Mathematics

Mathematics Project on Efficiency in Packaging

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Efficiency in Packaging

Hexagonal Packing

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.
    • © 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 = π R²
  • Area of square = 4 R²
  • Area of circle / area of square = π R² / 4 R²

This ratio will be evidently the same as the cross section of all the tins to the total base area.

Percentage efficiency = π / 4 × 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 × R.

To determine the other side, AB = 2 × R + 9 × 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 + 18R sin 60°

sin 60° = √3 / 2

Now AB = 2R + 18R × √3 / 2 = (2 + 9√3) R

Area of ABCD = 18R × (2 + 9√3) R = 18R² (2 + 9√3)

Percentage efficiency = 81π R² × 100 / 18R² (2 + 9√3) = 80.3%

Example 2 (64 Tins) - Calculation

Case 1: Square Packing

Each base circle is circumscribed by a square.

  • Area of one circle = π R²
  • Area of square = 4 R²
  • Area of circle / area of square = π R² / 4 R²

This ratio will be evidently the same as the cross section of all the tins to the total base area.

Percentage efficiency = π / 4 × 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 × R.

To determine the other side, AB = 2 × R + 8 × 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 + 16R sin 60°

but sin 60° = √3 / 2

Now AB = 2R + 16R × √3 / 2 = 2R + 8R√3 = (1 + 4√3) 2R

Area of ABCD = 16R × (1 + 4√3) 2R = 32R² (1 + 4√3)

Percentage efficiency = 64π R² × 100 / 32R² (1 + 4√3) = 79.3%

Example 3 (49 Tins) - Calculation

Case 1: Square Packing

Each base circle is circumscribed by a square.

  • Area of one circle = π R²
  • Area of square = 4 R²
  • Area of circle / area of square = π R² / 4 R²

This ratio will be evidently the same as the cross section of all the tins to the total base area.

Percentage efficiency = π / 4 × 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 × R.

To determine the other side, AB = 2 × R + 7 × 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 + 14R sin 60°

but sin 60° = √3 / 2

Now AB = 2R + 14R × √3 / 2 = 2R + 7R√3 = (2 + 7√3) R

Area of ABCD = 14R × (2 + 7√3) R = 14R² (2 + 7√3)

Percentage efficiency = 49π R² × 100 / 14R² (2 + 7√3) = 77.96%

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 = 43 π R³
    • Volume of cube = 8 R³
    • Percentage efficiency = 4π R³ / (3 × 8 R³) = π / 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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