Calculate Size of Pixel in X and Y Direction

Question

we have analytically calculated distance between the centers of a big circle and a small circle in mm.

We also have coordinates of the centers of the mentioned circles - yellow and blue - identified from an the image with image processing algorithm in pixels. Now we have to calculate the size of a pixel in mm in $x$-direction and $y$-direction based on the following information;

1. The analytical distance between center of the two circles in mm - $RDistance$
2. The center of the two circles identified with image processing in pixels, i.e. $(x_1, y_1)$ and $(x_2, y_2)$

I can calculate a single pixel size with the formula $$\frac{RDistance}{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}.$$

However, the requirement is to calculate the pixel size in both $x$- and $y$-direction differently. How can we calculate the pixel size $x$-direction and pixel size $y$-direction separately?

Answer 1

If the aspect ratio, which @MattiP mentioned in his comment above, is equal to 1, you have, $$\mathrm{pixel\space size\space in}\space y\mathrm{-direction} \space =\mathrm{pixel\space size\space in}\space x\mathrm{-direction} \space=\frac{RDistance}{\sqrt{2\left(\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2\right)}}.$$

If the aspect ratio is not equal to 1, then you have to use your image processing algorithm to find out,

$(A).\space$ the coordinates of the center of the $\mathrm{\color{blue}{blue\space circle}}$ directly above the $\mathrm{\color{brown}{yellow\space circle}}$ and
$(B).\space$ the coordinates of the center of the $\mathrm{\color{blue}{blue\space circle}}$ on the left hand side of the $\mathrm{\color{brown}{yellow\space circle}}.$

Denote the coordinates of the yellow circle and that of the blue circle above it as $\left(x_{yel}, y_{yel}\right)$ and $\left(x_{bluA}, y_{bluA}\right)$ respectively. Your image processing algorithm must give $x_{yel}=x_{bluA}$. We also assume that the analytical distance between these two circles is $RDistance_A$. Then, you shall write, $$\mathrm{pixel\space size\space in}\space y\mathrm{-direction}=\frac{RDistance_A}{\left|y_{bluA}-y_{yel}\right|}.$$

Now, denote the coordinates of the blue circle on the left hand side of the yellow circle as $\left(x_{bluB}, y_{bluB}\right)$. In this case, the image processing algorithm must give $y_{yel}=y_{bluB}$. We also assume that the analytical distance between these two circles is $RDistance_B$. Then, you shall write, $$\mathrm{pixel\space size\space in}\space x\mathrm{-direction}=\frac{RDistance_B}{\left|x_{bluB}-x_{yel}\right|}.$$

Answer 2

I would make an approximate solution like so: Let's assume that the number of pixels in each direction is large, so that the accuracy is sufficient. We don't know where the exact point lands within one pixel, so we have to approximate that the end points of the line are at the centre of a pixel.

Consider the line that connects the two points. It makes some angle with the horizontal axis, let's call that angle $\theta$. We get $$ \tan \theta \approx \frac{y_2-y_1 }{x_2-x_1} = Q $$ Calculate this auxiliary quantity $Q$, as it will be used later. If the width of one pixel is $\Delta x$ and height is $\Delta y$, we get $$ \cos \theta = \frac{1}{\sqrt{1+Q^2}} \approx \Delta x \cdot \frac{x_2 - x_1}{RDistance} $$ $$ \sin \theta= \frac{Q}{\sqrt{1+Q^2}} \approx \Delta y \cdot \frac{y_2 - y_1}{RDistance} $$ from which you can solve for $\Delta x$ and $\Delta y$. Can you continue from here?