Thanks Tyler Well. I've found that the readme of TRMM file (3B42):
The *grid* on which each field of values is presented is a 0.25°x0.25° lat./lon. (Cylindrical Equal Distance) global array of points. It is size 1440x400, with X (longitude) incrementing most rapidly West to East from the Dateline, and then Y (latitude) incrementing South to North from the southern edge. Quarter-degree latitude and longitude values are at grid edges: First point center (49.875°S,179.875°W) Second point center (49.875°S,179.625°W) Last point center (49.875°N,179.875°E) The reference datum is WGS84. It is the same datum used for gauges georeference. So I think that its ok to compare the x,y coordinates. I've not compared the gauges information against the satellite information yet, but I'll have this in mind. Thanks again for your help. On Aug 30, 2:18 pm, Tyler Erickson <[email protected]> wrote: > jpaulini, > > A few thoughts: > > - Precipitation is usually measured in units of length/time, so if the > area of the grid cell changes with latitute it doesn't matter. If it is > reported as a volume per grid cell (which I doubt it would be) then it does > matter. > - If you are comparing point gauge data to a gridded spatial variable, > you will need to reproject one of the other to match projections (and > datums). Because reprojecting a raster will alter the cell values (by some > interpolation method, whether explicitly known or not), I would suggest > reprojecting the point data. If the datums are not actually the same and > you assume they are, you could easily be misaligned by 10km (the error > varies by latitude). > > - Tyler > > > > > > > > On Tue, Aug 30, 2011 at 6:51 AM, jpaulini <[email protected]> wrote: > > Hi all, > > > This question goes mainly for Hanlie, but everyone is welcome.. > > > I'm thinking on doing some similar to you, Hanlie, comparing satellite > > information vs rain gauges. But I had the following doubt: > > > Do you need to change the projection of the satellite information to > > match the projection of the rain gauges georeferenced localization? I > > see that you are assuming 28 sq. km grid, but depending on the > > latitude that you are, it can be a different surface, right? > > > On Aug 15, 12:27 pm, Andy Wilson <[email protected]> wrote: > > > If you calculate Thiessen polygons for your 6 gauges and clip them to > > your > > > grid cell boundary then you'll be able to see how much each gauge would > > > contribute to the interpolated mean grid cell value. So you can just use > > the > > > areas of the polygons as weights for a weighted mean, and that should > > come > > > out the same as interpolating and then aggregating. If the position of > > the > > > gauges doesn't change, then the weights don't change so you only have to > > do > > > that once. Then you just apply your weighted mean function 730 times. > > > > Does that make sense or did I misunderstand something? > > > > On Mon, Aug 15, 2011 at 2:03 AM, Hanlie Pretorius < > > > > [email protected]> wrote: > > > > Hi, > > > > > I have rainfall measurements from 6 gauges that I want to interpolate > > > > to an areal value (a 'surface'), so that I can compare the > > > > interpolated gauge values to a satellite rainfall estimate that covers > > > > a grid cell of 28kmx28km. Two of the gauges are outside, but close to > > > > the border of the grid cell. Therefore, I also need to clip the > > > > interpolated surface to the grid cell and to get the average of the > > > > surface value in this clipped surface. > > > > > However, for each rain gauge I have 730 values representing a daily > > > > measurement over two years. As output, I need a text file > > > > with the interpolated rainfall values for each day in my time series. > > > > So, I was wondering if there is an 'easy' way to get my output without > > > > creating 730 GIS layers? > > > > > Regards > > > > Hanlie
