Just had a thought about pulling rings apart
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Joined: June 21, 2006
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Just had a thought about pulling rings apart
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Posted on Tue Mar 01, 2011 8:30 am
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I just had a thought, and I don't really know quite enough math or physics or whatever to figure it out. And I have no idea if this will make sense at all, but here goes:

Let's say I have a ring and I were to grab hold of either side with the (closed) opening at the top. Arranged thus, I could then pull in opposite directions and open the ring thereby deforming it.

So, suppose that X is the force needed to pull that ring apart.

Next thing is, suppose I were to weave a bunch of those rings together into a E4-1 sheet with none of the rings being welded or riveted and that the world is perfect.

If Y = (the number of rings in the sheet), would the force required to pull apart the weave thus be:

Y * X?

X^Y?

Or would it be something else?

Just a purely hypothetical question that's been bugging me for a while now especially since I don't have the know-how to figure it out.


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Being from the third world, BMR claims the right to speak in the third person.

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Posted on Tue Mar 01, 2011 9:35 am
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Speaking on ring-to-ring interactions...

Your point of failure is still a single ring, connected to another single ring.
In theory, the break point of the entire sheet shouldn't be much higher than the break point of a single ring.
(All it takes is one ring to buckle, and it's chain-reaction time, as the sheet stresses funny.)

In your 'perfect world' example, every ring holds weight and force completely evenly, and distributes it perfectly... That sheet could hold a lot more than a real sheet of mail, where a slightly lower AR ring, or a force applied incorrectly, can pop a ring as if it were made of... Something that pops really easily.

That being said, I dropped out of OAC Physics in highschool. (And didn't get a particularly stellar mark in Grade 12 Physics either...) So I could be horrendously mistaken.



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Posted on Tue Mar 01, 2011 1:21 pm
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Lotos does hit the mark with the failure point being at the weakest spot due to unequal distribution of load.
Also, where the other rings are placed and can apply forces will also affect the results. This would suggest that pulling apart as described would be a direct correlation for a 2:1 chain, which would be able to be pulled apart with force X.

When you move to sheet weaves, it would get a multiplier for the number of rings in a row that it could sustain. This is due to an increased cross sectional area normal (perpendicular) to the direction of force. The resulting ideal estimation would be X*(number of rings in a row). Of course there are also end effects and unequally distributed loadings to lower the actual load that can be applied.


while(!project.isFinished())
project.addRing();
// Maille Code V2.0 T7.1 R5.6 Eo.n Fper MFe.s Wsm Caws G0.8-1.6 I2.4-8.0 Pn Dcdejst Xw1 S07

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Posted on Tue Mar 01, 2011 7:54 pm
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Ack. There are reasons that I'm an agriculture scientist and burgeoning entomologist and NOT a physicist. This conversation just made my brain hurt!

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Posted on Thu Mar 03, 2011 7:00 am
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You forgot a variable = how the weight is being applied!

If applied to one edge ring it would be X

If applied to the edge it would be X times the number of rings to which you attached the weight.

if the weight is applied to the middle of the sheet (like a sling) it would be X times the number of rings used to support the sheet around the weight.
If the weight is dense (say lead or granite) the supporting rings would be fewer than to support the same amount of a lighter material (say an oak block or a brick).


Maille Code V2.0 T5.5 R4.1 Ef.o Fj12.1 MAu/Ag Whmi Cw G.5-2.5 I1-12.5 N10.10 Pj Dejt

Joined: June 21, 2006
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Posted on Thu Mar 03, 2011 7:13 am
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Ok, now my head is going all konky with all this physics stuff, hehe. But it's all rather fascinating. Hehe, makes me wish I had paid more attention in physics class.


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Posted on Fri Mar 04, 2011 4:30 am
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Hmm, well I'm guessing you're trying to think of this in very simple terms--like if you had a square sheet of maille and you were to put the sheet into some sort of contraption that pulled the sheet equally on all of the rings on the edges of the sheet... Let's just say that "B" equals the breaking point of one of the rings (and that all of the rings have this same exact breaking point [since the world here is perfect ;D]), and that "N" equals the number of rings across the sheet (and the sheet is square, so we don't have to deal with a second variable).

I would guess that the force would be distributed across each of the rings on the outer edge of each side of the sheet... and if you were pulling it only one way, I would guess that the maximum force you could pull on this sheet before it reached its breaking point would be B^N--suppose you have only one ring. its breaking point would be B ^ 1 = B. if you had a 2x2 sheet, the force would really be distributed across 4 rings.. so B^2.

this is just my guess, and honestly i'm tempted to come up with some sort of limit (calculus) that approaches the breaking point of a sheet with N number of rings across.

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Posted on Fri Mar 04, 2011 5:04 pm
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If you are going solely for pulling apart the weave at the sides...

Think of it as a rope: adding length to the rope does not add strength, but adding diameter does. A 2x2 sheet would have the same ideal resistance to force as a 2x400 sheet if you were pulling across the 2 dimension (length wise).

Perhaps a short diagram explanation (O's are rings, --'s are where you are pulling, F is force required to open one ring):
(1x2)
--OO--
Can support 1*F
(1x6)
--OOOOOO--
Can support 1*F
(3x2) (assume equally distributed along all 3 rings wide)
. OO
--OO--
. OO
Can support 3*F

Unfortunately, as it is in the field of strength of materials, with different loading setups there is a whole mess of solutions. The solution that I am posing is no where near the same if you wanted to study a sword hitting armor or something like that.


while(!project.isFinished())
project.addRing();
// Maille Code V2.0 T7.1 R5.6 Eo.n Fper MFe.s Wsm Caws G0.8-1.6 I2.4-8.0 Pn Dcdejst Xw1 S07

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