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When does a machine lose the object?

Meteorological visibility is one threshold. Your quantiser and your sensor impose two more, and they are shorter.

What this repairs

Reporting "trained with fog augmentation" as one knob, when β, airlight, illumination and camera response are four independent axes — and the model is only robust to the ones that were varied.

Assumes

The camera fights back
What the network actually predicts · Monocular Depth

Everything so far has been about how the image is formed. This is the question it was all for: at what distance does the object stop being recoverable — not to a human standing in the fog, but to whatever reads the delivered file?

The setup is the contrast law from the first concept. An object differing from its background by $\Delta J$ in scene radiance arrives at the sensor differing by $\Delta I = \Delta J\,e^{-\beta d}$ , because the airlight is common to both and cancels. That difference then has to survive being measured: it must be larger than the sensor noise at that signal level, and larger than the step between two adjacent code values. Two floors, both absolute, neither of which appears in the meteorological definition.

Three visibilities from one equation

$\Delta I(d) = \Delta J\,e^{-\beta d} \;\ge\; F$

Detection needs the arriving difference to clear a floor F. The floor is whatever is largest at that signal level: the sensor sigma, the quantisation step, or the compressor’s effective step.
$d_{\max} = \frac{1}{\beta}\ln\!\left(\frac{\Delta J}{F}\right)$

Solve for the distance. Note the shape: the reach is logarithmic in how much contrast you started with, so doubling an object’s albedo contrast buys the same distance as halving the noise.
$V = \frac{1}{\beta}\ln\!\left(\frac{1}{\varepsilon}\right) = \frac{\ln 20}{\beta}$

The meteorological visibility is the same formula with a relative floor — 5 % of the contrast you started with — rather than an absolute one. That is the only difference between the two numbers.
$\frac{d_{\max}}{V} = \frac{\ln(\Delta J / F)}{\ln 20}$

So the machine matches the meteorological range only when its contrast-to-floor ratio is exactly 20. With ΔJ = 0.25 and a floor of 0.02 the ratio is 12.5, and the machine loses the object at 84 % of the stated visibility. Each candidate floor gives its own distance, and the shortest of them is the one you actually have.
$\Delta d = -\frac{\ln 2}{\beta} = -0.23\,V \quad \text{per doubling of } F$

And the most useful form of all: every doubling of the noise floor costs 23 % of the meteorological visibility, at no change whatever to the fog.

Insight The camera settings move the visibility

Read that last step again with the previous concept in mind. Gloom raises the ISO; a higher ISO shrinks the well; a smaller well raises $\sigma$ ; and every doubling of $\sigma$ removes $23\%$ of the distance at which the object is detectable — while $\beta$ , the airlight and the visibility in the config all stay exactly where they were.

The same lever exists at the other end of the pipeline. Drop the JPEG quality and the effective step in a smooth region grows well beyond the 8-bit $1/255 \approx 0.4\%$ ; fog pushes the whole scene into a narrow band near the veil, where a compressor has least reason to spend bits and where $\log_2(1/0.2) \approx 2.3$ bits of range have already been given up.

So the machine's visibility is a property of the system, not of the weather. The fog sets $\beta$ . The camera sets the floor $F$ — and, through the local slope of the tone curve, how much of $\Delta J$ is still there by the time the floor is measured against it.

Pitfall Four axes, and robustness to only the ones you varied

"Trained with fog augmentation" names a knob that does not exist. There are at least four independent axes in this topic:

Extinction — $\beta$ , uniform or as a field.
Airlight — its brightness, its colour, its estimator, its dampening.
Illumination — the gloom EV and its near-field and fog-coupled terms.
Camera response — gain, ISO, noise, tone curve, compression.

A model is robust to the axes that moved during training and no others. Vary $\beta$ alone and you have a model that has seen exactly one airlight colour, one illumination level, and one noise level, all of which it may well have learned to rely on. The failure is not exotic: an airlight-only shift is what happens when the same fog is lit by sodium lamps instead of daylight, and the model has never seen it.

There is a connection back to monocular depth here that is worth making explicit. A depth network works because it has learned a prior over scenes, not because it solved the geometry — so what fog removes is not "a bit of contrast", it is the evidence the prior was keyed to: texture gradients, edge sharpness, apparent contrast falling with distance. The last of those is especially awkward, because contrast falling with distance is precisely the cue a monocular network uses for depth, and fog is a global multiplier on it. The network can read the fog as distance, which is right in a way that is not useful and wrong in a way that is hard to detect.

What to report, then, is not "fog augmentation: on". It is the range of each of the four axes, the camera profile the images were delivered through, and the metric evaluated on the delivered image rather than on the render. Anything less and the number is about a pipeline someone else cannot reconstruct.

Check yourself

To keep a foggy scene bright you raise the ISO. What happens to the distance at which a machine can still detect an object?

Show answer

It falls, logarithmically and predictably. Raising the ISO shrinks the well capacity, which raises the noise floor $F$ at the signal level of interest; the detection range is $\ln(\Delta J / F)/\beta$ , so each doubling of the floor costs $\ln 2/\beta$ , which is $23\%$ of the meteorological visibility. Note what has not changed: the fog, the config, and the visibility number anyone would quote for the dataset. This is the whole reason to distinguish the machine's visibility from the meteorological one.