Mathematic approach to anchoring scope

[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)

So, here is how I would reconcile these two 'different' views of Dockhead and barnakiel:

For any given, fixed anchor depth Y and provided there is enough chain, the heavier chain will outperform the lighter chain in terms of its ability to absorb energy. This is Dockhead's experience.

On the other hand, if I am ok to relocate to even deeper water, the lighter chain will eventually outperform the heavier chain, as for the same total weight of chain in the locker, the lighter chain, when fully paid out, can absorb more energy than the heavier chain, when this one is fully paid out. The heavier chain won't be able to get to those deeper grounds, as it has run out of chain, if both have same total weight. So, both will operate at different anchor depths and one needs to be prepared to accept that. Circumstances need to be such that this is possible (gust and swell-wise) and that is makes sense. This is perhaps barakiel's experience.

[MathiasW]

Originally Posted by slug

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

Yes, your observations are in perfect agreement with the model we have been discussing here. At that sort of anchor depth, even a smaller scope would work well.

The yawing I have not studied yet, but thanks for sharing this observation!

[MathiasW]

Originally Posted by NPCampbell

Both.

For including the elasticity of the steel itself, I can only recommend to use Bjarne's online calculator. It contains it all. For now, I do not intend to include it, as I do not think it is a major contributor.

Originally Posted by NPCampbell

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.

Not really. When the gust stops, the weight of the chain is pulling your vessel forward quite a bit, and this is precisely what can cause problems whilst at anchor. Over time, friction and what have you will dampen this oscillator, but it is not an over damped system by any means.

Originally Posted by NPCampbell

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

Precisely, and this is a very good example that the elasticity of the steel of the chain links are not storing an awful lot of energy. But, by the way, farmers that pull stumps out using a tractor all have a protection shield in their back...

Originally Posted by NPCampbell

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.

You are confusing me here. What do you mean by "100% of the energy does not get opposed as tension"? And yes, precisely: The flattening of the catenary is where the energy gets stored. It is called potential energy.

Originally Posted by NPCampbell

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.

OK, if you mean by tension the elastic stretching of the metal chain links, so, really pulling a steel rod to make it longer, then I am afraid I won't do that. As your example above already shows, the energy you can store in this is tiny compared to the other storage means.

I am afraid I still do not see where you are heading with all this. Can you perhaps be a bit more precise? Thanks.

[Dockhead]

Originally Posted by barnakiel

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..

Well, everyone on here knows how much I respect Dashew and his work. I've quoted him a million times on anchoring. This is not a contest about who's right and who's wrong -- I'm sure he would be the first to be delighted to learn something new. I am sure he didn't understand this. If you can find something in his work which shows otherwise, please quote it.

What you say about the same weight of chain but in a longer chain -- well, that is what Mathias just said. But Dashew NEVER said that (and if he did, I will be glad to be corrected, but I'm sure he didn't). In fact what Dashew says over and over again is that you should reduce the weight of the chain, and put the saved weight into the anchor -- not keep the weight of the chain the same and put it into a longer chain. Dashew says over and over again -- you often have limits on swing room, and even if you don't, it's desirable not to sweep too much of the seabed (something which is absolutely true). He shortens up his chain often to reduce the risk of snagging something. He can do this because he uses massive anchors. But this is not going to do anything to improve the catenary effect -- on the contrary.

Dashew on the contrary was of the opinion that catenary disappears just when you need it most (as Peter Smith said in the cited article), so you shouldn't bother trying to improve it -- put the weight into the anchor and use a good snubber.

I don't have time now to plow through all of Dashew's work to find that, but this view -- which I and many others took as gospel for decades -- is well summed up in Peter Smith's article "Catenary & Scope in Anchoring":

"It is then important to critically examine the mathematics of the catenary curve of a boat’s chain to investigate its true usefulness to a small boat. As the above 'old way of thinking' is commonly adhered to, it is a widespread myth that all boats should carry as much chain as possible – and the heavier the better.

"In fact, we shall see that catenary rarely offers much benefit which is truly worthwhile, and any unnecessary extra weight of the chain is often far better invested in other elements of the anchoring system."

https://www.petersmith.net.nz/boat-anchors/catenary.php

Peter Smith and Dashew were (are) friends, worked together on this as far as I know, and often quoted one another, so I believe this is harmonious with Dashew's views. I'll be glad to be corrected if I missed something in Dashew's work which is different from this.

Peter Smith reached this conclusion by showing, using Alain's models, that if you anchor in 6m of water with 48m of 12mm chain, the chain will be "bar tight" in 50 knots of wind, giving no benefit from catenary.

I guess his math is correct, and he DOES say that "benefit from catenary is even less in shallow water", so he does have an inkling of what Mathias is talking about.

But 6:1 in 8m of water with 48m of chain is not a case where catenary can "show itself". If you're going to be in 50 knot winds, you will want more chain out than 48m, and the effect of catenary -- as Mathias has shown us -- will increase dramatically if you put out say 100m of chain (like I surely would in such conditions) in say 25m of water.

This will work less well with lighter chain, but I guess 100m of 10mm chain in 25m of water will be pretty good too.


As to making the chain even longer than 100m, I come back to Dashew and question whether you ever have enough swing room to put out so much, or whether you would even want to. In my experience that's about the practical limit, anyway 90% of the time.


One P.S. for Barnie -- you say Dashew was writing all this prior to NG anchors -- no, that's not true! Dashew writes a lot about anchoring, all of it fascinating and valuable, in his blog, SetSail, or whatever it's called. He's been writing even recently. It's great stuff which I highly recommend.

[Dockhead]

For example:


"Steve Dashew Says: March 4th, 2010 at 9:15 am

1. Hi Chris:
   Our goal is to have the highest holding power, most reliable anchor system, for the lightest weight. Once you have sufficient strength in the chain to connect boat to anchor, the best system for the weight will always be light rode/big anchor. Taking this to extreme, we’d be better with a Spectra rode (much lighter and stronger than steel) connected to an even bigger anchor. The net result would be higher holding for less weight, but there are chafe and self-stowing issues.
   The other problem with heavy chain and increased catenary is fouling on bottom debris and/or damaging the sea bottom, which is a byproduct of the heavier chain."

https://setsail.com/chain-size-break...omment-page-1/

More good reading:


https://setsail.com/anchoring-system...true-approach/


This view is also echoed by another person I have tremendous respect for, John on Morgan's Cloud:


"I think that the idea that the catenary caused by heavy chain has benefit is wrong since on Morgan’s Cloud in winds of gale force, even with all 100 meters of our 7/16″ (about 11mm)"



https://www.morganscloud.com/2007/05/01/breaking-load/

[MathiasW]

Originally Posted by Dockhead

For example:[*]Hi Chris:
Our goal is to have the highest holding power, most reliable anchor system, for the lightest weight. Once you have sufficient strength in the chain to connect boat to anchor, the best system for the weight will always be light rode/big anchor. Taking this to extreme, we’d be better with a Spectra rode (much lighter and stronger than steel) connected to an even bigger anchor. The net result would be higher holding for less weight, but there are chafe and self-stowing issues.
The other problem with heavy chain and increased catenary is fouling on bottom debris and/or damaging the sea bottom, which is a byproduct of the heavier chain."

Hmm, Spectra. I am afraid he has lost me on that one. I would not tie up my vessel in a harbour slip / berth using Spectra. It kills the cleats. Why would I tie the vessel in such a way to the anchor then? With no means to absorb shock energy at all?

[Dockhead]

Originally Posted by MathiasW

Hmm, Spectra. I am afraid he has lost me on that one. I would not tie up my vessel in a harbour slip / berth using Spectra. It kills the cleats. Why would I tie the vessel in such a way to the anchor then? With no means to absorb shock energy at all?

Naturally you would use a snubber if you had a Spectra anchor rode.


Ships normally use Dyneema (or other UHMWPE like Acera) for their mooring lines by the way.

[MathiasW]

Originally Posted by Dockhead

Naturally you would use a snubber if you had a Spectra anchor rode.


Ships normally use Dyneema (or other UHMWPE like Acera) for their mooring lines by the way.

Ah, ok! Thanks!

[Dockhead]

https://www.dsm.com/dyneema/en_GB/ap...tow-ropes.html

[NPCampbell]

Originally Posted by MathiasW

I am afraid I still do not see where you are heading with all this. Can you perhaps be a bit more precise? Thanks.

Hmmm, how to explain ...

You state that the difference between Fw (wind force) and Fws (wind + swell), is totally accounted for by the difference in area between the two catenary positions of the chain. Let's call that force Fs.

To accommodate for the conservation of energy, Fs has to be completely opposed by other forces. You believe that Fs = sum of (m x g x h) for each chain link (ignoring the angle). However, I believe that Fs = T + sum of (m x g x h), where T is the tension in the chain. T is not potential energy because with chain, for all practical purposes, it is not elastic and therefore no work is performed and by definition it has no potential energy.

Are we sure that tension exists in this case? Yes.

If I pull horizontally on an anchor chain laying on the sea bed, with a force F, and the anchor does not move, 100% of the opposing force is tension, not potential energy. No distance is moved, no work is done, there is no potential energy, only tension. If I eliminate the force F, do I get a recoil in the chain equal to force F? No. Due to internal friction, plastic deformation, etc, there is little to no recoil because there is little to no potential energy.

Repeat the example above with an anchor where the chain extends vertically. If I pull on that chain vertically with a force F and the anchor holds, the only opposing force is tension, there is no potential energy.

Finally, take the example of an anchor chain where the chain extends from the anchor somewhere between horizontal and vertical. In this case, when I apply force to the chain, now there IS potential energy AS WELL AS tension in the chain. The potential energy is represented by summing up the mass of each chain link and multiplying it by gravity * the change in height of that link. Is there still tension in the chain? Absolutely. Did the tension change when the force changed? Absolutely.

As I apply an ever increasing force, F, to a chain, the catenary curve will flatten As the curve flattens, the tension forces opposing F will approach 100% of F and the potential energy due to gravity on the chain will become a tiny overall fraction of F.

Any clearer or am I missing something fundamental?

[rslifkin]

With the effectiveness of catenary thing, it scales based on size of boat I think. A bigger, heavier boat will carry more weight in chain. But that same larger boat will also typically be proportionally less impacted by gusts and wave action. So you have more energy absorption ability with the scaled-up ground tackle, but that effect scales up faster than the need for energy absorption. With a smaller, lighter boat, you typically won't be able to carry enough weight in chain to get enough absorption in heavy weather, so the key becomes matching up a snubber that's already starting to stretch noticeably by the time you've pulled out the useful catenary.

The Dashew giant anchor theory comes from 2 things in my understanding. First is ability to anchor in crappy bottoms and still have enough holding power. And the second is ability to anchor with short scope (even in shallower water where you'll pull the chain fairly tight) and still have enough holding power despite losing a lot to the worse angle of pull on the anchor. Realistically, if you can only carry, say 400 lbs of ground tackle, you'll probably have more anchoring options and better holding under bad conditions with a 200 lb anchor and 200 lbs of chain vs a 100 lb anchor and 300 lbs of chain.

[MathiasW]

Originally Posted by NPCampbell

Hmmm, how to explain ...

You state that the difference between Fw (wind force) and Fws (wind + swell), is totally accounted for by the difference in area between the two catenary positions of the chain. Let's call that force Fs.

To accommodate for the conservation of energy, Fs has to be completely opposed by other forces. You believe that Fs = sum of (m x g x h) for each chain link (ignoring the angle). However, I believe that Fs = T + sum of (m x g x h), where T is the tension in the chain. T is not potential energy because with chain, for all practical purposes, it is not elastic and therefore no work is performed and by definition it has no potential energy.

Are we sure that tension exists in this case? Yes.

If I pull horizontally on an anchor chain laying on the sea bed, with a force F, and the anchor does not move, 100% of the opposing force is tension, not potential energy. No distance is moved, no work is done, there is no potential energy, only tension. If I eliminate the force F, do I get a recoil in the chain equal to force F? No. Due to internal friction, plastic deformation, etc, there is little to no recoil because there is little to no potential energy.

Repeat the example above with an anchor where the chain extends vertically. If I pull on that chain vertically with a force F and the anchor holds, the only opposing force is tension, there is no potential energy.

Finally, take the example of an anchor chain where the chain extends from the anchor somewhere between horizontal and vertical. In this case, when I apply force to the chain, now there IS potential energy AS WELL AS tension in the chain. The potential energy is represented by summing up the mass of each chain link and multiplying it by gravity * the change in height of that link. Is there still tension in the chain? Absolutely. Did the tension change when the force changed? Absolutely.

As I apply an ever increasing force, F, to a chain, the catenary curve will flatten As the curve flattens, the tension forces opposing F will approach 100% of F and the potential energy due to gravity on the chain will become a tiny overall fraction of F.

Any clearer or am I missing something fundamental?

Yes, I have stated that somewhere. When the chain is flat on the beach, no potential energy can be gained. When it is hanging down vertically and I pull vertically, no potential energy can be gained either. In between, I can change the potential energy. And hence somewhere in-between, there must be a maximum, the sweet spot we have been talking about.

Originally Posted by NPCampbell

You believe that Fs = sum of (m x g x h) for each chain link (ignoring the angle). However, I believe that Fs = T + sum of (m x g x h), where T is the tension in the chain. T is not potential energy because with chain, for all practical purposes...

No, I do not believe anything, I just calculate stuff. If, by this formula, you mean that I say that the total force is strictly proportional to the area spanned in-between the two catenaries, one with swell, the other without, then please look again at my graph. I say it is related to it, not equal. One would indeed have to include the angle at each point when doing the summation. The error I am making here is related to the difference between a parabola and a cosh function. (When I ignore the angle of the chain and then work out what the chain's shape would be, I would end up with a parabola and not a cosh.) This diagram was only meant as an illustration that potential energy gain is related to the chain getting lifted and the more it gets lifted, the more gain there is. Perhaps it is a poor illustration to highlight the area in-between the two chains. Actually, I never bother to calculate that area. So, do not get hung up too much on that diagram, please.

And btw - please do not keep mixing forces and energies. If your T is tension, then it can never be energy, it is a force...

To conclude, my calculus is based on the exact equation for the potential energy of a catenary, which I had derived by calculating a line integral along the chain curve. I didn't feel, though, it was the right thing to explain the line integral in a high-level introduction to my approach. You can read up the details of the line integral in the long German paper. Sorry, it is in German, but the maths is trivial.

As said before, working with conservation laws (such as that for energy), is one of the most powerful tools we have in Physics. I can equally well work with forces, but it is generally more tedious. When working with energies, I convert to forces by differentiation, and when I work with forces, I convert to energies by integration. It is as simple as that. Two sides of the same coin.

Perhaps you can work out the potential energy of a catenary by integrating the anchor load from zero to its final value, keeping the anchor depth constant? You will find it matches my result, except possibly by a constant offset, as one is free to choose the zero point of energy as one sees fit.

[barnakiel]

Originally Posted by Dockhead

For example:Hi Chris:

1. Our goal is to have the highest holding power, most reliable anchor system, for the lightest weight. Once you have sufficient strength in the chain to connect boat to anchor, the best system for the weight will always be light rode/big anchor. Taking this to extreme, we’d be better with a Spectra rode (much lighter and stronger than steel) connected to an even bigger anchor. The net result would be higher holding for less weight, but there are chafe and self-stowing issues.
   The other problem with heavy chain and increased catenary is fouling on bottom debris and/or damaging the sea bottom, which is a byproduct of the heavier chain."

Nailed.


Very good detective work.


You may agree their context is vastly different though:


- no mention of swell, in any form,
- only a sub-part of the anchoring puzzle is being discussed, namely MAXIMISING THE HOLDING POWER,


So they are talking about a different thing, in a way. They are talking about how to get max holding. And they correctly conclude that max holding is achieved with max anchor. When anchor+chain total weight is fixed, max holding is when anchor weight is maxed out.


Now our question here seems to be - How do we connect that max holding point (assume there is no anchor but just a 1000 pounder concrete slab with point to shackle in your rode).


So, we are talking about the same thing just that we are looking at different parts of the whole as seen from the deck of a realistic boat (hardly anybody is carrying a 100o pounder concrete slab on their foredeck).




Yours,
b.

[barnakiel]

Originally Posted by Dockhead

(...)


What you say about the same weight of chain but in a longer chain -- well, that is what Mathias just said. But Dashew NEVER said that (and if he did, I will be glad to be corrected, but I'm sure he didn't). In fact what Dashew says over and over again is that you should reduce the weight of the chain, and put the saved weight into the anchor -- not keep the weight of the chain the same and put it into a longer chain. Dashew says over and over again -- you often have limits on swing room, and even if you don't, it's desirable not to sweep too much of the seabed (something which is absolutely true). He shortens up his chain often to reduce the risk of snagging something. He can do this because he uses massive anchors. But this is not going to do anything to improve the catenary effect -- on the contrary.

Dashew on the contrary was of the opinion that catenary disappears just when you need it most (as Peter Smith said in the cited article), so you shouldn't bother trying to improve it -- put the weight into the anchor and use a good snubber.

I don't have time now to plow through all of Dashew's work to find that, but this view -- which I and many others took as gospel for decades -- is well summed up in Peter Smith's article "Catenary & Scope in Anchoring":

"It is then important to critically examine the mathematics of the catenary curve of a boat’s chain to investigate its true usefulness to a small boat. As the above 'old way of thinking' is commonly adhered to, it is a widespread myth that all boats should carry as much chain as possible – and the heavier the better.

"In fact, we shall see that catenary rarely offers much benefit which is truly worthwhile, and any unnecessary extra weight of the chain is often far better invested in other elements of the anchoring system."

https://www.petersmith.net.nz/boat-anchors/catenary.php

Peter Smith and Dashew were (are) friends, worked together on this as far as I know, and often quoted one another, so I believe this is harmonious with Dashew's views. I'll be glad to be corrected if I missed something in Dashew's work which is different from this.

Peter Smith reached this conclusion by showing, using Alain's models, that if you anchor in 6m of water with 48m of 12mm chain, the chain will be "bar tight" in 50 knots of wind, giving no benefit from catenary.

I guess his math is correct, and he DOES say that "benefit from catenary is even less in shallow water", so he does have an inkling of what Mathias is talking about.

But 6:1 in 8m of water with 48m of chain is not a case where catenary can "show itself". If you're going to be in 50 knot winds, you will want more chain out than 48m, and the effect of catenary -- as Mathias has shown us -- will increase dramatically if you put out say 100m of chain (like I surely would in such conditions) in say 25m of water.

This will work less well with lighter chain, but I guess 100m of 10mm chain in 25m of water will be pretty good too.


As to making the chain even longer than 100m, I come back to Dashew and question whether you ever have enough swing room to put out so much, or whether you would even want to. In my experience that's about the practical limit, anyway 90% of the time.


One P.S. for Barnie -- you say Dashew was writing all this prior to NG anchors -- no, that's not true! Dashew writes a lot about anchoring, all of it fascinating and valuable, in his blog, SetSail, or whatever it's called. He's been writing even recently. It's great stuff which I highly recommend.

OK. I am with you. Thank you for setting this record straight.


Also when it comes to their blog, which I did not follow, thinking only about the context of their original books.


For the "100m" of chain thing, I have this comment:


Case A - a small boat - will not carry this much as it could not sail with this much,
Case B - can carry this much, but probably 'this much' is simply not enough for this boat size.



What I am trying to say is that in practical terms any cruising boat may in fact be "under-chained". It is possible that there is no way to get enough of chain.


Perhaps all we can do is just to strive and get as close to the 'enough chain' figure as we, with our specific boats, can.


b.

[barnakiel]

So, now.



PLS endure me, and if somebody can respond in a YES / NO (or close to this model) manner:


According to Math's math:


If all I can carry is 100 pounds of chain, should I :


- carry 40ft or 1/2 or,
- carry 100ft of 5/16,


Both are strong enough, both weight the same. I typically anchor in 6 to 10m (20-33ft of water).



Which option is better, by Math?


According to Alain's xls mentioned early on, the longer chain is recommended.



Thank you in advance of your feedback.



barnakiel