Exactly. And Lightroom and some image editors will allow you to export an image with a desired value for the "long edge" (rather than for desired width or height). It keeps your resize operations much more consistent, assuming you don't do a lot of cropping.
If you know the actual width or height and camera-to-object distant, you can easily calculate a ratio size-to-distance.
Say a object is 3 meters wide and a picture is taken with the camera positioned 12 meters from the subject. The ratio size-to-distance is 3÷ 12 = 0.250.
Inside the camera, the image forming rays from the lens, trace out the same ratio as ...
With conventional CMOS sensors the issue isn't so much the spacing between photosites (pixels). It's all of the circuitry on top of them (usually on the edges of each photosite) that blocks light from getting through to the semiconductor layer. What microlenses do is redirect light that would otherwise fall on the edges of photosites where the circuitry ...
I have no idea if the 1/4" description is valid for your camera, but in general, the tiny CCD digital sensors are often described by comparing their size to the video vacuum tubes used until about 1980. It represents the outer glass tube diameter, not the actual sensor portion of it.
Wikipedia at https://en.wikipedia.org/wiki/Image_sensor_format#...
This is not a mathematical aproach but an empiric one. Draw an example on a vector based program.
Figure 1 is a 24mpx representation of the object and text you describe. Download to see them in real size.
The text is 5/200 of the total length of the image. It is the same as 1/40. So if a 24 Mpx has 6000px width this text will have 6000/40=150px.
Figure 2 ...
I'm not very sure, but here goes
Scale factor=f/d //f is the focal length in metres
x'=x*f //x' length of image in metres
y'=y*f //y' breadth of image in metres
Convert dot per inch into dots per metres
Convert from metres to pixels