• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # Investigating different shapes of gutters.

Extracts from this document...

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.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Investigate different shapes of guttering for newly built houses.

Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 70? = 105.72cm� C = 80? Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 80? = 110.79cm� C = 90? Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 90? = 112.5cm� C = 100? Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 100?

2. ## The coursework problem set to us is to find the shape of a gutter ...

If we look at the trapezium we see that it is basically a triangle divided into 2 and putting them on either side of a rectangle. Contrarily to the triangle and the rectangles area being 112.5cm2 I predict that the trapeziums area will be 112.52 sin cos b As before

1. ## Investigating different shapes to see which gives the biggest perimeter

2: Area = 250.1 x 249.9 = 62499.99m2 I have now proved that the square has the biggest area out of all rectangles. Triangles: I am now going to carry on my investigation by investigating another different shape, triangles. I am going to use isosceles triangles because they have the biggest height out of all the triangles.

2. ## Geography Investigation: Residential Areas

1 which indicated whether or not there is a correlation in the data, positive or negative. Null Hypothesis: There is no correlation between intangible score and average time lived in the property in months To calculate Spearman's Rank I will use the formula: d = difference n = number of ranks Rs =1-(6?

1. ## GCSE Maths Coursework Growing Shapes

No. of lines No. of lines - 9n2 1 1 3 -6 3 2 21 -15 5 3 57 -24 7 4 111 -33 9 5 183 -42 D1 As there are all 9's in the D1 column, the formula contains9n2 - 9n.

2. ## A length of guttering is made from a rectangular sheet of plastic, 20cm wide. ...

the top and side y which was the height of the triangle. To find side x we had to to use 5*sin*the angle to find y we did 5*cos*the angle. We then had to find one parallel side to do this we did x*2+10 and the 2nd parallel side was the base which was 10.

1. ## The best shape of guttering

B30*E30 A30 B30 C30 D30-10 0.5*C31*C31*SIN(D31) B31*E31 A31 B31 C31 D31-10 0.5*C32*C32*SIN(D32) B32*E32 A32 B32 C32 D32-10 0.5*C33*C33*SIN(D33) B33*E33 A33 B33 C33 D33-10 0.5*C34*C34*SIN(D34) B34*E34 A34 B34 C34 D34-10 0.5*C35*C35*SIN(D35) B35*E35 A35 B35 C35 D35-10 0.5*C36*C36*SIN(D36) B36*E36 A36 B36 C36 D36-10 0.5*C37*C37*SIN(D37) B37*E37 Net Width Net Length Side 1 Side 2 Angle Area 24 400 13

2. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

It would analyse what I thought on aspects of the area. I would rate each of these aspects from 1 to 5 with 1 being the lowest score and 5 the highest. I carried this survey out in 10 different places across the study area to give my opinions on Sutton Harbour as a whole. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 