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Aug 12, 2007, 12:40:05 PM8/12/07

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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?

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

Aug 12, 2007, 1:56:18 PM8/12/07

to

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

Aug 12, 2007, 3:29:02 PM8/12/07

to

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>

Aug 12, 2007, 5:06:04 PM8/12/07

to

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

Aug 13, 2007, 6:00:36 PM8/13/07

to

there is a theorem,

all planar sections of any quadric surface

are conical sections;

therefore, all sections

ot the ellipsoid are ellipses;

eh?

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

Aug 14, 2007, 3:20:01 PM8/14/07

to

On Aug 12, 7:29 pm, "Driveby" <x...@x.invalid> wrote:

> On Sun, 12 Aug 2007 17:56:18 GMT, "W. Dale Hall"

>

>

>

> <mailtowdunderscorehallatpacbelldotnet@last> wrote:

> On Sun, 12 Aug 2007 17:56:18 GMT, "W. Dale Hall"

>

>

>

> <mailtowdunderscorehallatpacbelldotnet@last> wrote:

> 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>

> 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

Oct 6, 2007, 6:16:08 PM10/6/07

to

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?

> 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~

-----

Wikipedia—Contributor Home Page:

http://en.wikipedia.org/wiki/User:Kaimbridge

***** Void Where Permitted; Limit 0 Per Customer. *****

</pre>

Oct 6, 2007, 6:42:26 PM10/6/07

to

On Oct 6, 10:16 pm, "Kaimbridge M. GoldChild" <Kaimbri...@Gmail.com>

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

> oblate spheroid--all spheroids can be defined as triaxial: What

> creates the complication is the elliptical equator (hence, "scalene").

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~

-----

Wikipedia-Contributor Home Page:

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