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Spheroidal Isopathic Median

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Kaimbridge

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Apr 18, 2004, 2:36:11 PM4/18/04
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[ This was originally posted on a mathboard that allows formatting:

http://math2.org/cgi-bin/mmb/server?action=read&msg=29366 ]

In my elliptic explorations, I've come across an interesting
concept/property that I haven't seen addressed anywheres.
Let AP equal the ARC PATH (the transverse meridian, also known as
the great circle's azimuth at the equator) and TLAT be the
geographic latitude at the great circle's transverse equator (the
90° transverse latitude [90° "TvL"]: I.e., where Lat1 = 0 and
Long2 - Long1 = 90°, Lat2 = TLat): On a sphere, AP + TLat = 90°,
hence AP = 90° - TLat and TLat = 90° - AP.
But on a spheroid, however, AP + TLat > 90° (except for the 0/90°
cases).
Where a,b are the equatorial, polar radii and Oz--equaling
acos{b/a}--is the angle of oblate ellipticity, or angular
eccentricity: e^2 = sin{Oz}^2, e'^2 = tan{Oz}^2 and
f = ver{Oz} = 2 * sin{.5*Oz}^2 = 1 - cos{Oz}.
Denoting the spherical/graticular TLat, AP as TLats, APs and the
elliptically correct as TLate, APe; on a spheroid with minor
ellipticity (such as Earth), TLate ~=~ atan{tan{TLats}*sec{Oz}} and
APe ~=~ atan{tan{APs}*sec{Oz}}.
If, using TLats, one finds APe (via the standard geodetic distance
calculation, letting Lat1 = 0, Lat2 = TLats and Long2 - Long1 = 90°)
and labels the results TLatm:L1 = TLats, APm:L1 = APe, then finds
TLate using APs (this is a little more complicated--Lat1, Long1 = 0;
Azm1 = APe; Distance_0 = 10000; IC = 10; Distance = Distance + IC,
recurse until Long2 = 90°: When Long2 > 90°, then
Distance = Distance - IC; IC = .1 * IC, resume recursion) and labels
these results as TLatm:U1 = TLate, APm:U1 = APs, one will see that
(when TLats and APs are integers) the decimal of TLate will nearly
match that of APe's.
Using the average of one of the elements, find the second generation
of that set and do the same for the other element (using the above
routines): TLatm:L2 = .5 * [TLatm:L1 + TLatm:U1] = .5 * [TLats +
TLate], get APm:L2; APm:U2 = .5 * [APm:L1 + APm:U1] =
.5 * [APs + APe], get TLatm:U2. Repeat the process until like
elements converge to a common value: TLatm = TLatm:L = TLatm:U,
APm = APm:L = APm:U.
Again with minor ellipticities, TLatm ~=~ atan{tan{TLats}*sec{Oz}^.5}
and APm ~=~ atan{tan{APs}*sec{Oz}^.5}.
If one changed the TLat/AP relationship from APs + TLats = 90° to
APm + TLatm = 90° + [2 * ES] = [APs + ES] + [TLats + ES], then
TLatm = TLats + ES, APm = APs + ES and the ELLIPTIC SUPPLEMENT (ES)
is a single value, providing a common, offset pseudospherical merge
between TLat and AP, hence creating a "SPHEROIDAL ISOPATHIC
MEDIAN"--?:

http://math2.org/cgi-bin/mmb/server.pl?action=image&msgid=68282&fname=isopathicmedian_g.gif

While either the direct solution (finding a second set of geodetic
coordinates, given one set, a distance and azimuth) or inverse
solution (finding the distance and direction between two given
geodetic coordinate sets) can be utilized, the inverse approach is
considerably faster and easier:

--- Lat1 = 0: Long1 = 0: Lat2 = atan{tan{TLats}*sec{Oz}^.5}:
Long2 = 90°: APe = atan{cot{TLats}*sec{Oz}^.5}: ES = 0
------- ------- -------
---ES_o = ES: ES = .5 * [APe + Lat2 - 90°]
---Lat2 = TLats + ES
---Calculate APe
---Recurse until ES = ES_o
------- ------- -------
---TLatm = Lat2: APm = APe

Is this a known concept and, if so, does it have a more formal
name/identity? Is there a more direct equation for ES (such as
an--elliptic?--integral or series expansion)?
A likely question someone may have at this point is "yeah, so what do
you do with TLatm, APm once you have them?".
The immediate use this writer has, is as the object of differentiation
(the "differentiant" or "differentiand"?): Differentiating by
APs/TLate produces a different result than by APe/TLats.
With APm, one can find the theoretical, pseudospherical distance--or,
as is this writer's quest, the quadrantal arcradius--for a given APs,
from the equator out to TLats (remembering that APs --> TLate and
TLats --> APe).
Applying it further, APm can be used in finding the transverse surface
area--though it would seem that, as APe = APs at the equator, Qs goes
to Qe only in the "complete" case (i.e., 0 to 90°), hence ES needs to
be reduced as Lats goes from TLats to 0 (i.e., at the equator),
likely by the sine (squared?) of the geographic/spherical transverse
latitude ("TvL"):

Lats = asin{cos{APs}*sin{TvL}};
Latm = Lats + [sin{TvL} * ES] or [sin{TvL}^2 * ES];
(TLats = TLats + [sin{90°} * ES])

APm = APs + [sin{TvL} * ES] or [sin{TvL}^2 * ES];

Likewise, Longs = atan{sin{APs}*tan{TvL}}, so Long1 = Longs1,
Long2 = Longs2 and--instead of 90°--Long2 - Long1 = Longs2 - Longs1,
which, if finding Latm and APm via the direct solution, would be the
comparative measure.
This is presuming TvL is used exclusively, and not the parametric
TvL ("TpL")--or even "TvLe" or "TpLe"--in which case formulation
would likely become considerably more complex.
So, is there such a thing recognized as a "pseudospherical, spheroidal
isopathic median" (though likely identified under another name/term)?

~Kaimbridge~

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