Principle perimetres of triaxal ellipsoid
Nomen Nescio
semi-minor axis), what is the equation for finding the
principle (meridian) perimetres? Would you just substitute
"a" and "b" of an oblate spheroid with "a" and "c" for the
x-axis and "b" and "c" for the y-axis and find it the same way
you would spheroid?
A possible problem I see present with that idea is the polar
curvature! An oblate spheroid's radius of curvature at the
poles equals a^2/b. It would seem proper that the triaxal's
polar radius of curvature should be (a*b)/c throughout,
should it not?
But the meridional radius of curvature at the equator should
be c^2/a for the x-axis and c^2/b for the y-axis, while the
perpendicular should equal "a" and "b", respectively.
How should this discrepancy be addressed and resolved?
-Sir Thaddeus
W. Dale Hall
Aren't the sections equal to ellipses?
Isn't there a formula (I'd look it up, but this is, after
all, *your* problem) for the circumference of an ellipse
as a function of its major and minor axes?
Doesn't that just about solve the problem?
Or, maybe your ellipsoid doesn't have its principal axes aligned
along the coordinate axes.
Dale
...
For a 2D ellipse with semimajor axis a and semiminor axis b, the
perimeter is
p = 4 a E(e)
where E(e) is a complete elliptic integral of the second kind and
e = sqrt(1 - b^2/a^2) is the eccentricity.
See (63) at
<http://mathworld.wolfram.com/Ellipse.html>
George Orwell
I know how to do the perimetre of an ellipse and oblate apheroid, but I
am talking about a triaxal ellipsoid where a > b > c and since the
curvature merges at the poles the regular equation for an ellipse doesn't
seem proper.
Sorry if I wasn't clear enough. :(
-Sir Thaddeus
Il mittente di questo messaggio |The sender address of this
non corrisponde ad un utente |message is not related to a real
reale ma all'indirizzo fittizio |person but to a fake address of an
di un sistema anonimizzatore |anonymous system
Per maggiori informazioni |For more info
https://www.mixmaster.it
Major Quaternion Dirt Quantum
all planar sections of any quadric surface
are conical sections;
therefore, all sections
ot the ellipsoid are ellipses;
eh?
re
http://data.giss.nasa.gov/gistemp/graphs/Fig.D.txt
I did not look at the "corrected" data, but
I wonder, why it was said to be the result
of a "Y2K bug." (I'm sure that
the generic y2K feature was used
to extort lots of "emerging markets"
around the world, though.)
the UNIPCC has historically been trimmed
of anyone who did not go with the concensus,
with its primary agendum to be implimentation
of the Kyoto Protocol and successors (and
the precedent for Kyoto was Montreal,
after the Bronfman gang took-over DuPont;
their patents on Freons had expired).
another example,
the recent (?) Newsweek that I found,
coverstory on "the global warming hoax <wink>,"
abuses S. Fred Singer, as usual failing
to mention his bona fides, so that
he only appears to be an energy cartel shill.
why do people believe that
the cartels do not love the Kyoto Protocol, or that
they are against their own statistical data,
known as Peak Oil --
what's missing in the gestalt, here?
the glacier data that mosnieur Croppock linked,
is completely selective, as you can
tell by reading the encyclopedia-like article;
these glaciers are the ones
that are readily accessible!
http://nsidc.org/sotc/glacier_balance.html
thus:
having not reviewed Snelson's def.,
"tensegrity" as a term of art that happens
to ignore the neccesary coincidence
of compression is acceptable, due
to the well-known over-emphasis, since Rome,
on compression; does Snelson fall back
to that?
there is certainly no such thing
as a "pure tension structure" or
a pure compression one; per Bucky's dictat,
they are only & always conincident.
for instance, let's say that
we have an adequate model of the carbon atom
as a compressional element; but, see,
carbon does not exist in mono-atomic form!
thus:
is it true, that only LaRouchiacs know that
Leo Szilard was the foundational model for Dr. Strangleove?
if you don't believe that,
watch a few episodes of Sagan's PBS biopic,
"The Weaponeers" -- except that
Szilard was his mentor, so!
of course, there were probably compositional elements
of Sir Henry of Kiss.Ass., Macnamara, Chomsky et al
ad vomitorium from the Ayn RAND Corp.
thus quoth:
Only fragments of these remarks have been reported in the West, while
in Russia RIA Novosti gave its
readers a headline ("Clinton supports placement of BMD in Poland and
the Czech Republic") that was
directly opposite to what President Clinton said.
http://larouchepub.com/other/2007/3428clinton_at_yalta.html
--n~nerfman~n!
http://larouchepub.com/pr/2007/070730conyers_impeach_dick.html
14 Italian Senators Call for Cheney Impeachment
Aug. 1, 2007 (EIRNS)-
The Lyndon LaRouche Political Action Committee
(LPAC) issued the following release today.
Fourteen members of the Italian Senate have signed a call "to the
Members of Congress to support Rep. Kucinich's House Resolution 333
for the Impeachment of Dick Cheney."
http://larouchepub.com/pr/2007/070801italian_senators_call.html
Nomen Nescio
> On Sun, 12 Aug 2007 17:56:18 GMT, "W. Dale Hall"
>
>
>
> <mailtowdunderscorehallatpacbelldotnet@last> wrote:
> perimeter is
>
> p = 4 a E(e)
>
> where E(e) is a complete elliptic integral of the second kind and
>
> e = sqrt(1 - b^2/a^2) is the eccentricity.
>
> See (63) at
> <http://mathworld.wolfram.com/Ellipse.html>
As I replied to the other chap, I am competent with an ellipse
and oblate spheroid.
Allow me to illustrate with an example.
Let a = 6400, b = 5000 and c = 2500.
If we create an oblate spheroid where the semi-major axis
("a") = 6400 and the semi-minor axis ("b") = 2500,
the average radius of a meridian is ~ 4666.321 and its polar
radius of curvature is 16384.
Creating a second spheroid where a = 5000 and b = 2500, the
meridian's average radius ~ 3854.911 and its polar radius of
curvature is 10000.
So what would the average radius along the x-axis' and
y-axis' meridians be?
At first contemplation it would seem obvious that 4666.321
and 3854.911 (respectively) are the answers. But when the
distinctively differing polar radii of curvatures are
considered, these choices seem defective.
Shouldn't the triaxal spheroid's polar radius of curvature
equal (a*b)/c, in this case 12800?
Perhaps the better question to be asked is what the triaxal's
meridional radius of curvature recipe is?
-Sir Thaddeus
Kaimbridge M. GoldChild
> If a triaxal ellipsoid has x,y,z semi-axes a,b,c (c is the
> semi-minor axis), what is the equation for finding the
> principle (meridian) perimetres? Would you just substitute
> "a" and "b" of an oblate spheroid with "a" and "c" for the
> x-axis and "b" and "c" for the y-axis and find it the same way
> you would spheroid?
> A possible problem I see present with that idea is the polar
> curvature! An oblate spheroid's radius of curvature at the
> poles equals a^2/b. It would seem proper that the triaxal's
> polar radius of curvature should be (a*b)/c throughout,
> should it not?
> But the meridional radius of curvature at the equator should
> be c^2/a for the x-axis and c^2/b for the y-axis, while the
> perpendicular should equal "a" and "b", respectively.
> How should this discrepancy be addressed and resolved?
<pre>
Good observations! P=)
As you have surmised, a SCALENE ellipsoid is just a more complicated
oblate spheroid——all spheroids can be defined as triaxial: What creates
the complication is the elliptical equator (hence, "scalene").
Let's first examine an oblate spheroid.
The equatorial radius/semi-major axis is defined as "a", and the polar
radius/semi-minor axis is "b", like an ellipse. The various elliptic
parameters (e^2, e'^2, f, etc.) can be reduced to trigonometric
functions of the angular eccentricity, "oe" (actually, it is meant
to be the Greek "ethyl" ligature "omicron" + "epsilon", "oε",
"grethyl", but for simple ASCII text, "oe" is good enough), also known
as the "modular angle" and denoted as "alpha":
b a - b
oe = arccos(-) = 2 * arctan([-----]^.5);
a a + b
There are two principal elliptic integrands, "E'(P)" (and inverse,
"n'(P)"), and its complement, "C'(P)":
E'(P) = [1 - (sin(P)*sin(oe))^2]^.5,
= [cos(oe)^2 + (cos(P)*sin(oe))^2]^.5,
= [cos(P)^2 + (sin(P)*cos(oe))^2]^.5;
n'(P) = 1/E'(P);
C'(P) = [1 - (cos(P)*sin(oe))^2]^.5,
= [cos(oe)^2 + (sin(P)*sin(oe))^2]^.5,
= [(cos(P)*cos(oe))^2 + sin(P)^2]^.5;
While the planetographic (for Earth, read: "geographic") latitude, "LT",
is what is commonly known as *the* "latitude", a nore important form in
defining and working an ellipsoid is the parametric, or "reduced",
latitude, "RL" (symbolically, LT = "phi" and RL = "beta"):
sin(LT) = sin(RL) / C'(RL);
cos(LT) = (cos(oe)/C'(RL)) * cos(RL);
tan(LT) = sec(oe) * tan(RL);
sin(RL) = cos(oe) * n'(LT) * sin(LT);
cos(RL) = n'(LT) * cos(LT);
tan(RL) = cos(oe) * tan(LT);
There is an important relationship between E'(LT), C'(RL) and cos(oe):
cos(oe) = E'(LT) * C'(RL);
cos(oe) cos(oe)
E'(LT) = -------; C'(RL) = ------- = cos(oe) * n'(LT)
C'(RL) E'(LT)
Where "Lon" is a given (planetographic) longitude (symbolically "lambda"), let's now look at the spheroid's Cartesian parameterization, both triaxially and biaxially:
N(LT) = a * n'(LT);
X = a * cos(RL) * cos(Lon) = N(LT) * cos(LT) * cos(Lon);
Y = a * cos(RL) * sin(Lon) = N(LT) * cos(LT) * sin(Lon);
Z = b * sin(RL) = cos(oe)^2 * N(LT) * sin(LT);
or,
x = x(Lon) = [X^2 + Y^2]^.5,
= [(a*cos(Lon))^2 + (a*sin(Lon))^2]^.5 * cos(RL),
= N(LT) * [cos(Lon)^2 + sin(Lon)^2]^.5 * cos(LT);
y = Z;
X^2 + Y^2 Z^2 x^2 y^2
--------- + --- = --- + --- = 1,
a^2 b^2 a^2 b^2
= cos(RL)^2 + sin(RL)^2,
= n'(LT)^2 * [cos(LT)^2 + (sin(LT)*cos(oe))^2];
Thus the ellipsoidal 3D parameterization can be reduced to that of the
simpler 2D elliptical.
Isolating LT's unit integrand for radius,
r'(LT) = n'(LT) * [cos(LT)^2 + sin(LT)^2 * cos(oe)^4]^.5,
the radius at a given LT or RL can be defined:
R = [X^2 + Y^2 + Z^2]^.5 = [x^2 + y^2]^.5,
= R(LT) = a * r'(LT) = a * E'(RL);
While r'(LT) = E'(RL),
__ __
/ LT_f / RL_f
/ r'(LT)dLT ≠ / E'(RL)dRL.
__/ LT_s __/ RL_s
(where "V_s" is the lower boundary, or "standpoint",
and "V_f" is the upper boundary, or "forepoint")
To find the equivalent integral, RL is made a function of LT, "RL(LT)",
with its derivative used in the composite function form of E':
RL = RL(LT) = arctan(cos(oe)*tan(LT));
dRL(LT)
RL'(LT) = ------- = cos(oe) * n'(LT)^2;
dLT
Conversely,
dLT(RL) 1 cos(oe)
LT'(RL) = ------- = ------- = --------;
dRL RL'(LT) C'(RL)^2
With this, the composite function can be defined:
l'(LT) = E'(RL(LT)) * RL'(LT) = r'(LT) * RL'(LT),
= cos(oe) * r'(LT) * n'(LT)^2;
and
__ __
/ LT_f / LT_f
/ l'(LT)dLT = / r'(LT) * RL'(LT)dLT,
__/ LT_s __/ LT_s
__
/ RL_f
= / E'(RL)dRL.
__/ RL_s
But this presents a conflict: Does the unit radius at LT actually equal
r'(LT) or l'(LT)? As r'(LT) equals E'(RL), then r'(LT) *IS* the (unit)
radius——l'(LT) is considered "auxiliary", used to differentiate and
integrate with respect to LT instead of RL:
dRL
E'(RL)dRL = (E'(RL)---)dLT = E'(RL(LT)) * RL'(LT)dLT.
dLT
With an ellipsoid, there is also an auxiliary, complementary
parameterization:
(Note: Read V´ as "complement V", where the
acute accent, "´", is over the V)
a´= b; b´= a;
X´= b * cos(RL) * cos(Lon) = cos(oe) * N(LT) * cos(LT) * cos(Lon);
Y´= b * cos(RL) * sin(Lon) = cos(oe) * N(LT) * cos(LT) * sin(Lon);
Z´= a * sin(RL) = cos(oe) * N(LT) * sin(LT);
or,
x´= x´(Lon) = [X´^2 + Y´^2]^.5,
= [(b*cos(Lon))^2 + (b*sin(Lon))^2]^.5 * cos(RL),
= cos(oe) * N(LT)
* [cos(Lon)^2 + sin(Lon)^2]^.5 * cos(LT);
y´= Z´;
R´= R´(RL) = a * C'(RL) = [x´^2 + y´^2]^.5,
= a * cos(oe) * n'(LT) = b * n'(LT),
= a * r´'(LT);
Therefore, C'(RL) = r´'(LT) = cos(oe) * n'(LT).
While the primary parameterization deals with radius, coordinates,
positioning and the like, the complementary parameterization provides
for operations dealing with arc and curvature.
As such, most formulation dealing with surface——such as geodetic
applications——utilizes R´ in creating an auxiliary sphere, unique to
each calculation.
Since R´ is auxiliary, there must be something it is auxiliary to.
Limited to the parameters defined and presented within this discussion,
that "something" is the meridional radius of curvature and arc, "M":
m'(LT) = C´(RL(LT)) * RL'(LT)
= r´'(LT) * RL'(LT) = cos(oe) * n'(LT) * RL'(LT),
= cos(oe) * n'(LT) * (cos(oe) * n'(LT)^2),
= cos(oe)^2 * n'(LT)^3,
(and, since
cos(oe) * n'(LT)^2 = [cos(oe)^2 * n'(LT)^3 * n'(LT)]^.5,
= [m'(LT) * n'(LT)]^.5,
then RL'(LT) = [m'(LT) * n'(LT)]^.5!)
Thus,
M = M(LT) = R´(RL(LT)) * RL'(LT),
= a * m'(LT) = a * r´'(LT) * RL'(LT);
Or, more properly,
M(LT) m'(LT)
R´(RL) = ------- = a * [------]^.5,
RL'(LT) n'(LT)
= M(LT(RL)) * LT'(RL),
= a * (sec(oe) * C'(RL)^3) * (cos(oe) / C'(RL)^2),
= a * C'(RL);
(thus C'(RL)^2 also equals m'(LT)/n'(LT))
Therefore,
__ __
/ LT_f / RL_f
/ M(LT)dLT = / R'(RL)dRL.
__/ LT_s __/ RL_s
Now let's examine the scalene ellipsoid.
As expanded upon in a related discussion, "Secondary ellipsoid parmentation (sic)",
http://groups.google.com/group/sci.math/msg/8733012fc614ca70
http://groups.google.com/group/sci.math/msg/8733012fc614ca70?output=gplain
there are three conventionally defined semi-axes comprising a scalene
ellipsoid: The two equatorial boundaries, "a" and "b", denoting the
longer X and shorter Y horizontal axes, respectively——thus the equator
can be viewed and expressed as an ellipse——and the polar, "c",
representing the vertical Z axis.
Instead, keeping with the presentation of the ellipse and spheroid,
let's define the X semi-axis as "a_x", the Y semi-axis as "a_y" and the
polar Z semi-axis as "b".
Recalling that the x(Lon) axis of a spheroid equals
[X^2 + Y^2]^.5 = [(a*cos(Lon))^2 +
(a*sin(Lon))^2]^.5 * cos(RL),
converting to the scalene provides
[X^2 + Y^2]^.5 = [(a_x*cos(Lon))^2 +
(a_y*sin(Lon))^2]^.5 * cos(RL),
= a(Lon) * cos(RL);
Therefore, by changing a to a(Lon) and oe to oe(Lon), the primary
parameterization of a scalene ellipsoid at a given Lon can be reduced to
that of a spheroid:
a(Lon) = [(a_x*cos(Lon))^2 + (a_y*sin(Lon))^2]^.5;
(a_x = a(0); a_y = a(90°))
oe(Lon) = arccos(b/a(Lon));
E'(LT;Lon) = [1 - (sin(LT(Lon))*sin(oe(Lon)))^2]^.5;
C'(RL;Lon) = [1 - (cos(RL)*sin(oe(Lon)))^2]^.5;
LT = LT(RL;Lon) = arctan(sec(oe(Lon))*tan(RL)) = ...;
RL = RL(LT;Lon) = arctan(cos(oe(Lon))*tan(LT)) = ...;
n^(LT;Lon) = 1/E'(LT;Lon);
m'(LT;Lon) = cos(oe(Lon))^2 * n'(LT;Lon)^3;
r'(LT;Lon) = n'(LT;Lon) * [cos(LT(Lon))^2 +
sin(LT(Lon))^2 * cos(oe(Lon))^4]^.5,
= E'(RL;Lon);
M = M(LT;Lon) = a(Lon) * m'(LT;Lon);
N = N(LT;Lon) = a(Lon) * n'(LT;Lon);
X = a(0) * cos(RL) * cos(Lon),
= a(0) * n'(LT;Lon) * cos(LT(Lon)) * cos(Lon);
Y = a(90°) * cos(RL) * sin(Lon),
= a(90°) * n'(LT;Lon) * cos(LT(Lon)) * sin(Lon);
Z = b * sin(RL) = cos(oe(Lon))^2 * N(LT;Lon) * sin(LT(Lon));
x(Lon) = [X^2 + Y^2]^.5,
= a(Lon) * cos(RL) = N(LT;Lon) * cos(LT(Lon));
y = Z;
R = [X^2 + Y^2 + Z^2]^.5 = [x(Lon)^2 + y^2]^.5,
= R(LT;Lon) = a(Lon) * r'(LT;Lon) = a(Lon) * E'(RL;Lon);
While the scalene's primary parameterization mirrors that of the oblate
spheroid's——albeit, localized with respect to Lon——the complementary
parameterization and its extracts require the addition of both another
static value of "a" and a complex complement of a(Lon).
The simpler of the two is the constant, which is just a_x's and a_y's
geometric mean, "a_m":
a_m = [a_x * a_y]^.5; oe_m = arccos(b/a_m);
To identify the complement of a(Lon) and the accompanying
parameterization, let's examine the integrand for surface area, with
respect to RL.
For the oblate spheroid,
(RS(RL)^2)'= x * [x'^2 + y^2]^.5 = a^2 * cos(RL) * C'(RL),
= cos(RL) * [(a*b*cos(RL))^2 + (a*a*sin(RL))^2]^.5,
= cos(RL) * [b^2*((a*sin(Lon))^2 + (a*cos(Lon))^2)
*cos(RL)^2 + (a*a*sin(RL))^2)]^.5;
Converting it to the scalene case,
(RS(RL;Lon)^2)'= cos(RL) *
[b^2*((a_x*sin(Lon))^2 + (a_y*cos(Lon))^2)
*cos(RL)^2 + (a_x*a_y*sin(RL))^2)]^.5,
= cos(RL) * [(a(90°+/-Lon)*b*cos(RL))^2 +
(a_x*a_y*sin(RL))^2]^.5,
= a_m * cos(RL) *
[((a(90°+/-Lon)/a_m)*b*cos(RL))^2 +
(a_m*sin(RL))^2]^.5,
= a_m * cos(RL) * [x´(Lon)^2 + y´^2]^.5,
cos(oe(Lon))
= ------------ * x(Lon) * [x´(Lon)^2 + y´^2]^.5;
cos(oe_m)
From this configuration it can be deduced that
a(90°+/-Lon)
x´(Lon) = b * ------------ * cos(RL);
a_m
And, as the complementary radii for the spheroid is a´ = b and b´ = a,
the equivalent for the scalene ellipsoid is
a(90°+/-Lon)
a´(Lon) = b(Lon) = b * ------------ = cos(oe_m) * a(90°+/-Lon),
a_m
a_x a_y
= b * [---*sin(Lon)^2 + ---*cos(Lon)^2]^.5;
a_y a_x
b(Lon) a_m - b(Lon)
oe´(Lon) = arccos(-------) = 2 * arctan([------------]^.5),
a_m a_m + b(Lon)
a(90°+/-Lon)*b
= arccos(--------------),
a_x*a_y
a_x*a_y - a(90°+/-Lon)*b
= 2 * arctan([------------------------]^.5);
a_x*a_y + a(90°+/-Lon)*b
b´= a_m = [a_x * a_y]^.5;
The complementary auxiliary parameterization for the scalene ellipsoid
can now be defined:
X´= b(0) * cos(RL) * cos(Lon),
= b(0) * n'(LT;Lon) * cos(LT(Lon)) * cos(Lon);
Y´= b(90°) * cos(RL) * sin(Lon),
= b(90°) * n'(LT;Lon) * cos(LT(Lon)) * sin(Lon);
Z´= a_m * sin(RL),
= a_m * cos(oe(Lon)) * n'(LT;Lon) * sin(LT(Lon));
x´(Lon) = [X´^2 + Y´^2]^.5,
= b(Lon) * cos(RL) = b(Lon) * n'(LT;Lon) * cos(LT(Lon));
y´= Z´;
The next complication arising with the scalene ellipsoid is the
complementary integrand.
With an oblate spheroid,
[x^2 + y^2]^.5
E'(RL) = C'(90°-RL) = --------------;
a
[x´^2 + y´^2]^.5
C'(RL) = E'(90°-RL) = ----------------;
a
But, in the case of the scalene ellipsoid, although
[x(Lon)^2 + y^2]^.5
E'(RL;Lon) = C'(90°-RL;Lon) = -------------------,
a(Lon)
given the presence of a_m and b(Lon) in composing x´(Lon) and y´,
[x´(Lon)^2 + y´^2]^.5 [x´(Lon)^2 + y´^2]^.5
--------------------- ≠ ---------------------,
a(Lon) a_m
≠ C'(RL;Lon) = E´'(90°-RL;Lon);
Instead, a second complementary integrand, "D'(RL;Lon)", needs to be
introduced.
While:
C'(RL;Lon) = [(cos(RL)*cos(oe(Lon)))^2 + sin(RL)^2]^.5;
r´'(LT;Lon) = cos(oe(Lon)) * n'(LT;Lon);
for the actual auxiliary parameterization,
D'(RL;Lon) = C´'(RL;Lon) = E´´'(RL;Lon);
[x´(Lon)^2 + y´^2]^.5
= ---------------------,
a_m
= [(cos(RL)*cos(oe´(Lon)))^2 + sin(RL)^2]^.5;
The LT equivalent unit integrand, "r´´'(LT;Lon)", contains both oe´(Lon)
AND oe(Lon):
r´´'(LT;Lon) = n'(LT;Lon) * [(cos(LT(Lon))*cos(oe´(Lon)))^2 +
(sin(LT(Lon))*cos(oe(Lon)))^2]^.5;
The arc oriented auxiliary radius (R´) for the scalene ellipsoid thus
formulates as
R´= [X´^2 + Y´^2 + Z´^2]^.5 = [x´(Lon)^2 + y´^2]^.5,
= R´(RL;Lon) = a_m * D'(RL;Lon) = a_m * r´´'(LT;Lon),
= [(a_m*sin(RL))^2 + (b(Lon)*cos(RL))^2]^.5,
= n'(LT;Lon) * [(a_m*cos(oe(Lon))*sin(LT(Lon)))^2 +
(b(Lon)*cos(LT(Lon)))^2]^.5;
The original question can now be addressed: What about M (and N)?
Just as the oe(Lon) based E'(RL;Lon) and C'(RL;Lon) equate with those of
an oblate spheroid mirroring a(Lon) and b, so do M and N——for very
limited purposes (such as latitude conversion).
But for applications involving the complementary parameterization and D'
instead of C', a second M,N set ("M´","N´") is required.
Based on the relationships
m´'(LT;Lon) * r´'(LT;Lon) = m'(LT;Lon) * r´´'(LT;Lon);
n´'(LT;Lon) * r´'(LT;Lon) = n'(LT;Lon) * r´´'(LT;Lon);
m´'(LT;Lon) * n´'(LT;Lon) = m'(LT;Lon) * n'(LT;Lon)
r´´'(LT;Lon)
* (------------)^2;
r´'(LT;Lon)
m´'(LT;Lon) m'(LT;Lon) n´'(LT;Lon) n'(LT;Lon)
----------- = ----------; ----------- = ----------;
n´'(LT;Lon) n'(LT;Lon) m´'(LT;Lon) m'(LT;Lon)
m´'(LT;Lon) n´'(LT;Lon) r´´'(LT;Lon) D'(RL;Lon)
----------- = ----------- = ------------ = ----------,
m'(LT;Lon) n'(LT;Lon) r´'(LT;Lon) C'(RL;Lon)
cos(oe´(Lon))
= [(cos(LT(Lon)) * -------------)^2 +
cos(oe(Lon))
sin(LT(Lon))^2]^.5;
then
r´´'(LT;Lon)
m´'(LT;Lon) = m'(LT;Lon) * ------------;
r´'(LT;Lon)
r´´'(LT;Lon)
n´'(LT;Lon) = n'(LT;Lon) * ------------;
r´'(LT;Lon)
M´= M´(LT;Lon) = a_m * m´'(LT;Lon),
a_m * r´´'(LT;Lon)
= M(LT;Lon) * --------------------;
a(Lon) * r´'(LT;Lon)
N´= N´(LT;Lon) = a_m * n´'(LT;Lon),
a_m * r´´'(LT;Lon)
= N(LT;Lon) * --------------------;
a(Lon) * r´'(LT;Lon)
For a spheroid and scalene ellipsoid entry where a(Lon) equals the
spheroid's "a" (and the "b"s equate), the meridional radius of curvature
and arc at the equator and poles reduce to
M(0) = b^2/a = b * cos(oe); M(90°) = a^2/b = a * sec(oe);
With a scalene ellipsoid, using the complementary proper M´(LT;Lon),
b * b(Lon)
M´(0;Lon) = ---------- = b(Lon) * cos(oe(Lon));
a(Lon)
a_m * a(Lon)
M´(90°;Lon) = ------------ = a_m * sec(oe(Lon)),
b
= a(Lon) * sec(oe_m);
So now let's apply this to the scalene ellipsoid subsequently defined
by the OP:
> Allow me to illustrate with an example.
> Let a = 6400, b = 5000 and c = 2500.
a(0) = a_x = "a" = 6400; a(90°) = a_y = "b" = 5000;
a_m = [a_x * a_y]^.5 ≈ 5656.8542494924;
b = [ b(0°) * b(90°)]^.5 = "c" = 2500;
b(0) = 2209.7086912080; b(90°) = 2828.4271247462;
where
__
2 / 90°
Mr[Lon] = -- / M(LT;Lon)dLT;
pi __/ 0
__
2 / 90°
Mr[0,90°] = -- / Mr[Lon]dLon,
pi __/ 0
__ __
2 / 90° / 90°
= (--)^2 / / M(LT;Lon)dLTdLon;
pi __/ 0 __/ 0
(with the same integration/notation applying to M´)
> If we create an oblate spheroid where the semi-major axis
> ("a") = 6400 and the semi-minor axis ("b") = 2500,
> the average radius of a meridian is ~ 4666.321 and its polar
> radius of curvature is 16384.
> Creating a second spheroid where a = 5000 and b = 2500, the
> meridian's average radius ~ 3854.911 and its polar radius of
> curvature is 10000.
These values would be the results found with M:
M´/M´r (M/Mr)
---------------- -----------------
M(90°;0) 14481.5468787005 (16384.0)
M(0;0) 863.1674575031 (976.5625)
----------------------------------------------------
Mr[0] 4124.4837123015 (4666.3206429789)
/ [45°] 4247.1934767721 (4282.4874574569) \
\ [46.9366583°] 4255.1473189 (4255.1473185) /
M(90°;90°) 11313.7084989848 (10000.0)
M(0;90°) 1414.2135623731 (1250.0)
----------------------------------------------------
Mr[90°] 4361.3340056012 (3854.9110629751)
====================================================
Mr[0,90°] 4245.0574779647 (4271.5952873577)
> So what would the average radius along the x-axis' and
> y-axis' meridians be?
> At first contemplation it would seem obvious that 4666.321
> and 3854.911 (respectively) are the answers. But when the
> distinctively differing polar radii of curvatures are
> considered, these choices seem defective.
No, M(90°;Lon) does not need to be constant (just consider the equator
on an oblate spheroid——M ≠ N, though it is the same point), however
M and N *do* need to equate at the poles, since M and N merge into one.
The same properties apply to M´ and N´.
> Shouldn't the triaxal spheroid's polar radius of curvature
> equal (a*b)/c, in this case 12800?
This will occur when Lon = Lon_c = arctan((a_x/a_y)^.5) or
arctan((sec(oe_x)*cos(oe_y))^.5) (also noting that (a_x/a_y)^.5 =
a_x/a_m = a_m/a_y)——here, approximately 48.527065691°——resulting in
a(Lon_c) equaling a_m:
M(90°;Lon_c) = M´(90°;Lon_c) = N´(90°;Lon_c) = N(90°;Lon_c),
a(Lon_c)^2 a_m^2 a_x * a_y
= ---------- = ----- = --------- (here, 12800);
b b b
> Perhaps the better question to be asked is what the triaxal's
> meridional radius of curvature recipe is?
Melting it all down to a single equation:
Recalling that
a(Lon) = [(a_x*cos(Lon))^2 + (a_y*sin(Lon))^2]^.5;
a(90°+/-Lon) = [(a_x*sin(Lon))^2 + (a_y*cos(Lon))^2]^.5;
a_m = [a_x * a_y]^.5;
then
cos(oe(Lon)) *
[(cos(LT(Lon))*cos(oe´(Lon)))^2 +
(sin(LT(Lon))*cos(oe(Lon)))^2]^.5
M´(LT;Lon) = a_m * -------------------------------------,
[1-(sin(LT(Lon))*sin(oe(Lon)))^2]^1.5
a_m * a(Lon) * b^2 * [(sin(LT(Lon))^2 +
/a(Lon) * a(90°+/-Lon) \
(---------------------- * cos(LT(Lon)))^2]^.5
\ a_x * a_y /
= --------------------------------------------------;
[(a(Lon)*cos(LT(Lon)))^2 + (b*sin(LT(Lon)))^2]^1.5
For M(LT;Lon) (and M(LT) of an oblate spheroid), this greatly
simplifies to
cos(oe(Lon))^2
M(LT;Lom) = a(Lom) * -------------------------------------,
[1-(sin(LT(Lon))*sin(oe(Lon)))^2]^1.5
(a(Lon) * b)^2
= --------------------------------------------------;
[(a(Lon)*cos(LT(Lon)))^2 + (b*sin(LT(Lon)))^2]^1.5
If one wanted to approximate the scalene ellipsoid as an oblate
spheroid, "a" can be defined any number of ways——here, the most
efficient to providing equivalent scalene total averaging results
being either:
---The previously defined geometric mean, "a_m";
---The simple arithmetic mean of a_x and a_y, "a_s":
a_s = .5 * (a_x + a_y);
---The complete arithmetic mean or
divided difference, "A[0,90°]":
Where a(Lon) = A'(Lon),
__
2 / 90°
A[0,90°] = -- / a(Lon)dLon,
pi __/ 0
lim UT a(Lon_tn)
= UT->oo SUM ---------;
TN=1 UT
(Lon_1 = 0; Lon_ut = 90°)
---the "squared mean root" (i.e., the inverse
root mean square, which could also be considered
the "trapezoidal Heronian mean"), "a_i";
a_s + a_m a_x + 2*a_m + a_y
a_i = --------- = -----------------,
2 4
= [.5 * (a_x^.5 + a_y^.5)]^2;
Using the same process of finding M´r from M'(LT;Lon), the spheroidal
averages of M´/R´ and M/R (which, for whole quadrant average values, are
all one and the same), as well as the authalic surface area's "Ar",
where, for whole quadrant integration, RL = LT = L as variable,
(S(L)^2)'= (RS(RL)^2)'= a_m * cos(RL) * R´,
r´'
= (LS(LT(Lon)))^2)'= cos(LT(Lon)) * M´ * N´ ----,
r´´'
= cos(LT(Lon)) *
cos(oe(Lon)) r´´'
(------------)^2 * M * N ----;
cos(oe_m) r´'
__ __
2 / 90° / 90°
Ar = (--)^2 / / (S(L)^2)'dLdLon;
pi __/ 0 __/ 0
Keeping b at its scalene constant at 2500, the following valuations result:
Model a M´/R´ M/R Ar
------- --- ------- ----- ----
a(Lon): (Varying) 4245.057478 4271.595287 4601.246294
========= ============== =========== =========== ===========
A[0,90°]: 5721.511562485 4270.121208 4270.121208 4633.926912
a_s: 5700.0 4257.643527 4257.643527 4619.569683
a_i: 5678.427124746 4245.135883 4245.135883 4605.173736
a_m: 5656.854249492 4232.633889 4232.633889 4590.779997
If "a" is set to a_m and——using the same process for finding A[0,90°]——b
is set to B[0,90°] (here, equaling 2528.574765294), then the average
value of M/M´/R/R´ becomes 4243.576021 and Ar, 4600.124995: As all of
the other above valuations of "a" are significantly larger than a_m,
using B[0,90°] with any of these other "a" models will effect oversized
results.
If "a" is changed to a_s (5700), then the results become 4268.551671 and
4628.901538, respectively.
The choices for best scalene valuation thus depends on what is
attempted to be reflected——radius based operations (R/M) or
surface/arc/curvature oriented applications (Ar/M´/R´)?:
---For the radius based R/M, A[0,90°] and b reflect the
closest approximation: 4271.595287 ≈ 4270.121208;
---For Ar/M´/R´ surface referencing, either a_m and
B[0,90°] or a_i and b provide bet fit:
M´/R´ Ar
Scalene: 4245.057478 4601.246294
a_m,B[0,90°]: 4243.576021 4600.124995
a_i,b: 4245.135883 4605.173736
As the simple scalene average of M and M´ is 4258.3263825,
the valuation of 4257.643527 demonstrates that a_s and b is,
in fact, the obvious, most rudimentary, facile choice for
biaxial approximation of a scalene ellipsoid.
~Kaimbridge M. GoldChild~
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</pre>
Kaimbridge
wrote:
> On Aug 12, 4:40 pm, Nomen Nescio <nob...@dizum.com> wrote:
>
>> If a triaxal ellipsoid has x,y,z semi-axes a,b,c (c is the
>> semi-minor axis), what is the equation for finding the
>> principle (meridian) perimetres? Would you just substitute
>> "a" and "b" of an oblate spheroid with "a" and "c" for the
>> x-axis and "b" and "c" for the y-axis and find it the same way
>> you would spheroid?
>> A possible problem I see present with that idea is the polar
>> curvature! An oblate spheroid's radius of curvature at the
>> poles equals a^2/b. It would seem proper that the triaxal's
>> polar radius of curvature should be (a*b)/c throughout,
>> should it not?
>> But the meridional radius of curvature at the equator should
>> be c^2/a for the x-axis and c^2/b for the y-axis, while the
>> perpendicular should equal "a" and "b", respectively.
>> How should this discrepancy be addressed and resolved?
>
> <pre>
>
> Good observations! P=)
> As you have surmised, a SCALENE ellipsoid is just a more complicated
> creates the complication is the elliptical equator (hence, "scalene").
<snip>
A "clean" formatted copy can be found here:
http://groups.google.com/group/sci.math/msg/22b36d181b87df41
http://groups.google.com/group/sci.math/msg/22b36d181b87df41?output=gplain
~Kaimbridge~
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