Mathematic approach to anchoring scope

[Lodesman]

Originally Posted by valhalla360

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.

Can the engineer tell me how hard do you have to pull to make 130m of 3/8" chain, "for all practical purposes straight"?

[valhalla360]

Originally Posted by Lodesman

Can the engineer tell me how hard do you have to pull to make 130m of 3/8" chain, "for all practical purposes straight"?

Sure, tell us what your practical purposes are.

[StuM]

Originally Posted by Lodesman

Well first there's gravity, then added to that is elastic stretch - both temporary, but pull a bit more and you add deformity, which would come with some heating, that would dissipate to the surrounding water - lost energy, that you're not getting back

There's always lost energy that you're not getting back. I think that's what conachair meant.


In one word: Entropy

[Lodesman]

Originally Posted by valhalla360

Sure, tell us what your practical purposes are.

Duh! Keeping the anchor shank parallel to the seabed.

[StuM]

Originally Posted by thinwater

Less than 50 feet of chain in shallow water is a jackhammer. And you need more scope. Kind of simple.

Assuming all chain rode.
Rope can take the place of catenary as a shock absorber

[StuM]

Originally Posted by Lodesman

Can the engineer tell me how hard do you have to pull to make 130m of 3/8" chain, "for all practical purposes straight"?

I am not an engineer, but:


Take the simple case of a chain with the end points attached at the same height.
Decide what angle at the end points is acceptable for your "practical purposes".
Let's say you want it sufficiently straight that the chain departs from the anchor point at x°.
At either end point, with half the weight of the chain as the vertical component, the horizontal component is simple vector trigonometry.

[barnakiel]

Originally Posted by thinwater

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.

Thank you for reading into and for this answer. Can I please keep your attention one minute more?



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.


I would think when all catenary is gone, I would notice? (aka Sparrow's Paradigm)



(My guess: push forces from the wind are HUGELY overrated in extreme anchor pull calculations.) By maybe factor of 2. Hugely.



Maybe this is DUE TO thinking a lot about wind and swell BUT NOTHING about the friction of chain pulled horizontally thru water. E.g.



Good maths is good. Poor physics and limited real life experience is limiting good maths.



barnakiel

[StuM]

Originally Posted by Lodesman

Duh! Keeping the anchor shank parallel to the seabed.

That does not occur when the chain is straight. It occurs when the catenary lifts the last link. That's what this whole thread was about originally and we have all the theroretical answers in great detail at the link in the original post.

A "straight" chain is an entirely different situation. That is when the anchor shank is being pulled upwards by an amount determined by the depth of water and the distance between boat and anchor.

[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) ???

Perhaps your wind measurement device was completely out of whack? But more seriously, this seems like an awfully short chain for that kind of wind.

Any more information on what type of vessel it was, what the water depth was? What type of snubber / bridle used?

On that point of snubber / bridle - I see it often that they are way too short to be effective. A snubber of just 2 metres long is just micky mouse. A bridle of 4 metres the same. A snubber should be at least 10 metres long. Same holds for a bridle. As was said elsewhere, the energy is force times distance, and so you want to cover a good distance when elongating your snubber to keep the force in check.

Originally Posted by barnakiel

Maybe this is DUE TO thinking a lot about wind and swell BUT NOTHING about the friction of chain pulled horizontally thru water. E.g.
barnakiel

Have a look at post 30 by Panope in this thread, which I found most enlightening:

https://www.cruisersforum.com/forums...-215250-2.html

Seems like some serious amount of energy can be dissipated that way.

[Dockhead]

Originally Posted by Lodesman

Can the engineer tell me how hard do you have to pull to make 130m of 3/8" chain, "for all practical purposes straight"?

We discussed this in the old thread, in great detail. Are we going to rehash it all, here?


It takes 544daN to lift the last link of a 130m long, 3/8" chain, in 30 meters of water. That is a little less than the ABYC design wind load for a 14m boat in 28 knots of wind.


At that point, the pull on the anchor shaft is no longer horizontal. Alain's spreadsheet has calculators for all this, including the force required to get a given angulation on the anchor with a given combination of depth, chain length, chain size.


As to pure straightness of the chain -- that's the catenary sag formula. Here's a handy calculator: https://www.spaceagecontrol.com/calccabl.htm. I don't think we care about this, we care about the angulation on the anchor, and we care about the vanishing point of the damping effect of catenary. Reducing anchor angulation, and damping snatch loads on the ground tackle, are the two beneficial effects of catenary, so that's what we care about. Right?

[MathiasW]

Originally Posted by valhalla360

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.

Alright, what is your approach to the subject matter then? I mean concretely, no handwaving or pointing to some foggy statements. So far, apart from insults, your contribution to this topic has been underwhelming...

[MathiasW]

Originally Posted by Dockhead

It takes 544daN to lift the last link of a 130m long, 3/8" chain, in 30 meters of water. That is a little less than the ABYC design wind load for a 14m boat in 28 knots of wind.

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.

[valhalla360]

Originally Posted by Lodesman

Duh! Keeping the anchor shank parallel to the seabed.

If the chain is for practical purposes straight...that pretty much precludes the pull being parallel to the seabed.

Of course, this is where you have to understand that anchors don't need a perfectly parallel pull to stay dug in. It's nice but not necessary.

[valhalla360]

Originally Posted by MathiasW

Alright, what is your approach to the subject matter then? I mean concretely, no handwaving or pointing to some foggy statements. So far, apart from insults, your contribution to this topic has been underwhelming...

There's a reason it's a field that is mostly empirically driven.

There's a bit of theory and principals but trying to calculate it to the gnats ass doesn't really work.

[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.