I am currently at college study engineeering.As part of the course I am designing water system for cooling. The question that I am having trouble with is this
If i had a rectangle of 40 meters x 20.6 meters how many whole circles of 0.6 meters will fit into said rectangle. I am looking for a formula that will give me the answer intead of measuring. It saving time and using it for different sizes.
Not knowing whether you had any success with your problem, I thought I would take a look at it .
Since you did not clarify whether the .6 m was a radius or a diameter, I assumed it as the diameter of the pipe. If it is really the radius, you can work out the answer yourself.
The first packing assumes the pipes are stacked in rows and columns. I view the rectangle with the 40m horizontal. From 40m/.6m, we get 66.666...horizontal pipes, or 66 pipes with .666...(.6) = .40m by 20.4m vertical space empty.
From 20.6m/.6m , we get 34.333...vertical pipes, or 34 pipes with .666...(.6) = .20m by 40m horizontal empty space.
This results in 66x34 = 2244 pipes.
The next obvious approach would be to remove one pipe from the 2nd row and let the remaining 39 pipes roll down such that each pipe touches 2 pipes in the row below it, their centers forming an equilateral triangle, often referred to as hexagonal packing. This pattern repeats itslef to the top of the rectangle.
The vertical distance between pipe center becomes cos[arcsin(.3/.6)].6 = .5196m. The number of pipe rows, N, now derives from .5196(N - 1) + .6 = 20.6 from which N = 39.5. Therefore, the hexagonal packing allows us to increase the nuber of rows to 39, with 20 rows of 66 and 19 rows of 65 for a grand total of 2,555 pipes. However, we still have empty space of .40m by 20.6 and .3m by 40m or 16 m^2.
Our last possibility is to allow the first row to expand to the full 40m length with a gap between successive pairs and allow the row above to drop some distance, possible allowing us to fill the top row and have 40 rows.
Having .4m to exspand into, we would have .4/39 = .01025m between pipes. The half angle between adjacent vertical pipes now becomes arcsin(.3 + .01025/2) = 30.567º the cosine of which is .861035. The number of vertical rows now derives from .6cos(30.567)N .6cos(30.567) + .6 = 20.6 yyielding N = 39.71 rows. It was rather obvious that gaps of .01025 would not gain us much.
So we end up with 20(66) + 19(65) = 2,555 pipes and 17m^2 empty. Increasing the height to 20.774m would enable you to increase the number of pipes to 2,621.
Generally speaking, a square box would require the least enclosing material.
As for a formula for identifying the number of pipes that can be enclosed within a given rectangle or square, the following would be a possible start. The number of horizontal pipes derives from Nh = LhD, or the box length divided by the pipe diameter. The number of vertical rows derives from D(cos30º)N - D(cos30º) + .6 = H where Nv = (H - D + .866D)/(.866D). Checking our earlier results, For a rectangular container measuring 40m by 20.6 m, the maximum number of pipes enclosable derives from
.................N = LhD + (H - D + .866D)/(.866D)
.................N = 40/.6 + (H - D + .866D)/(.866D)
.................N = 66.666... + (20.6 - .6 + .866(.6))/(.866(.6)
.................N = 66.666... + 39.49 = 66 pipes in 20 rows and 19 pipes in 65 rows or 2,555 pipes. If possible, it would probably be logical to increase the box height to 20.774m and gain an additional 66 pipes.
