# [Proj] Re: libproj4 stmerc = French Gauss-Laborde projection

Martin Vermeer martin.vermeer at hut.fi
Thu Jun 22 03:37:46 EDT 2006

```On Thu, 2006-06-15 at 12:27 +0300, Martin Vermeer wrote:
> On Thu, 2006-06-15 at 05:05 -0400, Strebe at aol.com wrote:
> >
> > I'm not lost. I've analyzed the method and programmed it. It works
> > fine.
> >
> > Yes, p is the parametric co-latitude in the first formula. It's wrong
> > to use the same variable in the later formulae because p refers to
> > something else there -- in fact, it's a complex variable in the later
> > instances.
> >
> > The "trick" is this: Wallis uses the polar stereographic because it's
> > the simplest way to get a conformal mapping to the plane. Once the
> > ellipsoid is mapped, he treats the plane as the complex plane and
> > looks for a complex "co-latitude" which can be used with the polar
> > stereographic, but this time treating the polar stereographic as
> > function of a complex variable. The reason he does this is (a) to
> > preserve conformality; and (b) so that the central meridian (in which
> > the imaginary axis is 0) maps back to the parametric colatitude. At
> > this point the ellipsoid is mapped conformally in such a way that
> > leaves the central meridian effectively unmapped.
> >
> > Leaving the complex plane aside, using the colatitude as the parameter
> > to the elliptic integral of the second kind gives the distance from
> > the pole to the colatitude. Since this odd mapping Wallis contrived
> > effectively leaves the central meridian unmapped, and since any
> > analytic function applied to a conformal mapping results in a
> > conformal mapping, and since the elliptic integral has an analytic
> > form, all that is left is to push the mapping through the complex form
> > of the elliptic integral of the second kind. This "straightens out"
> > the central meridian to its true differential lengths whilst dragging
> > the whole complex plane with it in a conformal fashion. The result
> > must be the transverse Mercator, since the central meridian is
> > projected with correct scale and since a conformal projection is
> > unique except with respect to scale and rotation.
> >
> > Regards,
> > -- daan Strebe
> >
>
> Ah! But that's precisely what I have been doing numerically, using a
> polynomial expansion rather than an elliptic integral! (And starting
> from classical Mercator rather than stereographic, so it will run into
> problems at high latitudes.) It's essentially solving a boundary value
> problem, with the set of PDEs being the Cauchy-Riemann conformity
> conditions and the central meridian the boundary.
>
> I suppose I have to get it written up in english ;-)

Just found my Matlab routines for playing with this. Posted them here:

http://users.tkk.fi/~mvermeer/gk.zip

You run it as

>> gkmain(phi, lab, [phimin phimax delta maxk refphi]);

e.g.

>> gkmain(60, 5, [55 65 0.5 12 60])

(lab is the distance from the central meridian)

With this I have also obtained sub-millimeter precision (in typical
situations). Yes, it breaks down at the poles due to using Mercator as
the starting projection. Replacing this by stereographic as Dr Wallis
did, shouldn't be hard.

Using a polynomial expansion (instead of the elliptic integral /
Newton-Raphson thing) has the advantage of simplicity in implementation.
One disadvantage that I noticed is, that the normal matrix for
estimating the coefficients gets poorly conditioned for expansion degree
maxk = 15 or more.

I used this code in an exercise for my students; in a production
environment you would print out and hardwire the estimated coefficients
for the area of interest into your code.

The exercise instruction (in Finnish :) at

http://users.tkk.fi/~mvermeer/kotiharjB.pdf

- Martin

> >
> > In a message dated 6/15/06 01:18:47, martin.vermeer at hut.fi writes:
> >
> >
> > > On Wed, 2006-06-14 at 13:50 -0400, strebe at aol.com wrote:
> > > > Hm. I didn't know about that web page. Obviously it's wrong -- for
> > > some
> > > > reason "p" appears in several different roles. I tend to think
> > > that's
> > > > an error in conversion to a web page. (I see that the entire blurb
> > > is a
> > > > single graphic, not HTML mark-up.) Certainly he's been pedantic
> > > and
> > > > precise in all his communications with me.
> > > >
> > > > The p/2 exponent should read (e/2), where e is the eccentricity.
> > >
> > > Yes, I agree.
> > >
> > > > Use some other variable (perhaps p') in place of p in "Then, the
> > > > complex variable tan (p/2) can be obtained..." and "...yields the
> > > > argument p..."
> > >
> > >
> > > Actually the argument p is simply the (ellipsoidal) co-latitude
> > > 90d - phi.
> > >
> > > The common expression in u and v corresponds to exp(psi), where psi
> > > is
> > > the _isometric latitude_, i.e., essentially the "northing" in a
> > > traditional (non-transverse) Mercator map plane.
> > >
> > > Isometric latitude and longitude (psi, lambda) together as (x,y)
> > > co-ordinates in a plane define a conformal mapping from the curved
> > > Earth's surface. Using (psi, lambda) directly as rectangular
> > > co-ordinates produces classical Mercator. Using
> > >
> > > u + iv = exp(psi + i * lambda)
> > >
> > > i.e., polar co-ordinates, produces the stereographic projection.
> > > This is
> > > very much what Dr Wallis's formula looks like. Apparently for him it
> > > is
> > > only a trick leading somewhere... but then I also get lost.
> > >
> > > Regards Martin V
> > >
> > > PS you may want to look at
> > >
> > > http://users.tkk.fi/~mvermeer/geom.pdf
> > >
> > > pp 99-100 and around p. 90. Sorry it's in Fenno-ugrian formulese...
> > >
> >
> >
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