Mathemtic Formula for Tubes

[BMccarthy81]

I'm looking for a formula that I could use to find out how many .013" diameter wires can fit into a .75" ID pipe. I realize that we cannot divide the area of the pipe by the area of the wire, that give us the number we could fit in with the pieces over lapping. Is there a calculus or a algebraic formula out there that can be applied to this? Any suggestions would be greatful.

 

[rob46]

Per the National Electrical Code the total area of cross sections of conductors shall not exceed 40% of cross section of conduit. The NEC has tables and typical info on wires and conduits to make it easier.

 

[BMccarthy81]

i should have been more specific. we are trying to place as many of the wires into the pipe as possible. they are going to be held in place with wax and then have their ends ground down to a specific length. they are not to be used for anythiny electrical.

 

[SlideRuleEra]

BMccarthy81 - I had to solve this same problem for a project over 30 years ago! In our case, it was much simpler since we had to determine the maximum number of 1/8" OD dia. pipes that would fit inside a 1" ID dia. pipe. I solved it by trial and error, but co-worker discovered that there ARE formulas for such things - unfortunately I don't remember what they are called.

However - this web page "looks" like it may provide enough information to work out to your situation

http://mathworld.wolfram.com/CirclePacking.html

Best Wishes

p.s. When you find out, please let us know.

http://www.SlideRuleEra.net

 

[Lissandre]

You may try to solve the problem in parts.
Named X as the internal diameter of the pipe.
Named Y as the external diameter of the wire.
To form the first layer of threads leaned in the internal walls of the tube, you will need Z1 numbers of threads.
Z1= integer {[(X - Y/2) x 3.1416]/Y} = 179
You will now be able to imagine that possesses a new diameter tube X1 with value of X less 2 times Y.
Consider the integer number of the division of X for Y (call this as result of n). In this case is 28.
Repeat the procedure n times to obtaining the Z2,Z3... to Z28.
The sum of Z1 to Z28 plus 1 (the center) will be the wanted result.
Z = integration of (Z1 to Z28) + 1
Zn = integer {[(Xn - Y/2) x 3.1416]/Y} where Xn= X - n times Y.

 

[Lissandre]

Sorry !
The last correct sentece is:
Zn = integer {[(Xn - Y/2) x 3.1416]/Y} where Xn= X(n-1) - n times Y.

 

[Lissandre]

Layer D D-13 +(D-13)*Pi/13 Wires No.
1 750 737 178.10414 178
2 724 711 171.8209546 171
3 698 685 165.5377692 165
4 672 659 159.2545838 159
5 646 633 152.9713984 152
6 620 607 146.688213 146
7 594 581 140.4050276 140
8 568 555 134.1218422 134
9 542 529 127.8386568 127
10 516 503 121.5554714 121
11 490 477 115.272286 115
12 464 451 108.9891006 108
13 438 425 102.7059152 102
14 412 399 96.42272979 96
15 386 373 90.13954439 90
16 360 347 83.85635899 83
17 334 321 77.57317359 77
18 308 295 71.28998819 71
19 282 269 65.00680279 65
20 256 243 58.72361739 58
21 230 217 52.44043199 52
22 204 191 46.15724659 46
23 178 165 39.87406119 39
24 152 139 33.59087579 33
25 126 113 27.30769039 27
26 100 87 21.02450499 21
27 74 61 14.74131959 14
28 48 35 8.458134192 8

total 2598 wires of 0.013" in a 0.75" pipe.

 

[BMccarthy81]

Thank you all for getting back to me. I found out that the "circle packing" or "sphere packing" method is the best way to find out how many actual pieces of an OD wire into a specific ID of a pipe. The most efficient way to pack cirlces is a hexagonal pattern, six cirlces around on circle. The french mathematician found out this from studying bee hives, and investigating the shape of the honey-combs. His theorm was proven in the 1950's. It turns out that the cirlces use up about 90% of the actual area.
Wehave records of packing wire into tubes, the amount and their diameter. From the field tested method we found that we were using 85-87% of the pipes area. We were able to figure out a ball-park amount, and quoted the needed job.