Thanks a lot. It seems to be a very basic question, as you might relize, I'm making my firsts baby steps on GIS...
And the acronym pcs wasn't easy to search on the web... 2011/8/31, Robert Sanson <[email protected]>: > Hi Juan > > Projected Coordinate System. Usually means a cartesian system (2D > plane) rather than a spherical system (lat/long). > > Regards, > > Robert Sanson > >>>> Juan F Paulini <[email protected]> 31/08/2011 9:32 p.m. >>> > Thanks Hanlie, > > But I'm not sure about the meaning of the PCS acronym.. > > > > El 31/08/2011, a las 01:25, ping <[email protected]> > escribió: > >> Hi jpaulinin, >> >> I agree with Tyler that one can work in any coordinate system, as > long >> as you're using the same one for the gauge and the satellite data. >> >> I used a Transverse Mercator projected coordinate system for various >> reasons that were not connected with the rainfall. >> >> In the PCS, the TRMM grid cells are not all of equal size and they > are >> no longer squares. >> >> If you'll need the area of the grid cells for some reason, I think >> you'll have to work in a PCS. >> >> Regards >> Hanlie >> >> >> >> On Aug 30, 9:48 pm, jpaulini <[email protected]> wrote: >>> 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 > > > This email and any attachments are confidential and intended solely for the > addressee(s). If you are not the intended recipient, please notify us > immediately and then delete this email from your system. > > This message has been scanned for Malware and Viruses by Websense Hosted > Security. > www.websense.com >
