# Identify what projection this Australian map uses?

Im looking to put this map into google maps as custom tiles, but i now learn what projection is and that this map does not use the same format as google's default.

You can create custom projections i read in the API.

I suspect this map uses a common projection, likely the same as this one:

I believe it will likely be this one:

• You might find a way to solve this from an old question on this site HERE
– Dano
Commented Aug 31, 2013 at 12:45
• Well, not really. An unknown CRS must have been defined somewhere, while a map designer can use any projection he personally finds suitable for the subject of the map. Commented Aug 31, 2013 at 13:54

The graticule on the map suggests a conformal conic projection was used, because

• meridians and lines of latitude always meet at right angles, indicating conformality

• lines of latitude appear approximately circular and approximately equally spaced, characteristic of a conic projection.

I therefore attempted to register a screenshot of the map to a base feature projected using the "GDA 1994 Geoscience Australia Lambert" coordinate system (WKID 3112, EPSG). This is a Lambert Conformal Conic projection with central meridian at 134.0 degrees, standard parallels at -18 and -36 degrees, and meters for the units of measurement. The registration (an affine transformation) has a root mean square accuracy of 3 kilometers. How good is this? Some considerations are:

• The map appears to be a scan of a paper copy. This will introduce some relatively large distortions that cannot be corrected.

• It is an older map (1974) and therefore might not have been very accurate originally.

• My screen shot has a resolution of only 35 pixels per degree (because it had to include the entire map). The graticule is still several pixels thick at this resolution and therefore we should be lucky if it is accurate to better than about 1/35 degree vertically = 3 kilometers.

• Registration with a second-degree polynomial does not significantly decrease the root mean square residual (it decreases from 2720 meters to 2010 meters but requires an additional three parameters for the quadratic terms). This indicates the projection is either correct or close to correct (up to a simple change of origin as captured by a "false easting" and "false northing" value, or possibly a fixed rotation as reflected by the central meridian): using the wrong projection would cause the second-degree fit to be a significant improvement over the original affine fit.

I conclude provisionally that the original image likely can be registered to the EPSG 3112 coordinate system to within an acceptable error, given the qualities of the original map (and given that the essential features on it are only roughly drawn anyway: their boundaries are four pixels thick, or about 10 - 15 kilometers).

(I had difficulties with registration along the middle of the top of the map and attribute that to a local error in the graticule. Likewise, some of the graticule in the interior of the country may be grossly misplaced.)

This screenshot of the registration shows a five-degree graticule (black) and a polygon shape for Australia (gray with yellow outline) overlaid on the provisionally registered image of the original map. The red numbered crosses surrounding the figure are registration points: each one corresponds to a whole number of degrees in latitude and longitude and therefore should lie exactly on a crossing of the original map's graticule (which has a two degree spacing).

• wow you know your stuff, im very new to this and trying to get it into google maps using a custom projection as part of a project for the south australian museum, im stuck though trying to change the following example to my region (aus), are you interested in helping me, even being paid if necessary: facstaff.unca.edu/mcmcclur/GoogleMaps/Projections/Lambert.html Commented Sep 1, 2013 at 4:56
• @HaydenThring be sure to Accept this or another answer by hitting the tick symbol once you are satisfied that you have your best answer.
– PolyGeo
Commented Sep 2, 2013 at 0:35
• @PolyGeo ok did that , thanks, all awarded this one for initial detail/time given Commented Sep 2, 2013 at 1:09

My suggestion is

``````+proj=lcc +lat_1=-18 +lat_2=-36 +lat_0=-25 +lon_0=135 +ellps=aust_SA  +units=m +no_defs
``````

giving me this picture:

• this does look like it, but that code means nothing to me, i should have mentioned i have little experience in gis Commented Sep 1, 2013 at 5:01
• You can enter this code as a custom projection in QGIS. But whuber is right with lon_0=134 degrees as used in EPSG:3122. That is the meridian parallel to the right and left borders. Commented Sep 1, 2013 at 12:37
• Because the original image will have to be rectified anyway, its projection only needs to be determined modulo an affine transformation (a "re-rectification" if you will). In particular, the projection with the parameters in this answer is equivalent to the projection I proposed. (They differ only in their central meridian and therefore, because this is a true conic projection, are rotations of one another.) Thus both answers are equally valid. (I offered my answer to show more generally how one can go about finding a projection and checking its validity and accuracy.) Commented Sep 1, 2013 at 15:31
• i did get the impression both where correct and thus not sure who to award Commented Sep 1, 2013 at 22:41
• Sorry, no spare time on my side :-( Commented Sep 3, 2013 at 8:19