<HTML><FONT FACE=arial,helvetica><HTML><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"> gerald.evenden@verizon.net writes:<BR>
(Original e-mail in full at the end.)<BR>
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<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
The cusps at the ends of the equator seem unnatural. <BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">...<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">The cusps at the ends of the equator sure look like violations of conformality<BR>
to me.<BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
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?<BR>
<BR>
The position of the cusps is directly related to the eccentricity -- in fact, the relationship is surprisingly simple:<BR>
<BR>
Longitude of cusp = (pi/2) * (1 - e) radians<BR>
<BR>
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.<BR>
<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">Also, why does both of the other procedures that<BR>
we have looked at all contain discontinuities at the limits<BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
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.<BR>
<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">And a remaining note: this projection has not been published in an any<BR>
cartographic journal and has not been subject to normal peer review nor the<BR>
review of subsequent readers.<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
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.<BR>
<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">What magic twist<BR>
allows Wallis to come up with the above map when all others want to extend to<BR>
infinity?<BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
You are misundertanding the problem. The others don't "want" to extend to infinity; their accuracy simply degrades into uselessness.<BR>
<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">Is it truly a normal transverse mercator where the scale factor is<BR>
1. along the central meridian? <BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
Yes.<BR>
<BR>
<BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">Has that been checked? <BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
Yes, empirically, and yes, analytically.<BR>
<BR>
<BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">Sorry, I am still a<BR>
skeptic until I see the math and a functional program that can demonstrate<BR>
the conformal properties of the projection.<BR>
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</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
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:<BR>
<BR>
"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..."<BR>
<BR>
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.<BR>
<BR>
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.<BR>
<BR>
Regards,<BR>
-- daan Strebe<BR>
<BR>
<BR>
In a message dated 6/26/06 13:22:06, gerald.evenden@verizon.net writes:<BR>
<BR>
<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">And a remaining note: this projection has not been published in an any<BR>
cartographic journal and has not been subject to normal peer review nor the<BR>
review of subsequent readers.<BR>
</FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
</BLOCKQUOTE></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2"><BR>
<BR>
<BR>
<BLOCKQUOTE CITE STYLE="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px" TYPE="CITE"></FONT><FONT COLOR="#000000" FACE="Palatino" FAMILY="SERIF" SIZE="2">On Wednesday 14 June 2006 1:12 am, Strebe@aol.com wrote:<BR>
> You might contact Dr. David E. Wallis. He devised a much simpler method<BR>
> than Dozier's. I've implemented it for the full-ellipsoid. You can see a<BR>
> plot of an earth-like ellipsoid here:<BR>
><BR>
> http://mapthematics.com/Projection%20Images/Cylindrical/Transverse%20Mercat<BR>
>or. GIF<BR>
><BR>
> The method works for arbitrary eccentricities. Contact me privately if<BR>
> you're interested. Since it is Dr. Wallis's invention, I'll put you in<BR>
> contact with him.<BR>
<BR>
I wrote to the address on the web site but letter was returned undeliverable. <BR>
I suspect that the web page is several years old and not maintained.<BR>
<BR>
Not to beat a dead horse of several years ago, I have stared at the above gif<BR>
and it still bothers me and it does not seem real. The cusps at the ends of<BR>
the equator seem unnatural. Also, why does both of the other procedures that<BR>
we have looked at all contain discontinuities at the limits---most commonly,<BR>
they require isometric latitude which fails at 90 degrees. What magic twist<BR>
allows Wallis to come up with the above map when all others want to extend to<BR>
infinity? Is it truly a normal transverse mercator where the scale factor is<BR>
1. along the central meridian? Has that been checked? Sorry, I am still a<BR>
skeptic until I see the math and a functional program that can demonstrate<BR>
the conformal properties of the projection.<BR>
<BR>
The cusps at the ends of the equator sure look like violations of conformality<BR>
to me.<BR>
<BR>
As previously noted, the French TM has been added to libproj4 and the Dozier<BR>
procedure has also been pretty well conquered and will be add to<BR>
libproj4---probably as dtmerc. Neither of these routines will do |lat|=90<BR>
nor |lon|=90.<BR>
--<BR>
Jerry and the low-riders: Daisy Mae and Joshua<BR>
"Cogito cogito ergo cogito sum"<BR>
Ambrose Bierce, The Devil's Dictionary<BR>
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