Re: Secondary ellipsoid parmentation
Kaimbridge
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On Jul 11, 3:33 pm, Narasimham <mathm...@hotmail.com> wrote:
> On Jul 11, 5:42 pm, George Orwell <nob...@mixmaster.it> wrote:
> ----
>
>> Ellipsoid:
>> x=a*sin(theta)*cos(phi),
>> y=b*sin(theta)*sin(phi),
>> z=c*cos(theta).
>
>> Is it
>
>> x'=b*c*sin(theta)*cos(phi),
>> y'=a*c*sin(theta)*sin(phi),
>> z'=a*b*cos(theta)?
Yes, if "theta" equals the azimuth, using the spherical coordinate
system.
>> or
>> x'=b*c*cos(theta)*cos(phi),
>> y'=a*c*cos(theta)*sin(phi),
>> z'=a*b*sin(theta)?
Yes, if "theta" equals the *parametric* (or "reduced") latitude,
using
the geographic coordinate system.
> ---
> If you continue to use ' latitude and longitude" notation, these
> are same ellipsoids but with axes interchanged:
>
>> x'=bc*cos(theta)*cos(phi),
>> y'=ac*cos(theta)*sin(phi),
>> z'=ab*sin(theta)? -> lat,long = theta, phi
Right, but--again--it's the parametric "RLat" (usually denoted as
\beta), not the more recognized geographic "Lat" (usually denoted as
\phi).
There are a couple of different issues involved here.
--What are x', y', z'--derivatives or complements?
--While (AFAIK) it hasn't been recognized as such,
a scalene ellipsoid *CAN* be defined biaxially,
like a spheroid.
First, let's look at an oblate spheroid, which can be defined either
biaxially, like an ellipse, or triaxially:
Where RL = reduced/parametric latitude;
Lon = geographic longitude;
a, b = equatorial, polar radii;
Biaxial: x = a * cos(RL);
y = b * sin(RL);
Triaxial: X = a * cos(RL) * cos(Lon);
Y = a * cos(RL) * sin(Lon);
Z = b * sin(RL);
x = x(Lon) = [X^2 + Y^2]^.5,
= cos(RL) * [(a*cos(Lon))^2 + (a*sin(Lon))^2]^.5,
= cos(RL) * a(Lon);
R = [x^2 + y^2]^.5 = [X^2 + Y^2 + Z^2]^.5,
= [(a*cos(RL))^2 + (b*sin(RL))^2]^.5;
Now let's consider the complementary/auxiliary parameterization, first
by redefining the triaxial "a", "b", "c" radii:
Equatorial: a_x = "a"; a_y = "b";
Polar: b = "c";
Furthermore, the geometric mean of a_x and a_y, "a_m", the variable
equatorial radius, "a(Lon)", and its complement, "b(Lon)", will now be
introduced:
(Note: Read V' as "complement V", where the
acute accent, "'", is over the V)
a_m = [a_x * a_y]^.5;
a(Lon) = [(a_x*cos(Lon))^2 + (a_y*sin(Lon))^2]^.5;
a_y * b a_y
a_x'= ------- = b * [---]^.5;
a_m a_x
a_x * b a_x
a_y'= ------- = b * [---]^.5;
a_m a_y
a'(Lon) = b(Lon) = [(a_x'*cos(Lon))^2 + (a_y'*sin(Lon))^2]^.5,
a(90°+/-Lon)
= b * ------------,
a_m
a_y a_x
= b * [---*cos(Lon)^2 + ---*sin(Lon))^2]^.5;
a_x a_y
Thus, for an oblate spheroid,
a_x = a_y = a; a_x' = a_y' = a' = b;
The spheroid's complementary/auxiliary parameterization can now be
defined:
Biaxial: x'= a * cos'(RL) = a * -sin(RL);
y'= b * sin'(RL) = b * cos(RL);
or
x'= a' * cos(RL) = b * cos(RL);
y'= b' * sin(RL) = a * sin(RL);
Triaxial: X'= a * cos'(RL) * cos(Lon) = a * -sin(RL) * cos(Lon);
Y'= a * cos'(RL) * sin(Lon) = a * -sin(RL) * sin(Lon);
Z'= b * sin'(RL) = b * cos(RL);
or
X'= a' * cos(RL) * cos(Lon) = b * cos(RL) * cos(Lon);
Y'= a' * cos(RL) * sin(Lon) = b * cos(RL) * sin(Lon);
Z'= b' * sin(RL) = a * sin(RL);
Since the functions of Lon in X' and Y' are cofactors of sin(RL) (and
I believe they should still be cofactors of cos(RL)), it would seem
that the valid complementary parameterization is based on a' and b',
rather than cos'(RL) and -sin'(RL).
Thus,
R'= [x'^2 + y'^2]^.5 = [X'^2 + Y'^2 + Z'^2]^.5,
= [(b*cos(RL))^2 + (a*sin(RL))^2]^.5;
With a scalene ellipsoid, the same exact formulation as the oblate
spheroid applies to the primary parameterization, only substituting a
with a(Lon):
Triaxial: X = a_x * cos(RL) * cos(Lon);
Y = a_y * cos(RL) * sin(Lon);
Z = b * sin(RL);
Biaxial: x(Lon) = a(Lon) * cos(RL) = [X^2 + Y^2]^.5;
y = b * sin(RL) = Z;
R = [x(Lon)^2 + y^2]^.5 = [X^2 + Y^2 + Z^2]^.5,
= [(a(Lon)*cos(RL))^2 + (b*sin(RL))^2]^.5;
The presentation of the complementary/auxiliary parameterization
likewise mirrors that of the oblate spheroid's, with one important
caveat: Generally, the scalene's complementary parameterization is
presented in the context of its surface area integrand:
(RS(RL)^2)' = a_m * cos(RL) * [X'^2 + Y'^2 + Z'^2]^.5
Thus, the OP and follow-up both erroneously factored in a_m with
X', Y', Z'. The proper assignment works out as
X'= a_x' * cos(RL) * cos(Lon),
a_y * b
= b(0) * cos(RL) * cos(Lon) = ------- * cos(RL) * cos(Lon);
a_m
Y'= a_y' * cos(RL) * sin(Lon),
a_x * b
= b(90°) * cos(RL) * sin(Lon) = ------- * cos(RL) * sin(Lon);
a_m
a_x * a_y
Z'= b' * sin(RL) = a_m * sin(RL) = --------- * sin(RL);
a_m
x'(Lon) = a'(Lon) * cos(RL) = b(Lon) * cos(RL) = [X^2 + Y^2]^.5;
y'= b' * sin(RL) = a_m * sin(RL) = Z;
R'= [x'(Lon)^2 + y^2]^.5 = [X'^2 + Y'^2 + Z'^2]^.5,
= [(b(Lon)*cos(RL))^2 + (a_m*sin(RL))^2]^.5;;
A reply to another thread ("Principle perimetres of triaxal
ellipsoid") is being finished up, which apply these concepts.
~Kaimbridge~
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</pre>
Kaimbridge M. GoldChild
Right, but——again——it's the parametric "RLat" (usually denoted as
\beta), not the more recognized geographic "Lat" (usually denoted as \phi).
There are a couple of different issues involved here.
--What are x', y', z'——derivatives or complements?
~Kaimbridge~
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Wikipedia—Contributor Home Page:
Kaimbridge
> <pre>
> [ For coherent viewing, fixed-width font (such as
> "courier new") and "UTF-8" character encoding
> should be utilized ]
> (Note: Read V' as "complement V", where the
> acute accent, "'", is over the V)
> </pre>
http://groups.google.com/group/sci.math/msg/8733012fc614ca70?dmode=source&output=gplain
Reposted for notation clarification (the original Google posting
changed V' to just V'--thus muddying the distinction) and better
propagation.
Kaimbridge
wrote:
> There are a couple of different issues involved here.
>
> --What are x', y', z'--derivatives or complements?
>
> --While (AFAIK) it hasn't been recognized as such,
> a scalene ellipsoid *CAN* be defined biaxially,
> like a spheroid.
<snip>
> A reply to another thread ("Principle perimetres of triaxal ellipsoid")
> is being finished up, which apply these concepts.
Just posted:
http://groups.google.com/group/sci.math/msg/22b36d181b87df41
http://groups.google.com/group/sci.math/msg/22b36d181b87df41?output=gplain
> ~Kaimbridge~
>
> -----
> Wikipedia-Contributor Home Page: