Re: Discovery: libproj4 stmerc = French Gauss-Laborde projection

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Re: Discovery: libproj4 stmerc = French Gauss-Laborde projection

strebe
[hidden email] writes:
(Original e-mail in full at the end.)


The cusps at the ends of the equator seem unnatural. 

...

The cusps at the ends of the equator sure look like violations of conformality
to me.


Do the cusps on the August epicycloidal or the Eisenlohr seem unnatural? If so, do you believe those projections are not really conformal? If not, then what do you think the difference is?

The position of the cusps is directly related to the eccentricity -- in fact, the relationship is surprisingly simple:

Longitude of cusp = (pi/2) * (1 - e) radians

The spherical form is infinite because the cusp is at the pole... which projects to infinity. The greater the eccentricity, the closer to the prime meridian starts the cusps, and the stubbier the map.


Also, why does both of the other procedures that
we have looked at all contain discontinuities at the limits


Why is 1 - sqrt (1 - k) when k approaches zero ill-conditioned when computed as is, but marvelously well-behaved when computed as a series? The question is not the mathematics of the projection. We're talking about the same projection, whether Dozier, Kruger, or Wallis. The question is only how the projection is formulated. Wallis's formulation avoids poor conditioning. They all express the same underlying mathematics.


And a remaining note: this projection has not been published in an any
cartographic journal and has not been subject to normal peer review nor the
review of subsequent readers.


I think you mean to say that Wallis' formulation has not been published. Obviously the projection has, by any number of people over nearly two centuries.


What magic twist
allows Wallis to come up with the above map when all others want to extend to
infinity?


You are misundertanding the problem. The others don't "want" to extend to infinity; their accuracy simply degrades into uselessness.


Is it truly a normal transverse mercator where the scale factor is
1. along the central meridian? 


Yes.


Has that been checked? 


Yes, empirically, and yes, analytically.


Sorry, I am still a
skeptic until I see the math and a functional program that can demonstrate
the conformal properties of the projection.


You might avail yourself of a copy of L.P. Lee's monograph, "Conformal Projections Based on Elliptic Functions", Cartographica, Monograph Number 16, 1976. Quoting verbatim from p. 97:

"The positive y-axis represents part of the equator, extending from lambda = 0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly from a straight line to a curve... The projection of the entire spheroid is shown in Fig. 46, again using the eccentricity of the International (Hayford) Spheroid. It can be seen that the entire spheroid is represented withing the finite area without singular points..."

If you will take your attention to page 99, you will see a rendition of the (only true) transverse Mercator projection for the ellipsoid, generated by means of Thompson's 1945 formulation, succeeding across the entire ellipsoid because Lee (or someone) took the time to reformulate the problematic regions to make them suitable for calculation. The diagram is indistinguishable from the one I posted online and drew using Wallis's formulation.

Dozier seems to follow the formulation of Thompson, more or less, which is probably why it runs into trouble at the extremes and is certainly why it's more complicated than necessary. Wallis solves the boundary condition in a different way. It's not without its problems; you need to find a root which, for arbitrary eccentricities, defies any formulaic seed value. Nonetheless that can be solved efficiently in most cases and merely solved in all cases. There are one or two places where one must be careful of numeric techniques, like many other projections.

Regards,
-- daan Strebe


In a message dated 6/26/06 13:22:06, [hidden email] writes:



And a remaining note: this projection has not been published in an any
cartographic journal and has not been subject to normal peer review nor the
review of subsequent readers.




On Wednesday 14 June 2006 1:12 am, [hidden email] wrote:
> You might contact Dr. David E. Wallis. He devised a much simpler method
> than Dozier's. I've implemented it for the full-ellipsoid. You can see a
> plot of an earth-like ellipsoid here:
>
> http://mapthematics.com/Projection%20Images/Cylindrical/Transverse%20Mercat
>or. GIF
>
> The method works for arbitrary eccentricities. Contact me privately if
> you're interested. Since it is Dr. Wallis's invention, I'll put you in
> contact with him.

I wrote to the address on the web site but letter was returned undeliverable. 
I suspect that the web page is several years old and not maintained.

Not to beat a dead horse of several years ago, I have stared at the above gif
and it still bothers me and it does not seem real.  The cusps at the ends of
the equator seem unnatural.  Also, why does both of the other procedures that
we have looked at all contain discontinuities at the limits---most commonly,
they require isometric latitude which fails at 90 degrees.  What magic twist
allows Wallis to come up with the above map when all others want to extend to
infinity?  Is it truly a normal transverse mercator where the scale factor is
1. along the central meridian?  Has that been checked?  Sorry, I am still a
skeptic until I see the math and a functional program that can demonstrate
the conformal properties of the projection.

The cusps at the ends of the equator sure look like violations of conformality
to me.

As previously noted, the French TM has been added to libproj4 and the Dozier
procedure has also been pretty well conquered and will be add to
libproj4---probably as dtmerc.  Neither of these routines will do |lat|=90
nor |lon|=90.
--
Jerry and the low-riders: Daisy Mae and Joshua
"Cogito cogito ergo cogito sum"
   Ambrose Bierce, The Devil's Dictionary




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Re: Re: Discovery: libproj4 stmerc = French Gauss-Laborde projection

OvV_HN
From: Strebe-aol.com

>> What magic twist allows Wallis to come up with the above map when
>> all others want to extend to infinity?
> You are misunderstanding the problem. The others don't "want" to extend to
> infinity; their accuracy simply degrades into uselessness.

Recently I've been looking for a moment at the Transverse Mercator method
Michael Gendt et al. present in a seminar paper. I get the impression that
he is following Klotz from his 1993 publication.
The derived code is very lean, but fails in the same regions as Dozier's:
low latitude at high delta-longitude and above latitude 89.9999993 deg or
so. At low lat/high delta-long there can be convergence in the iterations,
but the answer is wrong.

Reference:
Geodätisches Seminar zum Thema Koordinatentransformationen mittels einer
analytischen Lösung der Gaußschen Abbildung
Michael Gendt, Frank Oheim, Stephan Gehrke
Technische Universität Berlin, 2000-12-21
<http://www.geodesy.tu-berlin.de/~rainer/seminararbeit/gendt/ut.pdf>
Equation 2.14c is wrong: d_n should be k_n.

Other methods - possibly Wallis' - perform a transformation and probably
reposition the difficult areas.
That means that under the finite precision of IEEE-754 math one has to use
two or three methods to cover really a complete (hemi)sphere with high
accuracy.

>From mr. Evenden:
> And a remaining note: this projection has not been published in an any
> cartographic journal and has not been subject to normal peer review nor the
> review of subsequent readers.
>From mr. Strebe:
> Wallis solves the boundary condition in a different way. It's not without its
> problems; you need to find a root which, for arbitrary eccentricities, defies
> any formulaic seed value. Nonetheless that can be solved efficiently in most
> cases and merely solved in all cases. There are one or two places where one
> must be careful of numeric techniques, like many other projections.
Hush hush, say no more - it's a secret!
Either big money or the NSA/CIA must be involved?
[One might want to grin or laugh at this point.]

By the way: will more discussions on this subject follow? Shouldn't we
change the subject title into "Complex Transverse Mercator" or so?




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Complex Transverse Mercator

OvV_HN
In reply to this post by strebe
From: Strebe-aol.com:

You might avail yourself of a copy of L.P. Lee's monograph, "Conformal
Projections Based on Elliptic Functions", Cartographica, Monograph Number
16, 1976. Quoting verbatim from p. 97:
"The positive y-axis represents part of the equator, extending from lambda =
0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly
from a straight line to a curve... The projection of the entire spheroid is
shown in Fig. 46, again using the eccentricity of the International
(Hayford) Spheroid. It can be seen that the entire spheroid is represented
withing the finite area without singular points..."

Reply:

Thanks for this explanation!
The numbers show it too:

International ellipsoid:
90*(1-eccentricity) = 82.62073 decimal deg

Tranverse Mercator:
lat0=0; lon0=0; x0=5e5; y0=0; k0=0.9996;
// International ellipsoid
lat=0; lon=82.50;  x,y = 18712722.276, 0 meters
lat=0; lon=82.60;  x,y = 18840409.942, 0
lat=0; lon=82.61;  x,y = 18853673.034, 0
lat=0; lon=82.62;  x,y = 18867090.964, 0
lat=0; lon=82.621; x,y = 18868446.553, 0.2947
lat=0; lon=82.63;  x,y = 18880722.285, 107.602
lat=0; lon=82.64;  x,y = 18894438.954, 366.186
lat=0; lon=82.65;  x,y = 18908216.295, 738.078
lat=0; lon=82.70;  x,y = 18977788.411, 3947.057
lat=0; lon=82.80;  x,y = 19119409.657, 15745.905

For what it's worth: alculated with my improved version of 'Dozier'.

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Re: Complex Transverse Mercator

Gerald I. Evenden-2
On Tuesday 27 June 2006 1:34 pm, Oscar van Vlijmen wrote:

> From: Strebe-aol.com:
>
> You might avail yourself of a copy of L.P. Lee's monograph, "Conformal
> Projections Based on Elliptic Functions", Cartographica, Monograph Number
> 16, 1976. Quoting verbatim from p. 97:
> "The positive y-axis represents part of the equator, extending from lambda
> = 0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly
> from a straight line to a curve... The projection of the entire spheroid is
> shown in Fig. 46, again using the eccentricity of the International
> (Hayford) Spheroid. It can be seen that the entire spheroid is represented
> withing the finite area without singular points..."
 Just a quick note as Tuesday is my "day off" from cartography.  What is the k
above---e or e^2?

What bothers me about the Wallis plot the curve is not a "smooth" transition
but suddenly takes off on an angle.  This may a graphic problem, but ... .

Yes, I need to look into Lee and a couple of other papers but getting them out
here in the boondocks of Cape Cod is not that easy.

Regarding your numbers, I am going to wait until I get proj_dtmerc going so I
can provide all the other support for the projection like ellipsoids, etc..  
What I was using before was just a skeleton main executing gkfor with input
lines of lat-lon.
--
Jerry and the low-riders: Daisy Mae and Joshua
"Cogito cogito ergo cogito sum"
   Ambrose Bierce, The Devil's Dictionary
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Re: Complex Transverse Mercator

Gerald I. Evenden-2
In reply to this post by OvV_HN
Sorry that I did not reply earlier.
On Tuesday 27 June 2006 1:34 pm, Oscar van Vlijmen wrote:

> From: Strebe-aol.com:
>
> You might avail yourself of a copy of L.P. Lee's monograph, "Conformal
> Projections Based on Elliptic Functions", Cartographica, Monograph Number
> 16, 1976. Quoting verbatim from p. 97:
> "The positive y-axis represents part of the equator, extending from lambda
> = 0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly
> from a straight line to a curve... The projection of the entire spheroid is
> shown in Fig. 46, again using the eccentricity of the International
> (Hayford) Spheroid. It can be seen that the entire spheroid is represented
> withing the finite area without singular points..."
>
> Reply:
>
> Thanks for this explanation!
> The numbers show it too:
>
> International ellipsoid:
> 90*(1-eccentricity) = 82.62073 decimal deg

I strongly do not believe any of these new extended TMs should be use for UTM
as the "standard UTM" is defined as the taylor expansion---warts and all.  
Anyone who has abused the limits of UTM will have trouble when using new
versions.  Secondly, UTM is bound by "law" to the limits of +-3.5 degrees so
these extensions are immaterial.

> Tranverse Mercator:
> lat0=0; lon0=0; x0=5e5; y0=0; k0=0.9996;
> // International ellipsoid
> lat=0; lon=82.50;  x,y = 18712722.276, 0 meters
> lat=0; lon=82.60;  x,y = 18840409.942, 0
> lat=0; lon=82.61;  x,y = 18853673.034, 0
> lat=0; lon=82.62;  x,y = 18867090.964, 0
> lat=0; lon=82.621; x,y = 18868446.553, 0.2947
> lat=0; lon=82.63;  x,y = 18880722.285, 107.602
> lat=0; lon=82.64;  x,y = 18894438.954, 366.186
> lat=0; lon=82.65;  x,y = 18908216.295, 738.078
> lat=0; lon=82.70;  x,y = 18977788.411, 3947.057
> lat=0; lon=82.80;  x,y = 19119409.657, 15745.905

As per  my comments three years ago, none of this make any intuitive sense.  
This probably explains why the German web page deviates as one approaches
pi/2.

BTW, how do these numbers stack up with the German web page.

I seem to recall the Lee article and must double check that I might have it
and forgotten about it.  If I do not have it, it is a pain to try and get a
copy.

The forward is functioning in libproj4 but I have not done any polishing other
than trimming Dozier's code.  I'll check out the inverse (which is coded)
this PM and then start looking at the details.
--
Jerry and the low-riders: Daisy Mae and Joshua
"Cogito cogito ergo cogito sum"
   Ambrose Bierce, The Devil's Dictionary
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Re: Complex Transverse Mercator

Clifford J Mugnier
In reply to this post by OvV_HN




"Secondly, UTM is bound by "law" to the limits of +-3.5 degrees so these
extensions are immaterial."

Actually, that's not exactly true.  Army TM-5-241-41 is a copious tome of
"Zone to Zone Extention Tables" that also document the Gauss-Krueger
Transverse Mercator to the (if I recall correctly), Fourteenth Derivative.
That extension of the series allows "round-trip" transformations to be made
to ±24° of the Central Meridian with an accuracy of 0.1mm.  That document
is the true "legal" extension, as far as the U.S. Army is concerned.

A long time ago, when I was a Topographic Engineer in the U.S. Army, I was
issued a COMPLETE set of the "TM" books of tables (all ellipsoids, etc.).
That was standard issue, and the stuff was carried around (about 60 lbs.),
from one duty station to another because of the truism, "All major battles
will be fought at UTM Zone Boundaries and/or Datum Boundaries!"  The
Topographic Engineers had to compute the zone extensions so that the
Artillery Corps could effectively supply indirect fire beyond the "normal"
zone limits.

Clifford J. Mugnier C.P.,C.M.S.
National Director (2006-2008)
Photogrammetric Applications Division
AMERICAN SOCIETY FOR PHOTOGRAMMETRY AND REMOTE SENSING
and
Chief of Geodesy
Center for GeoInformatics
LOUISIANA STATE UNIVERSITY
Dept. of Civil Engineering
CEBA 3223A
Baton Rouge, LA  70810
Voice:              (225) 578-8536
Facsimile:          (225) 578-8652
-----------------------------------------------------------
http://www.ASPRS.org/resources/GRIDS
http://www.cee.lsu.edu/facultyStaff/mugnier/index.html
-----------------------------------------------------------


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Re: Complex Transverse Mercator

OvV_HN
In reply to this post by Gerald I. Evenden-2
> From: "Gerald I. Evenden"

> I strongly do not believe any of these new extended TMs should be use for UTM
> as the "standard UTM" is defined as the taylor expansion---warts and all.
> Anyone who has abused the limits of UTM will have trouble when using new
> versions.  Secondly, UTM is bound by "law" to the limits of +-3.5 degrees so
> these extensions are immaterial.

>From my memory: wasn't the definition 3 degrees with an overlap of 40
kilometers? Is about the same, but legally different.


> As per  my comments three years ago, none of this make any intuitive sense.
> This probably explains why the German web page deviates as one approaches
> pi/2.
>
> BTW, how do these numbers stack up with the German web page.
If this is about Schuhr's web calculator after Klotz, located at:
<http://gauss.fb1.fh-frankfurt.de/cgi-bin/cgi_gk?modus=intro>
then I can only repeat:
This calculator suffers from the usual problems at low latitudes coupled
with large longitude differences with CM.

[more details including numbers replied to mr. Evenden personally. Let me
know if you're interested in the numbers].


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Re: Complex Transverse Mercator

Gerald I. Evenden-2
On Wednesday 28 June 2006 4:33 pm, Oscar van Vlijmen wrote:

> > From: "Gerald I. Evenden"
> >
> > I strongly do not believe any of these new extended TMs should be use for
> > UTM as the "standard UTM" is defined as the taylor expansion---warts and
> > all. Anyone who has abused the limits of UTM will have trouble when using
> > new versions.  Secondly, UTM is bound by "law" to the limits of +-3.5
> > degrees so these extensions are immaterial.
> >
> >From my memory: wasn't the definition 3 degrees with an overlap of 40
>
> kilometers? Is about the same, but legally different.

You're probably right about that.

Another note, at the equator one only has to go about 4.5 degrees from the CM
before the easting goes negative.  A no-no for grid applications.

--
Jerry and the low-riders: Daisy Mae and Joshua
"Cogito cogito ergo cogito sum"
   Ambrose Bierce, The Devil's Dictionary
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Re: Complex Transverse Mercator

strebe
In reply to this post by OvV_HN

The values in the table below agree with my calculations based on Wallis, down to millimeter precision. It's a fine validation of both methods.

Regards,
-- daan Strebe


In a message dated 6/27/06 10:40:04, [hidden email] writes:


From: Strebe-aol.com:

You might avail yourself of a copy of L.P. Lee's monograph, "Conformal
Projections Based on Elliptic Functions", Cartographica, Monograph Number
16, 1976. Quoting verbatim from p. 97:
"The positive y-axis represents part of the equator, extending from lambda =
0 to lambda = (pi/2)*(1-k)... At this point the equator changes smoothly
from a straight line to a curve... The projection of the entire spheroid is
shown in Fig. 46, again using the eccentricity of the International
(Hayford) Spheroid. It can be seen that the entire spheroid is represented
withing the finite area without singular points..."

Reply:

Thanks for this explanation!
The numbers show it too:

International ellipsoid:
90*(1-eccentricity) = 82.62073 decimal deg

Tranverse Mercator:
lat0=0; lon0=0; x0=5e5; y0=0; k0=0.9996;
// International ellipsoid
lat=0; lon=82.50;  x,y = 18712722.276, 0 meters
lat=0; lon=82.60;  x,y = 18840409.942, 0
lat=0; lon=82.61;  x,y = 18853673.034, 0
lat=0; lon=82.62;  x,y = 18867090.964, 0
lat=0; lon=82.621; x,y = 18868446.553, 0.2947
lat=0; lon=82.63;  x,y = 18880722.285, 107.602
lat=0; lon=82.64;  x,y = 18894438.954, 366.186
lat=0; lon=82.65;  x,y = 18908216.295, 738.078
lat=0; lon=82.70;  x,y = 18977788.411, 3947.057
lat=0; lon=82.80;  x,y = 19119409.657, 15745.905

For what it's worth: alculated with my improved version of 'Dozier'.

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