If the chart shows US location, it's not using a state plane zone, and I can't think of another US-based CRS that has those standard parallel values (46deg 40min and 49deg 20min). With the knowledge that the chart's area of interest is in Canada, I couldn't find any existing CRS that match. I think you'll have to create a custom definition. The central meridian may be the longitude at the center of the map.
Here's a link to the QGIS 1.8 help on projections which discusses creating a custom definition near the bottom of the page.
I don't have QGIS to check this but there have been several other questions here about adding a custom projected coordinate reference system. Cribbing from one, here's a PROJ.4 line for a North American Albers definition:
+proj=aea +lat_1=20 +lat_2=60 +lat_0=40 +lon_0=-96 +x_0=0 +y_0=0 +ellps=GRS80 +datum=NAD83 +units=m +no_defs
Now let's modify it to match yours:
+proj=aea +lat_1=46.66666667 +lat_2=49.33333333 +lat_0=?? +lon_0=-?? +x_0=0 +y_0=0 +ellps=GRS80 +datum=NAD83 +units=m +no_defs
You'll have to fill in the lat_0 and lon_0 (latitude and longitude of origin, try the map's center?).
It is basically possible to do what you want, but not the easy way.
I did the following with QGIS:
Restricting the manipulation to the northern hemisphere, I clipped the Natural Earth's land polygon and degree grid to positive values of latitude. Then I created a custom CRS for laea projection centered on the North Pole:
+proj=laea +lat_0=90 +lon_0=0 +x_0=0 +y_0=0 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs
and reprojected the data to it.
In a second step, I created a similiar laea projection at 0°E/0°N:
+proj=laea +lat_0=0 +lon_0=0 +x_0=0 +y_0=0 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs
and assigned it to the data from the last step. Note that I did not reproject the data to the new CRS in this step!
Then I reprojected that dataset back to WGS84, which puts the North pole into the middle of the Atlantic.
This is the picture I got:
I included the new degree grid in purple and the traditional grid in light green. The result is obviously useless around the equator, but should be usefull around the north pole.
EDIT
A very simple reprojection from Greenwich coordinates to North pole coordinates using a sphere would be:
En = (90°-Ng) * sin (Eg/180° * PI())
Nn = (Ng-90°) * cos (Eg/180° * PI())
With Eg and Ng degrees East and North in Greenwich coordinates (EPSG:4326)
En and Nn degrees of a geographic Coordiante system centered at the north pole.
This will give you concentric circles for the Greenwich latitudes.
Best Answer
You can add the matrices as ASCII Grid file to QGIS, and create contours from the lat and lon values. This will create a degree grid like this:
In this mailing list topic I described how to set up a suitable projection for it:
http://osgeo-org.1560.x6.nabble.com/Have-LCC-center-terms-need-PROJ-4-terms-td5102799.html
EDIT
Sample data can be found in the mailing list thread. The known projection parameters are:
This gives the following proj definition string:
No ellipsoid is given, so we take WGS84 as default.
The extent information inside the data reads:
which can be transformed to the GDAL-known AAIGrid header:
To get the x_0 and y_0 values, we take the lower left values from the data (last line,first value):
-86.433044 33.062443into a filell.txtand convert those to lcc using cs2cs:Resulting to
-228512.96 -694637.98 0.00.Note that the cells are 12km wide, and the cell value is for mid cell. So we have to invert the values and add 6000m to get the real lower left values:
so the complete proj string is:
Another approach is to look up the lon_0 value in the xlong table. It is in every line at column 20. So the x_0 is 19.5 * 12000 = 234000. Looking for lat_0 in the xlat table at the same column returns row 59 from the bottom, so the correct y_0 value is 58.5 * 12000 = 702000. The improved proj string is:
A more visual method is to translate the asc files into xyz files, and load those into QGIS as delimited text:
using the first lcc proj string with x_0=0 and y_0=0. Then look up the coordinates of lon_0 and lat_0, they are 234000 and 702000; which approves the last mentioned values for x_0 and y_0.