Terrain-Following Coordinate Transformation

From the point of view of the computational model, it is highly convenient to introduce a stretched vertical coordinate system which essentially "flattens out" the variable bottom at $z=-h(x,y)\!\,$. Such "$s\!\,$" coordinate systems have long been used, with slight appropriate modification, in both meteorology and oceanography [e.g., Phillips (1957) and Freeman et al. (1972)]. To proceed, we make the coordinate transformation:

${\hat {x}}=x$

${\hat {y}}=y$

$s=s(x,y,z)\!\,$

$z=z(x,y,s)\!\,$

${\hat {t}}=t$

See S-coordinate for the form of $s\!\,$ used here. In the stretched system, the vertical coordinate $s\!\,$ spans the range $-1\leq s\leq 0\!\,$; we are therefore left with level upper ($s=0\!\,$) and lower ($s=-1\!\,$) bounding surfaces. The chain rules for this transformation are:

$\left({\partial \over \partial x}\right)_{z}=\left({\partial \over \partial x}\right)_{s}-\left({1 \over H_{z}}\right)\left({\partial z \over \partial x}\right)_{s}{\partial \over \partial s}$

$\left({\partial \over \partial y}\right)_{z}=\left({\partial \over \partial y}\right)_{s}-\left({1 \over H_{z}}\right)\left({\partial z \over \partial y}\right)_{s}{\partial \over \partial s}$

${\partial \over \partial z}=\left({\partial s \over \partial z}\right){\partial \over \partial s}={1 \over H_{z}}{\partial \over \partial s}$

where

$H_{z}\equiv {\partial z \over \partial s}$

As a trade-off for this geometric simplification, the dynamic equations become somewhat more complicated. The resulting dynamic equations are, after dropping the carats:

${\partial u \over \partial t}-fv+{\vec {v}}\cdot \nabla u=-{\partial \phi \over \partial x}-\left({\frac {g\rho }{\rho _{o}}}\right){\frac {\partial z}{\partial x}}-g{\partial \zeta \over \partial x}+{F}_{u}+{D}_{u}$

${\frac {\partial v}{\partial t}}+fu+{\vec {v}}\cdot \nabla v=-{\frac {\partial \phi }{\partial y}}-\left({\frac {g\rho }{\rho _{o}}}\right){\frac {\partial z}{\partial y}}-g{\partial \zeta \over \partial y}+{F}_{v}+{D}_{v}$

${\frac {\partial T}{\partial t}}+{\vec {v}}\cdot \nabla T={F}_{T}+{D}_{T}$

${\frac {\partial S}{\partial t}}+{\vec {v}}\cdot \nabla S={F}_{S}+{D}_{S}$

$\rho =\rho (T,S,P)\!\,$

${\frac {\partial \phi }{\partial s}}=\left({\frac {-gH_{z}\rho }{\rho _{o}}}\right)$

${\partial H_{z} \over \partial t}+{\partial (H_{z}u) \over \partial x}+{\partial (H_{z}v) \over \partial y}+{\partial (H_{z}\Omega ) \over \partial s}=0$

where

${\vec {v}}=(u,v,\Omega )$

${\vec {v}}\cdot \nabla =u{\frac {\partial }{\partial x}}+v{\frac {\partial }{\partial y}}+\Omega {\frac {\partial }{\partial s}}$

The vertical velocity in $s\!\,$ coordinates is

$\Omega (x,y,s,t)={1 \over H_{z}}\left[w-(1+s){\partial \zeta \over \partial t}-u{\partial z \over \partial x}-v{\partial z \over \partial y}\right]$

and

$w={\partial z \over \partial t}+u{\partial z \over \partial x}+v{\partial z \over \partial y}+\Omega H_{z}$

Vertical Boundary Conditions

In the stretched coordinate system, the vertical boundary conditions become:

top ($s=0\!\,$):

$\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial u}{\partial s}}=\tau _{s}^{x}(x,y,t)$

$\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial v}{\partial s}}=\tau _{s}^{y}(x,y,t)$

$\left({\frac {K_{T}}{H_{z}}}\right){\frac {\partial T}{\partial s}}={Q_{T} \over \rho _{o}c_{P}}+{1 \over \rho _{o}c_{P}}{dQ \over dT}(T-T_{\rm {ref}})$

$\left({\frac {K_{S}}{H_{z}}}\right){\frac {\partial S}{\partial s}}={(E-P)S \over \rho _{o}}$

$\Omega =0\!\,$

and bottom ($s=-1\!\,$):

$\left({\frac {K_{m}}{H_{z}}}\right){\frac {\partial u}{\partial s}}=\tau _{b}^{x}(x,y,t)$

 & \left(\frac{K_m}{H_z}\right) \frac{\partial v}{\partial s} = \tau^y_b (x,y,t)</math>
 & \left(\frac{K_T}{H_z}\right) \frac{\partial T}{\partial s} = 0</math>

$\left({\frac {K_{S}}{H_{z}}}\right){\frac {\partial S}{\partial s}}=0$

$\Omega =0\!\,$

Note the simplification of the boundary conditions on vertical velocity that arises from the $s\!\,$ coordinate transformation.