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

[evm1024]

I'm late to the show and feel a little dopy having suggested this thread and it was already there.....

The image below is helpful. And it need an attribution so here it is:

By RDBury - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/inde...curid=16595012

I am no expert and have been trying to do the math and understand the physics. I've not done that kind of math in the last 3 or 4 decades and would rather do something else than come up to speed. (Um Beer)

Just some general thoughts.

First off the only time there is not a catenary in an anchor chain (in use off the bow) is when the chain hangs straight down. No force on the boat at all.

As for getting rid of a catenary the force would be impossible. The math that calculates the force would need to approach tan(0) which is undefined and more or less requires an infinite force.

For a working definition we could say that bar tight is the point where straightening the curve of the chain further results in an insignificant reduction in the force transmitted from one end of the chain to the other. Its ability to act as a shock absorber is gone.

Mostly when we think of a catenary we think of a chain suspended between 2 points. Not quite true in our case. We only look at half (or less) of this graph.

When we have any part of our anchor chain laying on the seabed that portion of the chain that is laying on the seabed is not part of the catenary. At the point where the chain just lifts off the sea bead ( 'c' in the diagram) the force transmitted by the chain (To) is parallel to the seabed. The length of chain laying on the seabed transmits the force To to the anchor. It is purely horizontal and thus the effective scope as seen by the anchor is infinite.

Laying out more chain will not change the scope and thus is of not use (ignoring any friction more chain on the bottom may cause).

As we increase the force on the rode the chain that is laying on the bottom is lifted. But so long at the force at the anchor remains horizontal no additional chain will increase holding.

If the force on the anchor is horizontal (as described above) and the anchor is dragging then the force To is greater than the holding power of the anchor. No further increase in scope will stop the anchor from dragging.

Once we increase the force to the point that we lift the last link off the seabed we have shifted up the catenary curve toward point 'r'. There is no longer a center of the catenary 'c' and the force on the anchor is no longer To (horizontal).

The forces at point r can be divided into an horizontal portion and a vertical portion. The sum of those forces is Tu. Straight from wikipedia: Tu = (T cos φ, T sin φ), where T is the magnitude of the force and φ is the angle between the curve at r and the x-axis.

At this point the vertical force is trying to break the anchor out and the horizontal force is trying to cause your anchor to drag. If the holding force of the anchor is greater than the horizontal portion of the chains pull then you will not drag. If the resistance to rotation is greater than the vertical force then the anchor will not rotate out.

Of course it is more complex than this. Tu is an vector and the resistance to drag and rotation also form a vector and the seabeds resistance to rotation and displacement (and many other factors that change the seabed) make this difficult.

Effective scope as felt by the anchor and geometric scope formed by the length of chain out and the water depth(plus height of chain attachment point) are not the same thing.

Letting out more chain in this case will increase the horizontal force felt by the anchor while decreasing the vertical force. In rare cases more scope will lay some of the chain on the seabed thus the force will go from Tu to To. I.E. only a horizontal force, no vertical force.

So how does increasing the length of chain deployed help stop dragging if in fact it increases the horizontal pull on the anchor?

In the most obvious case more horizontal pull reduces the chance of the anchor rotating out of the seabed.

Also, anchors are rated in holding power based on scope which really is not quite accurate and adds to the confusion. Just as the force applied to the anchor by the chain has vector force Tu (vertical and horizontal forces combined) the anchor has a vertical and horizontal holding vector with the vertical holding vector much less than the horizontal vector. Let's combine them and call it Hu (holding vector).

At low effective scopes Hu is much lower than Ho (horizontal holding) and thus Tu overcome it easily. As we increase the chain bent on we increase the effective scope and thus the horizontal component of Hu increases greatly.

Once We increase the effective scope to where Hu is greater than Tu the anchor will not drag.

Which brings up back to the beginning. If Tu is greater than Hu even if Hu = Ho then the anchor will drag regardless of the effective scope.

Scope is not the answer. Greater Hu is the answer. (I.E. Bigger, more effective anchor)

Whew, gotta lay off the espresso. The errors are mine. I claim no depth of knowledge on this subject. This is just what it appears to me to be likely. Correct away kind reader.

[evm1024]

An alternate name for the thread could have been (with apologies to Robert Asprin)

Catenary Mythtruths

[Dockhead]

Originally Posted by evm1024

I'm late to the show and feel a little dopy having suggested this thread and it was already there.....

The image below is helpful. And it need an attribution so here it is:

By RDBury - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/inde...curid=16595012

I am no expert and have been trying to do the math and understand the physics. I've not done that kind of math in the last 3 or 4 decades and would rather do something else than come up to speed. (Um Beer)

Just some general thoughts.

First off the only time there is not a catenary in an anchor chain (in use off the bow) is when the chain hangs straight down. No force on the boat at all.

As for getting rid of a catenary the force would be impossible. The math that calculates the force would need to approach tan(0) which is undefined and more or less requires an infinite force.

For a working definition we could say that bar tight is the point where straightening the curve of the chain further results in an insignificant reduction in the force transmitted from one end of the chain to the other. Its ability to act as a shock absorber is gone.

Mostly when we think of a catenary we think of a chain suspended between 2 points. Not quite true in our case. We only look at half (or less) of this graph.

When we have any part of our anchor chain laying on the seabed that portion of the chain that is laying on the seabed is not part of the catenary. At the point where the chain just lifts off the sea bead ( 'c' in the diagram) the force transmitted by the chain (To) is parallel to the seabed. The length of chain laying on the seabed transmits the force To to the anchor. It is purely horizontal and thus the effective scope as seen by the anchor is infinite.

Laying out more chain will not change the scope and thus is of not use (ignoring any friction more chain on the bottom may cause).

As we increase the force on the rode the chain that is laying on the bottom is lifted. But so long at the force at the anchor remains horizontal no additional chain will increase holding.

If the force on the anchor is horizontal (as described above) and the anchor is dragging then the force To is greater than the holding power of the anchor. No further increase in scope will stop the anchor from dragging.

Once we increase the force to the point that we lift the last link off the seabed we have shifted up the catenary curve toward point 'r'. There is no longer a center of the catenary 'c' and the force on the anchor is no longer To (horizontal).

The forces at point r can be divided into an horizontal portion and a vertical portion. The sum of those forces is Tu. Straight from wikipedia: Tu = (T cos φ, T sin φ), where T is the magnitude of the force and φ is the angle between the curve at r and the x-axis.

At this point the vertical force is trying to break the anchor out and the horizontal force is trying to cause your anchor to drag. If the holding force of the anchor is greater than the horizontal portion of the chains pull then you will not drag. If the resistance to rotation is greater than the vertical force then the anchor will not rotate out.

Of course it is more complex than this. Tu is an vector and the resistance to drag and rotation also form a vector and the seabeds resistance to rotation and displacement (and many other factors that change the seabed) make this difficult.

Effective scope as felt by the anchor and geometric scope formed by the length of chain out and the water depth(plus height of chain attachment point) are not the same thing.

Letting out more chain in this case will increase the horizontal force felt by the anchor while decreasing the vertical force. In rare cases more scope will lay some of the chain on the seabed thus the force will go from Tu to To. I.E. only a horizontal force, no vertical force.

So how does increasing the length of chain deployed help stop dragging if in fact it increases the horizontal pull on the anchor?

In the most obvious case more horizontal pull reduces the chance of the anchor rotating out of the seabed.

Also, anchors are rated in holding power based on scope which really is not quite accurate and adds to the confusion. Just as the force applied to the anchor by the chain has vector force Tu (vertical and horizontal forces combined) the anchor has a vertical and horizontal holding vector with the vertical holding vector much less than the horizontal vector. Let's combine them and call it Hu (holding vector).

At low effective scopes Hu is much lower than Ho (horizontal holding) and thus Tu overcome it easily. As we increase the chain bent on we increase the effective scope and thus the horizontal component of Hu increases greatly.

Once We increase the effective scope to where Hu is greater than Tu the anchor will not drag.

Which brings up back to the beginning. If Tu is greater than Hu even if Hu = Ho then the anchor will drag regardless of the effective scope.

Scope is not the answer. Greater Hu is the answer. (I.E. Bigger, more effective anchor)

Whew, gotta lay off the espresso. The errors are mine. I claim no depth of knowledge on this subject. This is just what it appears to me to be likely. Correct away kind reader.

Looks right to me.

For anyone who came late to this discussion - the "myth" does not appear to be a myth. Peter Smith and Dashew seem to have been right - the chain seems to actually go almost "bar tight" in strong conditions, even with quite heavy, long chains. We seem to approach the limits only with heavy (1/2") long (100m) chains in deep water, where in some situations we seem to get more than 1 tonne of holding force before the last link lifts, which may be enough to hold a largish boat in a real storm if dynamic loads are well controlled with snubbing and yaw control. But over 50 knots of wind, catenary is gone in most other cases, the data shows. So I was wrong.

[boden36]

There is no guarantee that catenary will remain positive. If you pull back on the chain rapidly, negative catenary and a whiplash effect may transfer through the chain. In this case inertia becomes your enemy!
This can happen on a strong initial gust which acts to rapidly straighten a chain that has been meandering along the seabed.


Regards,
Richard.

[Jim Cate]

Originally Posted by boden36

There is no guarantee that catenary will remain positive. If you pull back on the chain rapidly, negative catenary and a whiplash effect may transfer through the chain. In this case inertia becomes your enemy!
This can happen on a strong initial gust which acts to rapidly straighten a chain that has been meandering along the seabed.


Regards,
Richard.

I find this very hard to believe... the viscous drag of the chain lashing through the water added to the normal catenary drag would seem (to me at least) to prevent such an occurrence.

Any data to support the idea?

Jim

[dfelsent]

Ok so I’m too tired to do more than throw a (very small) Grenade into this discussion.
Stupid question I’m to lazy to model or find google refs in 30 seconds:
What is the actual elasticity of chain ignoring catenary? It’s steel. Steel stretches and bends elastically.
Ie everything is a spring. In the bar tight scenario the chain’s intrinsic elasticity may be comparable or more than the nominal catenary effects.
Numbers anyone?

[thinwater]

Originally Posted by dfelsent

Ok so I’m too tired to do more than throw a (very small) Grenade into this discussion.
Stupid question I’m to lazy to model or find google refs in 30 seconds:
What is the actual elasticity of chain ignoring catenary? It’s steel. Steel stretches and bends elastically.
Ie everything is a spring. In the bar tight scenario the chain’s intrinsic elasticity may be comparable or more than the nominal catenary effects.
Numbers anyone?

He brings up a point that you seldom hear. As you approach the WLL of the chain, the chain stretch become an important portion of the equation. I've run the numbers and measured it. Obviously, this is more true of high grade chain than BBB, since strain is proportional to load, not % WLL. This is just one of the reasons that transport chain must be grade 70 or higher.

[boden36]

Hi Jim,
No data, but lots of observation. I think as the caternary lessens, the velocity of the midpoint of the arc will increase. If you have drifted about so that you are over the anchor, then with a good gust, the boat will have some distance to run before she snubs. A 12 ton hull travelling at three knots when the chain tightens will overcome the effect of water drag on the chain.
So many variables ....



Regards,
Richard.

[Jim Cate]

Originally Posted by boden36

Hi Jim,
No data, but lots of observation. I think as the caternary lessens, the velocity of the midpoint of the arc will increase. If you have drifted about so that you are over the anchor, then with a good gust, the boat will have some distance to run before she snubs. A 12 ton hull travelling at three knots when the chain tightens will overcome the effect of water drag on the chain.
So many variables ....



Regards,
Richard.

Richard, it seems to me that even in your extreme example, as the boat moves backward the energy absorbed by reducing catenary will slow the boat so that it would not be going very fast as the chain tightens, and your proposed inertial energy would have been reduced to near nil by the time the chain straightens out. That is why catenary is so important in medium conditions. If the steady state load is high enough to keep the chain straight, then there is no velocity and no chain inertia to cause the "whiplash" effect that would raise the angulation at t he anchor.

At least that's how I see it, and I've never observed the whiplash effect... but I'm not sure how one could ever actually see that unless in the water in really severe conditions.

Anyone else with thoughts on this?

Jim

[evm1024]

We toss the term catenary around a lot but I think that we all know that the anchor chain only forms a true catenary under static conditions. I.E. The boat is not moving. Simply moving the bow around in wind and waves applies loading that "distorts" the catenary curve. I suppose if we applied more and more force gradually the catenary would retain its shape and flatten. But yanking it around will effect one end more than the other (simple inertia to start with).

I don't think that the actual curve or distortions of the curve matter much.

The plot below shows the difference between a catenary and a parabola As an example of how even a small change creates a different "named" curve.


When playing around with chain (small chain by hand as a kid for example) if you pull hard on it it will snap into what appears to be a straight line and oscillate close to a straight line.

In order to get the chain to snap up you had to pull the chain tight and then release the tension to allow the motion of the chain to carry it above the "straight" line (the catenary of course). It was all a matter of timing. If you did not release the tension the tension would dampen out the motion of the chain.

Without removing the tension the chain becomes "bar" tight and the oscillations die out shortly.

Air of course is much less viscous than sea water so I would expect sea water to dampen out any oscillations much more quickly than air. Whether this is a critically damped system or not I could not say.

The elongation of an anchor chain is in 2 parts. The structural stretch and elastic stretch.

Structural stretch is the settling in of the links to each other. Crushing crustraceans if you will. And squishing mud. This limits out pretty quickly I suspect.

Elastic stretch is the elongation of the wire used in the chain as a load is applied. If you keep the maximum force below the yield strength of the wire then the stretch is fully elastic and the wire of the chain will return to its unloaded length when the load is removed.

The general formula is:

stretch = (load * length of chain) / (area steel * young's modulus)

Basically this says that stronger (G70), larger (3/8") chain stretches less than weaker (g30), smaller (1/4") chain.

I suspect that at or below the WLL of an anchor chain that the elastic stretch of even G30 chain is less than 1% or 2%. And it would be less for stronger chain (G43, G70). In any case I suspect that whatever elastic stretch the chain has that the chain would still be "Bar tight" with or without the elastic stretch.

Elongation of G70 chain at fracture is listed at 20% or so. But looking at the strain-stress curve you can see that we are way past the elastic limits (and wll) of the chain.

N.B. I have no special knowledge of this subject. And I don't play a materials engineer on TV.

[RaymondR]

We are dealing with two types of forces, steady state for wind loading and cyclic loadings due to wave action.

Steady state is easy to handle and provided there is not enough fetch to allow the generation of a wave the vessel can be moored against a dock with fenders quiet safely.

Cycling loading from wave action is another matter as the vessel (In our cases less than or low multiples of wave length) will take on the rotational movement of the water as it passes the vessel.

The degree of horizontal and vertical positional translation in tri axial space is reasonable consistent and rather than attempt to hold a vessel over a fixed point, which will generate comparatively enormous forces, the strategy most often used is to restrain it over an average position with some sort of spring.

On occasions the offshore petroleum industry uses large gas springs to achieve this although they generally do not attempt to restrain vertical motion. The arrangements used constitute what are called "active motion compensation systems where the gas pressure on one side of a piston provides the restoring force and a responsive hydraulic system controlled by some sort of position sensing system on the other side provides the position correcting forces.

The cheapest and most practical systems use on drill rigs are the eight point chain and anchor mooring systems which exploits catenary behavior to provide the horizontal restoring force required. They are the same as we used on our boats except to a much larger scale.

Each of the eight winch/chain/anchor assemblies usually has a strain gauge and recorder fitted and I cannot recall ever having observed anything which might indicate a "whiplash" phenomenon by either direct observation or the recorder logs.

All wire and combination wire and chain systems are also used however wire tends to be much less robust than chain and require much larger winches, 3,000 ft of say 2 3/4" wire requires a fairly large drum and there are eight of them whereas chain can be stowed in lockers in the bottom of the hull in places fairly useless for anything else.

[AndyEss]

Uh, the anchor at the end of the chain is set in the sea floor soils - not fastened to an eyebolt set in an infinite plate of concrete.
While soil doesn't actually exhibit elastic behavior, to a first approximation it can be modeled as an elasto-plastic medium, of finite shear strength and zero tensile strength.
The deformations of the soil around the anchor will shed chain tension pretty rapidly because soil elastic moduli are orders of magnitude lower than steel.
And the soil will probably be deforming plastically at pretty low loads.

[Sojourner]

A lot of the maths cited are above my pay grade, being an MBA in my former life, but I would wonder re the argument against additional chain being of use vs a heavier anchor.....

I like using extreme examples to get my head around things. So if I have out 50 meters of chain on my anchor, could I pull it bar tight in x knots? 100 meters? 1000 meters? There is obviously a point at which my particular boat will not in the proverbial 100 year insurance storm be able to lift the actual weight of the chain from the seabed. So does the additional chain on the seabed not act as a 'heavier anchor' when combined with that weight, in addition to preventing the pull from going above horizontal? And then working backward, is there not a point in a real life scenario that my particular boat cannot lift the chain in an average storm? And if there is, then how can it be calculated that additional chain does nothing? Or is the thrust of y'alls arguments that in normal chain lengths, a decent storm will lift it all up?

[Dockhead]

Originally Posted by Sojourner

A lot of the maths cited are above my pay grade, being an MBA in my former life, but I would wonder re the argument against additional chain being of use vs a heavier anchor.....

I like using extreme examples to get my head around things. So if I have out 50 meters of chain on my anchor, could I pull it bar tight in x knots? 100 meters? 1000 meters? There is obviously a point at which my particular boat will not in the proverbial 100 year insurance storm be able to lift the actual weight of the chain from the seabed. So does the additional chain on the seabed not act as a 'heavier anchor' when combined with that weight, in addition to preventing the pull from going above horizontal? And then working backward, is there not a point in a real life scenario that my particular boat cannot lift the chain in an average storm? And if there is, then how can it be calculated that additional chain does nothing? Or is the thrust of y'alls arguments that in normal chain lengths, a decent storm will lift it all up?

Well, this is kind of the original question which started this thread. If you go back to the first dozen or so posts, you will find links to a calculator which will let you calculate the "last link" tension for various combinations of chain size, length, and depth.


You can find calculators and all the formulae and data you need for all of this on this brilliant site: Tuning an Anchor Rode


Working through a number of scenarios, I understood that, yes, "in normal chain lengths, a decent storm will lift it all up," if you define "decent storm" as something producing something like the ABYC design loads.



That doesn't mean that catenary is useless. The ABYC design loads are about 3 times the normal loads, for given wind strength, so if you have snubbing and yaw control well sorted, you might see, in a "decent storm", something more like those loads, where catenary will still be working a lot under a much wider range of conditions.



As to how catenary works -- it does not indeed work like a "heavier anchor on the seabed". Chain sliding along the bottom has very little friction, like orders of magnitude less, per unit of weight, than an anchor well set in decent bottom. Chain on the bottom doesn't do anything useful except form a reserve of catenary for gusts or increasing wind and wave forces.



So in trying to understand holding power in really bad weather, you should work back from scope and forget catenary. We have seen some people with a religious belief in scope, but in fact the relationship between scope and holding power of a given anchor is subject to science and is reasonably well understood:





OCIMF





Anchor


You get roughly double the holding power, from a given anchor, at 10:1 scope vs. 3:1 scope.


Whether you need all the potential holding power of your anchor or not depends on the anchor and the conditions, including bottom condition. Obviously the bigger the anchor you have, the more likely it is that part of the anchor's potential maximum holding power will be enough. You like extreme examples for testing concepts -- that of course is a standard analytic technique. Delfin has a 250 pound anchor which in a good bottom has holding force at 3:1 scope roughly equal to the breaking force of his chain. So he, like Steve Dashew, rarely have any need for more scope than that. YMMV.

[valhalla360]

I guess, the issue I see with this "thought experiment" is it can get a little misleading for typical cruisers...sure there are the rare cruisers who have 150m of chain and regularly anchor in 30-50m of water.

Most don't have nearly that much chain (I would bet 50-75m is more typical with rope used for rare extreme depths) and generally try to anchor in less than 10m of water. If these more typical cruisers don't grasp this key point, they may walk away thinking cantenary going "bar tight" and lifting the anchor shank vertically won't be an issue but in reality it very well may be in most common anchoring situations.

I did a quick and dirty using my traffic signal cantenary spreadsheet with bow roller to bottom of 33' and swing radius of 100' and and 1000lb horizontal force on the boat. The connection to the anchor is angling up at almost 45 degrees (visually assessed as my spreadsheet doesn't directly calculate it). Even down to 250lb of horizontal force, there was a substantial angle at the anchor shank.