Do I even need camera calibration for length measurements with one camera?

P.S. Regarding what I meant with:

This is an image out of a lecture. It shows columns placed parallel to the image plane. Interestingly the outer ones appear bigger, even if they are all equally sized.
[Picture 1] (I am only allowed to upload one picture to I stuffed all the grafics into one single image.)

Also a sketch regarding this “effect” I made my own:
[Picture 2]

Here is a model I made in a CAD Programm: (I turned on perspective projection.)
[Picture 3]

For one you may see that, even if the two rods are of the same diameter, the left one appears like it has a greater diameter. (At least slightly.)

Still the two rods are in the same plane. (At least they are touching the same plane. Maybe the issue is that they are “standing out” of that plane.) Did I get you right and mapping a homography to the four red corner points would solve this “issue”?
Definition of “issue”: The two rods of same diameter in the real world appear as having different diameters in an image.

In this one the “measuring plane” is placed parallel to the image plane:
[Picture 4]

It is already “rectified” (Turned into a rectangle :wink: ). So I guess this would be the expected output of applying a homography?

Still D_1 and D_2 are not equal. So to my question: If I know the diameter D_1. Am I able to calculate L_1? Or do I need an object of known length call it “L_reference” in the direction of L.
Since D_1 in the image doesnt equal D_2, measured in pixels. Even it the two rods are modeld with the same diameter. So calcualting e.g. 10 Pixels is one milimeter by counting the pixels along D_1, knwoing its 8 mm I would get a different result for D_2!
I guess measuring “3D” objects is harder than measuring flat objects like coins :confused:

Furthermore I succesfully confused myself now.

Thanks to all of you in advance. Especially @Steve_in_Denver and also to @crackwitz

That whole topic drifts away from my original question.