# Guttering Investigation

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Introduction

Guttering Coursework

Guttering Coursework

In my investigation I will be looking at guttering, where I will try and find the best type; that is the one which holds the greatest volume of water.

I was walking around my hometown of Antrim where I noticed how all the guttering was made into a semi circular open top shape. I wondered if this is because this is the shape which holds most water or is it because it is the cheapest to manufacture.

My hypothesis is that semi circular guttering will hold the greatest volume of water.

I will now investigate to see whether my hypothesis is true of whether there is another which can hold a larger volume of water. I will use a sheet of plastic and fold it into different shapes. I will find out the shapes’ cross sectional area rather than the volume as it is easier and the length will be the same so it is only the cross sectional area that varies.

I will fold a sheet of paper into different shapes to help me visualise the different shapes possible.

I will be investigating guttering which will have a perimeter of “B” cm.

Middle

Each side will be length “y” and the base will be B-2y

Area = x(B-2x)

A = Bx-2x2

Below is a sketch of the graph A= By-2y2

I realise from my maths knowledge that this is a quadratic expression. I also know that because it is a-x2.

Its shape will be . The graph will cross the axis where A=0.

∴Put A=0

Bx-2x2=0

x(Bx-2x)=0

x=0, B-2x=0

2x=B

x=B/2

Therefore the graph will cross the x axis at the values of 0 and B/2. Below is the sketch of my graph.

Due to the graph’s symmetry, the maximum area occurs half way between the two points 0 and B/2. This is the point B/4.

Maximum area = B/2 x B/4

= (B2)/8 = 0.125B2

I can now prove this by using differentiation.

A=Bx-2x2

dA/dx = B- 4x

At maximum point dA/dx = 0

∴B-4x=0

4x=0

x=B/4

To show this is maximum I will find d2A/dx2

d2A/dx2 = -4

Since d2A/dx2 is negative then the area is the maximum value when x =1/4. I have proved what I have found before. It is the same as triangular cross section but less that semi circular.

Conclusion

If the perimeter is B and there are n sides then the length of each side is B/n

I will divide the half regular polygon into n congruent triangles.

Area of 1 triangle = ½ bh

= ½ x B/n x h

= Bh/2n

∴ Area of half regular polygon = n x bh/2n

= Bh/2

= ½ Bh

I would like h to be in terms of n (number of sides).

Looking at one triangle:

tan θ = B/h

h tan θ= B/2n

∴h=B/2ntan θ

θ = 180/2n

θ=90/n

∴B/2ntan (90/n)

Area of ½ regular polygon= ½ Bh

= ½ B x B/(2n tan) (90/n)

= B2/ 4n tan (90/n)

I will double check my formula by using n=3 to find the area of a half regular hexagon which I already know from the cross sectional area of a trapezium.

A= B2/ (4(3) tan (90/3))

A= B2/(12tan 30)

A= 0.1443B2

This is the same area I got for the half regular hexagon so I know my formula works.

I will now use Microsoft Excel to record my results will see how the area increases as the number of sides increase for half regular polygons.

As I thought the area does get close to that of the semi circle as the sides increase but never exceeds it.

CONCLUSION: My hypothesis was correct, and the semi-circular guttering is the type which can hold the most water.

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