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Old 05-01-2024, 19:47   #1
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Anchoring equations

I understand that more weight, anchor and chain, on the bottom is better than less weight when it comes to holding.

that said, is there a way to calculate how much better one strategy might be than another?

Ie on a 35’ sailboat how would one measure the relative holding with the following:

140’ of rope, 35’ of chain ( 1 lb per foot), 15kg anchor( Bruce)
140’ of rope, 35’ of chain ( 1 lb per foot), 20kg anchor ( Bruce)
140’ of rope, 130’ of chain(1 lb per foot), 20kg anchor ( Bruce)

Obviously, the third one is what you’d prefer to ride out a storm with in 30’ of water with 180’ of rode deployed.

How much more holding power measured however one wishes does one scenario have over the others?

Over the past three years, I’ve moved from 1 to 2 to 3. I was much happier in a 30 knot wind with option three than I would have been with the first two.
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Old 05-01-2024, 22:22   #2
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Re: Anchoring equations

Dave - have a look at: Anchor

Loads of formulae to play with. However, if it was my boat I'd repurpose the Bruce as a garden ornament and replace it with a modern design. As per Fraysse's conclusions, I'd go for an all-chain rode with about 35-40' rope snubber.

The heavier the anchor the better the ultimate holding - but everything is a compromise. Put too much weight up in the bow and you'll get to bounce up and down a lot in a seaway. I'd think 20kg for a 35' Bennie is a good choice - provided its a Spade/Excel/Rocna/Mantus or similar.
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Old 05-01-2024, 23:33   #3
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Re: Anchoring equations

Assuming the angle of pull (at the anchor) remains the same, holding power increases with anchor weight at a rate of (roughly) 1/1 in "normal" seabeds (sands, muds)


The above does not work for special situations where a smaller anchor may not penetrate at all (thick weed, stones, hard sand), and a large version will. In this case the increase in holding power might be many times greater for the anchor that is just large enough to function properly (penetrate).
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Old 05-01-2024, 23:44   #4
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Re: Anchoring equations

Dave,

Worth having a read of some of the work Mathias has done on CF in a number of different threads.

https://www.cruisersforum.com/forums...se-274294.html
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Old 06-01-2024, 03:37   #5
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Re: Anchoring equations

Long story short:
The weight of the anchor does not affect holding, it is only the fluke size, the fluke to-shank angles, and the position of the crown, that does it, as well as the pulling direction (see aluminum fortress for example).
The weight of the chain does.
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Old 06-01-2024, 04:24   #6
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Re: Anchoring equations

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Originally Posted by Izikalvo View Post
The weight of the chain does.
Are you sure? I appreciate you manufacture some very interesting anchors.

However, this has been discussed at length on CF. The snyopsis I can remember is that when it really matters because its blowing a hooley, the anchor chain is bar tight and contributes very little. Conversely in flat calm conditions, the yacht may actually be lying to the chain and the anchor doing nothing, but that is a side issue.

The advice was therefore, to put the weight in the anchor and by association likely to be bigger with a larger surface area of the flukes which increases the holding. Or am I wrong
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Old 06-01-2024, 05:40   #7
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Re: Anchoring equations

Quote:
Originally Posted by Pete7 View Post
Are you sure? I appreciate you manufacture some very interesting anchors.

However, this has been discussed at length on CF. The snyopsis I can remember is that when it really matters because its blowing a hooley, the anchor chain is bar tight and contributes very little. Conversely in flat calm conditions, the yacht may actually be lying to the chain and the anchor doing nothing, but that is a side issue.

The advice was therefore, to put the weight in the anchor and by association likely to be bigger with a larger surface area of the flukes which increases the holding. Or am I wrong
I wouldn't say there is consensus on CF, on this matter. A lot of people equate "bar tight" with "straight line" - this is not the case with all chain rode at an effective scope. It is virtually impossible to completely eliminate the catenary, but at high loads, the "sponginess" of the system is gone.

In this matter, chain weight does contribute to the system, as it takes that much more load to lift the last link from the seabed (the point at which the anchor is pulled up from the horizontal).
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Old 06-01-2024, 05:46   #8
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Re: Anchoring equations

Quote:
Originally Posted by Pete7 View Post
Are you sure? I appreciate you manufacture some very interesting anchors.

However, this has been discussed at length on CF. The snyopsis I can remember is that when it really matters because its blowing a hooley, the anchor chain is bar tight and contributes very little. Conversely in flat calm conditions, the yacht may actually be lying to the chain and the anchor doing nothing, but that is a side issue.

The advice was therefore, to put the weight in the anchor and by association likely to be bigger with a larger surface area of the flukes which increases the holding. Or am I wrong
Though to be fair the dampening effect of the chain catinary doesn't just disappear as the load increases, just happens over a shorter & shorter distance so helps very little to lessen the load if the boat is moving & needs to be not moving. Still exists though. Little bit.
Very dependant on depth as well.
I use "20 + 2" nearly all the time unless a fierce blow is forecast. 20m chain plus 2 x depth of water in metres. Approximates roughly the scope needed so that 200Kg will just lift the last chain link with 10mm chain.
If this equation is right....



think it's hopefully close if not exact.

Y axis scope, x axis water depth, red line > equation, blue line > 20 * 2

https://www.desmos.com/calculator/qwcoew4w82

without even getting into snubbers, whole other can of fairly complex mechanics worms ��
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Old 06-01-2024, 06:05   #9
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Re: Anchoring equations

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Originally Posted by barcoMeCasa View Post
I use "20 + 2" nearly all the time unless a fierce blow is forecast. 20m chain plus 2 x depth of water in metres. Approximates roughly the scope needed so that 200Kg will just lift the last chain link with 10mm chain.
If this equation is right....
Oh that's quite similar to our formula we picked up from this thread:

https://www.cruisersforum.com/forums...ce-249475.html

We have a laminated card in the anchor locker with this image to remind me.

Pete
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Old 06-01-2024, 06:15   #10
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Re: Anchoring equations

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Originally Posted by Lodesman View Post
I wouldn't say there is consensus on CF, on this matter. A lot of people equate "bar tight" with "straight line" - this is not the case with all chain rode at an effective scope. It is virtually impossible to completely eliminate the catenary, but at high loads, the "sponginess" of the system is gone.

In this matter, chain weight does contribute to the system, as it takes that much more load to lift the last link from the seabed (the point at which the anchor is pulled up from the horizontal).
Agreed, I used bar tight rather than straight for the reasons you have pointed out. Also the catenary doesn't really work in very shallow depths. Lifting the last few metres of chain off the bottom takes little effort. Lifting vertically a long length of chain in deep water takes considerably more effort so has a real dampening effect. Maitias did some serious maths on this, I was just happy to benefit from his conclusions.

I did consider dropping from 8mm to 6mm chain. However, quite costly for a small 30-40kg saving in weight at the bow. Decided to leave it as is and spend the money on solar instead.

Pete
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Old 06-01-2024, 06:41   #11
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Re: Anchoring equations

Quote:
Originally Posted by Pete7 View Post
Also the catenary doesn't really work in very shallow depths. Lifting the last few metres of chain off the bottom takes little effort. Lifting vertically a long length of chain in deep water takes considerably more effort so has a real dampening effect.
I don't see it that way, unless of course you use a simple ratio (5:1 for instance) in calculating scope. If there is a sufficient length of chain out, it will still provide a catenary as required. At low load, most of the chain will just sit on the seabed, but then the potential energy in the system is partly provided by lifting the chain off the ground, and partly by the growing catenary.
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Old 06-01-2024, 07:18   #12
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Re: Anchoring equations

Quote:
Originally Posted by Pete7 View Post
Oh that's quite similar to our formula we picked up from this thread:

https://www.cruisersforum.com/forums...ce-249475.html

We have a laminated card in the anchor locker with this image to remind me.

Pete

Nice to see similar conclusions using the same logic & physics.

15m + 2 x depth seems closer to about 150Kg to lift last link with 10mm chain.

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Old 06-01-2024, 11:30   #13
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Re: Anchoring equations

This is a really interesting discussion. Thank you all.

Most of the discussion centers around all chain. What does a combination of say 2/3rds chain 1/3 rope do here? I had multiple occasions last summer where we were anchored in 30-50’ of water with a hard, to me and my crew anyway, wind blowing. I occasionally had 130’ of chain and another 60-70’ of rope out. How does that affect the forces?

I get the feeling that so long as the anchor had properly set and no one else broke free, I was in good shape with the tackle I have.
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Old 06-01-2024, 12:15   #14
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Re: Anchoring equations

I dunno what the diss is on the Bruce Anchor. I have almost exclusively used the Bruce in a variety of bottoms and depth in over 35 years without a hiccup.
My setup has likewise remained the same.
75' chain followed by 5/8" nylon rode, the length of which would vary according to the depth.
All three of my boats have been in the 40' range. Learning to anchor correctly is as important as anchor/rode selection.
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Old 06-01-2024, 12:51   #15
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Re: Anchoring equations

Quote:
Originally Posted by davefromoregon View Post
This is a really interesting discussion. Thank you all.

Most of the discussion centers around all chain. What does a combination of say 2/3rds chain 1/3 rope do here? I had multiple occasions last summer where we were anchored in 30-50’ of water with a hard, to me and my crew anyway, wind blowing. I occasionally had 130’ of chain and another 60-70’ of rope out. How does that affect the forces?

I get the feeling that so long as the anchor had properly set and no one else broke free, I was in good shape with the tackle I have.
If you looked at the 2nd thread mentioned by Pete, you'd have seen Boatman's 3xboat length + depth scope formula. Your chain length is not that far off from that, so that you could simply lay out all the chain and then an amount of rope equal to water depth. That would keep you securely anchored in anything short of storm-force winds, even with the Bruce. There would be little to be gained from putting out more rope than that - more rope than water depth and you add risk of chafe-damage or wrapping around your keel/rudder.
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