Sure, It is amazing. I wonder how did you do it.

Hans, glad to help. Thanks. ProMapper.

The meridians are straight lines. The parallels look plausibly like concentric arcs of circles. Going with that assumption, it is a conic projection. Which?

Three projections are most likely here: Lambert conformal conic, Albers equal-area conic, and equidistant conic. Many others have been described, but all others are rare and none has ever been used in association with weather.

The parallels are not equally spaced. Therefore it is not an equidistant conic. The parallels do not decrease in distance toward the extreme north or south. Therefore it is not an Albers. It must be Lambert.

I use Geocart to rapidly test hypotheses. The most important consideration in a conic projection is the “constant of cone”, which is the fraction of the circle comprised by the “pie wedge” of the cone. I quickly ascertained that this projection is idiosyncratic in that regard: It did not fit anything like the usual standard parallels for North America: The cone was more geared toward subtropical latitudes for unknown reasons. This is clear by how gradually the meridians converge. It is a poor choice for a map of North America.

Unlike the other two common conic projections, Lambert only really has one standard parallel. Yes, you can set two parallels to have “correct” scale, but on a conformal projection, “correct” scale depends entirely on the nominal scale that you arbitrarily assign to the map. This means that you can have the same constant of cone associated with an infinite number of Lambert projections, each with different standard parallels, and yet the only real difference between them is the scale. Therefore you are only obliged to determine a single standard parallel for a Lambert. With Geocart I was able to discern in short order that the standard parallel must be about 25°.

However, there was still a puzzling discrepancy: the parallels were too close together, no matter what standard parallel I chose. I noticed that they were too close together by the same proportion everywhere when the constant of cone closely matched. This means that the image had been compressed vertically, again for reasons unknown, and likely without any reasonable justification. With a few trials I homed in on the 0.97 vertical compression, and from there it was straightforward to home in on the 25° standard parallel, which I presume is exact. I also tried Lambert on WGS 84, but the fit was not as good as on the sphere.

With the parameters I discovered, the scale turned out to be 1:23,000,000 to six digits precision at my screen’s resolution of 101 dpi. Cute, but probably a coincidence, given that I did compare the projection overlay to the image or specify the vertical compression that precisely, as well as the fact that the original image was unlikely to have been specified at 101 dpi.

In all, I probably spent 20 minutes on the problem. If the scale had not been compressed idiosyncratically, it might have taken half that, and even less if the standard parallels had been meaningful for the situation.

— daan