geography rivers investigation

8/8/2019 Geography Rivers Investigation

1/22

Joe Blakey made this, if you do well in this exam, you should thank him for being such a nice guy

1 | P a g e

The Overall Aim of the Investigation

To investigate changing channel dynamics along Chew Brook.

Outline One Hypothesis That You Sought To Test

1. The hydraulic radius increases downstream2. Sediment size decreases downstream

3. Rocks become more smoothed and rounded downstream

4. The velocity of the river increases downstream

5. The discharge of the river increases downstream

6. The slope angle of the long profile decreases downstream

Outline the Sampling Scheme for Said Hypothesis

Long profile date - Data is obtained with a clinometer, a long tape measure and a

stick with a line that Mr Stewart has drawn on it (quality Audenshaw instruments)to use a reference when you look through the clinometers. Measured in non -

incremental points where theres a noticeable change in gradient. This is

quantitative data. (REMEMBERQuantitative is numerical, qualitative is descriptive

hence the lit)

Identify Risks & Mitigation Techniques

y Slipping / tripping > proper footwear + awareness of the danger

y The weather (wind, cold and rain) -> Waterproofs, layers, hat, scarf and gloves,

dry clothing

y Hypothermia-> Warm clothing, food, spare clothes

y Mist -> Stay in groups, awareness of location

y Getting lost maps, compasses, awareness, groups


2/22


2 | P a g e

Results and Analysis Of Graphs

I shall now analyse my predicted hypotheses with my collected data in order to validate them. This

refers to other work I did in my rivers investigation booklet for figures.

#1 The hydraulic radius increases downstream

(See Figure 10) The graph shows that the hydraulic radius falls after site 3, this could possibly be due to

the effects of the weir as can be seen in figure 1, after site 3. The correlation seen after site 3 is inverse toas expected, decreasing as the transition downstream takes place. However, despite this, there is no clear

overall trend, so the hypothesis cannot be verified.

#2 Sediment size decreases downstream

(See Figure 14) From this graph we can draw some conclusions which prove the hypothesis .. Site 1

has a wide range of sediment sizes, spread across the analysed spectrum of 0 100 mm. Site two

however, only has sediment sizes between 21 -80 cm, showing some support for the hypothesis as there

are now no larger sediment sizes, greater than 80m m. Site 3 however, provided definite support for the

hypothesis, as 7 out of the ten sediments analysed was in the smallest sediment range of 0 -20mm. The

most interesting of all the data shown in the histogram is that site 6 is entirely within the sedimen t range of

0-60mm, with most of these being featured in the 0 -40 range with the exception of one. Although sites 4and 5 are spread across most ranges, they both however, take up a larger area in the smaller sediment

ranges; hence the hypothesis can be dee med true as the collected data supports it.

#3 Rocks are more smoothed and rounded downstream

(See figure 17) The results collected shown in this figure do not show or suggest any possible

correlation. There is henceforth not enough evidence to prove the hypothesis above. However,

photographic evidence does show this, yet this alone is not extensive enough in my opinion to prove this. In

figure 18 below, it can be seen that the rocks are highly jagged and angular, where as figure 19 shows

them to be smoother and more rounded, through the processes of abrasion, hydraulic action and attrition

as the rock is carried downstream, so this is expected.

#4 The velocity of the river increases downstream

(See figure 15) The graph shows very little trend to suppo rt the hypothesis. It is, in fact, inverse to the

hypothesis between sites 2 and 5. But increases between sites 1 2 and 5 6, and these are the only

support for the theory. But the very fact that it is travelling at 80 m/s at site 2 and only half that a t site 4

totally contradicts the hypothesis. However, this may have been tainted due to the weir, which, as a type of

dam, would be expected to slow water down. But for these reasons cumulatively, there is not enough

evidence to draw any conclusion on this hypothesis.

#5 The discharge increases downstream

The discharge is defined as the cross sectional area multiplied by the velocity. However, the greater

the hydraulic radius, the greater the efficiency, meaning the greater the potential discharge (alth ough this is

also dependant on channel size and velocity still), this means that hydraulic radius and velocity should

increase proportionally if other variables are constrained (which when compared, no trend is shown, see

figure 9). Figure 16 shows the hy draulic radius; however this proves little, as it decreases downstream,

again, in my opinion, thanks to the weir. Hence the hydraulic radius does not support this theory. Nor does

the velocity graph, (figure 15), which as discussed previously, shows no tr end. Furthermore, conclusively,

by multiplying the cross sectional area by the velocity at respective sites and plotting (figure 20) it can be


3/22


3 | P a g e

seen the same anomaly results, there being a different trend after site 3, rising to and falling after. So once

more, the hypothesis cannot be proved.

#6The slope angle of the long profile decreases downstream

This is big as its the most important example of something ever.

(See Figure 11) This is easily provable with the collecting angle data;figure 11 shows a relatively perfect concave profile, steep at the top and

shallow at the bottom, matching perfectly to expectations. This theory is

further solidified by the Spearmans Rank Correlation Coefficient,

calculated in figure 12. This shows a strong negative value of-0.77419,

meaning that the trend is reliable and hence I can happily accept this

hypothesis.

Conclusion

In conclusion to my original hypotheses, I can only a ccept two of them to be true from the gathered

research.

The following hypotheses were ACCEPTED:

1. Sediment size decreases downstream2. The slope angle of the long profile decreases downstream

The following hypotheses were REJECTED:

1. Sediment size decreases

2. Rocks are more smoothed and rounded downstream*3. The velocity of the river increases downstream4. The discharge increases downstream

The rejected hypotheses remain undetermined as there is no clear alternative trend and so it can

therefore be deemed that they will be one of the following: True, yet unproven, thanks to anomalous results ,

There is no trend / correlation .

*This hypothesis however remains debatable, as photo evidence does strongly support the theory,

however collected data from the site contradicts this, and as of this discrepancy it cannot therefore be

accepted.

From the fieldwork we can therefore say that sediment size does increase downstream, such asshown by the fairly irrefutable evidence that at site 3 seven out of the ten sediments anal ysed was in the

smallest sediment range of 0 -20mm. Whereas site 1 had sediment sizes spread across the spectrum.

Similarly, it cannot be denied that the slope angle decreases downstream, as figure 11 clearly shows this,

and is strongly supported by figure 12, the Spearmans Rank Correlation Coefficient. Other hypotheses

investigated remain inconclusive.


4/22


4 | P a g e

Evaluation

If this investigation were to be repeated, I would suggest that firstly more sample points were taken, as the

room for anomalies being mistaken for real results is great, especially within a small part of river, so it

would also be desirable for a larger section of river to be observed. Possibly a better plan would be to

investigate a few sites, one at the source, one in middle course, one in the lower course and one at the

mouth. Furthermore, for data to be more reliable, such as that used only a few samples, such as load and

velocity, a greater number of samplings should have been used, again in an attempt to mitigate anomalies.

Techniques for collecting data were generally successful, although velocity was measured quite

primitively using a cork, ad two reference points, and timing the time it takes to move between them. The

margin for human error in this is very high, and so the use of a more reliable flow meter would be better.

Similarly, the judgement of smoothness of a rock is debatable as it is a matter of an individuals opinion, so

simply increasing the group size and taking the modal opinion would produce better results.

Strengths Weaknesses

y Reliable Date

y Teamwork and Co-operation

y Rushing

y Not enough sampling points

y Interference from a weir

y Time of year

Opportunities Threats

y Opportunity to see if theory iscorrect

y Opportunity to learn methodsof sampling

y Opportunity to collate primarydata

y Rain

y Wind

y Cold

y Hypothermia

y High Cliffs

y Uneven ground

y Large silt banks which can causesinking


5/22


6/22


6 | P a g e

The critical-value-shabam

(We NEVER went through this when we should have, but Ill do it

anyway, because I can)

Select significance level

Assign a NULL Hypothesis (what we expect) to Rs

e.g. There is no significant relationship betweenincome deprivation

and deaths from smoking

Assign an ALTERNATE Hypothesis (what we dont expect) to Rs

The is a significant relationship betweenincome deprivation and

deaths from

smoking

If Rs (greater

than) CV then

reject the null

hypothesis and

accept the

alternate

hypothesis

Peasy easy

squeezy lemons~


7/22


7 | P a g e

Chi Squared Sounds kinda oriental

We didnt learn this either. Faaaaabulooouuus.

Heres some stuff off the internet. Because I find this boring. Thats

why its in a boring font.

This is a statistical technique you will need for the AGS paper. It is pronounced to rhyme with

'sky' rather than 'tea'. (This man is funny)

We will use an example of corrie or cirque orientation.

Corries were identified from maps and the direction they face was recorded. Data was placed

into 4 categories, relating to the compass. Results shown below:

Orientation from N Frequency

0-89 30

90-179 5

180-269 6

270-359 11

Total number is 52.

Is this distribution random, or significant ?

Start by developing null hypothesis: The orientation of corries is random.

If this was correct, we would expect there to be how many corries in each category ?

52 / 4 = 13 in each.

This is obviously not the case, but the test will determine whether the differences are significant.


8/22


8 | P a g e

Formula is below:

O = observed frequency E = expected frequency

Corrie Data is set out as in table below. No 0 or E value should fall below 5.

0-89 90-179 180-269 270-359 TOTAL

0 30 5 6 11 52

E 13 13 13 13 52

0-E 17 -8 -7 -2

(O-E)2

289 64 49 4

(0-E)2/E 22.23 4.92 3.77 0.31 31.23

The value is 31.23

Has to be checked with significance tables.

Need to determine what are called the degrees of freedom.

This relates to the size of the sample, and is n-1 = 3.

For 3 d.o.f, value is 7.82 at the 5% significance level.

It is 11.34 at 1% level.

Since our value is greater than the value on the table we can reject the null hypothesis: there isless than 1% chance of the corrie orientation being random: there is some preferred orientation.

The example above tests one set of data against a theoretical frequency distribution. The 2 nd use

of Chi-squared is to compare 2 or more sets of data. This involves the production of a

contingency table.

For example: here is some humidity data of 2 stations: one near the sea and one far away.


9/22


9 | P a g e

Relative humidity % Near sea Away from sea

50-55 6 35 41

56-60 17 16 33

61-65 26 3 29

49 54 103

Note data is grouped. Sample size must be at least 20. There must be at least 1 observation in

each class.

Data need to be arranged into a contingency table. In this case, there are 2 columns, 3 rows, andtherefore 6 cells. The expected frequency for each cell needs to be worked out: using formula

below:

E value for cell = column total x row total

grand total

Can then work out the 0-E and square those for each cell to come up with the total which is the

chi squared value which we then compare to the values in the significance tables.

This time, degrees of freedom are calculated using the formula:

d.o.f = n(rows)-l x n(columns)-1

i.e 2 x 1 = 2

Calculated value = 38.6, so can reject the null hypothesis again.

Wasnt that fabulous?


10/22


10 | P a g e

And now introducing... Mann-Whitney U test

This test will tell you whether the medians of two sets of data are significantly different to oneanother. It works on unmatched, interval or ordinal data (see section on "Different kinds of data").It does not require that the data is normally distributed but it does require that both data sets arethe same shape.

Lets say you are investigating the effects of lifestyle on human body size . You feel it to be toomuch of an intrusion to measure people's weight or girth dire ctly, so you invent a way of assessingtheir size remotely.

You select two sites: Site one is outside the well known fast -food chain McBloaters. Site two isoutside the well known health -food chain McSmugs.

You stand outside both establishments and simply assess the body size of the punters patronisingthem on the following scale:

1 = Skeletalmmmmm2 = Thinmmmmm3 = Mediummmmmm4 = Plumpmmmmm5 = Fatmmmmm6= Bloater

You obtain the following results:

Body size scores of people patronising McBloaters and McSmugs

Sample 1 2 3 4 5 6 7 8

McBloaters

3 2 2 1 4 4 5 2

McSmugs

5 4 6 3 4 6 3 6

Firstly we state our null hypothesis (it's always the same for this test):

There is no significant difference between the medians of the two sets of data

Next we put the data in order from smallest to highest:

McBloaters

1 2 2 2 3 4 4 5

McSmugs

3 3 4 4 5 6 6 6


11/22


11 | P a g e

Next assign a rank to each piece of data:

McBloaters

1 2 2 2 3 4 4 5

RANK 1 3 3 3 6

9.5 9.5 12.5

McSmugs

3 3 4 4 5 6 6 6

RANK 6 6 9.5 9.5 12

Notice the lowest value is 1 (McBloaters data set), so this receives a rank of 1. Next we have threetied values (three values of 2 from the McBloater data set). These 3 items of data occupy threeranks but they are all of the same value, so we share out the ran ks thus: rank 2 + rank 3 + rank 4= 9. Divide by three and we end up with a rank of 3 for each piece of data. (See the red rank rowbelow McBloaters).

The next values come from the McBloaters and the McSmugs data set: A value of 3 fromMcBloaters and two tied values of 3 from McSmugs. We deal with these in the same way. Wehave used ranks up to 4 so the next three ranks that are available are rank 5, rank 6 and rank 7.

Add these together and share them out equally: 5 + 6 + 7 = 18/3 = 6.

Continue doing this to complete the table and the only moderately hard part of doing this test isover.

Add up the ranks for each data set:

Sum of McBloater ranks = 47.5

Sum of McSmug ranks = 88.5

Calculate the value U for each sample:


12/22


12 | P a g e

We can now check our calculations because U1 + U2 should equal n1 x n2. Happily, in ourcase this is indeed true.

We now take the smaller of the two values as our calculated test statistic. Our calulated value of U

is then 11.5

Our next task is to compare our calculated value with the critical value obtained from a table ofcritical values of U:

Table of Critical values of U (5% Significance)

n1/n2 1 2 3 4 5 6 7 8 9

1

2

3 1 1 2 2 2

4 1 2 3 4 4

5 1 2 3 5 6 7

6 1 2 3 5 68 10

7 1 3 5 68 10 12

8 2 4 6 8 10 13 15

9 2 4 7 10 12 15 17

10 3 58 11 14 17 20

We enter the table at the correct number of items of data for each data set (row 8 and column 8 in

our case). The critical value of U is therefore: 13

Our calculated value of U was 11.5

In a Mann-Whitney U test if the calculated value of U is less than the critical value we reject thenull hypothesis. In rejecting our hypothesis of no difference we are saying that the medians of thetwo sets of data are indeed significantly different. In doing this at the 5% significance level wewould expect to be correct in rejecting our null hypothesis 95% of the time.

We might also learn from this example of the dangers of jumping to conclusions. I would bet that

most people expected McBloaters to have the fatter people using it.It just goes to show, if you eattoo much of anything (even McSmugs lentil surprise) you will get fat. Let our maxim be "tolerance,

moderation, healthy exercise, wholesomeness, crispness and rhubarb".


13/22


13 | P a g e

Lets talk about graphs

line graphs simple, comparative, compound and divergent

bar graphs simple, comparative, compound and divergent

SEE PAGE 262 of AQA geog A2 Veryo importenteo (Ceebs with

notes on this)

scatter graphs and use of best fit line

pie charts and proportional divided circles

triangular graphs

kite and radial diagrams

logarithmic scales

dispersion diagrams (box and whisker plot to you and me)

1. Bar chart

The x-axis has labels, the y-axis may have numbers. There needs to be a gap between the bars.The example below, from a saltmarsh investigation, shows the Simpson Yule Diversity Index for12 sites in a saltmarsh. See investigation page.


14/22


14 | P a g e

2. Percentage bar charts

The x-axis has labels, the y-axis has percentages. There needs to be a gap between the bars.Each bar is divided up into coloured or shaded sections based on percentages. The examplebelow, from a downstream changes in a river investigation, shows the perc entage of stones in

each roundness category at 3 sites.

3. Divided bar charts

The x-axis has labels, the y-axis has numbers. There needs to be a gap between each of the bars.The size of each bar shows the total number. Each bar is divided up into coloured or shadedsections based on percentages. The example below, from a counter-urbanisation investigation,shows the results of a traffic survey carried out at four times of the day.


15/22


15 | P a g e

4. Histogram

The x-axis and y-axis both have numbers. The x-axis is divided into intervals. There are no gapsbetween bars. The example below, from a glaciation investigation, shows the number of stones ineach interval of the Cailleux Index.

5. Pie chart

These show the percentage of the total represented in each category. To work out the angle (indegrees) for each category, calculate percentage x 3 60. The example below shows employmentdata, derived from the census, for a rural ward in south -east England.


16/22


16 | P a g e

6. Scattergraph

Use the x-axis for the independent variable and the y -axis for the dependent variable. Theexample below, from a rural settlement hierarchy investigation, shows the total population size ofeach settlement (the independent variable) on the x -axis, and the number of services in eachsettlement (the dependent variable) on the y -axis.


17/22


17 | P a g e

7. Kite diagram

These show the change of a percentage over distance. They are most commonly used to showchanges in the percentage cover of plant species along an environmental gradient. The examplebelow, from a sand dunes investigation, shows changes in the percentage cover of a number ofspecies with distance inland in a sand dune system in south Wales.


18/22


18 | P a g e

Triangular Graphs Because square graphs are for peasants

Example: Service structuredata for selected urban areas can be plotted on a three-sided triangular graph. The

important features of a triangular graph are:

yEach axis is divided into 100, representing percentages.

yFrom each 100-0% axis, lines are drawn at angles of 60 degrees to carry the values.

yThe data used must be in the form of three components, each component representing a percentage value, andthe three component percentage values must add up to 100 per cent.

The position of the plots indicates the relative dominance of each of the three components and the value of the graph

arises in giving a quick visual comparison of contrasting component dominance for different areas. It is particularly

useful in identifying changes over time, since a position on the graph will change as the relative dominance of the

components change.

The graph can be used to show contrasting service structures for 4 locations in El Raval, an inner-city area of Barcelona

which has been the subject of radical urban reform. The choice of the three graph components is important and must

be in the context of the investigation. An example of data from one location (El Raval Site 2) is shown in map 1 below,

and this has been used along with data from three other sites (1,3 and 4) to compile the triangular graph.

Key

Gentrification

Immigrant

Services

Local Services

Professional

Services

Services ofPoverty

Training

Centres

Workshops

Map 1: Service Structure in El Raval, Site 2

Service Structure Data Summary Chart for Sites1-4

El Raval Service Structure

Service Site 1 Site 2 Site 3 Site 4

% Gentrification 60 11.4 3 0

% Immigrant Services 5 15.2 20 50

% Other Local Services 35 73.4 76 50

Total 100 100 100 100

Data example is for training purposes only. Its accuracy cannot be guaranteed.


19/22


19 | P a g e

Triangular graph to show the contrasting service structure for four areas of El Raval

Copy n paste:

Choropleth Maps

These are maps, where areas are shaded according to a prearranged key, each shading or colour type

representing a range of values. Population density information, expressed as 'per km,' is appropriately

represented using a choropleth map. Choropleth maps are also appropriate for indicating differences in

land use, like the amount of recreational land or type of forest cover.

An example from the Czech Republic is shown below.


20/22


20 | P a g e

Choropleth Map with Proportional Symbols

from ARCDATA PRAGUE, in GIS: Our Common Language, ESRI Map Book, Volume 12, 1997.

Disadvantages of Choropleth Maps

Although choropleths give a good visual impression of change over space there are certain disadvantages

to using them:

y They give a false impression of abrupt change at the boundaries of shaded units.y Choropleths are often not suitable for showing total values. Proportional symbols overlays (included

on the choropleth map above) are one solution to this problem.

y It can be difficult to distinguish between different shades.

y Variations within map units are hidden, and for this reason smaller units are better than large ones.

Isopleth maps

Isopleth maps differ from choropleth maps in that the data is not grouped to a pre-defined unit like a city

district. These maps can take two forms:

y

Lines of equal value are drawn such that all values on one side are higher than the "isoline" value andall values on the other side are lower, or

y Ranges of similar value are filled with similar colours or patterns.

This type of map is ideal for showing gradual change over space and avoids the abrupt changes whichboundary lines produce on choropleth maps. Temperature, for example, is a phenomenon that should be

mapped using isoplething, since temperature exists at every point (is continuous), yet does not change

abruptly at any point (like population density may do as you cross into another census zone). Relief maps

should always be in isopleth form for this reason.

Isopleth example: precipitation 10th June 2000 (mm)

The disadvantage of isopleths are that they are unsuitable for showing discontinuous or 'patchy'

distributions and a large amount of data is required for accurate drawing.

Proportional Symbol Maps


21/22


21 | P a g e

As the name implies, symbols (most often circles) are drawn proportional in size to the size of the variable

(e.g. employment change) being represented. Proportional symbol maps are not dependent on the size of

the area associated with the variable. In other words, on a proportional symbol map of Europe, tiny

Liechtenstein would have the same visual importance as Spain if their unemployment values were the

same. This would not be the case with a choropleth map.

An example of proportional circles is shown on the Czech Republic Voting Register map (above).

Scaling proportional symbols. Much research has gone into the optimal scaling for proportional symbols.

As a general rule, make sure that the area, rather than linear proportions like radius or length of a side, is

the scaled parameter. For example, if there are four times as many gentrified businesses in El Raval Site 1

than in Site 3, the area of the symbol should be four times greater for Site 1. If the symbol choice is a circle,the radius of the Site 1 symbol should thus be only twice as great (since area scales with the square of the

radius).

Dot maps

Used to show the distribution of phenomena where values and location are known. Dot maps create a

visual impression of density by placing a dot or some other symbol in the approximate location of the

variable being mapped. Dot maps should be used only for raw data, not for prearranged data or percentages.Appropriate themes for dot maps include the distribution of dairy farms, and population distribution in a

region.

Their limitations include the difficulty of counting large numbers of dots in order to get a precise value and

the need to have a large amount of initial information before drawing the map.

Dot map parameters. When constructing a dot map, two parameters must be considered: the graphical

size of each dot and the value associated with each dot. For example, you might stipulate that each dot be 2

pixels in diameter, and each represent 100 persons. In general, many small dots, each representing

relatively few instances of the attribute, is more effective than a few large dots, but is more tedious to

construct.

Random things of worth

Desire lines / paths. These are the imprints in the grass when a few

people have walked the route. AQA like to sound all hippy and

metaphorical and call lines that show where people go. You can draw

them proportionally, thicker for more people. Useful when drawing a

sphere of influence.


22/22


22 | P a g e

geography rivers investigation

Documents