This is the equation regulatory air quality runs on, worldwide. Before using it, say its assumptions out loud: steady wind, steady emission, flat terrain, no chemistry, no deposition, and a concentration profile that is exactly normal in both crosswind directions. None of them is true. The model survives because its errors are known and roughly conservative — a factor of two on an hourly average counts as good agreement, and anyone quoting it to three figures has misunderstood what they are holding.

The plume's width comes first: $\sigma = a\,x^{b}$, the Pasquill–Gifford power law. $x$ is the downwind distance in metres, $a$ and $b$ are bare fitted numbers read off a table for the hour's stability class (A very unstable, D neutral, F stable), and $\sigma$ is the spread in metres. Two spreads are needed and they have their own separate coefficient pairs: $\sigma_y$ — sigma-y — is the crosswind spread, and $\sigma_z$ — sigma-z — the vertical one.

Then the plume itself: $C = \dfrac{Q}{\pi \sigma_y \sigma_z u}\,e^{-H^{2}/(2\sigma_z^{2})}$ — C equals Q over pi sigma-y sigma-z u, times e to the minus H squared over two sigma-z squared. $C$ is the ground-level concentration on the centreline, $Q$ the emission rate in g/s, $u$ the wind at stack height in m/s, and $H$ the effective stack height in metres — steel plus rise, the number the last lesson built. Note the $\pi$ and not $2\pi$: the ground reflection is already folded in, because at ground level the real plume and its mirror image contribute equally and the 2 cancels.

The exponential is where the whole answer lives. It asks how many vertical spreads the plume is sitting ABOVE the ground, and it falls off ferociously: at $H = \sigma_z$ it is 0.61, at $H = 2\sigma_z$ it is 0.135, at $H = 3\sigma_z$ it is 0.011. Drop it and you have not made a small error — you have computed the concentration up on the plume centreline, hundreds of metres over the receptor's head.

And the closed form for the worst case, at whatever distance it happens: $C_{max} = \dfrac{2Q}{e\pi u H^{2}}\cdot\dfrac{\sigma_z}{\sigma_y}$. The $e$ is there because the maximum sits exactly where $\sigma_z = H/\sqrt{2}$, and the exponential is precisely $e^{-1}$ at that point. Read the $H^{2}$ and the economics of the whole industry falls out: double the effective height and the worst hour is quartered. That is why plume rise is worth more than steel, and why a scrubber that cools a plume can make the ground-level number worse while cleaning the gas.