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Discharge m³/s Rainfall mm 1 AQA GEO4B Geographical Issue Evaluation Skills Photocopiable/digital resources may only be copied by the purchasing institution.

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Presentation on theme: "Discharge m³/s Rainfall mm 1 AQA GEO4B Geographical Issue Evaluation Skills Photocopiable/digital resources may only be copied by the purchasing institution."— Presentation transcript:

1 Discharge m³/s Rainfall mm 1 AQA GEO4B Geographical Issue Evaluation Skills Photocopiable/digital resources may only be copied by the purchasing institution on a single site and for their own use © ZigZag Education, 2013

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3  You can be asked to undertake any one of the skills shown on the previous slide.  However, some are going to be more appropriate to the pre-release material than others.  Do make sure that you are familiar with all of the different calculation methods and that you take a calculator and full maths set (protractor, compass, ruler) into the exam. 3

4 4 This is a simple method of classifying the size of a stream based on the number of tributaries it has of certain sizes. When two tributaries of the same size meet, the stream moves up the hierarchy, e.g. when two first-order streams meet, they create a second-order stream. However, if a first-order and a second-order stream meet they simply remain a second-order stream. It needs the meeting of two second-order streams to create a third-order stream.

5 5 Top Tips:  Read ‘along the corridor and up the stairs’ to get the four-figure grid reference.  Use a ruler to measure exactly how far across the square you need to go (you need to subdivide each square into 10 segments).  On a 1:25,000 map, 4 mm is 1 segment across. These form your third and sixth numbers of the six-figure grid reference.  Don’t forget, if what you are looking for is on the line then the third (if going along) or sixth (if going up) number is 0.

6 6 1.Identify the cross- section on your OS map (if possible, draw a faint pencil line so you don’t lose your place). 2.Look at how the contour lines change across your profile; follow them round until you find out their height. 3.What would you expect the land to be doing either side of the river? © Crown copyright 2014 Ordnance Survey licence number 100040266

7 60 50 40 30 20 10 0 0 50 100 150 200 250 300 350 400 450 500 1.Use the scale on the OS map to help you understand the distance you will be travelling – 4 cm = 1 km, therefore 2 cm (the distance of the transect given in the question) = 500 m. 4 mm = 50 m. 2.Plot your axis labels on your graph – along the horizontal axis needs to be your distance (0– 500 m going up in 50 m intervals). Along the vertical axis needs to be the height of the land – check your transect – what’s the highest point shown? 3.Now make your first plot (0 m – the start of the transect should be 60 m high.) 4.Now measure 4 mm (50 m from the start of the transect). What is the height of the land? Plot this in the correct place on your graph. 7 x x 60m 6.Now continue working your way along the transect line and plotting the heights. 7.Finally, join the crosses up with a smooth curve.

8 8 River Distance in metres across the cross section Height of land (m)

9 9 How to plot it: 1.Choose a suitable scale for the axes – you can have one scale for rainfall and another for discharge. 2.Plot the rainfall as bars first. 3.Plot the discharge as a line. 4.Identify the peak rainfall (highest amount). 5.Identify the peak discharge (highest amount). 6.Calculate the lag time (time between peak rainfall and discharge). 7.Is the rising limb gentle or steep? 8.Is the falling limb gentle or steep? 9.What is unusual about the Wansbeck hydrograph? Can you suggest why? Peak Rainfall Peak Discharge Secondary discharge peak 1 st Peak Rainfall Lag time = 3 hours Steep rising limb Steep falling limb

10  This graph shows the distribution of data around the mean or median.  It is plotted as a line with a series of dots recorded if there are several pieces of data that are the same.  The points are NOT joined up. 10 Rainfall data for 5 th –7 th September Note how the number of hours when there was no rainfall recorded really skews the mean and median data. Mean Median

11 1.The mean is very straightforward; this is the average – add up all the data and then divide by the number of pieces of data you have. 2.For the median, you need to put all your data in order. Then find the middle number – this is the median. If you have an odd number of data sets this is easy, e.g. I have 51 sets of data so the median number will be the data found at number 26 as I count up. If you have an even number of data sets, you will need to split the difference between the two middle numbers, so if you have 50 sets of data you need to add the data you find at 25 to the data you find at 26 and then divide it by 2 to get your answer. 3.The mode is the most frequent number, so if in a list of numbers ‘15’ occurs five times and the other numbers only occur once or twice, then 15 is the mode. 11

12  This is a method used to show the spread (dispersion) of a set of data around the median. It is calculated by finding the difference between the upper and lower quartiles, which shows where 50% of all data is found.  The smaller the interquartile range, the more the data is grouped around the median showing that there is not much variation between the data. 12 n = number in the sample 3.2 mm 0.0 mm 3.2 – 0.0 = 3.2 mm What does this tell us about the spread of rainfall over these three days?

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14 TimeRiver flow (cumecs) 00.0057 01.0067.3 02.0078.5 03.0085.7 04.0093 05.0099.2 06.00107 07.00124 08.00145 09.00169 10.00191 11.00217 12.00243 13.00269 14.00297 15.00337 16.00357 17.00324 18.00304 19.00275 20.00241 21.00215 22.00206 23.00213 -139.57 -129.27 -118.07 -110.87 -103.57 -97.37 -89.57 -72.57 -51.57 -27.57 -5.57 20.43 46.43 72.43 100.43 140.43 160.43 127.43 107.43 78.43 44.43 18.43 9.43 16.43 19479.78 16710.73 13940.52 12292.16 10726.74 9480.92 8022.79 5266.41 2659.47 760.10 31.02 417.38 2155.75 5246.11 10086.18 19720.58 25737.78 16238.4 11541.2 6151.27 1974.03 339.66 88.92 269.94  = √8305.74  = 91.14 14 This standard deviation result shows a wide spread around the mean showing considerable fluctuation in discharge during the day.

15  Scatter graphs can be used to show the relationship between two data sets. However, it is only a ‘best fit’ relationship. For accuracy, Spearman Rank Correlation Coefficient can be used to prove the strength of a relationship.  Once points are plotted, a line of ‘best fit’ can be drawn.  The pattern of the scatter graph shows the relationship between the data.  With a positive relationship, as one set of data increases, so does the other.  With a negative relationship, an increase in one data set causes a decrease in the other. 15 Positive Relationship Negative Relationship Line of Best Fit

16  Cumulative frequency is when the data of one section is added to the data of the next section to give a running total.  This is particularly useful for giving an idea of how rainfall builds up over time and for showing why the ground would eventually become saturated leading to overland flow.  When plotted, cumulative frequency leads to a curve on a graph. 16

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