Digital Scanning: From Hard Copy to Soft Copy

Most papers and journals that we can gather are already in pdf form or printed form. When studying these papers, we take a look at the graphs and analyze the trends. We run into a problem when we need the actual values of the graph. This is where digital scanning and processing comes in.

The idea of digital scanning and processing is that if the graph printed is accurate and scaled, we can find the values of the data points we need in the graph just by using ratio and proportion. We can relate the pixels of the image to a specific scale on the actual graph, thus relating the pixel location to physical quantities represented by the graph.

Here we start by finding an old graph without the specific values of the data points. The graph I obtained comes from the Journal of Microbiology. By scanning this graph, we can now process the image to calculate the relationship of the pixel location to physical quantities.

Figure 1 Original Scanned Graph

Figure 2 Cropped Image of the Graph

First, we start by counting the pixels of the first tick mark to the last tick mark in the x and y axes. From the image and the use of Paint software, we find that the first tick mark in the x axis is at 18^th pixel and the last tick mark is in the 474^th pixel. These correspond to the 23 years and 100 years tick marks respectively. For the y axis, the first tick mark is at 799^th pixel corresponding to -1 R value and 0^th tick mark corresponding to the 6 R value.

From here, we compute for the unit/pixel constant.

77 years / (474 pixels -18 pixels) = 0.168859649 years / pixel (1)

7 R values / 799 pixels = 0.008760951 R values / pixel (2)

So given a data points x and y axis pixel count we can compute for the actual years and R values they represent.

X-axis

(data point(x) – 18 pixels) * constant_x-axis + 23 years = # of years (3)

Y-axis

(799 pixels – data point(y)) * constant_y-axis - 1 R value = R value (4)

For the x-axis, the subtraction of 18 pixels accounts for the offset of the origin and the addition of 23 years accounts for the starting year of the graph. While for the y-axis, the difference of 799 pixels and the data point y value accounts for the fact that the 0^th pixel along the y axis start from the top of the graph to the bottom. But subtracting the data point y value to the maximum number of pixels along the y-axis (799) we get the height of the data point from the base of the graph. The subtraction of negative 1 accounts for the fact that the graph starts at negative 1. The constants used are the values solved in equations 1 and 2.

Now, to find the values of the data points, we get the pixel locations of the data points and apply equations 3 and 4 to get the actual years and R values of the graphs.

Years	R		Years	R
18	200		23	4.24781
59	90		29.92325	5.211514
178	230		50.01754	3.984981
267	440		65.04605	2.145181
326	490		75.00877	1.707134
474	615		100	0.612015

(a) (b)

Table 1 Data Points of Ash or Triangular markers (a) is in pixels (b) in actual values

Years	R		Years	R
18	267		23	3.660826
59	420		29.92325	2.320401
178	410		50.01754	2.40801
267	627		65.04605	0.506884
326	550		75.00877	1.181477
474	630		100	0.480601

(a) (b)

Table 2 Data Points of Organic Matter or Circular markers (a) is in pixels (b) in actual values

Years	R		Years	R
18	276		23	3.581977
59	591		29.92325	0.822278
178	553		50.01754	1.155194
267	723		65.04605	-0.33417
326	672		75.00877	0.112641
474	673		100	0.10388

(a) (b)

Table 3 Data Points of Water or Square markers (a) is in pixels (b) in actual values

Years	R		Years	R
18	259		23	3.730914
59	460		29.92325	1.969962
178	440		50.01754	2.145181
267	610		65.04605	0.65582
326	585		75.00877	0.874844
474	643		100	0.366708

(a) (b)

Table 4 Data Points of Water or X markers (a) is in pixels (b) in actual values

From the solved actual values of the data points, we can now graph the data and compare to the original graph.

Figure 3 Reconstructed graph with superimposed original graph

We can see that the data points very closely match each other. Small deviations can be seen in position. This can be attributed to 2 things. One, the data points in the graph are large, hence the exact position of the data point can vary largely. Two, since the graph was initially photocopied a slight distortion can be seen. This distortion will correspond to a change in the actual value and the corresponding pixel location.

I then conclude that this method is effective in very closely approximating the values in a graph. It is only limited to the quality of the initial graph in terms of resolution, clarity of printing and precision. It is recommended that the graph to be scanned be on a separate sheet of paper rather than bound in a book or a journal.

For this exercise, I give myself a score of 10. I was able to complete the activity as well as repeat the process for the multiple data series available in the graph. I was also able to present the data in a complete form. Hence, I believe I deserve a perfect score.

I would also like to thank Arvin Mabilangan for the support and initial help in the understanding the methodology.