Possibly Original Thought About Chain Catenary - or - The Myth of the Bar Tight Chain

[evm1024]

Originally Posted by groundtackle

I'm struggling with this math. I see 15% elongation vs UBL, its regulated. Not inches but feet per hundred. 1 1/8" per 100' ?
Please correct me, I don't understand.
Chris

I could have easily blown the units....

This calculation was at 2000# load on the chain not UBS. Also consider that UBS is the point where the chain fractures having already elongated hugely and necked down. Elongation up to the yield point is linear and elastic. Past the yield point elongation is permanent and results in a decrease in the wires diameter.


(2 Kip force * 165 feet of chain) / (0.17 sq in of chain * 20300 ksi bulk modulus)

Gives us 0.096' elongation or 1.15".

G43 chain is made of astm 1022 steel which is where I got the 20300 ksi bulk modulus. 0.17 sq in is the diameter of the wire used in 5/16" chain converted to area (pie R squared) and times 2 for 2 links. pie * (0.329/2)^2 * 2

From the image the elongation from 0 force up to yield (2) is linear, elastic and "small". From yield force (2) to Fracture (3) there is a lot of elongation.

I'm just mucking about with these calculations and bow to observed results.

Legend for the image.

1: Ultimate strength
2: Yield strength (yield point)
3: Rupture
4: Strain hardening region
5: Necking region
A: Apparent stress (F/A0)
B: Actual stress (F/A)

(Ah, I transposed 20300 into 23000 in the prior calculation.

By [User:Slashme] (David Richfield) - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6039110

[Pelagic]

I find this fascinating, keep going..... but I just can't get this little scene out of my head

(Wife): Honey, I think we're dragging!
(CF'r): Wait, let me do the Math
[emoji6]

[Dockhead]

Originally Posted by Pelagic

I find this fascinating, keep going..... but I just can't get this little scene out of my head

(Wife): Honey, I think we're dragging!
(CF'r): Wait, let me do the Math
[emoji6]

Nah, you do the math beforehand, and then you don't drag

[Pelagic]

Originally Posted by Dockhead

Nah, you do the math beforehand, and then you don't drag

Wish that were true but if you have ever anchored in Repulse Bay (Hong Kong) where the bottom consists of 12 ft deep slurry on top of a hard shale bottom, Math doesn't help.[emoji20]

[Dockhead]

Originally Posted by Pelagic

Wish that were true but if you have ever anchored in Repulse Bay (Hong Kong) where the bottom consists of 12 ft deep slurry on top of a hard shale bottom, Math doesn't help.[emoji20]

Oh, sure it does. It tells you not to anchor there!




But seriously --


The math is extremely valuable. It tells you when catenary will stop working. It tells you the relationship between maximum anchoring force in different kinds of bottom. And it tells you how much of your maximum anchoring force remains, on what scope. It's extremely useful to understand this stuff. It really helps to connect the dots of experience.

[evm1024]

Originally Posted by Dockhead

SNIP

It really helps to connect the dots of experience.

That is perhaps one of the best quotes I've seen in a while. In other words Magic becomes Science.

[Delfin]

Originally Posted by Dockhead

Indeed, and that is a very complex calculation. As the chain gets longer and/or heavier, the point at which catenary no longer becomes effective, comes later. Eventually, even a force equal to the breaking strength of the chain is not enough to pull out the catenary. That is the whole point.

The sag you are referring to as not possible to "pull out" completely is a physical reality, but as Thinwater noted, the amount of sag in the chain that is left after you place a great deal of force on it is so minimal that for all intents and purposes one could say it doesn't exist. This is an asymptotic function, where as the angle at bow and anchor gets smaller, the amount of force required to lift the sag out of the chain increases. To get to zero angle, you need infinite force, which is what makes getting all the sag out impossible, even if the difference between the sag in real world conditions and no sag is essentially non existent. This is an explanation off the Interweb that says it better than I can:

[Dockhead]

Originally Posted by Delfin

The sag you are referring to as not possible to "pull out" completely is a physical reality, but as Thinwater noted, the amount of sag in the chain that is left after you place a great deal of force on it is so minimal that for all intents and purposes one could say it doesn't exist. This is an asymptotic function, where as the angle at bow and anchor gets smaller, the amount of force required to lift the sag out of the chain increases. To get to zero angle, you need infinite force, which is what makes getting all the sag out impossible, even if the difference between the sag in real world conditions and no sag is essentially non existent. This is an explanation off the Interweb that says it better than I can:

OK, thanks. I think I got that somewhere up the thread, in fact I knew it before, but this is a nice explanation. When catenary doesn't materially change the angle at the bow and anchor, and when it no longer absorbs energy materially, I think we can say that it's "bar tight".



The original idea of this thread was just that perhaps that happens later than we think, and that in some cases might happen later than the breaking strength of the chain. This idea of mine was BUSTED pretty quickly, for any normal size and quantity of chain we might carry on cruising boats.

[conachair]

Originally Posted by Dockhead

, and when it no longer absorbs energy materially, I think we can say that it's "bar tight".

This often comes up in (non physics ) forums (even in some magazines!) and maybe a little pedantic but.....

That's wrong!.

If the chain is really tight already from wind force alone and the boat gets blown onto the other tack and stops moving when the chain grabs it then a "bar tight" chain will absorb all the energy you throw at it until it breaks assuming the hook holds. So in the case of stopping a moving boat, if the boat speed goes from moving to zero then the kinetic energy has gone somewhere. All of it.
From a force point of view how quickly the energy transfers is the interesting bit. Energy = force x distance. So stick in snubber so the boat takes twice as far to stop and you've reduced the extra force down the chain to half.
Looking at it from an energy transfer point of view it also highlights how much the speed of the boat plays a part, kinetic energy also = ½mv². So with a snubber doubling the distance to stop the boat, some sort of drogue or riding sail to half the speed the boat gets up to then you've reduced the force by a further factor of 4. So in combination the force has reduced by a factor of 8. (bit more complex as the snubber won't stretch so much but the idea is the same).

For an anchored boat it's probably as good to be thinking of how fast the energy gets transferred rather than where it goes, it has to go somewhere which will be the chain stretching/reducing the catenary curve (what's left of it) if no snubber is involved. Just transferred a bit quicker than we would prefer
Similar to how hammers work.

[Dockhead]

Originally Posted by conachair

This often comes up in (non physics ) forums (even in some magazines!) and maybe a little pedantic but.....

That's wrong!.

If the chain is really tight already from wind force alone and the boat gets blown onto the other tack and stops moving when the chain grabs it then a "bar tight" chain will absorb all the energy you throw at it until it breaks assuming the hook holds. So in the case of stopping a moving boat, if the boat speed goes from moving to zero then the kinetic energy has gone somewhere. All of it.
From a force point of view how quickly the energy transfers is the interesting bit. Energy = force x distance. So stick in snubber so the boat takes twice as far to stop and you've reduced the extra force down the chain to half.
Looking at it from an energy transfer point of view it also highlights how much the speed of the boat plays a part, kinetic energy also = ½mv². So with a snubber doubling the distance to stop the boat, some sort of drogue or riding sail to half the speed the boat gets up to then you've reduced the force by a further factor of 4. So in combination the force has reduced by a factor of 8. (bit more complex as the snubber won't stretch so much but the idea is the same).

For an anchored boat it's probably as good to be thinking of how fast the energy gets transferred rather than where it goes, it has to go somewhere which will be the chain stretching/reducing the catenary curve (what's left of it) if no snubber is involved. Just transferred a bit quicker than we would prefer
Similar to how hammers work.

Well, I might be dense, or this might be above my physics pay-grade.


HOWEVER, why indeed can't you say that the chain has "stopped materially absorbing energy"? I don't see it.


When you have a nice catenary curve, and a wave crashes into the bow of the boat, the energy is aborbed by the catenary -- by the mass of chain being lifted as the curve is made shallower -- rather than yanking the anchor out of the seabed or smashing the bow roller. That's what I mean by "absorbing energy".


As the chain gets tighter, the amount of force required to change the distance to the anchor by a given amount, increases, and eventually, increases very steeply.



So less and less of the energy goes into lifting the sag out of the chain, and more and more gets transferred to the anchor and to the bow roller. Is this not so?


Until the chain starts behaving more like a bar ("bar tight"), and anchor-yanking-out and bow-roller-smashing forces start to appear. I said "materially" -- that doesn't mean that NO energy is being absorbed. As someone explained, catenary is not even the only actor -- the actual elasticity of the chain will absorb some energy. It just means that not enough is being absorbed to do much good.


Is that wrong?

Originally Posted by conachair

. . .
For an anchored boat it's probably as good to be thinking of how fast the energy gets transferred rather than where it goes, it has to go somewhere which will be the chain stretching/reducing the catenary curve (what's left of it) if no snubber is involved. Just transferred a bit quicker than we would prefer
Similar to how hammers work.

Well, yes. But surely "absorbing energy" can be reasonably understood also as "slowing down the transfer of energy", or is THIS the distinction that you are on about?


Naturally, all this is ALL about slowing down the transfer of energy -- neither catenary nor elasticity of the rode "absorbs" energy in the sense that it just disappears. It does of course eventually get transferred to the boat and to the anchor.

[conachair]

Originally Posted by Dockhead

Well, I might be dense, or this might be above my physics pay-grade.


HOWEVER, why indeed can't you say that the chain has "stopped materially absorbing energy"? I don't see it.

Cos it hasn't stopped absorbing energy And will continue to do so until it breaks if you apply more force.

When you have a nice catenary curve, and a wave crashes into the bow of the boat, the energy is aborbed by the catenary -- by the mass of chain being lifted as the curve is made shallower -- rather than yanking the anchor out of the seabed or smashing the bow roller. That's what I mean by "absorbing energy".

As the chain gets tighter, the amount of force required to change the distance to the anchor by a given amount, increases, and eventually, increases very steeply.

So less and less of the energy goes into lifting the sag out of the chain, and more and more gets transferred to the anchor and to the bow roller. Is this not so?

No. If the hook is snagged on a huge rock and the bow roller doesn't bend then it goes into the chain.

Until the chain starts behaving more like a bar ("bar tight"), and anchor-yanking-out and bow-roller-smashing forces start to appear. I said "materially" -- that doesn't mean that NO energy is being absorbed. As someone explained, catenary is not even the only actor -- the actual elasticity of the chain will absorb some energy. It just means that not enough is being absorbed to do much good.


Is that wrong?

Yup. Force x Distance. BIG force small distance is the same as small force big distance. The energy is still transferred , just really quickly over a short distance. If the connections each end don't distort then the chain will take the lot til it breaks, nowhere else for it to go.

Well, yes. But surely "absorbing energy" can be reasonably understood also as "slowing down the transfer of energy", or is THIS the distinction that you are on about?

Disagree, the rate it which the energy is absorbed is completely different to the fact that energy is getting transferred. Hit your thumb with a hammer to confirm

Naturally, all this is ALL about slowing down the transfer of energy -- neither catenary nor elasticity of the rode "absorbs" energy in the sense that it just disappears. It does of course eventually get transferred to the boat and to the anchor.

None goes to the anchor if it's a biggie and doesn't move. You need force and distance, just force doesn't transfer energy. Which is why your fridge magnets never need charging up, they don't need any energy to exert a force.
Probably back in the real world some will go into heating the water & other factors. But bottom line is, keep pulling a chain harder and harder it will continue to take all the energy you can throw at it until it breaks, a hammer doesn't work less the harder you hit with it.

All this from a very keen interest and the university of google & youtube, so if any physicists want to chime in please do, every day a school day

[transmitterdan]

Dockhead,

If something stops the boat the energy of 1/2mv^2 has to go somewhere. If not the chain then where did it go? Well, some was transferred to the water and some to the snubber and some to the anchor. That’s why snubbers get hot and melt in a storm. But a lot of energy goes into the chain. The metal heats up due to the application of energy and some energy is transferred to the water due to the motion. But all the energy has to go somewhere. Else the boat will go somewhere.

[conachair]

Might be a little easier to largely ignore the energy transfer bit and just focus on the deceleration and force.

[Dockhead]

Originally Posted by transmitterdan

Dockhead,

If something stops the boat the energy of 1/2mv^2 has to go somewhere. If not the chain then where did it go? Well, some was transferred to the water and some to the snubber and some to the anchor. That’s why snubbers get hot and melt in a storm. But a lot of energy goes into the chain. The metal heats up due to the application of energy and some energy is transferred to the water due to the motion. But all the energy has to go somewhere. Else the boat will go somewhere.

OK, OK, I understand now


Thanks for adding to my knowledge. One mustn't confuse FORCE, with ENERGY


Every day is school day!!

[Dockhead]

Originally Posted by conachair

Might be a little easier to largely ignore the energy transfer bit and just focus on the deceleration and force.

Exactly. That's exactly what I was confusing :headbang: