Interpolating 3D Model Data

QUESTION: I have an atmospheric model that produces a three-dimensional array. The first dimension is longitude, sampled at 3.75 degree increments. The second dimension is latitude, also sampled at 3.75 degree increments. The third array dimension is time. The array itself is a 96x48x1812 floating point array. It would be much more convenient for me to have this data sampled at either 2.5 or 5.0 degrees in latitude and longitude. How can I do this? I also need new latitude and longitude vectors that represent the center point of each grid square in the array. I'm not clear how to create these, either.

ANSWER: The Interpolate command can be used to resample the data on a different grid.

First, consider your longitude vector. There are 360 degrees of longitude, divided by 3.75 degrees:

   IDL> Print, 360/3.75
        96.000

Thus, your first dimension of your array is 96. Similarly, for the latitude vector:

   IDL> Print, 180/3.75
        48.000

If we want to sample at 5.0 degrees, or at 2.5 degrees, we will have to know how many samples we need. Consider first a 5.0 degree grid

   IDL> Print, 'Longitude samples: ', 360/5.0, '   Latitude samples: ', 180/5.0
        Longitude samples:       72.0000   Latitude samples:       36.0000

What this tells us is that we need to create location vectors for the Interpolate command that are 72 and 36 elements long. And they have to extend over 96 and 48 elements, respectively.

The easiest way to create such vectors is with the cgScaleVector command.

   lon_locations = cgScaleVector(Findgen(72), 0, 96)
   lat_locations = cgScaleVector(Findgen(36), 0, 48)
   time_locations = Findgen(1812)

These location vectors tell the Interpolate command where to sample the original array. Note that we have to have a time vector as well, but since we are not resampling time, this is easy to create. To create the resampled array, type this command:

   interpArray = Interpolate(array, lon_locations, lat_locations, time_locations, /Grid)

The result is a 72x36x1812 interpolated array.

If you need vectors to go along with this five degree array, centered on each grid square, you can create them like this. Assume the latitude grid goes from -90 to 90, and the longitude grid goes from 0 to 360.

   lon = Findgen(72) * 5.0 + (5.0/2.0)
   lat = Findgen(36) * 5.0 - (90 - 5.0/2.0)

To check:

   IDL> print, lon
       2.50000      7.50000      12.5000      17.5000      22.5000      27.5000      32.5000      37.5000
       42.5000      47.5000      52.5000      57.5000      62.5000      67.5000      72.5000      77.5000
       82.5000      87.5000      92.5000      97.5000      102.500      107.500      112.500      117.500
       122.500      127.500      132.500      137.500      142.500      147.500      152.500      157.500
       162.500      167.500      172.500      177.500      182.500      187.500      192.500      197.500
       202.500      207.500      212.500      217.500      222.500      227.500      232.500      237.500
       242.500      247.500      252.500      257.500      262.500      267.500      272.500      277.500 
       282.500      287.500      292.500      297.500      302.500      307.500      312.500      317.500
       322.500      327.500      332.500      337.500      342.500      347.500      352.500      357.500
   IDL> print, lat
      -87.5000     -82.5000     -77.5000     -72.5000     -67.5000     -62.5000     -57.5000     -52.5000
      -47.5000     -42.5000     -37.5000     -32.5000     -27.5000     -22.5000     -17.5000     -12.5000
      -7.50000     -2.50000      2.50000      7.50000      12.5000      17.5000      22.5000      27.5000
       32.5000      37.5000      42.5000      47.5000      52.5000      57.5000      62.5000      67.5000
       72.5000      77.5000      82.5000      87.5000