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Stage 4: Data analysis

Calculations and data presentation

Calculating the Cailleux Index

The raw data needed for each pebble are:

• the length of the longest axis (l)
• the radius of curvature of the sharpest angle (r)

For each stone, calculate Cailleux Index as follows

Ci = (2r/l)x1000.

Ci=100 for a perfectly spherical pebble. The lower Ci is, the more angular the pebble.

Calculating Krumbein's Index of Sphericity

The raw data needed for each pebble are the lengths of the a, b and c axes.

For each stone, calculate Krumbein's Index as follows

K = cube root of bc/a2

K = 1 for a perfectly spherical pebble. K must be between 0 and 1. The lower K is, the less spherical the pebble.

Zingg's shape classification

The raw data needed for each pebble are the lengths of the a, b and c axes.

Calculate the ratio b ÷ a

Calculate the ratio c ÷ b

Now classify each pebble into one of the four groups shown in the table

Type of pebble	b ÷ a	c ÷ b
Sphere	> 0.67	> 0.67
Disc	> 0.67	< 0.67
Rod	< 0.67	> 0.67
Blade	< 0.67	< 0.67

Analysing the extent of sediment sorting

Depending on the sieves that you used to sort sediment, you may have results for sediment size in millimetres or as phi sizes. Use the conversion table if you do not have the phi sizes already.

Sediment size
mm	phi
1.00	0
0.50	1
0.25	2
0.13	3
0.06	4
0.03	5
0.01	6

Use semi-logarithmic graph paper to plot a cumulative frequency graph of phi against mass. Plot phi on the linear x-axis. Plot the cumulative mass of sediment on the logarithmic y-axis.

On your finished graph, find the phi size values at 16% and 84% cumulative mass. Use these figures in the following formula

(phi at 84% mass - phi at 16% mass) ÷ 2

Use the following table to interpret the result

result	interpretation
<0.35	very well sorted
0.35 - 0.5	well sorted
0.5 - 0.7	moderately well sorted
0.7 - 1.0	moderately sorted
1.0 - 2.0	poorly sorted
2.0 - 4.0	very poorly sorted
> 4.0	extremely poorly sorted

Statistics

The roundness of pebbles can be compared between the two sample sites using the Mann-Whitney U test. Need more information about this test?

The following shows a worked example of how the test has been used for data collected for the Powers Index for two different sources of sediment. This is actual data collected at a fieldwork site in north Wales.

First state the null hypothesis: There is no significant difference between the medians of the two sets of data.

Powers Index	Site 1 (fluvio-glacial sediment)	Site 2 (glacial sediment)
very angular	1	5
angular	16	30
sub-angular	20	24
sub-rounded	29	18
rounded	16	5
very rounded	2	2

Next assign a rank to each piece of data.

Powers Index	Site 1 (fluvio-glacial sediment)	rank (R1)	Site 2 (glacial sediment)	rank (R2)
very angular	1	12	5	8.5
angular	16	6.5	30	1
sub-angular	20	4	24	3
sub-rounded	29	2	18	5
rounded	16	6.5	5	8.5
very rounded	2	10.5	2	10.5
TOTAL (∑)		41.5		36.5

Now calculate the U value for each sample.

U1 = 6.6 + [6 (7)] ÷ 2 - 41.5

U1 = 15.5

 

U2 = 6.6 + [6 (7)] ÷ 2 - 36.5

U2 = 20.5

Finally take the smaller of the two values as the calculated test statistic. In this case, the calculated value of U is 15.5.

If the calculated value of U is less than the critical value, then reject the null hypothesis.

In this example, the critical value of U is 5 at the 5% significance level. The calculated value of U is more than the critical value (as 15.5 is more than 5). Therefore the null hypothesis cannot be rejected. In conclusion, there is no significant difference between the median Powers Index for the glacial sediments and the median Powers Index for the fluvio-glacial sediments. This is probably not what we were expecting (or hoping for!). Just because the results don't agree with textbook models doesn't mean they are wrong.

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