Mathematics Project on Efficiency in Packaging

File Type: PDF
File Size: 581.72 KB

Submitted by Editor

Efficiency in Packaging

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 %

Blank

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.


Square Packing Hexagonal Packing Efficiency in Packing Efficiency in 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. (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 = ð R2 Area of square = 4 R2 Area of circle / area of square = ðR2 / 4 R2 This ratio will be evidently the same as the cross section of all the tins to the total base area. Percentage efficiency = ð / 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 + 18 R 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 = 81ðR2 ×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 = ð R2 Area of square = 4 R2 Area of circle / area of square = ðR2 / 4 R2 This ratio will be evidently the same as the cross section of all the tins to the total base area. Percentage efficiency = ð / 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 + 16 R 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 = 64ðR2 ×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 = ð R2 Area of square = 4 R2 Area of circle / area of square = ðR2 / 4 R2 This ratio will be evidently the same as the cross section of all the tins to the total base area. Percentage efficiency = ð / 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 + 14 R 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 = 49ðR2 ×100 / 14R2 (2 + 7√3) = 77.96 % Blank 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 ðR3 Volume of cube = 8R3 Percentage efficiency = 4 ð R3 / 3 × 8 R3 = ð / 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.

Add Your Comment - Guidelines
You can express your opinion or reaction in the form below!

100 characters required

0 Comments:

Be the first one to comment!

You may also want to see:

QUOTE OF THE DAY
The supreme art of war is to subdue the enemy without fighting. - Sun Tzu
September 23rd, 2019 - Monday
background

Sign in to continue