How to calculate the area of this object in an image?

[Walter Roberson]

Is the "object" everything that is black?

Is the "object" everything that is 8-connected to the largest black area -- so excluding the small stand-alone black areas?

Is the "object" the white area inside the black area?

[Muhammad Fiky]

the "object" is only the black-colored area surrounding the white circle. This means the "object" does not include the white circle area inside.

Answer 1

[Walter Roberson]

"area" could potentially refer to a physical quantity rather than just a pixel count.

In order to calculate as a physical quantity, either you have to have existing information about the physical sizes of the pixels, or else you need to have some reference information in the image that you can relate to physical quantities as a calibration. For example if there were a ruler in the image that was known to span 1 mm then the size of the ruler could be determined and knowing the associated physical size, the physical dimensions of the represented object could be calculated.

filename = "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1565304/image.jpeg";
img = imread(filename);
bw = im2bw(img);

object_pixels_count = nnz(~bw)

object_pixels_count = 142307

%for the purposes of illustration, we will not assume that individual
%pixels are square, and we will assume that **somehow** you know the actual
%physical size of each pixel
pixel_height = 2.54/80;
pixel_width = pixel_height * 4/3;

%then
object_area = object_pixels_count * pixel_height * pixel_width

object_area = 191.2725

[Muhammad Fiky]

Okay, I got the way to count the pixel dimension (area). Then I have few more questions that confuse me.

1. Does the pixel always come in square shape (in which height and width are the same size)?

2. In your code above, how do you measure the pixel height and width?

3. Why pixel height is 2.54/80 and pixel width is pixel height * 4/3 ? Is that just an illustration or assumption?

[Walter Roberson]

Your data might potentially have come from any of a number of different sources.

1) If it came from a standard digital camera, the underlying sensor locations are fairly likely square these days.

Hypothetically you might be using an anisotropic lens that caused concentrated external space more in one direction than in another. For example you might be using a spectrograph in which the useful information naturally is longer than it is wide, and rather than build a custom camera or use multiple cameras, the experimenters might choose to use a lens that squeezed the aspect ratio of the useable data into the aspect ratio supported by the camera.

This is not typically the case. More typically you would run into a different problem, which is that unless you are using a fairly short focal pane and a close lens, that what is being captured off-centre is probably further away than on-centre, so although the sensor pixels might be square, what is being recorded towards the edge might be gathering light from a larger physical area.

When I look at your image, I suspect "pupil" -- and pupil images are taken from quite a short range where the angular spread does not convert to much real-world difference in distance.

2) when you get to the point where you are not sure whether to call your recording device a "camera" or a "scientific instrument" then the assumption of square pixels for the sensors starts getting less doubtful. Ultra-high speed plane-shutter cameras do not necessarily use square sensor pixels for example.

3) devices such as MRI and CT and ultrasound do not record images directly: they record data that gets processed into images. The pixels for such device are not necessarily square. (When we were doing MRI work, we typically configured for x and y resolution equal and z twice as far apart -- and that meant that when slicing along the z, the pixels were not square.)

Why pixel height is 2.54/80 and pixel width is pixel height * 4/3

Those are made-up numbers.

72 pixels per inch is a common nominal resolution on Mac, with 100 pixels per inch being the more common nominal resolution on Windows. 72 arises out of historical typeface measurement systems that defined a "point" as 1/72 of an inch (and then a "12-point font" would be 12/72 of an inch according to a certain measurement.) But there has been some moves to redefine "points" in terms of 1/80 of an inch, at least for some purposes. And of course 2.54 is mm per inch -- so 2.54/80 would be mm per point. 2.54/72 or 2.54/100 might get used instead for similar reasons.

By comparison, a Canon EOS 5D Mark IV sensor is physically 36 mm x 24 mm with a resolution of 6720 x 4480, which is about 187 pixels per mm on the chip . But that's the chip-level resolution not taking into account optical path and lens: the field being recorded is probably not 36 mm x 24 mm (for example a camera recording pupils might have a 3 x magnifying lens, directing 12 mm physical width to the 36 mm sensor.)

Unless you are using a scientific instrument that has been pre-calibrated, or a scientific instrument specifically designed to record relevant lens and focal length information, then typically to find physical pixel sizes you would calibrate your system -- or you would have something in the field of view that would act as reference marks.

[Muhammad Fiky]

i use this code to obtain the number of pixels in a unit length.

d=imdistline;

the result is that line on the bottom left, showing 105.96 amount of pixels at 0.2 mm (actual length). With this in mind, and believing that my pixels are square-shaped, what are the changes on the MATLAB code above (pixel height and pixel width)?

The following is the actual picture taken from a digital microscope. I went to some lengths to transform the actual picture into black and white using Corel.

[Image Analyst]

See my attached spatial calibration demo to convert pixels into mm.