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The first-order combination of the Kelvin equation and Raoult's law for a chemical solution droplet with $A/r\ll 1$ and $B/r^3\ll 1$. Its maximum is at $r_c=\sqrt{3B/A}$, with $S_c-1=2A/(3r_c)$. For ambient saturation ratio $1<S<S_c$, the smaller equilibrium radius is stable and the larger is unstable: the sign of condensational growth is the sign of $S-S_{\mathrm{eq}}(r)$. In particular, exceeding $r_c$ alone need not activate continued growth at subcritical atmospheric supersaturation.

The equilibrium saturation ratio over a chemical solution droplet as a function of wet radius for fixed solute content. In the dilute radius approximation $S_{\mathrm{eq}}\simeq 1+A/r-B/r^3$, its maximum occurs at $r_c=\sqrt{3B/A}$. The corresponding critical supersaturation is $2A/(3r_c)$. Diameter and radius coefficients differ and must not be mixed.

$S=e/e_s^l$ is the ambient saturation ratio over a flat liquid-water surface; $S-1$ is the liquid atmospheric supersaturation. Curvature increases the surface equilibrium vapour pressure, producing the Kelvin penalty $A/r$, while dissolved solute lowers it and gives the opposing Raoult contribution $B/r^3$. Thus the dilute Köhler equilibrium curve is $S_{\mathrm{eq}}-1=A/r-B/r^3$, and the numerator is the atmospheric supersaturation available to drive growth beyond surface equilibrium.

$F_d^l$ describes the resistance of vapour diffusion through the surrounding air. $F_k^l$ describes the thermal-conduction resistance to removing latent heat released by condensation: a warmed droplet surface has a higher equilibrium vapour pressure and hence a weaker mass-transfer drive. Common continuum approximations are $F_d^l\simeq R_{v}T/(D_v{e}_s^l)$ and $F_k^l\simeq L_v[k_{\mathrm{air}}T{]}^{-1}[L_v/(R_{v}T)-1]$, showing the roles of diffusivity, conductivity, saturation pressure and latent heat. Both resistances have the supplied units $\mathrm{s}\mathrm{m}\mathrm{k}{\mathrm{g}}^{-1}$ and add for this coupled transport problem.

For a spherical droplet of constant density, $m=4\pi \rho r^3/3$. Multiplying the radius equation by $4\pi \rho r$ givesAt sufficiently large radius, curvature and solute corrections are negligible, soThis limit also requires sustained positive atmospheric supersaturation. It need not describe the earliest submicrometre stage or an unactivated haze particle.

Let $1<S<S_c$ be the ambient saturation ratio. There are two equilibrium radii: $r_{-}<r_c<r_{+}$. Condensation occurs if $S>S_{\mathrm{eq}}(r)$; evaporation occurs if $S<S_{\mathrm{eq}}(r)$. Therefore droplets with $r<r_{-}$ grow towards $r_{-}$, while those with $r_{-}<r<r_c$ shrink towards $r_{-}$. The smaller equilibrium is stable.

Above $r_c$, droplets with $r_c<r<r_{+}$ still undergo evaporation and ultimately approach $r_{-}$. Droplets with $r>r_{+}$ grow, and the larger equilibrium $r_{+}$ is unstable. Merely exceeding $r_c$ is therefore not sufficient for continued growth at subcritical ambient atmospheric supersaturation. For example, if $(S-1)/(S_c-1)=0.6$, the two intersections have approximately $r_{-}/r_c=0.676$ and $r_{+}/r_c=2.349$; a droplet of radius $1.5r_c$ undergoes evaporation. If instead $S>S_c$, there is no equilibrium barrier in this approximation and droplets activate into growing cloud droplets.

Köhler theory explains activation of cloud condensation nuclei, rather than rapid production of atmospheric precipitation. Diffusion-controlled droplet growth is limited by diffusion of water vapor and removal of released latent heat; approximately $d(r^2)/dt$ is proportional to the ambient atmospheric supersaturation. It slows in radius and tends to narrow a droplet size distribution. The available water vapor, changing temperature and shared depletion of atmospheric supersaturation also invalidate unlimited growth at prescribed $S$. Formation of rain-sized drops normally requires collision-coalescence, while colder clouds can additionally use cloud ice nucleation versus ice growth.

The ambient water vapor partial pressure divided by the equilibrium saturation pressure over a stated reference surface at the same temperature. Relative to flat pure water, atmospheric supersaturation is $s=S-1$. A curved solute-containing cloud droplet generally has a different equilibrium saturation ratio from unity.