[OSRS-PROJ] On the Ellipsoidal Transverse Mercator
Strebe at aol.com
Strebe at aol.com
Tue Oct 14 21:10:38 PDT 2003
Since Mr. Evenden has not successfully implemented Dozier's ellipsoidal
transverse Mercator, and since no one has come forward with any information about
Dozier's method or its reliability, I have decided to explain Wallis's method.
This method yields the ellipsoidal transverse Mercator to arbitrary accuracy,
covering the entire ellipsoid. For any eccentricity greater than zero, this
map is finite. The greater the eccentricity, the smaller the area required by
the map.
The map is bilaterally symmetrical. Each quadrant has the shape of a
tombstone. The base of each tombstone consists of half of the 0th and half of the
180th meridians. Two tombstones join base-to-base, and the combination of two then
join the other two side-by-side. The outer boundary consists of much of the
equator. Just how much depends on the eccentricity. The equator is also the
joint between the two sets of bĂȘche-bĂȘche tombstones.
I describe the method below. The method is due primarily to Dr. David E.
Wallis of Pasadena, California, USA. His innovation is to use the ellipsoidal
polar stereographic as the primary projection of the ellipsoid to the plane. He
treats the result as a complex space and applies a function to "straighten out"
the primary meridian such that it retains constant scale. That is the
definition of the transverse Mercator. My contribution is the simple iterative
technique to determine "z", as set forth below.
Regards,
daan Strebe
Geocart author
http://www.mapthematics.com
Method:
1) Project the ellipsoid to the ellipsoidal polar stereographic projection.
Use whatever formulation you like, but keep the scale factor at the center to
be 1.0. The result will be an x and y.
2) Construct a complex number z, using x as the real portion and y as the
imaginary.
3) Solve for psi:
z = tan (psi/2) * ((1 - e cos (psi))/(1 + e sin (psi)))^(e/2)
where e is the eccentricity, ^ is "raise to the power of" operator.
Solving for psi is the most difficult part of this method. For e < 0.32,
simply select 0.0 as your "seed" value for psi, and solve using Newton-Raphson. I
have no recommendations for larger e.
4) Pretend that psi is a meaningful geodetic latitude, even though it is
complex. Using the standard formulation but applied to complex arguments, compute
the parametric colatitude of psi. Call it psi'.
5) Using psi' as the amplitude, solve the complex elliptic integral of the
2nd kind. (Abramowitz & Stegun, p. 593 17.4.12)
Done. x is the real portion of the result; y is the complex.
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