Mean Radian For Spheroid

[Kaimbridge]

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I have come across a discrepancy regarding the mean radius/radian (i.e.,
the arc, surface radius) of an oblate spheroid.
First a clarification on formulation/notation ("S{" denotes integral
sign and boundaries):

S{0,.5*Pi} F'{p}dp = .5 * Pi * F'{HQ};
S{0,Pi} F'{p}dp = Pi * F'{HQ};
S{0,2*Pi} F'{p}dp = 2 * Pi * F'{HQ};

oo
F'{HQ} = sum F'{P_Ni} -:- oo,
Ni=1
where P_Ni = 90^o - [cos{AE_Ni} * 90^o] = Pi * sin{.5*AE_Ni}^2
and cos{AE_Ni} = [oo - Ni] -:- [oo - 1];
(of course, this is just the impractical, fundamental equation: For
actual solving, Gaussian Quadrature is used)

a,b = equatorial, polar radii; cos{Oz} = b -:- a;
sin{Oz}^2 = e^2 = [a^2 - b^2] -:- a^2;

I believe there are three different average radii that can be found:
Gaussian Mean Radius, volume derived radius and surface area derived
radius.

Gaussian Mean:

Mr = a * MR'{HQ}

oo
MR'{HQ} = cos{Oz} * sum [1 - (sin{P_Ni} * sin{Oz})^2]^-.5 -:- oo
Ni=1
----------------
Volume:

Volume = [4-:-3] * Pi * r^3 (sphere);
= [4-:-3] * Pi * Vr^3 (spheroid);

Vr = a * MV'{HQ} = [a^2 * b]^(1-:-3);

MV'{HQ} = cos{Oz}^(1-:-3) = [1 - sin{Oz}^2]^(1-:-6)
----------------
Surface Area:

Surface Area = 2 * Pi * r^2 (circle);
= 4 * Pi * r^2 (sphere);
= [2 * Pi * a^2]
+ [Pi * b^2 * csc{Oz} * LOGe{cot{.5 * [90^o - Oz^o]}}],
= 4 * Pi * Sr^2 (spheroid);

Sr = a * MS'{HQ};

MS'{HQ} = [.5 * (<atanh{sin{Oz}} * sin{Oz} * cot{Oz}^2> + 1)]^.5

The discrepancy involves the surface area radius/radian.
As area equals height (here, equaling Half-Pi/Quadrant/90^o--"HQ") times
width (radius) and--in the case of elliptical quantities--the width is
an integrated/averaged value along the height, it should follow that the
mean radian on a spheroid should equal the average of all geodetic
distances ("GEODx"s), measured from the equator to a latitudal sweep
from the pole to the equator at a longitudal/meridional distance ("M")
of 90^o:

M = Lon2 - Lon1 = 90^o, Lat1 = 0, Lat2 = P_Ni (90^o -> 0): Get
GEODx_Ni;

oo
MG'{HQ} = [sum GEODx_Ni -:- oo] -:- [.5 * Pi * a]
Ni=1

Gr = a * MG'{HQ}

Letting a = 6378.135 and b = 6356.75 (cos{Oz} = .9966471390):

n P_Ni Lr_n = a*MG'{P_Ni}
== ==== ==================
1 90 6367.446989
2 80 6367.771246
3 70 6368.704217
4 67.5 6369.020642
5 60 6370.131627
6 50 6371.879336
7 45 6372.807808
8 40 6373.735267
9 30 6375.475583
10 22.5 6376.578161
11 20 6376.891667
12 10 6377.814677
13 0 6378.135

Avg Gr_Avg
=== ===========================================================
1 6372.807808 = Lr_7
2 6372.790995 = [Lr_1 + Lr_13] -:- 2
3 6372.796599 = [Lr_1 + Lr_7 + Lr_13] -:- 3
4 6372.797300 = [Lr_1 + Lr_5 + Lr_9 + Lr_13] -:- 4
5 6372.797720 = [Lr_1 + Lr_4 + Lr_7 + Lr_10 + Lr_13] -:- 5
10 6372.798561 = [Lr_1 + Lr_2 + Lr_3 + Lr_5 + Lr_6
+ Lr_8 + Lr_9 + Lr_11 + Lr_12 + Lr_13] -:- 10
** ************************************
oo 6372.799401 = [Lr_1 + ... + Lr_7 + ... + Lr_13] -:- oo
= 6378.135 * .99916345475
Now compare:

MR'{HQ} = .99832145815: Mr = 6367.429033
MV'{HQ} = .99888112826: Vr = 6370.998685
MS'{HQ} = .99888213030: Sr = 6371.005076
MG'{HQ} = .99916345475: Gr = 6372.799401

The discrepancy becomes much greater as cos{Oz} is decreased:

a = 10000; b= a * cos{Oz};

<-- cos{Oz} -->
.75 .5 .25 .1
=========== =========== =========== ===========
Mr: 8615.669601 6864.402503 4458.257950 2352.715817
Vr: 9085.602964 7937.005260 6299.605249 4641.588834
Sr: 9155.311974 8307.144510 7527.264742 7176.639260
Gr: 9472.699551 9142.450665 8907.795824 8880.226988

Thus, is there a known reason why Sr and Gr are different (i.e., am I
missing some element or consideration in calculating Gr?). If they are
both equally valid, is one a better representative of a global mean
radius/radian?
As for calculating MG'{HQ}, is there a more direct equation, rather than
having to find the GEODx for each path?

~Kaimbridge~
--
UBasic Programming Forum: http://www.InsideTheWeb.com/mbs.cgi/mb426556

Global 2000 Spheroid [G2KS]: a = 6378.135, b = 6356.75, Gr = 6372.7994

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Before you buy.

[Kaimbridge]

In article <840nng$o9g$1...@nnrp1.deja.com>,
"Kaimbridge M. GoldChild" <kaimb...@my-deja.com> wrote:

<snip>

> ----------------
> Surface Area:
>
> Surface Area = 2 * Pi * r^2 (circle);
> = 4 * Pi * r^2 (sphere);
> = [2 * Pi * a^2]
> + [Pi * b^2 * csc{Oz} * LOGe{cot{.5 * [90^o - Oz^o]}}],
> = 4 * Pi * Sr^2 (spheroid);
>
> Sr = a * MS'{HQ};
>
> MS'{HQ} = [.5 * (<atanh{sin{Oz}} * sin{Oz} * cot{Oz}^2> + 1)]^.5

Typo, should be:

= [2 * Pi * a^2]

+ [2 * Pi * b^2 * csc{Oz} * LOGe{cot{.5 * [90^o - Oz^o]}}],

[Kaimbridge]

In article <84qr5i$jo7$1...@nnrp1.deja.com>,

Kaimbridge <kaimb...@my-deja.com> wrote:
> In article <840nng$o9g$1...@nnrp1.deja.com>,
> "Kaimbridge M. GoldChild" <kaimb...@my-deja.com> wrote:
>
> <snip>
>
> > ----------------
> > Surface Area:
> >
> > Surface Area = 2 * Pi * r^2 (circle);

As someone pointed out elswhere, that should be "Pi * r^2". P=(

[Mojca Miklavec]

I would like to ask if there exists a drawing tool for any parabolas and
hyperbolas and where could I get more information about it. I'm a
17-year-old student and have constructed a simple tool for it recently.

Mojca Miklavec

[Heidi Burgiel]

Dear Mojca,

There are numerous computer drawing tools for parabolas and
hyperbolas, but most of them do other things as well. I do not know
of any mechanical drawing tools (pencil, paper, and gadget) for this
purpose. What have you designed?

Heidi Burgiel