Entry 7: Applied Physics 186 - Activity 1 (Digital Scanning and Evaluation)

The rationale of this activity is to identify the actual physical values from digital graphs in scientific journals. The advantage of performing this method is that it gives you a very close approximation to what the real/physical values shown in the graphs. Also, with proper computational tools, one can derive the functional form of the curve itself; this further helps you in the analysis of the physical quantity you are concerned with.

The procedure starts by choosing a graph from a printed journal and followed by the digital conversion of the raw image (Fig. 1).

Fig. 1: Digitized copy of the raw image from a printed journal

To know the corresponding physical value of any point on the curve, it is essential to determine the scaling factor of the pixel dimensions in x- and y-axis. The scaling factor is equal to the span of two successive tick marks in pixel dimension relative to the absolute value of the difference between the physical values represented by the two tick marks.

$\color{white}SF_i\ =\ \frac{\Delta (pv)_i}{\Delta (pd)_i}$

where $\color{white}SF_i$ is the scaling factor, $\color{white}\Delta (pv)_i$ is the absolute value of the difference of the physical value at the tick marks and $\color{white}\Delta (pd )_i}$ is the absolute value of the difference of the pixel locations of the tick marks.

Example:
In the x-axis:
Two successive tick marks are located at 2740 and 2273 and the physical value associated to the two are 550 and 500, respectively. Therefore, the scaling factor $\color{white}SF_x$ = 0.107nm/px

Now to compute for the physical values in the graph, it must be considered that the location of the origin of the graph is not located at the origin of the image array. Thus, find a strategic pixel location which will be your reference for further computation of the values and note this as X and Y. In my case, I choose the location X = 876.946 px, and Y = 2851.77 px which represents the location for (350nm, -2).

Arbitrarily get sufficient number of points from the curve by noting their pixel locations (x, y). Use the equation below to generate the physical quantity of the point.

$\color{white}P_x\ =\ R_x\ + SF_x (x\ -\ X)$
$\color{white}P_y\ =\ R_y\ + SF_y (Y\ -\ y)$

where $\color{white}P_x$ and $\color{white}P_y$ are the computed physical value, $\color{white}R_x$ and $\color{white}R_y$ are the physical value of the reference point (X , Y). The x and y are the values of the pixel locations as noted beforehand.

Using Microsoft Excel, I plotted the results of the conversion of the point values and generated Fig. 2.

Fig. 2: Reconstruction of the derived physical quantities from pixel values

To verify the "validity" of the conversion used, I superimposed the derived curve with the actual digitized image. Ideally, when the two curves are nicely fitted; qualitatively, we can infer that the conversion of values is correct. The superimposed image is shown in Fig. 3.

Fig. 3: Nicely fitted curves of the raw image and the graph of the converted data

Basing from Fig. 3, the reconstruction made was good if not better, since it can be observed that the two curves almost overlap each other.

Things to do: Use interpolation methods to find the polynomial form of the curve.


Self Evaluation:

The activity was very refreshing and it made me realize that I have missed a lot of useful capabilities Microsoft Excel and OpenOffice Spreadsheet have to offer. Now I have learned that you can embed an image in a graph as a background and use it as a reference.
Again, I was reminded that reading the instructions first will help you a lot; especially in saving time.
After everything, I would bravely rate myself a 12 because of the new things that I have learned and that I was able to accomplish essential things to finish this activity.
If I would have enough time, I will try to derive the polynomial relationship of the curve. It could be done by using polynomial interpolation.