Questions for 25km East Asia domain experiment. – in #12: CCLM Starter Package Support

in #12: CCLM Starter Package Support

<p> The maximum time step allowed is limited by the Courant–Friedrichs–Lewy ( <span class="caps"> CFL </span> ) condition. This is the time step I am referring to. For 0.44 the maximum it is somewhere between 240s and 360s, with 240s on the save side. You can take this for your “rule of thumb” calculation for the maximum time step of higher resolutions. The minimum time limit is given by your computer resources. </p> <p> For the orographic correction you need to apply the moist adiabatic lapse rate (6.5 K/km is mostly used) during interpolation of one data set to the other. The difference between the two orographies is multiplied by the lapse rate and added (or subtracted) to one of the data sets. After that you may compare the data sets. This is a common praxis. However, one should be aware that taking a constant 6.5 K/km is also only a first order approximation. </p>

  @burkhardtrockel in #2d8324f

<p> The maximum time step allowed is limited by the Courant–Friedrichs–Lewy ( <span class="caps"> CFL </span> ) condition. This is the time step I am referring to. For 0.44 the maximum it is somewhere between 240s and 360s, with 240s on the save side. You can take this for your “rule of thumb” calculation for the maximum time step of higher resolutions. The minimum time limit is given by your computer resources. </p> <p> For the orographic correction you need to apply the moist adiabatic lapse rate (6.5 K/km is mostly used) during interpolation of one data set to the other. The difference between the two orographies is multiplied by the lapse rate and added (or subtracted) to one of the data sets. After that you may compare the data sets. This is a common praxis. However, one should be aware that taking a constant 6.5 K/km is also only a first order approximation. </p>
BR replied

The maximum time step allowed is limited by the Courant–Friedrichs–Lewy ( CFL ) condition. This is the time step I am referring to. For 0.44 the maximum it is somewhere between 240s and 360s, with 240s on the save side. You can take this for your “rule of thumb” calculation for the maximum time step of higher resolutions. The minimum time limit is given by your computer resources.

For the orographic correction you need to apply the moist adiabatic lapse rate (6.5 K/km is mostly used) during interpolation of one data set to the other. The difference between the two orographies is multiplied by the lapse rate and added (or subtracted) to one of the data sets. After that you may compare the data sets. This is a common praxis. However, one should be aware that taking a constant 6.5 K/km is also only a first order approximation.

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