16.1.3.2 Example 2: Two resolution domains

In the preceeding example, if it were desired to extend the latitudinal domain poleward of $10^\circ$N and $10^\circ$S to $30^\circ$N and $30^\circ$S where the meridional resolution was to be $\Delta_\phi=1^\circ$, then the two previous bounding latitudes would need to be replaced by four: one at $-30^\circ$ where $\Delta_\phi=1^\circ$, one at $-10^\circ$ where $\Delta_\phi={1/3}^\circ$, one at $+10^\circ$ where $\Delta_\phi={1/3}^\circ$ and one at $+30^\circ$ where $\Delta_\phi=1^\circ$. Poleward of 10 degrees, meridional resolution would telescope from $\Delta_\phi={1/3}^\circ$ to $\Delta_\phi=1^\circ$ over a span of $20^\circ$. The average meridional resolution in this region is calculated as the average of the bounding resolutions which is $\frac{1^\circ+{1/3}^\circ}{2} = {2/3}^\circ$. Therefore, there would be $\frac{20^\circ}{{2/3}^\circ}=30$ additional grid cells in each hemisphere between latitudes $10^\circ$ and $30^\circ$. This would be specified in the code as follows:

c
c     "nylats"  = number of bounding latitudes to define 3 region
c     "y_lat"   = bounding latitudes {-30, -10.0, 10.0, 30}
c     "dy_lat"  = resolution centered at "y_lat" {1, 1/3, 1/3, 1}
c     
       parameter (nylats=4)
       data (y_lat(j), j=1,nylats)  /-30.0, -10.0,  10.0,      30.0/
       data (dy_lat(j),j=1,nylats)  /1.0, 0.3333333, 0.3333333, 1.0/