 Just a very quick q. I'm looking for a mesh stuff sack to take a 10mm x 30m rope. What sort of capacity should I need? 15L, 10L? Cheers Marc
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 Do you already have any stuff sacs? It should be easy enough to figure out by putting the rope in one of those and estimating how much space could be spared.
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 Well, let's calculate the volume of the rope... V=pi*r^2*l r=10mm/2=0.5cm l=30m=3000cm V=pi*0.25*3000= 2356cm3= 2.4l Best case packing will be close hexagonal lay of parallel strands, which will increase the packed volume slightly. Looks like 10l should be plenty.
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| Edited: 07/07/09 13:10 |
CP...are you a mathematician? me...I'd have just grabbed a likely looking size and thrust the rope in to see....
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> CP...are you a mathematician? Engineer. But they try to teach us stuff at school that might just be useful in our later lives, if only we'd remember to actually use it. The volume of a cylinder isn't exactly rocket science...
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 Just as well or we'd need John Burley for that! 
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.jpg) Captain paranoia wrote (see)
> CP...are you a mathematician? The volume of a cylinder isn't exactly rocket science...
School was a while ago and whilst I still use a calculator less than younger colleagues for addition, subtraction, multiplication and division, working out the volume of a cylinder might as well be rocket science as far as I'm concerned. Use it or lose it!Problem is I never really got it in the first place.
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> CP...are you a mathematician? Engineer. But they try to teach us stuff at school that might just be useful in our later lives, if only we'd remember to actually use it. The volume of a cylinder isn't exactly rocket science...
Erm...it is if you are trying to design a rocket.... Also explains the different approach....I started as a farmer...learnt some engineering design......then worked on a farm....if it aint broke, then hit it harder!
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 um... I'm more of a spacecraft engineer than rocket scientist myself... though it was thinking like CP's above that gave me a crap answer to the alpkit Christmas contest question about how many beers would fit in their fridge!!! I'd say that you'd be a long way from the ideal closed-packed hexagonal in any rope bag... all those loops of empty space at each end of the coil. But still the maths does give a minimum volume value so you know it will need plenty more than that. That is to say, the hexagonal cylindrical packing that CP suggests has an efficiency of (pi / (2*(3^0.5))), or 0.906. But I'm certain that no amount of judicious coiling could get close to 90% of the space packed with rope...
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| Edited: 07/07/09 14:44 |
Yup...thought as much..... stuff it in and see what happens. Always knew my expensive agriculutural training would come in handy...
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Simon, Can you stop talking about AI!
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Now....where did I put that long glove....oh...Waldo has it, he uses them to climb over electric fences...
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LOL!
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space is outdoors last time I looked
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> That is to say, the hexagonal cylindrical packing that CP suggests has an efficiency of (pi / (2*(3^0.5))), or 0.906. Over what cross-section, and into what enclosing space, though...? 1/0.906 ~+10% ~'slightly increase'... ;-) > But still the maths does give a minimum volume value so you know it will need plenty more than that. That was the approach I took; as an engineer, I calculated the minimum volume, compared it with the candidate volume, shrugged and said "yeah, that's more than 100% margin; it'll do"... I confess that my mental model was a nicely coiled rope, not just random stuffing into a bag...
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| Edited: 07/07/09 17:49 |
CP, the 0.906 threshold is the maximum efficiency of circular packing in two dimensions presuming the cylinders are incompressible, no matter what bounding shape you choose. Not that anyone wants the details of how I got there... but... draw the basic unit cell of a closed packed hexagonal 2D lattice; you get four circles whose centres form a parallelogram. Looking inside this parallelogram, you see that you have four segments; two equal to one sixth of the circle... two equal to one third. In other words, the parallelogram contains one unit circle. The efficiency of packing over infinite 2D space is therefore the ratio of the area of the circle to the area of the parallelogram (since this in turn will tessellate perfectly) The circle (diameter d) has area pi.(d/2)^2 = (pi/4).d^2 the parallelogram has area = base x height = d.d sin60 = d^2.(3^0.5)/2 ratio simplifies out to (pi / (2*(3^0.5))) as above. I can't believe I just posted that in a outoors forum thread about packing ropes! I should get out more... Personally I like to chain coil so the packing is far less efficient and I need a nice big bag. Now that's better isn't it!!!
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| Edited: 07/07/09 18:04 |
> the 0.906 threshold is the maximum efficiency of circular packing in two dimensions presuming the cylinders are incompressible I know, John; I was joking... > The efficiency of packing over infinite 2D space ...and what I was alluding to: the boundary conditions for a finite space will determine the actual packing efficiency... Oh, us engineers; we're such wags. It's a laugh a minute...
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 I love this thread!!!!
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> the 0.906 threshold is the maximum efficiency of circular packing in two dimensions presuming the cylinders are incompressible I know, John; I was joking...
... wow... he thinks I've got a sense of humour... .... but then he's another engineer so that doesn't really count...  ...... but Trev likes it so that must be good, right  ?
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| Edited: 08/07/09 10:17 |
