



 
Hi John.john2
Think about the crosssectional areas. Consider hexagonal circle packing (http://mathworld.wolfram.com/CirclePacking.html). If you pack them such that the centers of three adjacent circles form an equilateral triangle, then within the equilateral triangle there is some space in the middle. The trigonometry works out this way: An equilateral triangle has angles of 60 degrees, one sixth of a circle. So the area of the portion of the triangle which is cable is three times 1/6 of a cable crosssectional area. That totals (1/2) π × r^{2}. The area of the whole equilateral triangle (whose side length is 2r) is r^{2} × √3. Thus the ratio of the cable area to the total triangular area is π / (2 √3), which is about 90.7%. The reason we calculated this is because the whole surface of the pipe can be tiled using such equilateral triangles (there is a margin of error along the edge, but with a large number of triangles fitting into the middle, it becomes fairly insignificant). Thus, the percentage of area in the triangle is the same as the percentage of area in the whole pipe. The area of the pipe is π 19.05^{2} which is about 1140.1 sq mm. 90.7% is cable, so that is 1033.5 sq mm. The crosssectional area of each cable is π 0.28^{2} mm (which makes 0.2463 sq mm). So we divide the area per cable into the area covered by cable. That means 1033.5 / 0.2463 = 4196 cables. Hope this helps.  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 