Mathematic approach to anchoring scope

[Dockhead]

So, to sum up what we've discussed and figured out here.


I think Mathias has a real breakthrough in thought about anchoring and chain catenary.


I'd like to see the math peer reviewed.


I presume that everything is correct.


And if so, it really changes our thinking and practice in profound ways.



It contradicts some things which Peter Smith and Dashew came up with decades ago, which many of us have taken as gospel.


I think it seriously undermines the idea that you should make the chain as light as possible and put the weight in the anchor. Because Dashew didn't understand what Mathias has described here -- which is the large effect of catenary in deep water on short-ish scope.


We were groping around these issues in the thread I started, "Myth of the Bar Tight Chain", but neither I nor anyone else had the math to really crack the issue.

[rslifkin]

Originally Posted by Dockhead

I think it seriously undermines the idea that you should make the chain as light as possible and put the weight in the anchor. Because Dashew didn't understand what Mathias has described here -- which is the large effect of catenary in deep water on short-ish scope.

That idea is probably still valid if you have limited capacity for total anchor+chain weight, but if you aren't weight limited, might as well get the biggest anchor you can fit and then the heaviest chain you can carry enough of in the locker.

[Dockhead]

Originally Posted by rslifkin

That idea is probably still valid if you have limited capacity for total anchor+chain weight, but if you aren't weight limited, might as well get the biggest anchor you can fit and then the heaviest chain you can carry enough of in the locker.

Sure.


But a big issue with heavy chain is weight in the bow, which hurts performance.



330kg in the very bow of the boat is very significant.


Dashew on some of his boats had the chain locker just ahead of the mast, a great setup I think. I would do it that way if I were building a boat from scratch. Then I might go with even heavier chain -- maybe 14mm. 100 or even 120m of it. If you can get the chain locker under the waterline, then the anchor chain is even ballast.

[MathiasW]

Originally Posted by Dockhead

So, to sum up what we've discussed and figured out here.


I think Mathias has a real breakthrough in thought about anchoring and chain catenary.


I'd like to see the math peer reviewed.


I presume that everything is correct.


And if so, it really changes our thinking and practice in profound ways.



It contradicts some things which Peter Smith and Dashew came up with decades ago, which many of us have taken as gospel.


I think it seriously undermines the idea that you should make the chain as light as possible and put the weight in the anchor. Because Dashew didn't understand what Mathias has described here -- which is the large effect of catenary in deep water on short-ish scope.


We were groping around these issues in the thread I started, "Myth of the Bar Tight Chain", but neither I nor anyone else had the math to really crack the issue.

Thank you Dockhead! :blushing:

As far as peer review of the maths is concerned, I have been exchanging emails with Bjarne on this topic as a result of this thread. His maths is using a different coordinate system as I do, but he has verified that he gets the same result for the potential energy of a chain as I do. And because of this I believe the engine behind his calculator is based on the same maths as my work is - except that he has included many more things than I do. And that is why I am advertising it as the first link on my web page...

What still needs to be verified independently is the next step of doing an analytical derivative of the chain's energy with respect to anchor load. This is more tricky than one might think, as one needs to get the boundary conditions right. Without keeping 'something' constant, this derivative is without any bound, which was initially very confusing to me. But it simply reflects the fact that when I have an unlimited amount of unbreakable chain and unlimited water depth, I can store as much energy as I want in that chain...

I have worked out this derivative with two different boundary conditions. One whilst keeping the anchor depth constant, the other whilst keeping the chain length constant. In the later case I needed one more iteration as I had screwed up there initially. In that case it is also needed to keep the upper point of the chain fixed at a reference point like the vessel's bow. I am still a bit struggling where this second scenario really applies, though. Both calculations can be found in my long German essay, if you have version 20 or higher. Below that version number there was a mistake in this derivation.

In any case, had you asked me a year ago what my anchoring tactics are for a storm, I would have answered - fleeing into as shallow as possible water. I guess the intuitive reckoning behind that was: A) water is generally calmer there, and I guess that still holds, and B) I need less chain in shallow water to establish a full catenary. So I thought, I have much more chain reserve left, compared to deeper water, where I might be forced to pay out most of my chain and get very close to the limit. And I do not like to be close to limits. In shallow water, if I pay out all chain, way more than needed, surely, I thought, I would add a great margin to the bare minimum. That made me feel safe.

Now, if you ask me again, I would do the opposite. The point A) still holds, and so fleeing into a mess of a whirlpool in deeper water does not make sense, but as long as the gusts and swell are not a lot different at the place of deeper water, that would be my preferred place to go to. And the reason is simply that in the above reckoning I had not appreciated that the chain works so poorly at absorbing additional energy (i.e. gusts and swell), when the scope is very large. And in shallow water, in a storm, a fully developed catenary will have a large scope. It has no choice.

[Dockhead]

Originally Posted by MathiasW

Thank you Dockhead! :blushing:

As far as peer review of the maths is concerned, I have been exchanging emails with Bjarne on this topic as a result of this thread. His maths is using a different coordinate system as I do, but he has verified that he gets the same result for the potential energy of a chain as I do. And because of this I believe the engine behind his calculator is based on the same maths as my work is - except that he has included many more things than I do. And that is why I am advertising it as the first link on my web page...

What still needs to be verified independently is the next step of doing an analytical derivative of the chain's energy with respect to anchor load. This is more tricky than one might think, as one needs to get the boundary conditions right. Without keeping 'something' constant, this derivative is without any bound, which was initially very confusing to me. But it simply reflects the fact that when I have an unlimited amount of unbreakable chain and unlimited water depth, I can store as much energy as I want in that chain...

I have worked out this derivative with two different boundary conditions. One whilst keeping the anchor depth constant, the other whilst keeping the chain length constant. In the later case I needed one more iteration as I had screwed up there initially. In that case it is also needed to keep the upper point of the chain fixed at a reference point like the vessel's bow. I am still a bit struggling where this second scenario really applies, though. Both calculations can be found in my long German essay, if you have version 20 or higher. Below that version number there was a mistake in this derivation.

In any case, had you asked me a year ago what my anchoring tactics are for a storm, I would have answered - fleeing into as shallow as possible water. I guess the intuitive reckoning behind that was: A) water is generally calmer there, and I guess that still holds, and B) I need less chain in shallow water to establish a full catenary. So I thought, I have much more chain reserve left, compared to deeper water, where I might be forced to pay out most of my chain and get very close to the limit. And I do not like to be close to limits. In shallow water, if I pay out all chain, way more than needed, surely, I thought, I would add a great margin to the bare minimum. That made me feel safe.

Now, if you ask me again, I would do the opposite. The point A) still holds, and so fleeing into a mess of a whirlpool in deeper water does not make sense, but as long as the gusts and swell are not a lot different at the place of deeper water, that would be my preferred place to go to. And the reason is simply that in the above reckoning I had not appreciated that the chain works so poorly at absorbing additional energy (i.e. gusts and swell), when the scope is very large. And in shallow water, in a storm, a fully developed catenary will have a large scope. It has no choice.

In short, you well busted a myth. Well done.


Thanks to you -- we now know that scope is very much not necessarily a good thing. More anchor chain is a good thing, but not scope. To get the benefit from putting out a lot of anchor chain, you need to be in deep enough water to keep the scope down to a moderate level.


A new chapter in anchoring knowledge has been written. Maximum kudos.

[MathiasW]

Originally Posted by Dockhead

Sure.


But a big issue with heavy chain is weight in the bow, which hurts performance.



330kg in the very bow of the boat is very significant.


Dashew on some of his boats had the chain locker just ahead of the mast, a great setup I think. I would do it that way if I were building a boat from scratch. Then I might go with even heavier chain -- maybe 14mm. 100 or even 120m of it. If you can get the chain locker under the waterline, then the anchor chain is even ballast.

My analysis shows that when you are only concerned with how deep you can anchor in, then a longer thinner chain is better than a shorter thicker chain, both having the same total weight. These two chains only differ in their spread of their weight along the chain.

But what is missing in that approach is the analysis of their elasticity. Whether the thinner chain will still win here remains to be seen. At the same anchor depth, the thinner chain has a larger scope than the thicker chain has, and according to the elasticity analysis that is bad - the chain's elasticity is the product of the anchor depth times a universal function depending solely on effective scope. So, the thinner chain will have to venture out into even deeper water to compensate for that and match the elasticity of the thicker chain. The question is whether - it being longer than the thicker chain - it has enough chain to compensate for its disadvantage on the elasticity side - currently I do not know. And gusts and swell might well be even larger further out there.

So, if I know what the deepest water is I have to anchor in - or I am comfortable anchoring in - I would then pick the thickest chain that is still long enough to give me a full catenary at that depth and is with its total weight still within my weight budget at the bow. All this is of course also assuming a certain maximum wind / swell load I expect to have to face, as this determines the chain length to begin with.

[MathiasW]

Originally Posted by MathiasW

My analysis shows that when you are only concerned with how deep you can anchor in, then a longer thinner chain is better than a shorter thicker chain, both having the same total weight. These two chains only differ in their spread of their weight along the chain.

But what is missing in that approach is the analysis of their elasticity. Whether the thinner chain will still win here remains to be seen. At the same anchor depth, the thinner chain has a larger scope than the thicker chain has, and according to the elasticity analysis that is bad - the chain's elasticity is the product of the anchor depth times a universal function depending solely on effective scope. So, the thinner chain will have to venture out into even deeper water to compensate for that and match the elasticity of the thicker chain. The question is whether - it being longer than the thicker chain - it has enough chain to compensate for its disadvantage on the elasticity side - currently I do not know. And gusts and swell might well be even larger further out there.

So, if I know what the deepest water is I have to anchor in - or I am comfortable anchoring in - I would then pick the thickest chain that is still long enough to give me a full catenary at that depth and is with its total weight still within my weight budget at the bow. All this is of course also assuming a certain maximum wind / swell load I expect to have to face, as this determines the chain length to begin with.

So, here comes the analysis. I have again started from my expression for the elasticity of the chain, derived by differentiating the potential energy of the chain with respect to the anchor load F, making sure to keep the anchor depth Y constant.

So, it is exactly the same universal plot as before, but I am now parametrising the x-axis slightly differently. I do not use scope L/Y, but rather the quantity a/Y, where a = F/(m g) is the catenary parameter. Small values of 'a' mean little wind, and large values mean storm.

Again, the chain's elasticity is best at a certain sweet spot, where a/Y = 0.474 holds, with a = F/(m g). As before, the height of the peak itself scales with the anchor depth Y. Now, let's compare two chains, one chain twice as heavy as the other one per metre. Clearly, when going for the lighter chain to twice the anchor depth, 2Y, I have exactly the same ratio a/Y as with the heavier chain, and therefore exactly the same value in this graph. But, as the graph scales with Y, in real terms, the elasticity of the lighter chain is twice as high as that of the heavier chain. It requires precisely a twice as long chain (so in terms of scope, we end up at the very same point in the previous graph!), and the total weight of these two chains is thus the same. But as the elasticity is so much better for the thinner chain I could even save on the total weight of this chain, if I wanted to, and still have more elasticity.

So, in total, I gain a lot when using a lighter chain. But this does come at the expense of having to anchor in even deeper water, and I have to face the wrath of the Gods of the Seas out there. (And the chain's breaking load being reduced...)

But the rule is simple - halving the chain weight per metre means a chain of same total weight (i.e twice as long) lets me anchor in twice as deep water, with twice the elasticity of the chain.

[barnakiel]

Originally Posted by Dockhead

(...)



I think it seriously undermines the idea that you should make the chain as light as possible and put the weight in the anchor. Because Dashew didn't understand what Mathias has described here -- which is the large effect of catenary in deep water on short-ish scope.


(...)

Well, this is not exactly what the Dashews said.


If I remember well, their attitude was not to "make the chain as light as possible". I think they recommended LIGHTER and LONGER chain of the same or better strength. So say rather than 100' of 12mm to get 150' of 10mm - while moving from plain chain to high grade. This gives the same weight of chain, more of it, and stronger.


Now, as for heavier anchors - PLS look at the date the wrote their books. It was the pre-Mason, pre-Rocna, actually pre-Brugel (as it was used in EU but not in the US then) era. So, where they recommended a bigger anchor, they meant something more allong: if your Bruce is 10 pounder, get a 15 pounder Bruce. And they were 100% correct as this as with anchors we want AREA, as long as we have an anchor heavy enough to dig in easy into most bottoms. Once it, digs in well, bigger anchor only improves holding and minimizes dragging speeds.


They recommended bigger anchors as too many cruisers had TOO SMALL ONES back then. This is still, surprisingly, the case today.



So. My summary: it is my strong belief that the Dashews DID UNDERSTAND, and very well, what math Mathias tells us, just they had different limitations back then and so they applied different solutions.


- to improve penetration and holding - get a bigger anchor,
- chosing between 100' and 150' of chain, go for 150'.


I do not think what they said contradicted nor differed from what Math says.


Or am I wrong and Math say we should switch back to shorter and heavier chain?


If so, I bark my words back, BUT I am most surprised: as from my years of anchoring experience (about 15 years) in all bottoms and conditions (rtw), I have concluded that LONGER chain always improves how the boat behaves at anchor. I did play with both, and I found LONGER, lighter (lighter per meter, not per lockerful) chain the better solution (in a small boat, perhaps in a bigger boat the opposite is true).



If in doubt, drop them a message, they might have something to add to our discussions here.


BTW It is always good to remember where Steve comes from. I would be very careful assuming he could 'not understand' something.


About Steve's dad here:

https://en.wikipedia.org/wiki/Stanley_Dashew


Cheers,
b.

[MathiasW]

PS: This result is also easily seen in the original formula for the chain's potential energy:

E_pot = 1/2 m g (L Y - a (L - X))

Now, when I half the chain mass m per metre, but double the length L and double the anchor depth Y, then this doubles the potential energy, and this doubling then also translates to the derivative with regards to the anchor load...

[NPCampbell]

Originally Posted by MathiasW

PS: This result is also easily seen in the original formula for the chain's potential energy:

E_pot = 1/2 m g (L Y - a (L - X))

Now, when I half the chain mass m per metre, but double the length L and double the anchor depth Y, then this doubles the potential energy, and this doubling then also translates to the derivative with regards to the anchor load...

Forgive me for not reading the entire post. This may have been answered already. I got to this statement on your web page and I believe this is a false premise:

"It is done by representing swell as the kinetic energy of the vessel and then requiring that this energy is temporarily absorbed by an increase of the potential energy of the anchor chain. "

I believe it's a false premise because any force on the vessel that translates to a force at the anchor in the X direction is represented by an increase in tension, which is not considered potential energy.

Consider a slack chain stretched between two fixed points with an empty light weight water bucket hanging from the mid point. The chain will assume a catenary curve. If I add water to the bucket (F=mg), some of the force will be used to increase potential energy by "straightening" the chain from a catenary curve to a "V" shape. However, most of the force is just increasing tension in the chain. As I add more water to the bucket, the percentage of that force converted to tension will approach 100%.

Without accounting for tension, the formulas for potential energy will be incorrect.

[MathiasW]

Originally Posted by NPCampbell

Forgive me for not reading the entire post. This may have been answered already. I got to this statement on your web page and I believe this is a false premise:

"It is done by representing swell as the kinetic energy of the vessel and then requiring that this energy is temporarily absorbed by an increase of the potential energy of the anchor chain. "

I believe it's a false premise because any force on the vessel that translates to a force at the anchor in the X direction is represented by an increase in tension, which is not considered potential energy.

Consider a slack chain stretched between two fixed points with an empty light weight water bucket hanging from the mid point. The chain will assume a catenary curve. If I add water to the bucket (F=mg), some of the force will be used to increase potential energy by "straightening" the chain from a catenary curve to a "V" shape. However, most of the force is just increasing tension in the chain. As I add more water to the bucket, the percentage of that force converted to tension will approach 100%.

Without accounting for tension, the formulas for potential energy will be incorrect.

Hello NPCampbell, welcome to this discussion thread!

May I ask what you mean when you say 'accounting for tension'? Are you talking about the metal links of the chain stretching (ever so slightly) and storing some energy that way?

Or are you talking about the force along the chain, as it gets passed from the bow along the chain?

With regards to the former - no, I have not included that, I am afraid.

But when you talk about the tension as simply being the load in the chain, then the good news is that this is inherently taken care of in the catenary approach.

In fact, the very existence of the catenary shape is a consequence of the intricate play between wind load on the vessel (or any other load), and the chain weight. It is all balanced out very neatly such that at any point along the chain the load / or tension vector is pointing precisely along the direction of the chain. It has to, as the chain cannot support any forces perpendicular to it, without a kink in it, as your example also illustrates.

So, in approaching this problem, you have at least two different possible paths: You can work out all the forces as you suggest and then derive everything from there. I have seen others doing it, and it can be done, but it is painful. My preference has been to look instead at the energy balance and how these forces change it. After all, if the catenary gets pulled harder, it will have less slack. So, the increase in tension is equivalent to an increase in potential energy. It is simply looking at two sides of the same coin. Change of tension, change of potential energy, and vice versa. You will find formulas for the resulting tension in my pdf digests.

With regards to swell energy: There is the notion of conservation of energy. So, when the vessel initially had some kinetic energy and then got stopped by the anchor gear, this energy had to go somewhere. Some will have boiled the ocean, but the bulk of it will have had to be converted to additional potential energy in the chain. And so that's what I calculate.

Looking at energies has the huge benefit that I do not have to work in the time domain, like Alain has done it.

This is a well-trodden path in physics of applying conservation laws and I am very confident it works.

[MathiasW]

PS: As an addendum to this point: The catenary shape can be derived by requiring this system to have the lowest possible potential energy given the boundary conditions of a fixed chain length and fixed supports on either side.

[MathiasW]

Originally Posted by barnakiel

If so, I bark my words back, BUT I am most surprised: as from my years of anchoring experience (about 15 years) in all bottoms and conditions (rtw), I have concluded that LONGER chain always improves how the boat behaves at anchor. I did play with both, and I found LONGER, lighter (lighter per meter, not per lockerful) chain the better solution (in a small boat, perhaps in a bigger boat the opposite is true).

Our responses seem to have crossed mid-air...

My analysis supports this view - at least, when you are prepared to anchor in deeper water and have the same total weight of chain out there, just more stretched...

[slug]

Originally Posted by MathiasW

Our responses seem to have crossed mid-air...

My analysis supports this view - at least, when you are prepared to anchor in deeper water and have the same total weight of chain out there, just more stretched...

I didn’t have the energy to read the entire thread

From experience I observe that short scope in deep water works well .

By short scope I mean one hundred meters of chain laid in twenty five meter depth water ..4 to 1

Another observation is YAW and chain weight

For years I sailed with 12 mm chain

It was to much weight in the bow and didn’t stack correctly in the chain locker so when time came to replace the chain I went with 10 mm

The first thing I noticed was much less YAW with the light 10 mm chain

The 10 mm chain stored less energy when compared to heavier 12 min. Chain

[NPCampbell]

Originally Posted by MathiasW

Hello NPCampbell, welcome to this discussion thread!

May I ask what you mean when you say 'accounting for tension'? Are you talking about the metal links of the chain stretching (ever so slightly) and storing some energy that way?

Or are you talking about the force along the chain, as it gets passed from the bow along the chain?

Both.

Yes, the chain is stretching and storing the energy but due to internal friction, friction with the water, plastic deformation, and forces released in a non-axial manner you only get a small fraction of your applied energy back in the direction that the force was applied.

In another example, if I hang a 1000 kg weight from a chain, it exerts a force of 1000k x g downwards. Since the weight won't accelerate, there must be an equal and opposite force of 1000k x g. 100% of this opposing force is represented by tension in the chain. If the weight is instantly released, the chain should bounce back but it won't be anything approaching 1000k x g due to the reasons I mentioned above. If it did bounce back with a force of 1000k x g then I would stop pulling stumps out using a chain because it would kill me every time the stump broke

In your original example, the anchor chain has a force applied to it due to wind loads on the boat, etc. If the chain was vertical, and the anchor held, 100% of the force would be opposed by the tension in the chain. However, when the chain has a catenary curve, and a force is applied, 100% of the energy does not get opposed as tension. A portion of it is used to raise the the chain against gravity and flatten the catenary curve.

This brings me back to my original point. You must account for the percentage of the applied force on the chain that is represented by tension in the chain as not all of it gets translated to the potential energy of raising the chain against gravity.