How can i find the resolution of an image?

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Carole 2012-11-16

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评论： Palguna Gopireddy 2022-11-27

Hi, how i can find the resolution of an image with matlab (number of pixels in millimeter). thanks

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Image Analyst 2019-8-24

在 MATLAB Online 中打开

See solutions below. What imfinfo gives you might not always contain xresolution, and when it does, it may not be correct or what you think. For example

fi = imfinfo('cameraman.tif')

gives 72 inches. Obviously not correct or even applicable. However, it might be correct depending on the image capture system.

Walter Roberson 2019-8-24

In my experience, XResolution and YResolution information returned by imfinfo() is wrong much much more often than it is correct.

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Thomas 2012-11-16

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编辑：Thomas 2012-11-16

在 MATLAB Online 中打开

Use 'imfinfo'

E.g.

 info = imfinfo('new1.jpg')
info = 
         Filename: [1x95 char]
      FileModDate: '01-Oct-2001 17:19:44'
         FileSize: 27387
           Format: 'jpg'
    FormatVersion: ''
            Width: 600
           Height: 650
         BitDepth: 24
        ColorType: 'truecolor'
  FormatSignature: ''
  NumberOfSamples: 3
     CodingMethod: 'Huffman'
    CodingProcess: 'Sequential'
          Comment: {[1x69 char]}

There you can fine the resolution widthxheight bit depth etc..

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Walter Roberson 2012-11-16

Good point, Eric. Carole, is the image of a single spot or is it of a larger area that you moved along?

Carole, XResolution is an optional EXIF tag. In most cases it is not known or not meaningful.

Carole 2012-11-16

Sorry for the stupid questions but I like to understand. So what's the solution to convert a diameter of a region in pixel to millimeter. is there a trick using a regle of three for example? or what is the parameter necessary to make this conversion?

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Walter Roberson 2012-11-16

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Most often, you cannot figure out what the resolution of an image is.

If you are using a better-quality camera in auto-focus mode, the EXIF information might include the distance to the focus. If you have that and the lens aperture, then by knowing the height and width in pixels, you can calculate the angle subtended by the pixel range, and thus the angular resolution. (Perhaps you do not need the aperture; I have forgotten the details of the calculation.)

If you are using a medical device such as CT or MRI, then the images for those are usually created as DICOM images, for which there is DICOM meta-data that includes the resolution in the PixelWidth tag.

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Carole 2012-11-16

编辑：Carole 2012-12-20

I do not know the resolution. I want to calculate the resolution from the image. Is this possible??

Eric 2012-11-16

编辑：Walter Roberson 2019-8-24

It looks to me like the device has no imaging capability ( http://www.delasco.com/pcat/1/Diagnostic/Delta20/Delta20/ ). You need to pair it with another camera. I would say your best bet would be to do your own calibration of the systems. You could buy a resolution target (something like an Air Force 1951 resolution target from Edmund Optics) that has small features of known sizes. Image them and determine what the resolution is that way.

-Eric

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Carole 2012-11-17

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编辑：Carole 2012-11-17

I'm looking for the solution to convert a diameter of a region in pixel to millimeter. is there a trick using a regle of three for example? or what is the parameter necessary to make this conversion? I have these informations but I don't know how can I use it:

-the resolution of the device is 41.1 megapixel

-1 megapixel = 0.264 mm

-1 dpi = 3.937 pixels/mm

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Walter Roberson 2012-11-18

How about you tell us the make and model of what you are using, so we can see if we can understand the manual?

Palguna Gopireddy 2022-11-27

@Carole.

If you have studied the Digital Image Processing by Gonzalez Book. You understand that 'Resolution can't be explicitely mentioned using the pixels. It has to be mentioned along with dpi value'. For exmaple 512*512 image with 72dpi has better resolution compared to 512*512 image with 36dpi

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Image Analyst 2012-11-18

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编辑：Image Analyst 2012-11-18

Carole, Sure, go for it. So multiply any lengths in pixels by 0.022 to get length in mm, and multiply and area in pixels by 0.022^2 to get area in mm^2. Of if you know the length and height of the total image field of view then you can use that too. For example, if the height = 960 pixels and it's 21 mm. Then the spatial calibration factor is 21/960 = 0.021875 mm per pixel. So let's say an object is 342 pixels long, well that's 342 * 0.021875 = 7.48125 mm long. If the area = 20,000 pixels, it's 20,000 * 0.021875^2 = 9.57 mm^2 in area.

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Image Analyst 2012-11-19

No, that's not. You can't use any number that you look up somewhere. They have no idea what you imaged or the spatial calibration that may be needed. You can't trust the camera either unless it's a calibrated system like CT, MRI, or Microscopy where they make sure to embed the accurate spatial calibration in the header or EXIF info. Let's say that you took a picture of me standing in front of a bunch of trees with the moon overhead. Your camera will tell you the resolution in terms of pixels - how many pixels tall and wide your image is, but the number of millimeters one pixel corresponds to depends on what you're looking at. Even though you have the same camera with the same focal length and same CCD size, one pixel on me might be 2 mm, while one pixel on the tree in the distance behind me might be 200 mm, and one pixel on the moon might be 200 kilometers if you were measuring something on the moon. That's why you need to measure some known length (like a ruler) and that's why that spatial calibration only applies in the imaging plane where the thing was measured, not for anything closer or farther away. Does that help illustrate the issue to you better?

Walter Roberson 2012-11-20

Especially a picture of something of known size, such as a ruler.

Image Analyst 2012-11-20

The how did you get the pictures? Did you get them from that dermascope or not? If not, where did they come from and what does the dermascope have to do with anything?

And the print size has nothing at all to do with any kind of distance in your image. I could print a picture of a galaxy and a picture of a virus both on the same size paper printout with the same size image pixel dimensions. But of course their spatial calibration is vastly different.

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TIAN YUAN WANG 2017-6-30

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I'm sure you can get the FOV size from the Dicominfo.

If the FOV is 400 mm, matrix is 512 by 512; then the Resolution = FOV/matrix = 400/512 = 0.7813 mm/pixel.

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Walter Roberson 2017-6-30

Unfortunately the person was working with a png not with a dicom image.

Looking at the fov is typically the wrong way to get sizing from dicom images. The best way depends upon what kind of image it is. I do not recall the best way for CT. MRI images have a pixel centre to centre for x and y that can be used. (The tricky part is determining the z resolution)

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Antonis L 2020-6-16

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Ιs it possible to find the real image dimensions of a .jpeg, through the XResolution and the YResolution given from the imfinfo function, without the pixel/mm reconstruction factor? Τhank you in advance

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Image Analyst 2020-8-4

spatial_calibration_demo.m

No. It's not possible. There is no such conversion that works for all or most jpg images and all/most fields of view. You have to calibrate your specific image. See attached spatial calibration demo.

Walter Roberson 2020-8-5

You would need to know the distance between the image sensor and the lens:

https://www.physicsclassroom.com/class/refrn/Lesson-5/The-Mathematics-of-Lenses

https://www.nayuki.io/page/absolute-and-relative-lens-apertures

When f is the focal length and

is the distance between the sensor and the lens, and

is the distance between the lens and the object being photographed, then

By knowing the image sensor physical size, and its resolution, and the portion of the image that the object of interest occupies, you can determine the physical size of the sensor portion covered by the projection of the image. Then you can use the recorded readings such as aperture and distance to the target together with the optical formulas, to figure out what the physical size of the distant object must have been. You then combine the physical size of the object with the number of pixels needed to represent it to find the resolution.

Alternately if you know the physical size of the object and the number of pixels needed to represent that size, you can directly calculate resolution.

The further you get from the object, the more imprecise the calculations get. It doesn't take much in casual photography before