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My question is as follows: if a (calibrated) camera is facing a perfectly flat surface (i.e. the optical axis is perpendicular to the surface), is the area represented by a pixel independent of the position of the pixel on the sensor?

I simplified the problem (1D instead of 2D) and made a figure of what I think models the issue:

enter image description here

The first horizontal line represents the camera sensor and the bottom one represents the flat surface. They are parallel and the camera optical axis (vertical line) is perpendicular to the two because of the hypotheses. The variable $c$ is a constant representing the width of a pixel, and $y$ is the projected length. The variable $x$ represents the position of the pixel on the sensor.

The question is now formulated as: does $y$ depends on $x$?

I applied two Thales theorems and found that $y = \frac{b c}{a}$, so $y$ seem to be independent of $x$:

Since $\frac{x}{d} = \frac{a}{b} = \frac{a_1}{b_1}$ and $\frac{c}{y} = \frac{a_1}{b_1}$,

then $\frac{c}{y} = \frac{x}{d}$ and $\frac{x}{d} = \frac{a}{b}$

so $y = \frac{c d}{x}$ and $d = \frac{b x}{a}$

finally $y = \frac{b c x}{a x}$ = $\frac{b c}{a}$

Are my modeling and my reasoning valid?

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For an ideal pinhole camera model, your geometric reasoning is correct: the projected pixel area on an orthogonal flat world-surface will be independent of sensor pixel position on the flat image-plane, and only depends on the object distance (b in your equation) and focal length (a in your equation).

Stated in other words, the crosssection slice of the 3D viewing-cone at a given depth z is scaled linearly on to the image projection surface.

The fact that this mapping scale depends on the object distance (vs focal distance) is the primary reason of the perspective projection distortion on the captured image with multiple layers of depths.

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  • $\begingroup$ Thanks for the answer, to state it in yet other words this means that any pixel represents say 1 cm2 of ground surface, that's it ? $\endgroup$
    – Louis Lac
    Commented Dec 7, 2021 at 13:40
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    $\begingroup$ @LouisLac yes all pixels repsents the same amount of area. Note that a "finite" area of object-plane is mapped into a finite area image-plane. $\endgroup$
    – Fat32
    Commented Dec 7, 2021 at 14:03
  • $\begingroup$ I would add that most real lenses do not fit the pinhole camera ideal, and that as the field of view gets wider, the chances for a good fit go down. $\endgroup$
    – TimWescott
    Commented Dec 7, 2021 at 16:12
  • $\begingroup$ @TimWescott indeed so for all practical lenses... $\endgroup$
    – Fat32
    Commented Dec 9, 2021 at 11:04
  • $\begingroup$ Well, I'm not sure about the "all practical lenses" statement. It kind of depends on how you define "practical" -- if you pull "serious art photography" into the definition of "practical" there are a few serious landscape photographers who use literal "tiny hole in a metal diaphragm" pinhole cameras in their work. $\endgroup$
    – TimWescott
    Commented Dec 9, 2021 at 15:46

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