How to figure out the resolution and KB size of an image

I am no photographer or photo editor. But when cropping or saving images of observations, I have been wondering about the target image size or resolution of image files.
So, for example, given this observation (not mine). The EXIF info of the first image says its width is 4032 pixel, the height 3024, and no other file specific info. How to calculate the resolution, and the its file size?
Thanks

The resolutoin you already mentioned (4032x3024). The file size depends on the compression used and other factors. There is no general formula.

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iNat resizes images, to 2048 pixels in the biggest dimension. The EXIF data will only give you the original dimensions on upload.

As far as sizes go, when you convert a picture to a non-lossless format (like JPEG), it is compressed, thus the final size doesn’t just depend on how large the image is (a picture where all pixels are the exact same colour will be much lighter, for an extreme example).

Once you have the size and compression, on screen what you see is what you get. The resolution in DPI/PPI (dots/pixels per inch) is simply your pixel size divided by the size of your display/print, so if you want to maintain for example 300 DPI, you can work out the maximum physical size at that resolution from the pixel size.

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these are the dimensions of image. I remember that resolution uses the “dpi” suffix.

I think I understand your explanation. Though I am still confused when editing images.
I have just used MS Paint to create a rectangle of 400x300 px. The size in inches is 3.15x4.22. The program says that the image resolution is 96 dpi.
How is the pixel size relate to the dpi then?

The default system-wide scaling Microsoft uses for most displays on Windows is 96 dpi. You can tweak it, for example if you have a screen that has a higher pixel density, like some phones or laptops, or the opposite if you want everything to be bigger for accessibility reasons for example.

If you do the maths, 400 px / 4.22 inches = 94.79 dpi and 300 px / 3.15 inches = 95.24 dpi, close enough for government work as they say.

The display resolution for a Samsung S22 is 425 dpi/ppi, so that same 400x300px image on that screen would be in the default settings 400/425=0.94 inches by 300/425=0.71 inches. If you or the software you’re using force that phone to display your image at 96dpi, you’ll get back the previous dimensions in inches by stretching the image, “cloning” pixels to fill up the gaps.

The main context where dpi matter is when printing, since stretching the image will result in a loss of quality. The standard for printing is 300 dpi, so your 400x300px image could only be 1.33x1 inches while meeting that standard. If you go bigger you’d get jagged pixels.

The dpi value MS Paint displays isn’t a property of the image itself, this is what I meant when I said “what you see is what you get”, a 400x300px image which MS Paint calls 96 dpi and a 400x300px which it calls 300 dpi are the exact same, the value (as well as any size in inches it might say) is only given so that other programs that work in inches or both pixels and inches can figure out what to do with that image in different contexts (MS Word when you’re working with a “real page size” like A4 or US letter, for example).

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