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breaking ALDI climbing gear


evanbb

6-Aug-2009
12:55:33 PM

On 6/08/2009 gfdonc wrote:
>I'm no engineer, but I wager that's not the case.

>Now time for the real engineers to join the debate ..
>

It's flattering that you think an engineer might be able to contribute to this, but in my experience, very few engineers do any engineering. They pay a drafter and modeller to do the maths, then pay a manufacturing company to make stuff. Sometimes we might approve invoices? Or offer some comment in passing on the colour of a design. But no, I've got no real idea of how the forces might play out in a fall. I'd just design a test rig and see what happens.


ajfclark

6-Aug-2009
1:05:59 PM

What about railway engineers? ;-)

yankinoz

6-Aug-2009
1:19:26 PM

On 6/08/2009 gfdonc wrote:
>YankinOz wrote:
>>If a leader falls on the first piece of pro I would think the forces
>on the belay device would be the same as that on the leader.
>
>I'm no engineer, but I wager that's not the case.

not on the device itself. yeah, i agree with you.

>Firstly the force on the belay isn't the same as that on the leader, because
>of friction through the runner system which reduces the force, as well
>as the dynamic nature of the rope.

In the case of a fall from the first piece of protection I disagree - there's only one biner between the leader and the belay so i would expect equal forces on both sides of the rope. Add more pro, more biners, more angles, more friction then the force on the belay side would reduce.
I've been lifted off the ground and I'm sure it takes a fair bit of force to pull 70kg off the ground - greater than 7Kn? is that how the works? I've pulled belayers bigger than me off the ground. there's still a lot of force hitting the belay.

>Secondly the force on the belay is mostly on the carabiner that the rope
>is passing around. The belay device is being subjected to enough force
>to introduce friction into the rope, not directly being pulled by the fall
>- assuming of course an ATC-style plate, rather than a figure-8 where the
>device is directly being loaded.
>
>The force on the exit side of the plate is going to be somewhat limited
>by how hard you are holding the rope. Reckon you can pull 10kN one-handed?

no way I can hold the rope without some mechanical aid! The system locks the rope but the force is still there - we take it on our harness. So my question to the engineers - what's the difference between the forces on the leader's harness compared to the belayer's?

>Now time for the real engineers to join the debate ..

I hope so :)

devlin66

6-Aug-2009
1:27:49 PM

edited as requested to remove very bad maths. see below.


wallwombat

6-Aug-2009
1:42:35 PM

I'm not really up on all this technical stuff, so I tried googling killernewtons.

All I got was this.

yankinoz

6-Aug-2009
1:55:41 PM

noice graphic. (and I got the hint)

found this:
http://splitterclimbinggear.com/Fall_Forces.html

it's in pounds but interesting none the less and answers my question.

doesn't answer how much force an ATC would take.

tas alex

6-Aug-2009
2:24:29 PM

On 6/08/2009 devlin66 wrote:
>1kN is approx. to 980kg weight. so 7kN = approx. 6860kg weight. You would
>need 0.072kN to start pulling you off the ground.

force(N) = ma, as a=g(9.8)
9.8N = 1kg
1kN = ~100kg

hero

6-Aug-2009
3:13:23 PM

is it just me, or are there a lot more geeks climbing now days?


evanbb

6-Aug-2009
3:16:38 PM

On 6/08/2009 hero wrote:
>is it just me, or are there a lot more geeks climbing now days?

I've wondered about that too; not necessarily now-a-days, but climbing in general. I've concluded that I encounter most climbers through the interpipes, which is a filter that encourages nerds.

I do think there's more geeky trad climbers than sport, just through the need to think about this sort of stuff.

devlin66

6-Aug-2009
3:28:44 PM

On 6/08/2009 tas alex wrote:
>On 6/08/2009 devlin66 wrote:
>>1kN is approx. to 980kg weight. so 7kN = approx. 6860kg weight. You would
>>need 0.072kN to start pulling you off the ground.
>
>force(N) = ma, as a=g(9.8)
>9.8N = 1kg
>1kN = ~100kg

Yes, your right. I don't where my head is today. Probably buried in the 37 drawings they wanted yesterday as always. Maybe if I spent less time starung at Chockstone and working i'd get it done :-)

My comment with correction should read....

1kN is approx. to 102kg weight. so 7kN = approx. 714kg weight. You would
need 0.01kN to start pulling you off the ground.

tas alex

6-Aug-2009
5:15:21 PM

On 6/08/2009 hero wrote:
>is it just me, or are there a lot more geeks climbing now days?

Hardly. That's pretty basic stuff


wallwombat

6-Aug-2009
5:25:44 PM

On 6/08/2009 hero wrote:
>is it just me, or are there a lot more geeks climbing now days?

It's not just you.

slackjaw

6-Aug-2009
5:32:59 PM

Just an observation on my part, I'd suggest that belay devices are built reasonably solidly to dissapate the heat from long abseils/lower offs rather than the need for strength?


evanbb

6-Aug-2009
5:37:42 PM

On 6/08/2009 slackjaw wrote:
>Just an observation on my part, I'd suggest that belay devices are built
>reasonably solidly to dissapate the heat from long abseils/lower offs rather
>than the need for strength?

Hooray! Some common sense. Makes perfect sense to me.

devlin66

6-Aug-2009
5:58:10 PM

I would also suggest that the shape req'd to produce the mechanical effect has an inherent volume and therefore you end up with way more material than needed for a pure strength requirement.

Olbert

6-Aug-2009
6:28:28 PM

There is a lot of dodgy maths going on here. To lift a 70kg person you need more then 700N or 0.7KN. This is the force acting directly on the belayer, not at any other point in the whole system.

Also there is friction through the first runner, so if a leader falls onto it then the belayer does not experience all of the force generated by the leaders fall. There is actually a great deal of friction taking up a great deal of the force generated.

Im not sure exactly how much but a good way to test it would be for two people of equal weight to loop a rope through a single runner and then tie in to both sides and hang. Then to measure the amount of friction from that piece all you would have to do is find out how much weight you would have to add to one of those people before one person went up and the other went down.

Obviously in this situation if there was zero friction from the runner then an infinitely small weight differential between the two people would cause them to move. In the same way if there was an infinite amount of friction then it doesnt matter how much weight is added they people arent moving.

Considering it takes a hefty leader to lift a small belayer even if the first piece is fallen on i would suggest there is a decent amount of friction.

rob668

6-Aug-2009
7:22:01 PM

On 6/08/2009 devlin66 wrote:
>On 6/08/2009 tas alex wrote:
>>On 6/08/2009 devlin66 wrote:
>>>1kN is approx. to 980kg weight. so 7kN = approx. 6860kg weight. You
>would
>>>need 0.072kN to start pulling you off the ground.
>>
>>force(N) = ma, as a=g(9.8)
>>9.8N = 1kg
>>1kN = ~100kg
>
>Yes, your right. I don't where my head is today. Probably buried in the
>37 drawings they wanted yesterday as always. Maybe if I spent less time
>starung at Chockstone and working i'd get it done :-)
>
>My comment with correction should read....
>
>1kN is approx. to 102kg weight. so 7kN = approx. 714kg weight. You would
>need 0.01kN to start pulling you off the ground.
>
Just for the general public's safety can you please EDIT your original message and stipulate there very simply to not trust what you'd written there.

This will help minimise the chance of some poor punting taking that a 1 kN rated biner (or whatever) will support them, with their 9 best mates, each lugging a slab of beer.

Cheers,

Rob

devlin66

6-Aug-2009
9:58:21 PM

On 6/08/2009 Olbert wrote:
>There is a lot of dodgy maths going on here. To lift a 70kg person you
>need more then 700N or 0.7KN. This is the force acting directly on the
>belayer, not at any other point in the whole system.

Exactly so at the belayer you would only need a fraction more than the weight of said belayer to begin lifting him up. Add in whatever friction and other energy absorbers are present to then get the force needed by the falling climber to arrive at that point.

mikl law

7-Aug-2009
12:24:59 PM

The most common situation would be where a belayer is holding a big fall, and is being lifted. Lets say that he's fat and being accelerated, maybe the equivalent of 100kg. The forces on the biner are higher than this (100kg on the upwards rope, and probably 60kg on the anchor rope{{ this force is mostly generated over the back edge of the belay plate}}) The sideways forces on the belay device are pretty small, probably in the order of 40kg (0.4 kN)

In a worst case, the rope is experienceing its maximum impact force of (lets say 10kN or 1000kg ), and the other end is tied off, there is no significant force on the belay biner. the side force on the belay plate could now be 2000kg.

I tested pretty much in this mode and it was starting to look a bit wonky at 32 kN (3265kg)


ajfclark

7-Aug-2009
2:15:19 PM

On 7/08/2009 mikl law wrote:
>I tested pretty much in this mode and it was starting to look a bit wonky at 32 kN (3265kg)

That's basically trying to pull the belay biner through the belay device isn't it?

There are 44 messages in this topic.