Your calculation is absolutely correct. An image with 1616 pixels for example 101x16 has a surface of  2.39053 square mm if 1mm corresponds to the length of 26 pixels. So either this last assumption has to be false or your manual measurement.

if one square mm would correspond to 26 pixels then 5.1 pixel-lengths would correspond to 1 mm which would lead to a total area of about 62 square mm

Best regards,

Volker

The first answer makes perfect sense. Indeed, if you have 26 linear pixels per 1mm, even if it is a FULL HD screen (whose diagonal should be about 2203 pixels), it would have a diagonal of about 3.34 inch, which I would not expect from even high-end mobile phone screens.

However, if it were 26 square pixels per 1 mm^2, then, e.g., a FULL HD screen would have a diagonal of 43.2 cm, or about 17.009 inches, which is plausible. However, I do not understand where the error by a factor of 2 may come from. Maybe, aspect ratio, if, e.g., you manage to switch to 1920x1024 on an "almost-square" monitor, then the width of a pixel is just a little bit longer than 1/2 of its height.

You can look at iruler.net for an on-screen ruler, but it only shows it horizontally, because the original purpose was different. But it should still give you a hint.

Upd: Unless, of course you meant PRINTER's ratio, which would then be about 660DPI, which again is plausible (but then I do not have the answer to your original question)

What I tried was in an Image I kept a reference object(Scale) of 3cm. I calculated number of pixels for 3cm using euclidean distance formula. Using which I found the first measurement i.e. 1mm=26pixels

For second measurement I did segmentation of a region in the image and then calculated the pixels lying in that segmented region. It had 1616 pixels.

I was trying to find out a relation using first calibration for converting the segmented area in Sq.mm

Your calculation is correct given that the pixels are square, i.e, the pixel density is same in both dimensions also related to aspect ratio.

Still when i tried to measure the segmented area i was getting an approximate length of 5.77 and approximate width of 5.80 mm. Making the area around 33.466 Sqmm. But as per the above calculation it just showed 2.34

The surprising part is that your calculated answer diverges from the experimental one by such a large factor... Ok, so maybe your millimeters correspond to a photo or something like that. Clearly, the size of pixels in mm could vary depending on the angle, but surely not nearly by so much. I would verify the calculations, e.g., that you did not forget to take a square root in the Euclidean distance formula, and that you did not confuse any units (sorry, just the factor of the error suggests that something should be off by a multiple).

Just to check your first data, a 3cm object at 26 pixels per mm should be 780 pixels long, a rather close or magnified shot. And, since you mentioned Euclidian distance, this distance COULD be, e.g., a diagonal of a rectangle of 720 by 300 pixels.

To check your second data, 1616 pixels is, e.g., about 40 by 40 pixels. However, if the ratio of 26 pixels per mm is correct, 40 pixels would be only about 1.4mm, which is far less than 5.77 or 5.8mm. Alternatively, 5.77mm by 5.80mm at 26 pixels per mm would occupy about 150 by 150 square pixels (22500 is far more than 1616). Another alternative is if the dimensions of the segmented area and the corresponding number of square pixels is correct, then 1mm would be about 7 pixels long (not 26). Maybe this would help you find the source of discrepancy.

I have gone through all the replies and find that It's all depends on picture capturing system or imagery system's aspect ration. To calibrate your image with object size, you can perform a experiment.

Take a circular object's image and major its radius in all the 360 degree angle from center. Image of circular object is also circular then the multiplication factor is same in all direction otherwise calibrate it in all directions depends on resolution angle and number of pixels.

I would like to know the experiment results. 

I don't understand very well what you mean by

"I calculated number of pixels for 3cm using euclidean distance formula.".

If you have a reference object of a known distance you would decide for one line for example the middle or a border to measure.  If it is parallel to one of the image axis just count the pixels and then you know what number of pixels corresponds to 3cm. If it is not parallel, count one for each horizontal or vertical step and sqrt(2) for each diagonal step.

Can you explain what you mean by: "I calculated number of pixels..."? Or maybe you could make an image with the reference object available.

Best regards,

Volker

Sir just as you said I kept the reference object parallel to the image axis just and counted the pixels. I kept a same reference for all the images (3cm*3cm square). To measure distance in pixels i tried checking distance between end points of the border.

So what do you mean by

"But the actual answer when measured manually comes approximately around 31-32"

is this the imaged object that you measure or is this a measurement on the image?

Sir its an irregular shape object. I tried to find its max height and width and then approximated the area. Max height was around 5.8 ans width was 5.9 so the answer should be less than that. So i approximated it to be 31-32

If you do not have any reference inside the image (see attached file), it is practically impossible to convert pixel in mm because the pixel has no size. 

Firstly, calculate diagonal resolution in pixels by following formula:

  dp= sqrt (wp^2+hp^2) 

Now calculate Pixels per unit length by following equation:

 pixels per unit length= dp/diagonal length of screen

Diagonal length of screen means, length in mm, cm, etc of for example 1024*512-pixels screen.

Now by using formula of area of particular shape equivalent to circle you will be able to calculate area in pixels.

Where:

dp is diagonal resolution in pixels

wp is width resolution in pixels; here 512

hp is height resolution in pixels; here 1024

--

Thanks