Drum/Spool Capacity

There is alot of info about this but I am trying to find out the size drum I need for a specific length and thickness.

My math is not what it used to be. Can anyone tell me if I have rearranged the equation right or wrong? I am trying to rearrange for A but the results are not giving me what I expected. (its driving me nuts!)

If anyone can prove me right or wrong, I will post a working excel spreadsheet here for all too use.

For anyone wondering. It came about because I was talking with some non nautical friends about transocean cable laying ships and how big their spools must be, and how often they change them. It tweaked my curiosity.

 

INGERSOLL-RAND Drum Capacity Estimator http://www.learneasy.info/MDME/MEMmods/MEM30009A/lifting_systems/Ingersoll_Drum_Calculator.htm

 

It depends how you stack the windings. Are they one on top or do they stagger like the calculator?

 

Those calculations are for wire rope drums, not cables and especially not transoceanic ones. Transoceanic cable tanks (not spools) are on a whole other level. Think of a cargo ship where all the cargo holds are circular tanks the width of the ship and the entire depth of the hold. Cables cannot be stored under tension or twist. There are absolute minimum bend radius, minimum bend radius at load, maximum install tension, lifetime to installed tension, not-to-exceed tensions, etc...lots of stuff to consider.

Edit,
Here is a slide show that has a brief discussion and a good cutaway of a cable ship and its tanks.
http://dls.virginia.gov/commission/materials/subseacables.pdf

 

If you want a simplified answer, then you calculate the internal volume of the cylindrical vessel. Since any cable or rope probably has a minimum bend radius, you can treat the volume like a hollow cylinder. Thus: Vol = π x(D^2 -d^2)/4 x L where L is the axial length. All units the same m for example.

Once you have the volume, you imagine you extrude that volume through a square opening equal to the diameter of the cable. Vol= A^2 x L2 where A is the cable diameter in same as previous units and L2 is the length of the cable. So solving for L2 = Vol/A^2. This approximation assumes that the cable nests in radial and axial direction in it's diameter. The truth is that each row has a partial turn at the begining and end so in reality slightly less material will fit. If the material will not nest as tight you can increase the dimension A to compensate.

 

Thanks for all your input. The link you have sent jehardiman is particularity interesting.

I don't think I laid out the question very clearly so

"What is the equation to work out, for a given length of cable/rope/spaghetti- how big a spool is needed when the material can stack horizontally and vertically?"

The question framed in a different light might be; How do we work out the size of spool needed to hold the 50m (est) of cable required to pull this welsh lifeboat back up the ramp?

For the record I will not be using the info for any practical purpose apart from passing it on in the pub as trivia on a Friday night!

 

This is a link to the US Navy's handbook on "the design and engineering of wire rope systems"

Interesting bedtime reading!

https://apps.dtic.mil/dtic/tr/fulltext/u2/a955305.pdf