Find Area of the object in mm^2 from binary image

Hi All, I have two questions related to each other. I have a binary image and I want to calculate the area of the object in mm^2. I have calculated the area in pixel by using this code

   [B,L] = bwboundaries(Bw,'noholes');
   stats = regionprops(L,'Area','perimeter');
   area = [stats.Area];

but I don't know how I can convert this in mm^2.

here is my original image in cartesian coordinate and size is 512 x 512.

I have converted the image in polar coordinate , its size is 256x256 . here it shown

can I calculate the length of the line in mm^2. can anybody please help me . Thanks in Advance

[jonas]

The length of the line in mm^2?

Answer 1

[Guillaume]

Unless you know the scale of your image (pixels/mm) it is impossible to convert pixel measurements such as area or perimeter to rea world physical units such as mm^2.

You would know the scale of your image either by knowning carefully measuring or calculating the magnification of your imaging system or by capturing an image of an object of known dimensions (under the same imaging conditions).

If you have done neither of these, you're out of luck.

[Alina tom]

How to calculate the scale of the binary image ? I have x and y coordinates of the line and just plot them . how I can calculate the line length in the lower image.

[Alina tom]

if i have the image size 512x512 and know the screen dpi is 96 then can i use this information to calculate area in mm^2?

[jonas]

You have to understand that you cannot calculate the length in any other unit than number of pixels, because you need a scale for the conversion between pixel size and real world size [m]. I mean, for all we know your picture could be anything from the death star (several miles in diameter) to a grain of sand (a few microns). If not even a human can tell the difference, how are you supposed to calculate it?

The dpi is a measure of resolution, not real world scale. With the dpi you could calculate the length of your segment if you print it on a piece of paper.

[Alina tom]

Thanks Sir for you replay. Actually i have calculated these images from coronary artery OCT images . this is the original image

 and this is the calculated inner area of the artery in binary

i know the axial resolution of OCT light which is 20 µm and lateral resulation which is 30 µm and scan diameter is 8.3 mm . can this information uses to get area in mm^2?

[Alina tom]

Here is the scale for OCT image . is this helpful

[Guillaume]

"i know the axial resolution of OCT light which is 20 µm and lateral resulation which is 30 µm and scan diameter is 8.3 mm . can this information uses to get area in mm^2"

I know nothing about optical coherence tomography and have no idea what scan diameter refers to. I also don't know what axial and lateral mean in the context of your image.

If the measurements you give are indeed the resolution of your whole imaging system, then yes you could use that information since you now know that one pixel correspond to a physical area of 20x30 μm² (maybe! depending on what axial/lateral mean).

[Alina tom]

if we consider this scale is true . then how i can use this to calculate the area in mm^2? can you please help me

[Guillaume]

Well, if a pixel correspond to a physical size of 20x30 μm², then simply multiply your area in pixels by 600 to get the area in μm².

However, note that I strongly suspect that I've not understood your description correctly and that the scale of the pixels is not 20x30 μm².

Bottom line: you need to understand your imaging system and find out the mapping between pixel size and real world size. It's not something we can figure out for you.

[Alina tom]

I have check it again . the scale for our imaging system is

1024 pixels = 9.5 mm

now the size of image is 512x512.

I can calculate the area in pixels like this and then multiply area with 4.75. like this

   [B,L] = bwboundaries(Bw,'noholes');
   stats = regionprops(L,'Area','perimeter');
   area = [stats.Area];
   AreaMM = area*4.75;

is this correct?

[Walter Roberson]

one pixel is 1/1024 * 9.5 = 0.00927734375 mm then, at least in that direction. That would make one square pixel about 8.607E-5 mm^2 .

[Angelo]

thank you! very helpful for my case as well