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Deleted User YDKXO
Guest
Absolutely right, V. It cannot dig through the weed like the Delta or the Fisherman's anchor can. Its best suited to mud or clay bottoms
Anchors, cables and catenarys.... or not?
- Thread starter Bajansailor
- Start date
Absolutely right, V. It cannot dig through the weed like the Delta or the Fisherman's anchor can. Its best suited to mud or clay bottoms
benjenbav
Well-Known Member
If the effect of the anchor is excluded it should be simple maths* to work out how much weight is required to be heaved onto the seabed in order to hold a boat of a given mass in a fixed place in predictable wind and water conditions. The friction should be straightforward to include if the digging in effect of the hook is not a factor.
I suspect that the mech eng wizards amongst us will quickly demonstrate that, without a mass of chain which just simply could not be carried, the boat would start to drift away in wind conditions in which those of us with anchors on the end of our chains would expect it to stay put.
* Obviously not for a poncy arts graduate like me. But I might be able to write a little poem about it afterwards.
I suspect that the mech eng wizards amongst us will quickly demonstrate that, without a mass of chain which just simply could not be carried, the boat would start to drift away in wind conditions in which those of us with anchors on the end of our chains would expect it to stay put.
* Obviously not for a poncy arts graduate like me. But I might be able to write a little poem about it afterwards.
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jfm
Well-Known Member
If the effect of the anchor is excluded it should be simple maths* to work out how much weight is required to be heaved onto the seabed in order to hold a boat of a given mass in a fixed place in predictable wind and water conditions. The friction should be straightforward to include if the digging in effect of the hook is not a factor.
I suspect that the mech eng wizards amongst us will quickly demonstrate that, without a mass of chain which just simply could not be carried, the boat would start to drift away in wind conditions in which those of us with anchors on the end of our chains would expect it to stay put.
* Obviously not for a poncy arts graduate like me. But I might be able to write a little poem about it afterwards.
The maths is hard. Mech eng graduates can do all sorts of calcs for you either with friction data (empirical) or by assuming coulomb friction (which is a defined model, and quite representative of smooth surfaces on each other in everyday life). But I doubt anyone has data for chain dragged along sandy seabed.
However, I fully agree your hunch that with data it would be shown that the chain weight needed couldn't practicably be carried. Also the chain length would foul the whole anchorage or in some enclosed anchorages you'd run out out of space
Looking forward to the sonnet!
Hurricane
Well-Known Member
I think he needs some suggestions of technical words to include in the poem.
jfm
Well-Known Member
Switched to here from the other fred that was overdrifting (geddit?)
Yup but those trainers aren't engineers/physicists/whatever. As discussed on here, in flying school all pilots are taught that planes fly because of Bernouille, with no mention of Newton, which is also completely incorrect
MapisM
Well-Known Member
LOL, I suspect that both the "chain holds the boat" and "Bernoulli flies the planes" are simply quick and easy ways to make helmsmen aware that they should use some chain when anchoring, and pilots aware that they should try to avoid stalling...
benjenbav
Well-Known Member
Too busy and bereft of talent for a sonnet so you'll have to make do with a limerick:
There once was a man from Bahrain,
Who got in a pickle with chain,
Three times is the scope,
But more with a rope,
Was the thought going round in his brain.
or, given that clean ones are less fun:
The popular crooner, Paul Anka,
Bought a boat, some chain and an anchor,
It drifted; then a bang,
And out a voice rang,
Be off with you right now, you ******.
There once was a man from Bahrain,
Who got in a pickle with chain,
Three times is the scope,
But more with a rope,
Was the thought going round in his brain.
or, given that clean ones are less fun:
The popular crooner, Paul Anka,
Bought a boat, some chain and an anchor,
It drifted; then a bang,
And out a voice rang,
Be off with you right now, you ******.
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Deleted User YDKXO
Guest
If the effect of the anchor is excluded it should be simple maths* to work out how much weight is required to be heaved onto the seabed in order to hold a boat of a given mass in a fixed place in predictable wind and water conditions. The friction should be straightforward to include if the digging in effect of the hook is not a factor.
I suspect that the mech eng wizards amongst us will quickly demonstrate that, without a mass of chain which just simply could not be carried, the boat would start to drift away in wind conditions in which those of us with anchors on the end of our chains would expect it to stay put.
* Obviously not for a poncy arts graduate like me. But I might be able to write a little poem about it afterwards.
If it was easy somebody would have done it by now. It would be relatively simple to calculate the wind loading on a particular boat and thus the load applied to the anchor chain but I don't know how you'd start to calculate the friction between the seabed and the chain because the seabed material is so variable and the chain itself is not uniform. However, I shall buy a couple of my civil engineering chums a beer or two to see if they can help
AndieMac
Well-Known Member
I think you made the right decision Mike, we had a disappointing result from our experience in both setting/holding and the galvanizing was very poor.
Thankfully though their returns policy was sound!
guernseyman
Well-Known Member
Not too difficult to find some data:
Line type....Seabed Type....Starting Friction....Sliding Friction
.....................................Coefficient............Coefficient
Chain.........Sand...............0.98....................0.74
................Mud with sand...0.92...................0.69
................Mud/clay..........0.90...................0.56
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Deleted User YDKXO
Guest
Thats interesting. I searched the net and couldn't find anything like that. Have you got the link?
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Deleted User YDKXO
Guest
That Craig guy is very persuasive!
kid's inheritance
Well-Known Member
I wonder if anyone ever played with increasing the friction of the chain. If it was (for example) spiked , the possibility of avoiding an anchor (with the benefits of no cable grabbing, deformation etc) would be increased . Bugger on the windlass though 
DAKA
Well-Known Member
I dont think your pleasure boat anchors in sandy bays can be related to ships anchors, not because of chain size but more to do with shallow water and lack of growth on the seabed.
A chain sat on sand will not have much to grab on to and each chain link can slide as easily as the next.
out in deeper water the sea bed will have seaweed which will tangle between each link.
imagine the chain as a long narrow bit of velcro.
Your 1/2 inch dinghy chain will hold nicely in blanket weed.
A chain sat on sand will not have much to grab on to and each chain link can slide as easily as the next.
out in deeper water the sea bed will have seaweed which will tangle between each link.
imagine the chain as a long narrow bit of velcro.
Your 1/2 inch dinghy chain will hold nicely in blanket weed.
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Deleted User YDKXO
Guest
Actually I think quite the opposite as IMHO the chain will slide over the weed more easily but hey, what do I know
DAKA
Well-Known Member
Actually I think quite the opposite as IMHO the chain will slide over the weed more easily but hey, what do I know
I dont know enough to argue Mike and agree I should have qualified my comment similarly.
Just thinking out loud now but I think you are in dry land mode thinking sand vers wet slippery seaweed.
the seabed is saturated sand/mud and very slippery as soon as it is agitated (liquefaction ), each bit of weed is well anchored to the bed, add together and you have a conjoined strength velcro effect ( I think ) deduced from the **** I haul up after anchoring in deep water compared to shallow sandy bay.
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jfm
Well-Known Member
1. Well if we use those figures (whose relevance is questionable - I think they are coulomb figures, but never mind about that) then the maths is easy. I'll pick the highest coeff so as to work in favour of the "the chain not the anchor holds the boat" faction which I'll call "BICH" (= boat is chain held)
2. The HIGHEST coeff there is 0.98. The weight of 10mm anchor chain is quoted here as 1.85kg/metre
3. In a typical 6m deep nice anchorage let's say you go wild and dump 40m of chain out, of which 30m lies flat on the seabed. The holding force is simple 30 x 1.85 x coefficient of 0.98 = 55kgf.
4. What is 55kgf? Well a man can reasonably lift 55kg, with a bit of effort. Two an a half of those big metal 20litre jerry cans filled with water. The little bowthruster fitted on the new Princess v39 is the Sleipner Sidepower SE60, rated at 60kg force. So the chain has roughly the same pull as these things. Clearly that will hold the boat in modest wind, but not in 20kts of wind, say. Remember of course that as soon as there is any snatch load or sailing sideways about the anchor, some or all of the chain moves and the 0.74 max coeff will apply not the 0.98, so the holding force of the chain reduces
5. As for tide, a 40 foot mobo (as a guess; I can do the maths if you want) probably does 1 knot with 55kgf and no wind. Certainly not 2-3 knots (it's a cube relationship between kgf and speed at these speeds)
6. Remeebr, all this is with loads of chain out - 40m in 6m of water, which is nearly 8x scope, waaaaay more than RYA teaches
7. All of which is what we generally agreed upon above, I think, ie that the BICHes are correct in light winds (or <1knot tide), but once you get above (say) 15kts of wind, or >1knot tide, or in a lower coeff ground (the lowest being 0.56) the anchor starts to contribute and in say 25kts wind or 3 knots of tide is doing most of the work and the BICHes are wrong.
8. For those who can still stay awake with the maths, let's calc the catenary loads. Imagine a 40 footer boat in 6m of water, with a bow roller 2m above the water surface, with the "RYA taught 3x method" ie 20m of chain paid out and the wind is blowing such that 5m of that chain is flat on the seabed, after which the upper 15m of chain rises in a catenary to the bow roller which is 8m up in the sky (so the catenary is less steep than 45 deg - the "triangle" of the catenary is 8m high and 12m long, if you get my drift). The chain is 8mm, so 1.35kg/metre. Reasonably normal mediumish winds scenario I think?
9. The horizontal pull that is being resisted by the anchor plus the 5m of chain flat on the seabed is
Catenary Force in kg (from here) = Specific chain weight x (catenary chain length squared)/(2 x Depth)ie 1.35 x 15x15/2x8 = 19kg
10. To get a 19kg pull from the chain alone (ie ignore the anchor) needs a coeff of friction of 19/(1.35x5) which is about 3, ie way higher than the data above, so the anchor must be doing most of the work. Now if you put out 35m of chain, ie 20m chain flat on the seabed and a 4.5x scope, then the coeff to get 19kg of force is 0.7 which is do-able using the above data. This however is light winds (19kg pull) and virtually no tide, and you have paid out 35m of chain in a 6m ancorage, and the chain is now just holding you (ignoring snatch loads) and the anchor is doing nothing.
11. All a bit crazy maths, sorry, and i dont suggest anyone does these clacs when parking their boat but the results are plausible imho. An interesting thing from these maths is that those (many) people who say the RYA 3x scope is too little are right. If you go to say 5x scope you are puting out 6% more chain length but you perhaps triple the length of chain lying flat on the seabed and that gives you useful extra holding force
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BartW
Well-Known Member
are you sure about this formula ?
shouldn't you reduce the chain weight by the Archimedes force (upwards) from the water on the chain ?
now even if I'm right, it will only make your case stronger (that its mainly the anker that holds the boat)
benjenbav
Well-Known Member
1. Well if we use those figures (whose relevance is questionable - I think they are coulomb figures, but never mind about that) then the maths is easy. I'll pick the highest coeff so as to work in favour of the "the chain not the anchor holds the boat" faction which I'll call "BICH" (= boat is chain held)
2. The HIGHEST coeff there is 0.98. The weight of 10mm anchor chain is quoted here as 1.85kg/metre
3. In a typical 6m deep nice anchorage let's say you go wild and dump 40m of chain out, of which 30m lies flat on the seabed. The holding force is simple 30 x 1.85 x coefficient of 0.98 = 55kgf.
4. What is 55kgf? Well a man can reasonably lift 55kg, with a bit of effort. Two an a half of those big metal 20litre jerry cans filled with water. The little bowthruster fitted on the new Princess v39 is the Sleipner Sidepower SE60, rated at 60kg force. So the chain has roughly the same pull as these things. Clearly that will hold the boat in modest wind, but not in 20kts of wind, say. Remember of course that as soon as there is any snatch load or sailing sideways about the anchor, some or all of the chain moves and the 0.74 max coeff will apply not the 0.98, so the holding force of the chain reduces
5. As for tide, a 40 foot mobo (as a guess; I can do the maths if you want) probably does 1 knot with 55kgf and no wind. Certainly not 2-3 knots (it's a cube relationship between kgf and speed at these speeds)
6. Remeebr, all this is with loads of chain out - 40m in 6m of water, which is nearly 8x scope, waaaaay more than RYA teaches
7. All of which is what we generally agreed upon above, I think, ie that the BICHes are correct in light winds (or <1knot tide), but once you get above (say) 15kts of wind, or >1knot tide, or in a lower coeff ground (the lowest being 0.56) the anchor starts to contribute and in say 25kts wind or 3 knots of tide is doing most of the work and the BICHes are wrong.
8. For those who can still stay awake with the maths, let's calc the catenary loads. Imagine a 40 footer boat in 6m of water, with a bow roller 2m above the water surface, with the "RYA taught 3x method" ie 20m of chain paid out and the wind is blowing such that 5m of that chain is flat on the seabed, after which the upper 15m of chain rises in a catenary to the bow roller which is 8m up in the sky (so the catenary is less steep than 45 deg - the "triangle" of the catenary is 8m high and 12m long, if you get my drift). The chain is 8mm, so 1.35kg/metre. Reasonably normal mediumish winds scenario I think?
9. The horizontal pull that is being resisted by the anchor plus the 5m of chain flat on the seabed is
Catenary Force in kg (from here) = Specific chain weight x (catenary chain length squared)/(2 x Depth)ie 1.35 x 15x15/2x8 = 19kg
10. To get a 19kg pull from the chain alone (ie ignore the anchor) needs a coeff of friction of 19/(1.35x5) which is about 3, ie way higher than the data above, so the anchor must be doing most of the work. Now if you put out 35m of chain, ie 20m chain flat on the seabed and a 4.5x scope, then the coeff to get 19kg of force is 0.7 which is do-able using the above data. This however is light winds (19kg pull) and virtually no tide, and you have paid out 35m of chain in a 6m ancorage, and the chain is now just holding you (ignoring snatch loads) and the anchor is doing nothing.
11. All a bit crazy maths, sorry, and i dont suggest anyone does these clacs when parking their boat but the results are plausible imho. An interesting thing from these maths is that those (many) people who say the RYA 3x scope is too little are right. If you go to say 5x scope you are puting out 6% more chain length but you perhaps triple the length of chain lying flat on the seabed and that gives you useful extra holding force
Wow. Not sure I followed it all but yo' bich sure has a nice feel to her.
I guess if you put one of those Waterbuoy keyring things on the end of the chain the whole thing would float?
jfm
Well-Known Member
are you sure about this formula ?
shouldn't you reduce the chain weight by the Archimedes force (upwards) from the water on the chain ?
now even if I'm right, it will only make your case stronger (that its mainly the anker that holds the boat)
Yeah I thought of that but I left that out Bart becuase it's only 1/8th. I mean the specific gravity of steel is about 8. So it doesn't really change the result and by ignoring it I am helping the BICHes