Submitted by Editor
Hexagonal Packing
Efficiency in packin
Objective
To investigate the efficiency of packing of objects of different shapes in a cuboid box.
Efficiency is the percentage of box space occupied by the objects.
Description
1. Took a certain number of cylindrical tins and packed them in a cuboid container.
(a) For illustration I took 81 tins.
(b) Second time I took 64 tins.
(c) Third time I took 49 tins.
2. The cylindrical tins can be placed in two different ways These are
(a) Square packing
(b) Hexagonal packing
3. I wished to study which packing out of two is more efficient.
4. To understand the difference between the two packing I have drawn figures on the left side pages.
Example 1 (81 tins) Calculation
Case 1
Square packing
Each base circle is circumscribed by a square. Area of one circle = n R2
Area of square = 4 R2
Area of circle / area of square = nR2 / 4 R2
This ratio will be evidently the same as the cross section of all the tins to the total base area.
Percentage efficiency = n / 4 x 100
= 78.5 %
Therefore the efficiency in case of square packing is 78.5%
Case 2
Hexagonal packing
Here we determine the sides of the base of the container in terms of the radius of the cylindrical tin.
One side of the rectangular base i.e. BC = 18 x R.
To determine the other side, AB = 2 x R + 9 x h, where h is the altitude of the equilateral triangle formed by joining the centres of three adjacent circles.
h = 2R sin 60°
AB = 2R + 18 R sin 60°
sin 60° = V3 / 2
Now AB = 2R + 18R x V3 / 2 = (2 + 9V3) R
Area of ABCD = 18R x (2 + 9V3) R = 18R2 (2 + 9V3)
Percentage efficiency = 81nR2 x100 / 18R2 (2 + 9V3)
= 80.3 %
Example 2 (64 tins) Calculation
Case 1
Square packing
Each base circle is circumscribed by a square. Area of one circle = n R2
Area of square = 4 R2
Area of circle / area of square = nR2 / 4 R2
This ratio will be evidently the same as the cross section of all the tins to the total base area.
Percentage efficiency = n / 4 x 100
= 78.5 %
Therefore the efficiency in case of square packing is 78.5%
Case 2
Hexagonal packing
Here we determine the sides of the base of the container in terms of the radius of the cylindrical tin.
One side of the rectangular base i.e. BC = 16 x R.
To determine the other side, AB = 2 x R + 8 x h, where h is the altitude of the equilateral triangle formed by joining the centres of three adjacent circles.
h = 2R sin 60°
AB = 2R + 16 R sin 60°
but sin 60° = V3 / 2
Now AB = 2R + 16R x V3 / 2 = 2R + 8RV3 = (1 + 4V3) 2R
Area of ABCD = 16R x (1 + 4V3) 2R = 32R2 (1 + 4V3)
Percentage efficiency = 64nR2 x100 / 32R2 (1 + 4V3)
= 79.3 %
Example 3 (49 tins) Calculation
Case 1
Square packing
Each base circle is circumscribed by a square. Area of one circle = n R2
Area of square = 4 R2
Area of circle / area of square = nR2 / 4 R2
This ratio will be evidently the same as the cross section of all the tins to the total base area.
Percentage efficiency = n / 4 x 100
= 78.5 %
Therefore the efficiency in case of square packing is 78.5%
Case 2
Hexagonal packing
Here we determine the sides of the base of the container in terms of the radius of the cylindrical tin.
One side of the rectangular base i.e. BC = 14 x R.
To determine the other side, AB = 2 x R + 7 x h, where h is the altitude of the equilateral triangle formed by joining the centres of three adjacent circles.
h = 2R sin 60°
AB = 2R + 14 R sin 60°
but sin 60° = V3 / 2
Now AB = 2R + 14R x V3 / 2 = 2R + 7RV3 = (2 + 7V3) R
Area of ABCD = 14R x (2 + 7V3) R = 14R2 (2 + 7V3)
Percentage efficiency = 49nR2 x100 / 14R2 (2 + 7V3)
= 77.96 %
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Remarks
1. In the calculations here the number of tins was fixed and the cuboid dimensions are variable.
A similar exercise may be done with fixed cuboid dimensions and variable number of tins.
2. We can also determine the efficiency for packing of spheres in a cuboid.
Volume of sphere = 4/3 nR3
Volume of cube = 8R3
Percentage efficiency = 4 n R3 / 3 x 8 R3
= n / 6 = 52 %
Note
When 81 and 64 tins were taken Hexagonal packing was more efficient but in case of 49 tins Square packing was more efficient.
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