, not a bull. If it ain't a cow, it ain't milk...
Rob
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bigwallgumbie
Apr 7, 2004, 7:48 PM
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Shank, good attempt, but no cigar:
In reply to: 1 kg (kilogram)=2.2046 lbs (pounds)
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
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drunkencabanaboy
Apr 7, 2004, 7:54 PM
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Mike is oh-so-right about the t=1.4 s - not the 1s previously quoted - but he forgot the most important remaining question -
what type of harness would you use for our chocolate milk cow? BD Alpine Bob maybe? I Think such a diaper-style harness would fit around her utters.
Also, to make the physics easier, treat the rope like a spring. Spring physics are taught in HS physics classes.
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bigwallgumbie
Apr 7, 2004, 8:05 PM
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Drunken texan (W?):
The diaper harness sounds like a good idea (I know home much it hurts to get a nad up under a leg loop, never mind an udder) but I'm thinking she might like a chest harness to keep her upright, them cows don't have really big hips ya know. Shit, we're talking about a spherical cow ain't we? Hmmm... Haulbag? Being as it is a brown cow, how do you tell if shes get s the "milk squeezed outta her" or if its the shit-scared-outta-her (esp if said cow just got back from the potrero....) Could be a nasty suprise when you go for that glass of free chocolate milk...
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...
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robmcc
Apr 7, 2004, 8:10 PM
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Maybe we should go with a rigid, spherical, chocolate cow of uniform density. Then we could just bolt it. Otherwise, I'd go with the haulbag idea, or maybe a cargo net.
Rob
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bigwallgumbie
Apr 7, 2004, 8:12 PM
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Any idea what the spring constant for a rope is? Shank, where'ld ya get that 7%? To be a little more accurate we could do a spring/damper thing but I ain't doing the math... I'm procrastinating hard, but not that hard
Wait, doesn't petzl have a program thing for all this stuff?
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jason1
Apr 7, 2004, 8:14 PM
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shouldn't someone be arguing that it's really an elastic collision...
really force diagram the on side of the top piece... then account for the other... then take away for friction determined by how much rope runs over the top biner(with rope stretch and distance belayer is pulled up)...
you guys should use one of those OSHA harnesses.... the big uncomfortable ones with the bungee on back...
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bigwallgumbie
Apr 7, 2004, 8:15 PM
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Mmm, solid cholacte cow, its like an easter bunny but 10000 times bigger!!! We'll need a lotta weed to eat that sucker.
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robmcc
Apr 7, 2004, 8:16 PM
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In reply to: Any idea what the spring constant for a rope is?
Well, there's the problem. It isn't constant for a climbing rope.
Rob
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robmcc
Apr 7, 2004, 8:19 PM
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I have to reluctantly concede that I was wrong about something very important. It's hard to admit, but I really have no choice.
This thread is actually MORE fun than a mouth full of lemon juice after sucking on the business end of a blender.
Rob
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drunkencabanaboy
Apr 7, 2004, 8:30 PM
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As you can see from my photo - chest harnesses are not needed - and they don't release their milk any easier than you do. Let's not be ridiculous people.
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drunkencabanaboy
Apr 7, 2004, 8:35 PM
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In reply to: In reply to: Any idea what the spring constant for a rope is? Well, there's the problem. It isn't constant for a climbing rope. Rob
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
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