Mariner69
Well-Known Member
My apologies, it is the number of shackles to be veered.
Each shackle is 15 fathoms long: 90 feet traditionally. In metric they are 27.5 metres long.
So to anchor in 9 metres we use 4.5 shackles, 123.75 metres of chain.
Anchor wally
- Thread starter halb
- Start date
My apologies, it is the number of shackles to be veered.
Each shackle is 15 fathoms long: 90 feet traditionally. In metric they are 27.5 metres long.
So to anchor in 9 metres we use 4.5 shackles, 123.75 metres of chain.
thinwater
Well-Known Member
The ambiguity in this statement is that any anchor that is well buried takes some of the rode into the bed with it, and the angle of that cable forms a sort of reverse catenary. Thus, so long as the angle of the cable is less than the angle of the chain pulled into the seabed, the anchor feels no change. This angle is VERY dependent on anchor design and bottom type.
Of course, if your anchor does not bury well, then the original statement is more plain; the chain needs to stay on the bottom, which in a blow in shallow waters require substantial scope.
(Several posters have use ~ 250 kg as a reasonable force for calculations. That's really not much of a blow--only about 25 knots assuming a 35-40' boat and a short snubber. 50-60 knots is a blow, at which time the chain is coming off the bottom with any practical scope. No arguing, just sayin'.)
Lucky Duck
Well-Known Member
A ratio of almost 14:1???
RAP77
Well-Known Member
I use 3 x depth plus 10 meters. his works for the sort of depths we anchor in.
If it is forecast to be windy I will veer as much as I can
If it is forecast to be windy I will veer as much as I can
skipmac
Well-Known Member
Why make it complicated? Why on earth use all these formulas with square roots, shackles, fathoms and phases of the moon?
I use a minimum scope of 5:1 with chain. S0, if the water is 5 meters deep I add about one meter for bow roller to the water (chose the appropriate addition to match your boat) so consider the depth 6 meters. Then do the very, very simple math of 5 (scope) times 6 (depth) to get 30 meters of chain. In a blow I prefer 7:1 scope so would multiply 7 times 6 to 42 meters of chain.
If you prefer a different scope ratio then use a different multiplier.
I use a minimum scope of 5:1 with chain. S0, if the water is 5 meters deep I add about one meter for bow roller to the water (chose the appropriate addition to match your boat) so consider the depth 6 meters. Then do the very, very simple math of 5 (scope) times 6 (depth) to get 30 meters of chain. In a blow I prefer 7:1 scope so would multiply 7 times 6 to 42 meters of chain.
If you prefer a different scope ratio then use a different multiplier.
GHA
Well-Known Member
Because it gives you an idea of what's actually going on below the water. 2 X Depth plus 23m is close to the catinery/scope graph for10mm chain & 230kg.
In your example, 6m X 2 +23 gives 35m which, for 10mm chain, will just lift the last link from the seabed if you have a horizontal force of 230kg.
You don't have to go that route, but it is closer to how the chain behaves in the real world than just scope ignoring depth.
pmagowan
Well-Known Member
I use a wetted finger and a suck through the teeth! Never failed. My anchor rode socpe depends on too many variables to be determined by a formula. How do you calculate for numbers of rums consumed or if you had port and cheese after supper? Then there is the question over how often the ship will be abandoned in favour of the pub. This is before we get to such banalities as the phase of the moon, forecast wind and alignment of the planets. Such questions as, 'is the shore all jagged?' or 'will I need a pee in the middle of the night?' determine how much extra scope I might dump out. In general once my complex calculations are complete and I have the exact, perfect amount of scope out, I just toss a bit more in for luck. Then I wind a bit more in when we are too close to that boat behind, and throw it back out again because I am being silly we are miles away! I put a hand on the chain to feel the fish talking and then announce that we are well stuck for the night, wheres that rum gone!!
vyv_cox
Well-Known Member
I take it we are not talking about yachts here?
I carry 60 metres of chain and quite often anchor in 10 metres depth, although I prefer less. 4:1 scope is quite adequate for a new generation anchor, as I have found many times when diving on it. More than 4:1 is nice but in crowded anchorages often impossible.
RichardS
N/A
I've heard that repeated as a literal statement several times on here over the years.
That Haydn chappie has a lot to answer for.
Richard
JumbleDuck
Well-Known Member
Square root x 12 is a terrible idea because it's not length-invariant: in 36 feet / 18 metres / 6 fathoms one would put out 72 feet / 50 metres / 30 fathoms of chain. I like the 144 <anything> example, though!
oldbilbo
...
I've heard that repeated as a literal statement several times on here over the years.
That Haydn chappie has a lot to answer for.
Then there's this....
I believe I know from whence those words come, but I'l leave others to find them as they will.
PaulRainbow
Well-Known Member
I can't remember any of the theory's, so i just bung a load over the bow. If i can motor astern i bung some more over. Sometimes 2:1 is enough, other times i need 10:1. I've anchored in 2m and i've anchored in over 100m 30odd miles offshore, never dragged yet (once i've decided it's set).
JumbleDuck
Well-Known Member
I don't understand why that would happen. Can you explain?
vyv_cox
Well-Known Member
Not sure about the 250 kg = 25 knots. I know that I can pull a 30 ft boat forward by the anchor chain in a force 6 because I used to do it. Even in the strength of my youth I doubt that I could pull 250 kg, or even half that. A user of Anchorwatch wrote an article in YM a couple of years ago using data gathered over a whole season and his values were nowhere near that.
RichardS
N/A
Lucky Duck
Well-Known Member
Maybe spinnaker sheets, halyards, etc had been pressed into service 
oldbilbo
...
In certain circumstances, 'deep water kedging' has its advantages - if one is prepared for it - as Adlard Coles detailed in one of his race-winning accounts 'back in the day'. He carried a roll of piano wire. I/we carried a spool of that Kevlar-cored lightweight cordage used to pull fibre-optic cables through u/ground piping.... several hundred feet of it.... during the 2003 Fastnet Race.
We used it twice to good effect, once in 40-odd metres SE of The Shambles Bank - where we had the satisfaction of seeing a well-sailed 40-foot hot racing trimaran sail past us backwards at 3 knots OTG, and disappear into the murk astern - and also in 70-odd metres south of Start Point.
We gained more than a few places with that ancient ploy, won our Class ( Double Points ) - and the Class Series.
Many thanks, old Adlard!
sarabande
Well-Known Member
Haydn is a sort of Delphic oracle. I think I know what he means, viz:
An anchor directly connected to the boat, without any chain on the seabed, is almost guaranteed to move; it is being pulled upwards.. Enable a good length of chain to lie on the seabed and make the anchor pull horizontally, keeping enough slack in the chain so that the cyclical snubbing forces natural to a boat's motion do not allow the chain to become taut, and the anchor will continue to hold as much as the design and seabed allow.
It's the horizontal pull, and avoiding a catenary direct between the boat and the anchor that are important.
An anchor directly connected to the boat, without any chain on the seabed, is almost guaranteed to move; it is being pulled upwards.. Enable a good length of chain to lie on the seabed and make the anchor pull horizontally, keeping enough slack in the chain so that the cyclical snubbing forces natural to a boat's motion do not allow the chain to become taut, and the anchor will continue to hold as much as the design and seabed allow.
It's the horizontal pull, and avoiding a catenary direct between the boat and the anchor that are important.
vyv_cox
Well-Known Member
If that was true it would not be possible for an anchor with a rope rode, or short chain and rope, to function. Last week I anchored on my Fortress with a scope of only 3:1, 15 metres of rode in 5 metres depth, an angle of 17 degrees. It held well in 35 knots of wind.
I have dived on my anchors many times in fresh winds, watched all the chain lift off the bottom and the shank move vertically. Both Delta and Rocna coped with this perfectly well, not dragging after days of winds in the 30 knot area.
skipmac
Well-Known Member
Using a formula that adds a constant to the calculation is going to give odd results at the extremes and that formula doesn't deal well with what happens to the catenary in more extreme conditions.
For example, if one anchors in water 2 M deep then add 1 M for the distance bow roller to the water by my formula one would need only 15 M of chain for a 5/1 scope. By your formula if adding the 1 M for the bow/water distance (which should be done to get the correct effective water depth) one would put out 29 M of chain, a scope of almost 10/1 which is I think is excessive and in a crowded harbor impractical.
If anchoring in 30 M by my method one would put out 155 M of chain, by your method 87. 87 M would be a scope of 2.6/1. That would be sufficient in mild to moderate conditions but if it comes a blow the catenary will decrease dramatically leaving the effective angle of pull on the anchor too acute.
Since my past cruising was primarily in the Bahamas and Caribbean where tides are minimal and depths usually moderate I generally adhered fairly close to the X5 formula but like any formula or rule of thumb it should be used with consideration of the specific situation and conditions. For example, if anchoring in deeper waters I would tend to a lower multiplier unless conditions became extreme and of course in any depth with strong winds would add a bit more scope.