conical equal-area projection, secant case has e.g.

+proj=aea +lat_1=... +lat_2=... +lon_0=... +lat_0=... +x_0=... +y_0=... etc...

I would ask if there is such thing, for transverse cylindrical equal-area, secant case? e.g:

+proj=tcea +lon_1=... +lon_2=... +lon_0=... +lat_0=... +x_0=... +y_0=... etc...

and what is for the azimuthal equal area, secant case, I know just this for tangent case

+proj=laea +lat_0=... +lon_0=... +x_0=... +y_0=... etc...

what should one add to get secant case not only tangent at a point?

Edit: image


The transverse cylindrical equal area projection can be made into a secant case by specifying a scale factor with the +k_0= parameter. The secant lines (standard circles) of such a projection do not follow meridians, there are no "standard meridians", and using +lon_1 +lon_2 would be ambiguous, so they are not used. Therefore we indicate a scale factor that will allow two small circles to be standard (no distortion).

The Lambert azimuthal equal area projection is tangent, there is no secant case. In fact, if we tried to create a secant version of an azimuthal equal area projection, we would get very odd results, where the center of the projection would become a sort of singularity and directions would be undetermined. That would somehow destroy the concept of an "azimuthal" projection. Such projections are not used in practice.

  • Ok, so the azimath case is tangent and there is no secant case, and for transverse cylindrical, one specify k_0 to adjust the secant line, is there a relation for e.g. to adjust the secant lines at equator to be 1/6 of width of the zone before the secant line rightside and automatic the same will be leftside?
    – Khaled
    Dec 6 '18 at 9:08
  • I added an image to show what I mean
    – Khaled
    Dec 6 '18 at 9:11
  • In Transverse Mercator, the scale factor generally used is 0.9996, which makes the standard circles at about 1.62° either side of the central meridian. Transverse Mercator is different from cylindrical equal-area.
    – FSimardGIS
    Dec 6 '18 at 12:44
  • The relationship would be k_0 = acos(dLon) where dLon is the angular separation at equator between the central meridian and the standard circles (for the spherical case).
    – FSimardGIS
    Dec 6 '18 at 12:46
  • 1
    Oops, my bad, it's cos, not acos, so k_0 = cos(dLon)
    – FSimardGIS
    Dec 6 '18 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.