# Investigating different shapes of gutters.

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Introduction

## Introduction

A gutter is an object, which catches rainwater and carries it along a channel. I am going to investigate different designs of gutters to see which one can hold the most water. To do this I will vary the shapes and dimensions of the gutter, which will enable me to find which one can hold the greatest quantity of water.

## Hypothesis

After looking at guttering on the tops of houses I found that they had mainly semi-circular cross-sectional areas. I then asked myself was this because it looked better or was it because semi-circular cross-sectional areas carried the greatest amount of water? And from this I get my hypothesis. My hypothesis is that semi-circular guttering is the best shape for guttering because it carries the most water.

### Investigation

I found a piece of guttering at home and I measured it to be 21cm.

For each shape I am investigating I will find the general case using Lcm as the perimeter then I will use 21cm as the perimeter so that I can compare them with the semi-circle to verify or nullify my hypothesis. As the gutters are used to catch water there will be no top on them, which is shown in the diagrams below.

### I intend to investigate the following prisms:

Semi-circular cross- section Rectangular cross-section

Triangular cross-section Trapezoidal cross-section

Middle

overflow

15cm (example)

6cm (example)

The two sides add up to L in a general case therefore each side is L/2

L/2 L/2

C

The area of a triangle is ½ absinC a=L/2 b=L/2 C=angle between the two sides

A= ½ * (1/2L)*(1/2L) sinC

A=1/8LsinC

SinC is the only variable and it has to lie between a range of angles therefore in this case it has to be 0<sinC<180

As you can see the area depends on the angle C so to find the best angle I will use the sine graph shown below:

The maximum angle is 90° so if C=90° Therefore A= 1/8Lsin90°

From the graph you saw that the maximum area was when sin90° = 1

The maximum area = 1/8L(1)

=1/8L

If I use 21cm to = L

The Maximum area =1/8*(21*21)

A=

Triangular Rectangular

cross-section cross-section

a

h

b

To find the area of a trapezium the formula is ½(a+b)*h

A= ½ (sum of the parallel sides)* perpendicular distance between

I am going to use the values x and L-2x

Conclusion

To find the area of 1 triangle I will use the formula A= ½ bh = ½ base*height

The base =2L/n

360/n To find the height I will use trigonometry and half a

triangle as shown.

180/n°

L n/2 h

2L/n

90°

L/n

To find the area of one triangle I will fill the height(h) and the base(b) into the formula A= ½ bh

A= ½

A=

Area of half a polygon =

=

I will use a length of 21cm for L for a various number of sides (n). This is shown on the following table:

As you can see from the tables and the graph the more sides there are in a polygon, the closer the areas are to the semi-circle area but no matter how close the values are to the semicircle they will never exceed it. This proves once again that the best area is the semi-circular cross section.

### Summary

Area=1/8L Area=1/8L

L=21cm L=21cm

Area=55.125cm Area=55.125

Area=(L-2x+xsinC)xcosC Area=

L=21cm x=21/3=7 C=30° L=21cm

Area= 63.65 Area=70.2

## Conclusion

I investigated different shapes of gutters. I put forward my hypothesis that a semi-circular gutter would hold the most water. I examined different shapes of gutters and found as the number of sides increased for a regular polygon they tended towards that of a semi-circle but never exceeded it. My hypothesis is proved to be correct and I can conclude that semi-circular gutters are most efficient since they carry the most water.

Ciara Mc Kay 4 Layde

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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