[Proj] Re: libproj4 stmerc = French Gauss-Laborde projection
Martin Vermeer
martin.vermeer at hut.fi
Thu Jun 15 02:27:38 PDT 2006
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 ;-)
Thanks
Martin V
>
> 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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