How much space does 6mm chain occupy?

Greenheart

Greenheart

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I definitely once saw a table showing how much chain (of different sizes) would fit into a 'well' of given proportions...but it was probably fifteen or twenty years ago. :(

Granted it's pretty dense stuff, but does anybody know approximately what length of chain fits into a cubic foot of space, (or whatever unit of measurement you prefer)?
 
Dan, your answer is to take a cardboard box down to the local chandlers and speak nicely to them explaining once you have worked it out you will be buying a long length of chain.

Pete
 
Is that the 'sensible' way to look at the question. Is it more a case of how much chain can 'useably' fit into a particular chain locker. Mine forms a precarious mountain with lots of wasted space around the sides. I recall an old discussion where one idea was to fix a small traffic cone under the hole so that the chain 'laid' itself around it. This helped avoid the chain 'toppling' and getting tangled and also used the room more effectively. Even after allowing for the space used by the cone there would still be wasted space in the locker towards the top?
 
dancrane said:
I definitely once saw a table showing how much chain (of different sizes) would fit into a 'well' of given proportions...but it was probably fifteen or twenty years ago. :(

Granted it's pretty dense stuff, but does anybody know approximately what length of chain fits into a cubic foot of space, (or whatever unit of measurement you prefer)?
Click to expand...

My50M of 8mm chain forms a cone about 300mm across and 600mm high is it pi r squared h over 3 ?? !!
 
pcatterall said:
My50M of 8mm chain forms a cone about 300mm across and 600mm high is it pi r squared h over 3 ?? !!
Click to expand...

I'm not sure of the maths...but thanks for the description! It's evidently dense enough for its weight to be more of an issue than the space it occupies. :)
 
nimbusgb said:
50m of 10mm chain fit into a plastic bin 400mm x 320mm x 230mm
Click to expand...

That would make about 12 litres for the 20 m if my maths is correct.
I once asked the same question here, but regarding 8mm chain. Answers varied, but several said approx. 10 litres, which I think is about right.
So my guess for the OP:s 6mm chain would be around 8 litres.
 
Thanks very much for these answers. I was actually mainly wondering whether five meters or ten would suffice (with a line as well) for my dinghy's danforth next year...

...in my head, I was visualising sackloads of chain squirming all over the cockpit floor, but I reckon 10 metres of 6mm will fit snugly inside the hand-bailer!
 
The Following looks good but there are so many other factors, is it stowed wet or dry, flat water or at sea (boat rolling), short link long link studded or not. I would recommend trial and error...

Borrowed from here:
http://www.boatdesign.net/forums/boat-design/rig-anchor-chain-stowage-44690.html

V = 0.85*L*D^2 (where * means multiply & D^2 means D*D)
V = volume in cubic feet of the stowed cable
L = chain cable length in fathoms and
D = chain cable dia in inches

or V = 0.1417*L*D^2
using L in feet.

If you prefer metric use
V = L*D^2/49000
V = volume in cubic metres
L = length in metres
D = dia in millimetres.

But be careful of the basic shape etc. of the chain locker.

If possible it should be circular in plan view. If it must be rectangular make it as square as possible. If it is rectangular with more than 1.5 to 1 proportion of long to short sides the cable may not lie properly.
There must be no side stiffeners or obstructions to the cable. If this is not possible then angle bar stiffeners should be welded mouth-on i.e. to form a V shape on the bulkhead or use half-round bars. If possible fill up the mouth-on angles with wood and plenty pitch. The same with a pipe-pipe but preferably use a solid half-round.
If the chain locker is high do not fit a ladder, cut semi-circular holes in the locker bulkhead to form hand and foot holes. Pitch them about about a foot apart horizontally and about 9 or 10 inches max vertically. Remember you'll be very close to the bulkhead and you won't be able to see too well. Grind the edges smooth to prevent injury to your hands.
The height of the stowed cable (assumed stowing flat or horizontal) should be at least 3 times the diameter of the locker with an allowance of half the diameter to the lip of the chain pipe. If not circular use the diagonal instead of the diameter.
The chain pipe should be vertical and situated at the middle i.e. at the centre of the circle or at the intesection of the diagonals if not circular. The lower edge of the pipe, the lip, should be flared and very strong and it should be on level with or just below any beams or longitudinals under the deck that is the crown of the chain locker.
Remember that the cable is only the length stowed in the locker, theoretically you can reduce the actual length by the distance from the anchor shackle to the lip of of the chain pipe. If there are two anchors you'll only stow half the cable. It is obvious but you would not believe how many people have used the entire length on two for one locker.
You should arrange a drainage space at the base of the locker to allow water to run off.
Arrange the bitter end to have the attachment outside of the chain locker. If you have to release the cable while at sea you don't want to be inside the locker when the cale flies and takes you with it.

The sizes quoted are the minimum and you should increase them if this does not impinge on your design.

I trust I am not teaching granny how to suck eggs but it might help others who are also wondering.
Michael

Just in case;
Circular
Using the minimum height of stowed cable as 3 times the locker diameter then
pi/4*dia^2 is the area and 3*dia is the stowed flat cable height

and pi/4*dia^2*3*dia = Vol = 2.356*dia^3
so that dia = 0.75*Vol^(1/3) or 0.75 times the cube root of Vol

Rectangular
Say the width of the wide side is W and the width of the narrow side is N. Call the proportion W/N f so f =1 for a square and f = 1.5 for the limit as mentioned above.

The floor area = N*fN = fN^2
The diagonal = (N^2+f^2*N^2)^0.5 = N*(f^2+1)^0.5

Vol = The flat, stowed volume = area*3*diag
Vol = fN^2*3*N*(f^2+1)^0.5 = 3*f*N^3*(f^2+1)^0.5

still with me? Good!

so then N = [Vol/(3*f*(f^2+1)^0.5]^(1/3)
simplify this to N = Vol^(1/3)*a coefficient that depends on on f

Prop'n 1.00 to 1; N = 0.6177, W = 0.6177, diag = 0.8736
Prop'n 1.25 to 1; N = 0.5502, W = 0.6878, diag = 0.8808
Prop'n 1.50 to 1; N = 0.4977, W = 0.7466, diag = 0.8972

So there it is!
 
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