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Length of Cable on a Reel

Date: 05/07/2003 at 13:04:24
From: Joe
Subject: Length of cable on a reel

What is the formula used to determine the length of cable on a reel?
Specifically the way to determine the "k" factor if I use the formula

   L = A*(A+B)*C*k

where 

   L = length of wire (in feet)
   A = diameter of wire (in inches)
   B = diameter of drum (in inches)
   C = width of drum (in inches)
   k = some empirical factor.

If I were going to just run a bunch of iterations (can't find k 
value), and if I were using a 30-inch drum 30 inches wide and winding 
2.5 inch cable, would I determine the length of the first layer using 
a diameter of 30 inches or 31.25 inches (half of the wire diameter)?

What do I use for a k value? (My reference book only goes to 1.5" 
diameter.) What do I use for the initial diameter?

If I use a 30"x30" drum and 2.5" cable and run three iterations 
starting with the first "layer" diameter of 30", I can get 329.87' of 
cable in three layers. But if I start with a diameter of 31.25" I can 
get 353.43' of cable in three layers.

Which method is correct?


Date: 05/07/2003 at 15:42:18
From: Doctor Peterson
Subject: Re: Length of cable on a reel

Hi, Joe.

Your formula seems very odd; for example, if you increase the diameter 
of the cable, that should decrease the amount that will fit on the 
reel, yet your formula shows an increase. Did you copy the formula and 
definitions right?

Ah! Looking closer at your discussion, I realize that your "diameter 
of wire" is apparently not the diameter of the cable itself, but the 
outside diameter of the full reel, while "diameter of reel" is the 
size of the core around which it is wound. Your "k" apparently 
incorporates the diameter of the cable, so you have to look it up in 
a table rather than include the cable diameter in the formula.

Here is an answer I have given to a similar question:

   Calculating the Length of String on a Reel
   http://mathforum.org/library/drmath/view/56488.html 

The formula I came up with there is

  L = pi(Do^2 - Di^2)W/(4Dw^2)

where

  L = length of wire (all measurements in the same unit)
  W = width of reel
  Di = diameter of core
  Do = diameter of full reel
  Dw = diameter of wire

As I suggest there, you will probably have to multiply by an 
arbitrary factor that depends on the nature of the wire and the way 
it is wound, but will not vary (much) for different sizes of reel and 
perhaps of wire.

Using your variables, plus "d" for the diameter of the wire itself, 
this would be

  L = length of wire (in feet, requiring division by 12 below)
  A = diameter of full reel (in inches)
  B = diameter of drum [core] (in inches)
  C = width of drum (in inches)

  L = pi/12 (A^2 - B^2)C/(4d^2)

    = pi/(48d^2) (A - B)(A + B)C

Comparing this to your

  L = A*(A + B)*C*k

it seems that you are not subtracting B from A (or else I am still 
misunderstanding what A is), and that k = pi/(48d^2) times whatever 
adjustment factor is necessary to account for space between turns.

Using your example, my formula would give, for a 30-inch by 30-inch 
reel holding three layers of 2.5-inch cable (so that the outer 
diameter A is 30+6*2.5 = 45),

  L = pi/(48*2.5^2) (45-30)(45+30)30 = 0.01047*15*75*30 = 353.4 ft

If that is not reasonably close to what you actually measure, I would 
recommend winding several layers, and then use the ratio of the 
actual length to the calculated length as an additional factor in the 
equation, in order to make it more realistic.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/08/2003 at 10:32:19
From: Joe
Subject: Length of cable on a reel

Thanks for your help. 

To clear up what A is. A = 1/2 thickness of all of the wire from the 
outside of the hub, or the radius of thickness of the wire wound on 
the drum, or the outside radius minus the drum radius.

I got this formula from a pocket reference book; however, the k values 
they use don't go high enough for my application.

I understand your formula and agree with your conclusions. However, I 
have one more question. When you set the volume of the wire equal to 
the square of the wire thickness dw^2, (assuming a square cross 
section) the formula works. But if you use the actual circular cross 
sectional area (pi*r^2)*length of wire and set that volume equation 
equal to the first equation the answer is wrong.

Why do you assume a square cross-section when in actuality it isn't 
square but rather a circle?


Date: 05/08/2003 at 10:58:35
From: Doctor Peterson
Subject: Re: Length of cable on a reel

Hi, Joe.

I assumed a square cross-section in order to include both the wire 
itself and the air around it, supposing that the wires lie in this 
pattern:

         *****       *****       *****
       **     **   **     **   **     **
      *         * *         * *         *
     *           *           *           *
      *         * *         * *         *
       **     **   **     **   **     **
         *****       *****       *****
       **     **   **     **   **     **
      *         * *         * *         *
     *           *           *           *
      *         * *         * *         *
       **     **   **     **   **     **
         *****       *****       *****
       **     **   **     **   **     **
      *         * *         * *         *
     *           *           *           *
      *         * *         * *         *
       **     **   **     **   **     **
         *****       *****       *****

and not, say, this:

         *****       *****       *****
       **     **   **     **   **     **
      *         * *         * *         *
     *           *           *           *
      *         * *         * *         *
       **     *******     *******     **
         ******     *******     ******
            *         * *         *
           *           *           *
            *         * *         *
         ******     *******     ******
       **     *******     *******     **
      *         * *         * *         *
     *           *           *           *
      *         * *         * *         *
       **     **   **     **   **     **
         *****       *****       *****

In the former case, each circle can be thought of as included in a 
square that fits with those around it, filling the whole space. In 
the second picture, we would have to use hexagons, and then we would 
still be missing some air at the top, bottom, and sides.

In any case, apart from edge effects that would make our formula 
inaccurate for small amounts of wire, we are really just multiplying 
the circular cross-section by some factor related to how the wires 
lie, which we really have to adjust at the end anyway, since nothing 
works as well in reality as in the mathematical approximations we're 
using. The square assumption just keeps things simple.

I was wondering if your A might be my (Do - Di); but your calling it 
a "diameter" kept me from considering that it might be, as you now 
say it is, (Do - Di)/2. Then my Do - Di is your 2A and my Do + Di is 
your 2(A+B) [that is, Do + Di = (Do-Di)+2Di = 2A + 2B], so my formula 
using your variables is

  L = length of wire (in feet, requiring division by 12 below)
  A = thickness of wire on reel (in inches)
  B = diameter of drum [core] (in inches)
  C = width of drum (in inches)

  L = pi/12 (Do - Di)(Do + Di)W/(4Dw^2)

    = pi/12 2A 2(A+B) C/(4d^2)

    = pi/(12d^2) A(A+B)C

and your k is pi/(12d^2), times any adjustment factor you need. I'm 
curious to see how their k's agree with this, and whether they adjust 
it using more than just a constant factor.

In any case, it sounds like we've got it figured out.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/08/2003 at 13:05:35
From: Joe
Subject: Thank you (Length of cable on a reel)

Thanks for all your help, I agree with your version of using a square 
cross section to include the air spaces.  If you're still interested, 
the k values from the table are

 .25" = 3.29
  .5" = 0.925
 .75" = 0.428
   1" = 0.239
1.25" = 0.152
 1.5" = 0.107
1.75" = 0.0770

Thanks.


Date: 05/08/2003 at 13:17:37
From: Doctor Peterson
Subject: Re: Thank you (Length of cable on a reel)

Hi, Joe.

Thanks. It looks as if their k is about 0.9 times mine, slightly more 
for the smaller wire sizes. It makes sense that this is less than 1, 
since you would want to allow for non-optimal layering, which would 
fit less cable in the same space. You may want to multiply my formula 
by about 0.9 to be consistent, and then check the results against 
reality.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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