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Quirky Math Problem for Engineers / Math Whiz's
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So I've been stuck with doing the inventory clerk's job and have to inventory a pile of hydraulic hose and copper cable on reels. Seems easy enough. Pull the hose / cable off and measure it right? Wrong - 500' spools of 1" 2 braid 3000 PSI hose weighs a lot and there isn't a winder in the building. The obvious answer is to use math. I've checked a couple of calculators on line and the answers don't match with the result. In one case the error was around 30% under the actual. (actual was 235', estimated was 165 or something like that)

The basics -
determination of the amount of material for a single layer of material is pretty straight forward,
info needed is diameter and number of wraps that the hose/cable or whatever makes on a single pass across the reel.

The complications start on the subsequent passes.
the diameter changes & therefore so does the length per layer, the material may or may not nest within itself & therefore require a factor to account for the nesting.

This one is way off...
http://www.ingersollrandproducts.com/...ing/winches/drum.htm

Basic brief of the problem
https://www.quabbin.com/...pacity-reel-or-spool


This one says it's doing one thing, but really it's doing another.
http://www.handymath.com/...len.cgi?submit=Entry
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Re: Quirky Math Problem for Engineers / Math Whiz's [racin_rusty] [ In reply to ]
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If you ask me how to estimate length in the spirit of your links, this is what I would do, using the notation of your middle link:

The gross volume of the spool that is filled with cable is:

(Pi*T/4)*[(B+H)^2-B^2] = Pi*T*H*(2B+H)

Not all of this space is filled; let F be the fraction of volume that is actually occupied. Then, F times the above is the occupied volume. This equals the cable volume, which is:

(Pi*D^2/4)*L

Setting these equal and solving for L gives:

L = F*T*H*(2B+H)/D^2

A best case value for F is hexagonal packing of circles in a plane:

F = Pi/(4*sin60) = .907

This, however is where uncertainty creeps is, especially since packing will be less efficient near the boundaries, and the consequences of this will depend on the relative sizes of the spool and the cable diameter. I think the equations above would give an upper bound on the actual length.

Having said all this, if I were in your shoes, I would go by weight: divide the weight of cable on a spool by its weight per unit length. Obviously you would need to know and subtract the weight of the empty spool.
Last edited by: wesley: Jun 25, 16 1:56
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Re: Quirky Math Problem for Engineers / Math Whiz's [racin_rusty] [ In reply to ]
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Have you thought about doing it by weight. The hose should weigh the same per foot no matter how tightly it is wound, right?
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Re: Quirky Math Problem for Engineers / Math Whiz's [JoeB] [ In reply to ]
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Just take a guess. If no one else is doing inventory, who's to know.
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Re: Quirky Math Problem for Engineers / Math Whiz's [racin_rusty] [ In reply to ]
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Weight is the easiest. Weigh a foot of the product. Weigh an empty spool. Then weigh the whole package. Its basic subtraction and division after that.

Or, if you can, tare the scale with an empty spool on it so that you can remove the spool weight from the equation. Then you only need to divide the gross weight by the weight of one foot.

who's smarter than you're? i'm!
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Re: Quirky Math Problem for Engineers / Math Whiz's [JoeB] [ In reply to ]
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The problem with weight is that we do not have a scale with high enough capacity. I'm guessing, but some of the reels have to be in the 1000 lb range.
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