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I'm trying to compute something called the "Drying Ratio" following: Eidhammer, 2018

The drying ratio is the ratio of the precipitation to the incoming vapor flux. So,

$DR = \frac{F_{in} - F_{out}}{F_{in}} = \frac{P}{F_{in}}$

This makes some assumptions about sources/sinks (minimal evaporation for example). The vapor flux can be expressed as:

$\vec{F} = \frac{-1}{g} \int_{P_{sfc}}^{P_{top}} q\vec{V}dp$

Integrating the vapor transport field along a particular control volume grid-box, and over time yields, length units (same as precipitation).

When the wind component U and V are always coming from the same direction (same sign), the incoming flux into a particular control volume is found by summing up the V components for i,j and i+1, and the U components for i,j and i,j+1.

How can this be done when the vector field changes direction? In this case the opposite side of the control volume would then be the location of the incoming flux.

One could go grid-cell by grid cell and find the sign of the wind direction, and then sum the incoming flux from the upwind cell using an "if" to find the direction.

Is there another method? I wonder if there is a vector calc identity I am missing. This also seems like a much more general problem and I know it must be addressed in many other situations.

Thanks

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  • $\begingroup$ I believe for each grid point you'd basically want to calculate the movement across each of the 4 (or 8) surrounding grid cells in a circle around it and sum them... because the flux isn't just what is coming in but what is going out. And then the sign of u and v will automatically account for which directions are going in and which are coming out in aggregate. $\endgroup$
    – JeopardyTempest
    Commented May 26, 2022 at 21:21
  • $\begingroup$ Seems (unless I took too cursory a look) to be part of the much larger topic in general of advection across gridpoints, which you can probably find quite a bit about how to do. $\endgroup$
    – JeopardyTempest
    Commented May 26, 2022 at 21:24
  • $\begingroup$ So basically the four directly adjacent squares would be $q_{i-1, j}u_{i-1,j} - q_{i+1, j}u_{i-1,j} + q_{i, j-1}u_{i,j-1} + q_{i, j+1}u_{i,j+1}$ $\endgroup$
    – JeopardyTempest
    Commented May 26, 2022 at 21:37
  • $\begingroup$ And if you went on to include the four kitty-corner squares it would be $q_{i-1,j-1}(u_{i-1,j-1}+v_{i-1,j-1})\frac{\sqrt{2}}{2} + q_{i-1,j+1}(u_{i-1,j+1}-v_{i-1,j+1})\frac{\sqrt{2}}{2} + q_{i+1,j+1}(-u_{i+1,j+1}-v_{i+1,j+1})\sqrt{2} + q_{i+1,j-1}(-u_{i+1,j-1}+v_{i+1,j-1})\frac{\sqrt{2}}{2}$ $\endgroup$
    – JeopardyTempest
    Commented May 26, 2022 at 21:38
  • $\begingroup$ If I make no mistakes, as a rough idea! Basically the sign before the u/v are opposite the sign of the grid cell offset in that direction (so for example you add the u term when doing i-1, and subtract it when doing i+1). Of course in reality, things advect more than one grid point unless using a very small time step, so you need more complex calculations? I would think there's packages for gradient/advection matrix calculations in most any programming languages? But it's certainly not my expertise area! $\endgroup$
    – JeopardyTempest
    Commented May 26, 2022 at 21:43

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