Image analysis: Gravel Sieve

Question

In this link: http://broadwaysand.com/?page_id=165, you can find different gravel pictures.

A quarter dollar coin gives the scale, because its diameter is 0.955 inches (24.26 mm)

I want to get the Particle Size Distribution and plot the cummulative granulometry curve, this is the percentage of the material less than a given size.

I'm not very experienced on image analysis and I don't know where to start: How to calculate the picture size identifying the coin, and afterwards find the particle sizes.

I don't think this can be do accurately. you need to know that Sieve does not depend on the surface layer of the Gravel only. Sieve takes the whole mass of gravel. The picture shows only the first layer of the gravel. — Basheer Algohi, Aug 21 '15 at 21:48
You could start by taking a look at the following: How to Count Cells, Optimal Binarize value for counting experiment, and Count Elements in Image. — MarcoB, Aug 21 '15 at 21:55
Also, related Image Processing: Concrete test blocks http://mathematica.stackexchange.com/questions/27075/image-processing-concrete-test-blocks — s.s.o, Aug 24 '15 at 09:24

score 12 · Answer 1 · edited Aug 22 '15 at 04:27

12

It will only be good for a rough estimate, but you can segment the image using WatershedComponents and then visualize the results

img = ColorConvert[Import["https://i.stack.imgur.com/t3u7G.jpg"], "Grayscale"]; 
seg = WatershedComponents[img, Method -> {"MinimumSaliency", 0.5}];
seg // Colorize

Now you can use ComponentMeasurements to get the area of each colored region and then plot the results. Below I've plotted a histogram showing the number of detected particles (vertical axis) of each size (on the horizontal axis).

areaComp = ComponentMeasurements[seg, "Area"];
areas = areaComp[[All, 2]];
Histogram[areas]

Histogram[areas, "Log", "CDF"]

Mathematica graphics

edited Aug 22 '15 at 04:27

Dr. belisarius

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answered Aug 22 '15 at 00:58

bill s

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thanks to @belisarius for the log histogram... that looks more like the plot the OP was expecting! – bill s Aug 22 '15 at 12:56
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I thought exactly the same :) – Dr. belisarius Aug 22 '15 at 14:18
Area[D_] := (Pi/4)*D^2 so Area[24.26 mm] = 462.244 mm^2, as the coin looks larger than any other pebble the histogram should reach the 100% before 462 mm. – Enrique Pérez Herrero Aug 22 '15 at 21:05
Nice job bill s! – Enrique Pérez Herrero Aug 22 '15 at 21:09
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In fact, the largest segment is the coin, which has area 47949 pixels (it's the first element of areas). – bill s Aug 22 '15 at 22:36

Image analysis: Gravel Sieve

1 Answers1