Mathematic approach to anchoring scope

[Lodesman]

Originally Posted by Dockhead

At that point, the pull on the anchor shaft is no longer horizontal.

At that exact mathematical point, the pull is still horizontal - you have to start pulling the chain straight in order to change the angulation - that is the point of my question. The discussion about X angle of the anchor reduces holding by a couple percent is nonsensical, as it would take an enormous amount of force to get the anchor chain into a straight line.

And yes we did hash this out previously, but it seems to have come back to life.

[dfelsent]

Originally Posted by noelex 77

Mathematically it is impossible to eliminate catenary (while gravity is present), but in practice there are many anchoring situations where the chain visually appears dead straight.


It is difficult to capture the whole length of chain on camera, but these photos are examples of the appearance of the last few metres of chain.

Great photos.
BTW there is a way to eliminate catenary. Use zero scope so the anchor chain is vertical. Not particularly helpful I know. [emoji3]
I often think these threads come from someone’s desire to have control of ones situation by pure analysis and other’s explanations that many real world problems are messy enough that analysis must be coupled with judgement and practice.
Seamanship is an art informed by experience, engineering, and science.

[noelex 77]

Originally Posted by dfelsent

Great photos.
BTW there is a way to eliminate catenary. Use zero scope so the anchor chain is vertical. Not particularly helpful I know. [emoji3]

Ah, true. I had forgotten about that exception. It is more applicable than anchoring without gravity .

[Lodesman]

Originally Posted by dfelsent

Great photos.
BTW there is a way to eliminate catenary. Use zero scope so the anchor chain is vertical. Not particularly helpful I know. [emoji3]
I often think these threads come from someone’s desire to have control of ones situation by pure analysis and other’s explanations that many real world problems are messy enough that analysis must be coupled with judgement and practice.
Seamanship is an art informed by experience, engineering, and science.

That would be 1:1 scope

[Dockhead]

Originally Posted by MathiasW

Sorry for my ignorance, but are these 544 daN the wind force, or the force at the bow?

In the following I will ignore the difference between daN and kp, as it is rather late for me...

If it is wind force, then 544daN at 28 kn of wind correspond to an effective windage area of 43 square metres, which is rather a lot for such a small vessel.

If it is force at the bow, one first has to subtract the weight of the imaginary vertical chain to get to the wind load, so for a 10 mm chain roughly 30 m x 2 kp/m = 60 daN. And 484daN correspond to an effective windage area of 39 square metres, which is still rather large.

So, to me the issue seems to be that the windage area is assumed to be way too high for such a small vessel. A vessel of that size has more likely an effective windage area closer to 10-15 square metres. I mean, if the beam is 4.5 metres, say, the coach roof 2.5 metres above the water, that makes 11.25 square metres. Plus lazy jack, mast, Genova, rig, perhaps another 2 square metres. Then this has to be reduced by some factor < 1 as it is not a flat square surface of a brick, but the wind can slip around it more easily. I would find it hard to believe that this can get anything close to a windage area of 40 square metres...

A friend of mine measured the windage area of an Etap 39 quite accurately with a gauge and found it to be 7 square metres. If I apply scaling law here, so 14 m = 46 ft and the Etap 39 being 39 ft, then I arrive at

7 square metres x (46/39)^2 = 10 square metres

for a vessel of 14 m, which is consistent with my very rough estimate above.

The ABYC is a design load which considers the likely peak forces on the ground tackle. Not only wind force but some dynamic loads from sea motion.


The wind load by itself would not be the right value to use -- the ground tackle is subject to other loads than that.


Thinwater measured about half of the ABYC values in real life, but I think he agrees that the ABYC values are good to design to -- right, Thinwater?

[Dockhead]

Originally Posted by Lodesman

At that exact mathematical point, the pull is still horizontal - you have to start pulling the chain straight in order to change the angulation - that is the point of my question. The discussion about X angle of the anchor reduces holding by a couple percent is nonsensical, as it would take an enormous amount of force to get the anchor chain into a straight line.

And yes we did hash this out previously, but it seems to have come back to life.

To make it completely straight is impossible. To approach straightness is not impossible. Depending on how close you approach straight, yes, the forces become enormous. It's easy to play around with using the catenary sag calculator which I linked to.


As to anchor angulation -- yes, no disagreement -- at the Critical Tension, as Alain calls it, pull is still just horizontal. But from then on all additional tension adds angulation. In some combinations of depth and scope and chain size, it doesn't take much more force to pull the anchor at close to the same angle as a rope rode would. Alain gives a calculator which allows you to play around with it.


If you're using 8mm chain, 16m of it, in 4m of water (including bow roller height) -- a very typical real life situation for many boats -- then the values are:


Angulation without catenary (like a rope rode): 14.48 degrees

Critical force: 39daN
Force to get to 5 degrees: 60daN
Force to get to 10 degrees: 126daN
Force to get to 12 degrees: 228daN
Force to get to 13 degrees: 381daN


So you're within a degree and a half of no catenary already at 381daN of force, which is less than 27 knots of wind at ABYC norms.


But it will FEEL like catenary is gone already at 10 degrees, I'm sure, of not sooner. 10 degrees of angulation happens at 15 knots of wind (ABYC) in the given case.


That's why anchoring like that, in our old boat, which had 8mm chain, and typically in that kind of depth, we ALWAYS used a snubber, even in light weather.

[barnakiel]

In flatline scenario, it is very hard to get catenary to zero. Anybody who worked with their chain on a dock can say.


In deep water scenario, however, it is bang easy to zero the catenary (the catenary effect, the spring effect).


Water depth 10, bow 1 m, chain 11 meter, catenary effect ZERO.


Water depth 1 meter and you still need quite some pull to zero the catenary.


So a bit against the grain of better catenary in deep water.


b.

[MathiasW]

Originally Posted by noelex 77

Mathematically it is impossible to eliminate catenary (while gravity is present), but in practice there are many anchoring situations where the chain visually appears dead straight.


It is difficult to capture the whole length of chain on camera, but these photos are examples of the appearance of the last few metres of chain.

Thanks for sharing these pictures! I completely agree that a fully developed catenary will not always be present, but I am still striving for achieving one, if I can manage...

[MathiasW]

Originally Posted by dfelsent

Great photos.
BTW there is a way to eliminate catenary. Use zero scope so the anchor chain is vertical. Not particularly helpful I know. [emoji3]
I often think these threads come from someone’s desire to have control of ones situation by pure analysis and other’s explanations that many real world problems are messy enough that analysis must be coupled with judgement and practice.
Seamanship is an art informed by experience, engineering, and science.

Indeed, very true!

My hope was that my work would help beginners to avoid some of the basic mistakes. Clearly, the old buffs know it all by experience and do not need the maths.

I am not such an old buff yet and I do like to understand why things are behaving as they do, knowing that the models have all their limitations. It helps me to make better informed decisions and hopefully avoid serious accidents.

Along this line, I do appreciate all the feedback that has come through this thread. I have tried to include what I have distilled out of this on my web page. But more feedback is always welcome!

So, thanks again to all!

[MathiasW]

Originally Posted by Dockhead

The ABYC is a design load which considers the likely peak forces on the ground tackle. Not only wind force but some dynamic loads from sea motion.


The wind load by itself would not be the right value to use -- the ground tackle is subject to other loads than that.


Thinwater measured about half of the ABYC values in real life, but I think he agrees that the ABYC values are good to design to -- right, Thinwater?

OK, thanks for sharing this info. I am waiting to get a gauge to make my own measurements of the forces involved.

I have tried to deal with energy bursts (i.e. swell) separately, so I need to avoid accounting twice for things. But again, I need to make measurements...

[MathiasW]

So, I am trying to include my first ever graph in such a thread. Hopefully it works... The preview sucks...

The graph shows how a partial catenary is extended by a virtual chain below the seabed to complete the catenary. It also shows the forces involved.

What I wanted to point out in particular is that the argument is flawed that one cannot have a catenary, because it would get too close to the breaking load of the chain, or its working load.

A partial catenary like the one shown here will ALWAYS have a higher chain load at the bow than a fully developed catenary (at the same anchor depth ) has.

The full graph has been added to and discussed on my web page.

Note that at any point along the chain the down-ward pointing force Fg is equal to the weight of the chain to its left. After a little calculus one finds that the load in the chain at that point is equal to Fw plus the weight of a virtual chain hanging vertically down at this point, ending at the depth Y2 where the catenary becomes horizontal. This is kind of neat and easy to remember.

Finally, Fw may not necessarily be wind load only. It is just the horizontal component of the load at the anchor, whatever its origin may be.

[MathiasW]

Originally Posted by barnakiel

So how come with just 17 meters of chain and in wind gusting 45 for well over a day, we never ever felt the catenary gone flatline - anchored in Eastern Australia (behind the reef) ???


Swell was max 2 ft. We were right behind a flat island.

Perhaps you were lucky in that your anchor was located in a small ditch / trench with a slope, and so as far as the anchor was concerned, the pulling angle at the anchor shaft was still horizontal?

[Lodesman]

Originally Posted by Dockhead

If you're using 8mm chain, 16m of it, in 4m of water (including bow roller height) -- a very typical real life situation for many boats -- then the values are:


Angulation without catenary (like a rope rode): 14.48 degrees

Critical force: 39daN
Force to get to 5 degrees: 60daN
Force to get to 10 degrees: 126daN
Force to get to 12 degrees: 228daN
Force to get to 13 degrees: 381daN


So you're within a degree and a half of no catenary already at 381daN of force, which is less than 27 knots of wind at ABYC norms.

IMO, that is inadequate scope; I'd use twice that on a calm day. Expecting F6/7, I'd put out another 15m. In that case, maximum possible angulation is 5º and it takes (according to Alain's spreadsheet) 16748 daN to do it. Don't think the chain will hold.

[RaymondR]

MathiasW, you might have a look at how the American Bureau of Shipping and Det Norske Veritas handle wind loading calculations. There's also a mountain of work on anchoring offshore floating structures for the oil industry.

[MathiasW]

Hi Doghead,

It is perhaps me, but I am not getting the point of this:

Originally Posted by Dockhead

If you're using 8mm chain, 16m of it, in 4m of water (including bow roller height) -- a very typical real life situation for many boats -- then the values are:

Angulation without catenary (like a rope rode): 14.48 degrees

Critical force: 39daN
Force to get to 5 degrees: 60daN
Force to get to 10 degrees: 126daN
Force to get to 12 degrees: 228daN
Force to get to 13 degrees: 381daN

So you're within a degree and a half of no catenary already at 381daN of force, which is less than 27 knots of wind at ABYC norms.

What are you trying to prove here? If you are using too short a chain, certainly, the catenary will be spoilt. I won't argue on that point.

Or am I missing something?