Core 11 min 3 of 4

Intrinsics: from metres to pixels

K is an affine map applied after the division. Each entry is a fact about a sensor.

What this repairs

Reading $K$ as "the camera matrix" without being able to say which entry changes when you crop, resize, or switch to a different lens.

Assumes

The pinhole model

The pinhole equations produce a position on the image plane in metres, measured from the optical axis. A file on disk stores pixel indices from the top-left corner. The intrinsic matrix is exactly the conversion between those two, and nothing else:

$K = \begin{bmatrix} f_x & s & c_x \\ 0 & f_y & c_y \\ 0 & 0 & 1 \end{bmatrix}, \qquad \begin{pmatrix} u \\ v \\ 1 \end{pmatrix} = K \begin{pmatrix} x_n \\ y_n \\ 1 \end{pmatrix}, \qquad x_n = \frac{X}{Z},\; y_n = \frac{Y}{Z}.$

Note the order. The division happens first, producing normalised coordinates $(x_n, y_n)$ ; then $K$ acts, and it is a plain 2D affine map — scale, shear, translate. Nothing about $K$ is perspective. It is a units conversion that happens to be written as a matrix.

Entry by entry:

$f_x, f_y$ — focal length in pixels. This is the physical focal length divided by the physical pixel pitch, which is why " $f = 520$ " is meaningless without an image size and why the number changes when you resize an image. $f_x \neq f_y$ means non-square pixels; on modern sensors the ratio is $1$ to within a fraction of a percent, and a calibration that returns $1.04$ is telling you about your calibration, not your sensor.
$c_x, c_y$ — the principal point, where the optical axis pierces the sensor. Nominally the image centre; in practice off by a few to a few tens of pixels from manufacturing tolerance. It moves when you crop.
$s$ — skew, non-rectangular pixels. Zero. It has been zero for thirty years. Its only use is as a sanity check: if your calibration returns a large skew, one of your other parameters is absorbing an error.

Pitfall Resizing an image changes K, and not by a scalar

Scale an image by $\alpha$ in width and $\beta$ in height and the correct new intrinsics are $f_x' = \alpha f_x$ , $f_y' = \beta f_y$ , $c_x' = \alpha c_x$ , $c_y' = \beta c_y$ . Cropping is different again: focal lengths are unchanged, and only the principal point moves, by the crop offset. Getting this wrong is the single most common cause of a reconstruction that is subtly, uniformly wrong in a way no residual plot makes obvious — because the reprojection error can stay small while the geometry is bent.

There is a half-pixel question hiding here too: whether pixel $(0,0)$ is centred at $0$ or at $0.5$ . Conventions differ, the discrepancy is half a pixel, and it matters exactly when you are chasing the last half pixel.

Focal length and field of view

$\tan\left(\frac{\theta_h}{2}\right) = \frac{W/2}{f_x}$

Half the image width subtends half the horizontal field of view, at distance f_x. Both in pixels, so the units cancel.
$\theta_h = 2\arctan\!\left(\frac{W}{2 f_x}\right)$

The usable form. Note it depends on the image width — a focal length alone never determines a field of view.
$f_x = \frac{W}{2\tan(\theta_h/2)}$

And back, for when a datasheet gives you degrees and your code wants pixels.

Check yourself

You centre-crop a 1920×1080 image to 1280×720. What are the new intrinsics?

Show answer

$f_x$ and $f_y$ are unchanged — cropping removes pixels, it does not change the angle any remaining pixel subtends. The principal point moves by the crop offset: $c_x' = c_x - 320$ , $c_y' = c_y - 180$ . The field of view is smaller, but that is because $W$ and $H$ shrank in the formula, not because $f$ changed.