Core 11 min 2 of 5

Airlight is an estimate

A is scattered light with a path length behind it, an estimator in front of it, and a dependence on β.

What this repairs

Treating airlight as "white", or as a constant that has nothing to do with β, the sky, or where the camera happens to be pointing.

Assumes

The Koschmieder model, honestly
Intrinsics: from metres to pixels · Projective Geometry & the Camera Model

Airlight is not a colour someone picked. It is the ambient light scattered into the line of sight along its path, accumulated over that path — which is why it appears in the equation weighted by $1 - t$ rather than as an additive constant.

That gives it two properties an opacity knob does not have. It saturates: only a path long enough to be optically thick reaches the full ambient radiance $L_s$ . And it is a colour, inherited from the light in the scene — overcast daylight, sodium street lighting, a sunlit fog bank — so a fog render whose airlight is pure white has quietly asserted a specific and usually wrong illuminant.

Why thin fog must not inherit the full sky

$L_A(d) = L_s\big(1 - e^{-\beta d}\big)$

The veil term on its own: the airlight actually accumulated by a ray of length d. It is the part of the fog equation that does not depend on the scene at all.
$\beta d \ll 1 \;\Longrightarrow\; L_A(d) \approx L_s\,\beta d$

Expand for a short optical path. In thin fog over a scene a few tens of metres deep, the veil is linear in β and nowhere near the ambient radiance — the horizon simply has not had enough path to reach it.
$f(\beta) = f_{\min} + \frac{f_{\max} - f_{\min}}{1 + \kappa\,\beta/\beta_{\text{ref}}}$

The pipeline cannot integrate that per scene, so it applies a monotone dampening factor to the estimated airlight instead — a stand-in for the saturation above that needs only β.
$f(0) = 1, \qquad f(\beta_{\text{ref}}) = 0.45 + \tfrac{0.55}{2} = 0.725, \qquad f(\infty) = 0.45$

The shape a figure should be able to assert: full ambient brightness in clear air, exactly halfway at the reference β (80 m visibility) with strength 1, and a floor it approaches but never reaches. Note that it is applied to estimated airlight only — a colour an author typed is left exact.

Pitfall Three estimators, three ways to be wrong

The pipeline can estimate the airlight three ways, and each fails on a recognisable class of image:

From the sky. The mean of the pixels a sky mask marks. Correct when the mask is correct, and its fallback when there is no sky in frame — or no mask — is white, which is the single most common source of a whiteout render.
Dark channel prior. Take the per-pixel minimum over channels, find the brightest of those, and read the airlight there. It is looking for the most haze-dominated region, and it is fooled by anything large, bright and achromatic: a white truck, a lit billboard, a snowbank.
Dark channel with a heuristic. The same estimate blended toward a sky-like prior, which trades a little accuracy on clean images for not falling over on the pathological ones.

None of them can be checked from the estimate alone — one RGB triple looks as plausible as another. The only audit is to go back to the pixels it was computed from: which pixels the mask marked, which ones the dark channel picked, how much weight the prior got. An airlight estimate you cannot attribute to specific pixels is a number you are trusting for no reason.

Then there is the sky, which has no depth at all. Nothing in a sky pixel is at a finite range, so $t = e^{-\beta d}$ has nothing to evaluate — and leaving those pixels at the far clip renders a sky of uniform airlight with a hard line where the geometry ends.

The fix is geometric rather than cosmetic. Treat the fog as a layer of finite height, build the ray direction for the pixel from $K^{-1}$ , take the camera's up-vector from its pitch and roll, and use the length of that ray's segment inside the layer as $d$ . Rays that climb steeply leave the layer quickly and get a short path; rays near the horizon stay inside it and are capped. The sky gradient is then a consequence of the geometry instead of a texture.

Insight The sky path in three numbers

Take a camera on the ground under a fog layer $80$ m deep, with a cap of $1000$ m on the path any single ray may accumulate:

Straight up, uniform density: exactly $80$ m — the layer height, since the ray crosses it perpendicularly.
Straight up, with the density tapering linearly over the top $10$ m: $75$ m. The taper removes $10^2/(2 \cdot 10) = 5$ m of vertical optical depth, and that quadratic bite is what stops the horizon having a hard edge across it.
Horizontal: the ray never exits the layer, so it takes the $1000$ m cap. That cap is the reason the horizon saturates to a constant instead of running to infinity.

Note what the camera's own attitude does here. The up-vector at zero pitch is $-Y$ , because the camera frame is OpenCV's and $+Y$ points down; pitch the camera up and the pixel that looks vertically upward moves, so the entire sky gradient tilts with it. A fog sky that does not move when the camera pitches is not a fog sky.

Check yourself

You render fog with the from-sky estimator on a frame that has no sky in it — a tunnel approach, or a camera pitched at the road. What comes out?

Show answer

A white airlight, and therefore a whiteout. With no sky pixels to average, the estimator falls back to white, and the render composites toward pure white rather than toward the actual ambient colour, which in a tunnel approach is dim and strongly tinted. The dampening factor does not save it: dampening scales a colour's brightness, so it makes a wrong colour darker rather than making it right. Either supply a mask that covers the real ambient source, use the heuristic estimator, or set the airlight explicitly — in which case, note that the dampening no longer applies at all by default.