Exercise #3


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  • Problem #1 - The HYSPLIT GUI menu noted that the user generated meteorology file has a resolution of 1 km while the text in the Tutorial section indicated a resolution of 10 km. Which is correct?

  • Hint - Use the Display Data / Grid Domain utility program

  • Solution - showgrid.ps



  • Problem #2 - Use the profile program to compare the meteorological data at Dayton, Ohio (25th 1800 UTC) between the WRF27uw and WRF27. Both have the same spatial resolution but use different boundary layer parameterizations.

  • Hint - Note that because the WRF27uw and the WRF27 files have different starting times, they require different offset times for the 1800 UTC sounding to appear in the output file.

  • Solution - At around 925 hPa (mid-boundary layer) the WRF27uw and the WRF27 data are very similar with respect to temperature, but the wind direction difference is about 6 degrees. This may be in part due to the 350 m difference in boundary layer heights (PBLH) between the two models. The actual sounding at Dayton has been extracted from the CAPTEX data archive.



  • Problem #3 - For HYSPLIT calculations meteorological data are saved on orthogonal grids. Various conformal projections are used to convert latitude‐longitude to X‐Y coordinates. Approximately what angle would a west wind (vector = 270°) in the wrf27uw data appear to have along the 75°W meridian when projected in the orthognal X‐Y framework?

  • Hint - Use the Profile program with the true wind displayed as a vector to determine the grid rotation. Note that if the projection were Polar Stereographic rather than Lambert the rotation would simply be the difference (23°) between the Pole or Reference longitude, where the meridians are parallel to the Y coordinate (98°W), and the longitude requested (75°W).

  • Solution - Although the grid and true vectors could be examined at any level, finding a west wind where the V component is near zero, ensures that we can use the small angle assumption (the arc tangent in radians = V/U). At level 120 hPa, the true angle is 0.5/17.8 (=1.6°) while the grid angle is 4.9/17.1 (=16.4°) which means that a west wind would show as a 255° vector.