Core 10 min 3 of 5

Fog is not one number

Two fields with the same mean β are not the same fog, and the mean does not determine the mean transmission.

What this repairs

Believing fog density is a scalar, so a model trained on uniform fog meets a fog bank and every metric was computed against a homogeneity assumption the world does not have.

Assumes

The Koschmieder model, honestly

Real fog comes in banks. It pools in dips, thins over warm tarmac, and drifts — so the density along one ray is not the density along the ray next to it, and a single $\beta$ is a claim about the atmosphere that is almost never true outdoors.

The pipeline models this by multiplying a base $\beta_0$ by a smooth random factor field: fractal Perlin noise, remapped into a range between a minimum and a maximum factor, with a contrast parameter and a correlation length quoted as a fraction of the image. The same machinery can modulate the airlight, together or independently. Optionally the field is normalised to hold its mean at $1$ , so that $\overline{\beta} = \beta_0$ regardless of how the field is shaped — which is the only way to compare a uniform fog and a patchy one at the same nominal visibility.

Same mean β, more light through

$t(u,v) = e^{-\beta(u,v)\,d}$

Transmission is exponential in the local extinction, so the map from β to t is convex — this one fact drives everything below.
$\mathbb{E}\!\left[e^{-\beta d}\right] \;\ge\; e^{-\mathbb{E}[\beta]\,d}$

Jensen’s inequality for a convex function, with equality only when β is constant. Spreading β around a fixed mean can only raise the average transmission, never lower it.
$\overline{\beta} = 0.0375\ \mathrm{m^{-1}},\; d = 60\ \mathrm{m} \;\Longrightarrow\; \overline{\beta} d = 2.25$

A concrete case: 80 m visibility, a surface at 60 m. Uniform fog gives a transmission of 0.106.
$\tfrac{1}{2}\big(e^{-0.5 \cdot 2.25} + e^{-1.5 \cdot 2.25}\big) = 0.180$

Now let half the frame sit at half the mean β and half at one and a half times it. The mean transmission is 70 % higher than the uniform case — same nominal visibility, measurably clearer picture, and that is before anyone looks at the variance.

Insight What "the same MOR" survives, and what it does not

Normalising the field to a fixed mean makes the parameter comparable. It does not make the images comparable, and the derivation above is why: a mean-preserving spread raises the mean transmission, so the patchy render is on average easier than the uniform one at the same stated visibility — while its densest patches are considerably harder than anything the uniform render contains.

Two consequences for evaluation. A frame-averaged metric will move in the optimistic direction when you add heterogeneity at fixed mean $\beta$ , which looks like the model coping and is not. And a per-region metric will show the opposite, because the worst region got worse. If a robustness claim rests on one number per image, heterogeneity is where it stops meaning anything.

Pitfall A 2D field is not a volume

The factor field is defined over the image, not over the world. Two things follow, and both are cheap to state and expensive to discover later.

The optical depth becomes $\beta(u,v)\,d$ rather than the line integral $\int_0^d \beta(s)\,ds$ that the physics asks for. Those agree only if the density is constant along each ray, which is exactly the assumption the heterogeneity was introduced to remove.

And the field is tied to the frame, so it is not view-consistent. Move the camera and the patches stay where they were on the sensor rather than where they were in the world; a stereo pair gets two different fogs; a video sequence gets patchiness that swims with the camera. A model trained on that can learn the swimming as a cue, which is a cue that exists in no real dataset.

Check yourself

Two datasets both report a meteorological visibility of 80 m; one is uniform, one is a mean-normalised fBm field. Which has the better average contrast, and does that make it easier?

Show answer

The heterogeneous one has the better average transmission, by Jensen's inequality — the convexity of $e^{-\beta d}$ guarantees it, and the example above puts the gap at $70\%$ for a surface at $60$ m. So on a frame-averaged metric it looks easier. Whether it is easier depends on where the objects are: the same field contains patches denser than the uniform fog, and an object behind one of those is harder than anything in the uniform dataset. The averages and the worst cases move in opposite directions, which is the whole reason to report both.