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  • Level: GCSE
  • Subject: Maths
  • Word count: 1504

Investigating different shapes of gutters.

Extracts from this document...

Introduction

image00.png

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:

image01.png

image66.png

Semi-circular cross- section                                  Rectangular cross-section

image10.pngimage19.png

image36.png

image42.png

Triangular cross-section                                 Trapezoidal cross-section

...read more.

Middle

image11.pngimage09.png

      overflow                 image14.pngimage13.png

        15cm (example)

6cm (example)

image16.pngimage15.png

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

                                                                                                   L/2        L/2image17.pngimage18.png

                                                                                                                       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=

image20.png

image21.png

image22.pngimage23.pngimage23.png

image25.pngimage24.png

image26.png

image27.png

Triangular        Rectangular

 cross-section                  cross-section

image28.png

        aimage29.png

image31.pngimage30.png

image32.png

                                  h

                             b                image33.png

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                               image34.png

...read more.

Conclusion

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

image53.pngimage54.pngimage55.png

        The base =2L/n

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

        triangle as shown.

        180/n°image56.pngimage54.pngimage57.png

               L n/2                 himage58.png

               2L/n

90°

                                                                    L/nimage59.png

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

image63.pngimage61.pngimage62.png

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

                  L=21cm                                                                   L=21cm

                  Area=55.125cm                                                               Area=55.125

                       Area=(L-2x+xsinC)xcosC                Area=image64.png

        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

...read more.

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