Mathematic approach to anchoring scope

[MathiasW]

Hi Dockhead,

It took me quite a while to read through the entire thread that you had started a few years ago...

Originally Posted by Dockhead

I used to think this, too, until others proved otherwise beyond a doubt, in this thread: https://www.cruisersforum.com/forums...in-215250.html, "Myth of the Bar-Tight Chain".

Catenary DOES go away at a certain force. We found the limits of this -- 200m of 1/2" chain will still have meaningful catenary at its breaking strength; but 100m will not.

I am still struggling with this. I am using the same equations as Alain does, but now also including swell, as I found an analytical formula for the potential energy of a catenary curve, so I can compare two different catenaries and take the difference. That is the main thing I am doing when working out what I call dynamic anchoring. (This is, btw, perhaps what you had been asking for in your earlier thread. )

But in all these calculations, I never got anywhere close to the breaking point of a chain, unless you try to anchor in extremely shallow water with a huge swell head on.

For instance, look at my results for Aeff = 12 square metres and a 12 mm chain, so half an inch, and only minimal swell. Then, in the absence of swell, at gusts of 9 BFT and 25 meters of depth, I get 92 metres of chain are minimally required, with the load at the anchor being ~600 kp. The load at the bow will be higher than that as it also has to support part of the weight of the chain. It comes out as 650 kp for this case. This is far away from the breaking load such a chain has, which is perhaps 7-8 times of this. BTW - For that very reason, I am using a Duplex chain, 10 mm, with a breaking load of 7200 kp.

Now, when paying out more chain, you do not increase the load at the anchor, to the contrary. The additional chain weight is simply lying on the seabed or, if not, has to be supported by the two end points of the catenary. One being the bow roller, the other where the chain touches the ground. For a catenary curve that is ending horizontally at the seabed, it turns out the force at the bow roller is simply the load at the anchor plus the weight of a chain of the same type hanging down vertically at the anchor. Hence, in the above example 25 meters x 2 kp/m = 50 kp. It is not so that the entire weight of the catenary has to be supported by the bow roller.

So, if anything, this argument of breaking load being a problem, can only support trying to achieve a catenary. The load forced upon the chain by a scope rule generally will be higher than what a catenary does. You need a very high load to get to a bar-tight chain to begin with.

So, I really fail to see why the breaking load of the chain should limit me in obtaining a catenary. If we ever were to get close to the breaking point of a new chain, the anchor would yield, unless it is hopelessly stuck somewhere. And yes, not having enough chain will possibly stop me , but the breaking load I just do not see.

What am I overlooking here?

[MathiasW]

Originally Posted by Lodesman

That's clipped from the 1995 edition. I don't think earlier editions mentioned it, but OTOH they stated something to the effect that the resultant scope would hold to 60 kts (iirc).

Thank you Lodesman! Much appreciated. I do not have access to that manual where I am currently located and only got indirect pointers to this BA rule.

[barnakiel]

My good people.


You know I mostly blah rather than hear ya. Da animal with small ears and huge mouth. My bad.



Yet.


What exactly are the conditions that would remove all catenary from my chain ???


Are they not the same conditions that tell me I should not be in these conditions, in this anchorage. The conditions that pulled the cleats from my deck one hour ago?


Boat Bavaria 30 ft long. Chain 10mm / 100 meters. Achorage depth 6 meters.


???


Thank you in advance.


PS Please respond swiftly, we are dragging slowly now and expect to be ashore in about half an hour. I will collect your responses from the beach bar - that is just next to the wreckage... !!!


Now. I mean it. How much wind and swell does it take?


b.

[Dockhead]

Originally Posted by conachair

Completely disagree, like 11/10 - it's a very fundamental mistake in some fairly basic physics and doesn't help anyone getting a rough handle in what's going on. Nothing to do with terminology at all.

You should be focusing on how quickly the energy of a moving boat transfers , that's why a snubber helps.
Saying no more energy can be absorbed by the chain is just plain wrong and will confuse anyone trying to get and handle of the physics of what's happening, it has to go somewhere. Though to be fair this basic fundamental mistake even makes it into published articles. But if you're going to talk physics you really should get it right, it makes so much more sense that way and you look better than getting it wrong... .....

Trying to understand where this vitriol comes from. What, exactly, are you disagreeing with?

If the physics are badly expressed, why don't you improve the description?

What about this specifically do you disagree with:

". . . to be more precise (but not actually more correct), is reduction of the peak forces by spreading out the transfer of energy between boat and anchor in time. Of course the energy doesn't disappear, but it is absorbed temporarily and transferred over a longer period of time. OK, are you happy now? The point is that as the chain tightens, the useful reduction of peak forces by the action of catenary fades and soon disappears. It does start to behave like a bar. THAT is why we need snubbers."

If you actually know the physics, rather than just sneering, let's have them. I will be glad for the correction.

[thinwater]

Originally Posted by valhalla360

...In shallow water, you often are relying on friction between the chain and seabed as much as cantenary to absorb loads....

Yes and no. Straight pull, no. That has been covered.


However, if the boat is yawing (and it is), then this is an important factor until the wind causes the chain to lift. It reduces yawing, which reduces chain tension in two ways.

[Dockhead]

Originally Posted by MathiasW

Hi Dockhead,

It took me quite a while to read through the entire thread that you had started a few years ago...

I am still struggling with this. I am using the same equations as Alain does, but now also including swell, as I found an analytical formula for the potential energy of a catenary curve, so I can compare two different catenaries and take the difference. That is the main thing I am doing when working out what I call dynamic anchoring. (This is, btw, perhaps what you had been asking for in your earlier thread. )

But in all these calculations, I never got anywhere close to the breaking point of a chain, unless you try to anchor in extremely shallow water with a huge swell head on.

For instance, look at my results for Aeff = 12 square metres and a 12 mm chain, so half an inch, and only minimal swell. Then, in the absence of swell, at gusts of 9 BFT and 25 meters of depth, I get 92 metres of chain are minimally required, with the load at the anchor being ~600 kp. The load at the bow will be higher than that as it also has to support part of the weight of the chain. It comes out as 650 kp for this case. This is far away from the breaking load such a chain has, which is perhaps 7-8 times of this. BTW - For that very reason, I am using a Duplex chain, 10 mm, with a breaking load of 7200 kp.

Now, when paying out more chain, you do not increase the load at the anchor, to the contrary. The additional chain weight is simply lying on the seabed or, if not, has to be supported by the two end points of the catenary. One being the bow roller, the other where the chain touches the ground. For a catenary curve that is ending horizontally at the seabed, it turns out the force at the bow roller is simply the load at the anchor plus the weight of a chain of the same type hanging down vertically at the anchor. Hence, in the above example 25 meters x 2 kp/m = 50 kp. It is not so that the entire weight of the catenary has to be supported by the bow roller.

So, if anything, this argument of breaking load being a problem, can only support trying to achieve a catenary. The load forced upon the chain by a scope rule generally will be higher than what a catenary does. You need a very high load to get to a bar-tight chain to begin with.

So, I really fail to see why the breaking load of the chain should limit me in obtaining a catenary. If we ever were to get close to the breaking point of a new chain, the anchor would yield, unless it is hopelessly stuck somewhere. And yes, not having enough chain will possibly stop me , but the breaking load I just do not see.

What am I overlooking here?

The question we explored in that thread was -- is the chain ever really "bar tight"? What if there is still useful catenary at the breaking strength of the chain?


And we figured out that at some point with heavier or longer chain, it might be, but not at typical values we experience.

[thinwater]

Originally Posted by Dockhead

...Catenary DOES go away at a certain force. We found the limits of this -- 200m of 1/2" chain will still have meaningful catenary at its breaking strength; but 100m will not....

I think you would have gotten fewer arguments with "WLL" than "BS."



And so long as the chain is sized for the boat, the diameter doesn't not actually matter in the math (the tension ~ weight ~ strength). Just the ratio of tension to WLL.



But you clearly have the essence of the lesson. It is the length of chain. Once there is a lot of chain out (let's say over 250 feet) you have catenary under all common conditions, regardless of the water depth. The obvious corollary is that less scope is needed. With less chain out you have less weight out and thus less catenary. Less than 50 feet of chain in shallow water is a jackhammer. And you need more scope. Kind of simple.

[thinwater]

Originally Posted by barnakiel

My good people.


You know I mostly blah rather than hear ya. Da animal with small ears and huge mouth. My bad.



Yet.


What exactly are the conditions that would remove all catenary from my chain ???


Are they not the same conditions that tell me I should not be in these conditions, in this anchorage. The conditions that pulled the cleats from my deck one hour ago?


Boat Bavaria 30 ft long. Chain 10mm / 100 meters. Achorage depth 6 meters.


???


Thank you in advance.


PS Please respond swiftly, we are dragging slowly now and expect to be ashore in about half an hour. I will collect your responses from the beach bar - that is just next to the wreckage... !!!


Now. I mean it. How much wind and swell does it take?


b.

I'll go slow. The---answer--is--in--the--first--post.


Seriously, you heard the list of variables. No one answer. Anywhere from 15 knots to 45 knots, depending.

[MathiasW]

Originally Posted by thinwater

I think you would have gotten fewer arguments with "WLL" than "BS."

And so long as the chain is sized for the boat, the diameter doesn't not actually matter in the math (the tension ~ weight ~ strength). Just the ratio of tension to WLL.

We are in complete agreement on both accounts!

[MathiasW]

Originally Posted by Dockhead

The question we explored in that thread was -- is the chain ever really "bar tight"? What if there is still useful catenary at the breaking strength of the chain?


And we figured out that at some point with heavier or longer chain, it might be, but not at typical values we experience.

Hmm, ok, but this is not what I see with essentially the same Ansatz as Alain. But then again, I have a chain longer than 100 metres and when I do the maths for my boat and such a chain, I should be good to achieve a decent catenary up to 9 BFT, depending on water depth and swell.

[Dockhead]

Originally Posted by MathiasW

Hmm, ok, but this is not what I see with essentially the same Ansatz as Alain. But then again, I have a chain longer than 100 metres and when I do the maths for my boat and such a chain, I should be good to achieve a decent catenary up to 9 BFT, depending on water depth and swell.

Without lifting the last link? It would be interesting to see the math. I think Alain's calculator will do that in a jiffy.



I have 100m of chain and I've definitely lost all catenary long before 9 Bf.



I believe the size of the chain is another critical variable, along with the depth.

[MathiasW]

Originally Posted by Dockhead

Without lifting the last link? It would be interesting to see the math. I think Alain's calculator will do that in a jiffy.

Sure!

If I ignore swell completely, I use the same formula as Alain does for the length of a catenary with a perfectly horizontal attachment to the anchor shank:

L = square_root(Y(Y+2a))

where a = F_wind/(m g)

m being the mass of the chain in water per meter, and g = 9.81 m/s^2

This page is in German, but there are others for the wind pressure:

https://de.wikipedia.org/wiki/Winddruck

(
I found an online calculator just now that seems to give similar numbers:

https://www.engineeringtoolbox.com/w...ad-d_1775.html
)

According to this, for 9 BFT, I have 367 N/m^2 as wind force per unit area. If I assume an effective windage area of my trimaran to be 20 square meters, then this makes

F_wind = 7.348 kN

With a chain of 10 mm, which has m = 2 kg/m in water, I thus get

a = 374 m

Plugging this into the above formula yields

for Y = 15 m water depth

L = 107 m

for Y = 12 m water depth

L = 95.5 m

So, I seem to be ok as long as it does not get too deep.

As a note aside, the British Admiralty rule would give 159 m and 143 m, respectively. And with the rules for the heavier chains that Lodsman mentioned, those BA values would have to be divided by a factor of 1.5 So very much the same ball park as my results.

[valhalla360]

Originally Posted by Lodesman

On a separate note, I cringe at peoples' notions of catenary disappearing completely. At the point that the last link comes off the seabed it is still pulling horizontally, and there is still a catenary in the chain, even though it will appear bar-taught and straight from your vantage on the bow. I think Alain's site has a formula to calculate the force required to get to that point. To pull the chain to a straight line requires considerably more force. This is of course dependent on the length and weight of the chain, but it is certainly possible for many cruising boats to have sufficient chain, such that catenary would never disappear. IIRC, this is around 350 feet of chain - the force required to pull this into a straight line would part the chain - assuming the anchor held fast and the chain held fast to the bow.

The physicist will tell you cantenary never goes away and technically they would be right. It is impossible to pull a chain hard enough that it is perfectly straight (exception being if it's hanging straight down or in a weightless environment).

The engineer will tell you for all practical purposes yes, it does go away. Once a 100m length of chain is within a few centimeters of being in a straight line, for all practical purposes, it's straight.

[valhalla360]

Originally Posted by thinwater

Yes and no. Straight pull, no. That has been covered.


However, if the boat is yawing (and it is), then this is an important factor until the wind causes the chain to lift. It reduces yawing, which reduces chain tension in two ways.

That is what I was refering to...if you can reduce the momentum of the boat, you reduce a lot of it's ability to pull an anchor free from the bottom.

If you just have a steady state system where the boat pulls straight back, settles in with no motion, it really can't generate nearly as much force on the anchor as when it's bucking and yawing about.

[valhalla360]

Originally Posted by MathiasW

Now, that is a myth! If you simply work out the force a chain lying on the seabed can withstand, you will be disappointed. It is friction coefficient times weight in water. The friction coefficient is less than one. So, the force required to pull this chain on the seabed is less than the weight of the chain.

Assuming a 10 mm chain, which is 2 kp/m weight in water, and 30 m of chain on the seabed, this is just 60 kp, which is a small fraction of the holding power a, say, 30 kg anchor could develop. And it is only twice the weight of the anchor. So, you seem to feel safe if three anchors are just lying on the ground, not dug in at all???

So, this chain lying on the seabed really does not improve holding power by a lot.

What it does, though, is to add room for the chain to gain more potential energy when needed and thereby reduce shock loads. But this is an effect due to potential energy, and not due to friction.

And this is precisely, what I had calculated with my formulas...

You are back to laboratory conditions with the chain neatly laid out and the boat sitting nice a still as the wind holds it in place. In reality, it's digging in pulling sand back and forth and absorbing a lot of motion, just not in a direct straight line back along the wind direction.

Why should the maths not scale? The physics does. Some variables will scale differently than others, but all scale. If that were not the case, why would small models in a water tank be of any use to study? But professional vessel designers do this all over the world! Some folks claim catenaries only work for small vessels, but not for large vessels, others claim the opposite. I happen to have a bulletin of the Japanese Loss Prevention and Ship Inspection Department, which works with the catenary formula. So, at least they seem to believe in it. And for sure there is a lot of money at stake there, so they had better get it right...

Have you ever done scale modeling? I was referring to your simplistic analogy that implies a 30ft cruising boat and a 1000ft oil tanker react in the same way.

It's not as simple as double the length of the boat and double the force on the anchor. Yes, you can scale up but what happens is as you get into large ships, things happen differently. This is one of the tricky things with doing scale models. If you don't fully understand how size impacts a set of motions and forces, it's easy to make false assumptions.

A light catamaran with boards up, might yaw back and forth every 10 seconds. On a 1000ft oil tanker, it might take several minutes and before it really gets any good momentum going, the wind is already slowing it down for the return swing. The result can be the catamaran rode is going thru loading that varies by 50-100% in rapid fashion vs a tanker that might go thru a 5-10% variation in a very slow and controlled manner. This makes the catamaran far more likely to break the anchor out.

And yes, on our vessels we can and must use snubbers / bridles. They help a great deal. But they will not absorb 100%, it is always a sharing of the load between the snubber and the chain, as otherwise the snubber would not engage. Think of it like a voltage divider in an electrical circuit created by two resistors in series. Consequently, it is good to know how well the chain can cope with its part in the deal...

By spreading out the loading a snubber drastically reduces the force on the anchor shank, so ignoring it ignores a major component.

Again, the catenary formula will ALWAYS yield a chain length where a horizontal pull at the anchor is achieved, no matter how hard the wind blows. This is not an assumption, but simply physics. If you pay out enough chain, the pull will be parallel. Period. Now, of course, it can happen that you do not have enough chain for doing that, but it is still interesting to know how much chain would have been needed. And, funnily enough, it turns out it is less chain than one might have thought, at least when anchoring in deep water. There, it is often less than a scope 9:1 would suggest.

Some say they have SEEN the catenary to be gone. I doubt that. In a serious storm, you may have 100 m of chain out. With the anchor perhaps at 10 m. So, this is already a scope that is hard to tell with the naked eye. A difference of as little as one degree angle at the bow, with which the chain is pointing into the water, would make about 15-20 of meters difference were the chain hits the seabed. So, it may still be lying on the seabed and you just don't know it. And how can one tell as little as one degree angle difference on a rocking boat?

You are looking at the theoretical vs the practical. Theoretically, there is always cantenary but I've been diving on the anchor in moderate conditions and watched the chain lift straight a bit at the anchor shank. I'm sure with even more out in storm conditions, you could still see it go bar tight and for practical purposes eliminate the cantenary.

I have said it before: I am not trying to imply my work has solved all problems and gives answers to everything. But it does show what chain length is minimally needed when using certain models for the real world. It helps at least me to make an informed decision when anchoring. I will adapt my formulas to account for factors at my anchorage that are not included in the model as best as I can, but at least I have a reference point to start from. And if I have more than one model, I can see how compatible their results are. And yes, I will pay out a little more chain than what the formula suggests.

Have to disagree that you've proven anything for the "real world".

What you've done is look at a "theoretical set of conditions" which has some correlation to the real world.