mPixelSize in LayerInfo

Qiao

Hi Mike,

When we use Global Mapper to load a GeoTiff file with meter as projection planar unit, it shows correctly the spacing of the image in Pixel_Scale (Pixel_Width & Pixel_Height too except the fourth digit is truncated, right?) of the metadata window. However, when we use SDK's GM_GetLayerInfo(), the returned mPixelSizeX and mPixelSizeY are slightly different than the correct ones (e.g., 2.000000 becomes 2.0007579857619753).

Is that true that in SDK, even the input projection is in meters, it will re-calculate the mPixelSizeX/mPixelSizeY based on the goegraphical (degrees) extents? Do you have any comments on that?

Thanks,

Qiao

global_mapper

Qiao,

Yes, the mPixelSizeX and mPixelSizeY values are calculated based on the actual size of the image using the Vincenty distance calculation and not the less accurate projected distance.

Thanks,

Mike
Global Mapper Support
support@globalmapper.com

Yann DUPUY

Hi Mike,

sorry for digging up an old thread but I need some explanations.
I'm trying to export LANDSAT rasters with the "best" pixel resolution. I was expecting to find such information within the GM_LayerInfo_t structure, in the mPixelSizeX and mPixelSizeY parameters. But I really don't understand the value I can find there!

For instance, loading a [1; 1.5] [43.5; 44] rectangle with the SDK, I have
mPixelSizeX 7.9482874971255688
mPixelSizeY 7.9482874971255688

I was expecting something like 30 meters which is Landsat common precision (without panchromatique data).

If I load the same data with GlobalMapper application and display the metadata, the pixel width and height are
PIXEL WIDTH=0.0004883 arc degrees
PIXEL HEIGHT=0.0004883 arc degrees

If I consider the minute of arc length being 1852 meters, it would give something like
mPixelSizeX = 0.0004883 * 60 * 1852 * cos(latitude=44°) = 39 meters
mPixelSizeY = 0.0004883 * 60 * 1852 = 54 meters

And this the result I have if I measure the drawn pixels in GlobalMapper view.

So where do these nearly 8 meters come from ?

Thanks for your help,
Yann

global_mapper

Yann,

Which version of the SDK are you using? Are you loading the Landsat imagery from a local file or from an online source? The pixel size reported in mPixelSizeX and mPixelSizeY should basically just be the distance in meters along a line crossing through the center of the image in the horizontal or vertical direction divided by the pixel width or height as appropriate.

Your other calculations do seem correct though.

Thanks,

Mike
Global Mapper Guru
gmsupport@bluemarblegeo.com
http://www.globalmapper.com

Yann DUPUY

Mike,

I am using SDK 13.1 and loading Landsat Imagery from provided online source (I tried 13.2 and noticed the difference in Landsat resolution, as mentionned in the release notes).
I am using Global Mapper 11.02 to check things.

Pixel width and height are set to 0 after a call to

status = GM_LoadOnlineLayer(server, &boundingRect, &layer, 0);
const GM_LayerInfo_t *layerInfo = GM_GetLayerInfo(layer);

and I want to do something like

GM_ExportRaster(filename, GM_Export_GeoTIFF, layer, 0, sizeX, sizeY, GM_ExportFlags_HideProgress);

to use my "filename" texture as intended. That's why I thought I could use the mPixelSize and mGlobalRectangle to deal with it.

Thanks,
Yann

global_mapper

Yann,

The mPixelSizeX and mPixelSizeY will be the resolution of the most detailed layer in the online Landsat layer. Online layers don't have a native pixel width and height as the pixel dimensions depend on the zoom scale you are accessing the layer at.

For the Landsat layer in Global Mapper v13.2 application I get the following metadata:

FILENAME=
DESCRIPTION=WMS Global Mosaic, pan sharpened
UPPER LEFT X=-94.3460738171
UPPER LEFT Y=39.0442623607
LOWER RIGHT X=-94.3280226967
LOWER RIGHT Y=39.0377210524
WEST LONGITUDE=94° 20' 45.4937" W
NORTH LATITUDE=39° 02' 40.0680" N
EAST LONGITUDE=94° 19' 39.6239" W
SOUTH LATITUDE=39° 02' 15.0291" N
PROJ_DESC=Geographic (Latitude/Longitude) / WGS84 / arc degrees
PROJ_DATUM=WGS84
PROJ_UNITS=arc degrees
EPSG_CODE=4326
COVERED AREA=0.4381 sq mi
PIXEL WIDTH=0.0001221 arc degrees
PIXEL HEIGHT=0.0001221 arc degrees
BIT_DEPTH=24

As you can see the pixel width and height are 0.0001221 arc degrees, which is consistent with the 7.9 meters per pixel reading you are getting. This is what I would expect as the Landsat data is natively 15m resolution, so the highest zoom level on the server is the 7.9m zoom scale as the next one up (about 16 meters) can't quite capture the full resolution of the data.

Thanks,

Mike
Global Mapper Guru
gmsupport@bluemarblegeo.com
http://www.globalmapper.com

sisocano

Hi
Global Mapper Guru

I have a simple question about calculating width and lenght in km of DEM .*flt files using header data .*hdr

so i have :

ncols         1240
nrows         760
xllcorner     9,4371914430
yllcorner     34,4300096770
xdim          0,00032722957580642
ydim          0,00026934061315784

Area dimension calculated by globalMapper in Projection GeoGraphic (Lat/Lon) WGS84  is 37.29 x 22.07 km

if i use this rule :

Lenght/Width = PixelSize * ncols/nrows   that give me worng value.

1240 * 0,00026934061315784 = 0,3339823603157216

760  * 0,00032722957580642 = 0,2486944776128792

i want to know if there is any way to calculate distances ?

Mykle

Length = ncols * xdim = 0,40576

Are you calculating Length = ncols * Ydim ??

If your units are angular (degrees), it's more complicated than that.