Mathematic approach to anchoring scope

[MathiasW]

Originally Posted by 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

Knowing the approximate windage area would be most helpful before giving any guidance.

[rslifkin]

Originally Posted by 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

Personally, I'd go for the 100 ft of 5/16 in that situation. Not from a catenary performance perspective, but because 40 feet is a bit short from the perspective of being able to avoid bottom chafe in less than perfect bottoms. 100 feet gives you more ability to keep line off the bottom.

[barnakiel]

Originally Posted by rslifkin

Personally, I'd go for the 100 ft of 5/16 in that situation. Not from a catenary performance perspective, but because 40 feet is a bit short from the perspective of being able to avoid bottom chafe in less than perfect bottoms. 100 feet gives you more ability to keep line off the bottom.

However, this is exactly what I already have. I am trying to learn how the solutions provided by Mathias bear on the practical solutions I arrived at from experience and from Alain's work expressed in his xls file.


barnakiel

[barnakiel]

Originally Posted by MathiasW

Knowing the approximate windage area would be most helpful before giving any guidance.

face: 4.5 sq m,
side: 12 sq m,


weight 4t


I do not the D / Cd but the boat is not boxy, rather plenty of very oblique areas.


b.

[MathiasW]

Originally Posted by barnakiel

face: 4.5 sq m,
side: 12 sq m,


weight 4t


I do not the D / Cd but the boat is not boxy, rather plenty of very oblique areas.


b.

OK, if this is the brick size, if you will, then accounting for streamlining effect, perhaps the effective windage area is as little as 3 square metres.

For the thinner chain, which is close to 8 mm, at 7 BFT, the catenary parameter would be a = 44.1 m.

with y = sqrt(L^2 + a^2) - a

this yields for L = 100 ft = 30.5 m: Y = 9.5 m

as the maximum depth you could anchor at whilst still maintaining a full catenary. This is not accounting for swell, so just the wind force only.

In the elasticity graph you are with a/Y = 4.6 well beyond the sweet spot.

But a chain shorter by 40/100 and heavier by the same amount, would only allow you to anchor in up to 9.5 m * 40/100 = 3.8 m, so that would not be enough.

8 BFT would reduce these numbers to 6.8 m and 2.7 m, respectively.

If you increase the effective windage area, these numbers would get worse.

So, this is not really new insights, just using the same formulas that Alain did many years ago.

Now, when you include swell, my Aeff = 3 sqm digest says the following: With a 4t boat and a velocity of 0.5 kn at anchor due to swell, I have to absorb an energy of 130 J. Looking at the graph for 100 J and 8 mm chain tells me the maximum anchor depth - for a chain of 30.5 m and 7 BFT - is about 7.5 metres, so 2 metres less to what I worked out above in the absence of swell. For 8 BFT that has gone to something like 4.7 m, and at 9 BFT your chain is not long enough anymore to establish a full catenary when such a swell hits your vessel. At these small water depths, the swell causes a significant increase of the load on the anchor, and for that reason alone you may want to relocate to deeper water. But a thick chain would prevent you from doing that, if you insist on a limit of its total weight. It would not be long enough.

And I believe with this point we are touching for the first time what I had initially thought was my new contribution to this field: Swell is more difficult to absorb in shallow water and leads to an increase in anchor load that you may want to reduce by relocating to deeper water.

So, with that total weight of the chain, the thinner chain is the only option, really. You are rather chain-weight limited in your configuration. If you expect more than 8 BFT in gusts, or large swell, or the effective windage area being larger, you may even want to consider a slightly thinner chain of higher quality.

[barnakiel]

Originally Posted by MathiasW

OK, if this is the brick size, if you will, then accounting for streamlining effect, perhaps the effective windage area is as little as 3 square metres.

For the thinner chain, which is close to 8 mm, at 7 BFT, the catenary parameter would be a = 44.1 m.

with y = sqrt(L^2 + a^2) - a

this yields for L = 100 ft = 30.5 m: Y = 9.5 m

as the maximum depth you could anchor at whilst still maintaining a full catenary. This is not accounting for swell, so just the wind force only.

In the elasticity graph you are with a/Y = 4.6 well beyond the sweet spot.

But a chain shorter by 40/100 and heavier by the same amount, would only allow you to anchor in up to 9.5 m * 40/100 = 3.8 m, so that would not be enough.

8 BFT would reduce these numbers to 6.8 m and 2.7 m, respectively.

If you increase the effective windage area, these numbers would get worse.

So, this is not really new insights, just using the same formulas that Alain did many years ago.

Now, when you include swell, my Aeff = 3 sqm digest says the following: With a 4t boat and a velocity of 0.5 kn at anchor due to swell, I have to absorb an energy of 130 J. Looking at the graph for 100 J and 8 mm chain tells me the maximum anchor depth - for a chain of 30.5 m and 7 BFT - is about 7.5 metres, so 2 metres less to what I worked out above in the absence of swell. For 8 BFT that has gone to something like 4.7 m, and at 9 BFT your chain is not long enough anymore to establish a full catenary when such a swell hits your vessel. At these small water depths, the swell causes a significant increase of the load on the anchor, and for that reason alone you may want to relocate to deeper water. But a thick chain would prevent you from doing that, if you insist on a limit of its total weight. It would not be long enough.

And I believe with this point we are touching for the first time what I had initially thought was my new contribution to this field: Swell is more difficult to absorb in shallow water and leads to an increase in anchor load that you may want to reduce by relocating to deeper water.

So, with that total weight of the chain, the thinner chain is the only option, really. You are rather chain-weight limited in your configuration. If you expect more than 8 BFT in gusts, or large swell, or the effective windage area being larger, you may even want to consider a slightly thinner chain of higher quality.

Thank you, Mathias!


barnakiel

[barnakiel]

BTW If your formulas and algorithms are open source then it may be possible to build a neat Android App to input the boat and swell data and get outputs and visualize them.


b.

[rslifkin]

The swell in deep vs shallow water thing makes a lot of sense. Especially if you're reaching the point where your rode is stretched pretty much tight. At the same rode angle, the change in angle from a 3 foot swell will be much less in deeper water where its relative height change of boat vs anchor is much smaller. Because of that, the rode won't be trying to pull the boat forward as far while it's rising up on the swell, which should reduce the load spikes on the system (particularly if there's enough load already that catenary and stretch can't readily smooth out all of the swell load).

[MathiasW]

Originally Posted by barnakiel

BTW If your formulas and algorithms are open source then it may be possible to build a neat Android App to input the boat and swell data and get outputs and visualize them.


b.

Surely all can be made open source. I had tried to tempt my son into building such an app, but so far have not had any luck with it...

Still, I have not given up on that idea quite yet...

[barnakiel]

Originally Posted by MathiasW

Surely all can be made open source. I had tried to tempt my son into building such an app, but so far have not had any luck with it...

Still, I have not given up on that idea quite yet...

Building an Android app is bloody easy. The limitation is in finding the required time to build, test and debug.


Also, most people are afraid of side loading, so such an App would have very few users.


b.

[MathiasW]

Originally Posted by barnakiel

Building an Android app is bloody easy. The limitation is in finding the required time to build, test and debug.


Also, most people are afraid of side loading, so such an App would have very few users.


b.

My son told me one would need to buy tools, but I could not quite believe that. Some open source tools should exist. I guess I need to look myself...

Perhaps it is less effort to piggy-back on an existing app.

[barnakiel]

Originally Posted by MathiasW

My son told me one would need to buy tools, but I could not quite believe that. Some open source tools should exist. I guess I need to look myself...

Perhaps it is less effort to piggy-back on an existing app.

Your son just wanted his dad off his back. we all did!


All that is required to build is free and is available online.


'To piggy-back on an existing app' I do not understand the meaning. This would be doing what exactly?


barnakiel

[MathiasW]

Originally Posted by barnakiel

Your son just wanted his dad off his back. we all did!


All that is required to build is free and is available online.


'To piggy-back on an existing app' I do not understand the meaning. This would be doing what exactly?


barnakiel

I guess you may have nailed down the explanation...

With piggy-backing I meant to find somebody who is selling or giving for free an app like an anchor watch, and see whether there is interest to add this functionality jointly.

[barnakiel]

Originally Posted by MathiasW

I guess you may have nailed down the explanation...

With piggy-backing I meant to find somebody who is selling or giving for free an app like an anchor watch, and see whether there is interest to add this functionality jointly.

This may be a good idea. I wrote my own AIS app as well as a number of non sailing Apps, but here at CF there are people who wrote anchor alarm apps - I believe this popped up in threads some time back.Dig into older anchoring App threads and send them PMs.


barnakiel

[Pete7]

Originally Posted by NPCampbell

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.

A minor point but I think you mean elastic deformation. Plastic deformation occurs once the elastic deformation is exceeded past the yield point and means the material won't return to its original dimensions. Failure is likely to occur in short order with on a small increase in force, not something we want to do with an anchor chain.

Pete