Talk:Spherical Earth/Archive 1

Archive 1

Archive 2

Archive 3

Stade

Do you happen to have a source for the length of the "stade". I recall reading somewhere that figure for the stade was derived from that very same measurement by erastothenes. In other words, a pure case of circular reasoning: "We know the correct circumference of the earth, Erastothenes measured it in stades, hence we can assume that the stade was approximately such and such long. Now, let's see how close Erastothenes got when he measured the earth... WOW! Almost on the nose!!!". -- Cimon Avaro on a pogo stick 14:07 25 May 2003 (UTC)

Against merge

I vote against merging the articles. —Preceding unsigned comment added by 24.17.76.161 (talk) 10:25, 20 August 2005 (UTC)

Please use modern dating

Meaning BCE and CE. Outdated models like BC and AD should not be part of the encyclopedia. [written by 69.181.136.51 on 23:44, 14 April 2007]

So we object to B-BCE/CE dating systems. Here I was thinking this was all just fashion. Sorry the traditional method offends you (and that I had nothing to do with its use). :OP 209.149.99.2 (talk) 13:47, 27 February 2008 (UTC)

Chinese and others

This article doesn't discuss the views of Chinese and other non-Europeans much at all (except a brief mention of Persians). AFAIK, the Chinese largely believed the world was spherical from early on, before 0 AD I believe. In fact, I'm not even sure whether the idea the world is flat was ever a popular phenomenon at least in recorded history. Someone who knows more about research on the historical views of non-Europeans should add to this and the world is round articles. Nil Einne 15:29, 13 April 2006 (UTC)

As I wrote in Talk:Flat Earth... This article (in Portuguese) says that Chinese only started to discuss the possibility of an spherical earth in the 17th century. I've read other sources agreeing that, during the Antiquity and the Middle Ages, the Chinese believed in a flat Earth... But I agree that the article should elaborate on the views of non-Europeans. --Leinad ¬ »saudações! 21:16, 13 April 2006 (UTC)

The article on Abubakari II of the Mali Empire alleges that scholars in medieval Timbuktu speculated about a "gourd-shaped world"... I'm not sure how true the claim is.

Plato

Um... the quote from Plato seems to imply that he thought the world was a dodecahedron - only approximately spherical. Remove or support?

Climate change

At the end of "In fact, the Earth is reasonably well-approximated by an oblate spheroid,", an editor keeps trying to add: "and some believe that it is due to climate changes (NASA-Most Changes in Earth's Shape Are Due to Changes in Climate)."

Climate affects DEVIATIONS from the ellipsoid. It does not contribute in any way to the basic ellipsoid. Therefore it is a complete misunderstanding of the NASA information to claim, with reference to the ellipsoidal shape, "and some believe that it is due to climate changes". Further, this information is completely out of context and irrelevant to the paragraph and to the entire article. The article does not deal with nuances of the geoid. It deals with how we approximate the earth as a sphere. The influence of climate does not affect that approximation at all. If it did, that information should be included somewhere else in the article, along with HOW it influences the spherical approximation. But it does not affect it and therefore doesn't belong anywhere in this article. There are other articles in which it MIGHT belong, such as "geoid".

Strebe 19:17, 29 May 2007 (UTC)

Ok fine, the article can then be added under 'external links' I guess.--S3000 11:38, 30 May 2007 (UTC)

I question its relevance even there. What does it have to do with theories that the Earth was round? -Elmer Clark 09:39, 18 August 2007 (UTC)

Whaaaat?

For fun (and because I needed to calculate the surface area of a slightly smaller planet for a game), I fired up Excel and plugged in the surface area equation, but I got '46,515.04 km' as the return value....not 6,371.007 km. The equation I used is below:

=SQRT(E1*E2/2)
E1=(C1+(B1*C2))/SQRT(C1-C2)
E2=LN(B1+(SQRT(C1-C2))/B2)

Where B1 = a (6,378.14), B2 = b (6,356.75), C1 = a^2 (40680631.59) and C2 = b^2 (40408295.99).

Could someone find where I mucked it up? Blast ^{[improve me]} 04.06.07 1931 (UTC)

Nevermind...bloody misplaced parenthesis. Blast ^{[improve me]} 04.06.07 1947 (UTC)

Discussion of ellipsoidal model, syntax, &c

Pol098, thanks for your contributions. A few notes: the parenthetical text you add in the notes concerning the ellipsoid should go somewhere else, if anywhere. In the form it was in, it associated Pythagoras with the ellipsoid. It also made for extremely ungainly prose.

A sentence fragment after a colon should not be capitalized, but an otherwise complete sentence after a colon should.

It is best to keep numbers consistent throughout, which is one reason for reverting the sub-millimeter accuracy of your numbers. The other reason is as the disclaimer states, which you also tried to delete. The disclaimer is useful, since, like you, people have (a) supplied precision inconsistent with other values; and (b) wondered why we don't provide 11 digits (or whatever their favorite number is). Since geodesy has certain expectations for precision, the article distances itself from those expectations and gives a cogent reason. Strebe (talk) 20:32, 16 January 2008 (UTC)

Pythagoras and Spherical Earth

Okay . . . just a few seconds ago I read something really scholarly on google books that said Pythagoras believed the Earth was plate shaped. Another site said we don't know what he believed at all. Whom am I to believe? —Preceding unsigned comment added by 209.149.99.2 (talk) 13:42, 27 February 2008 (UTC)

Formula for ellipsoidal quadratic mean

These comments relate to the section entitled "Spherical models".

Shouldn't the formula for the "ellipsoidal quadratic mean" be

$Q_{r}={\sqrt {\frac {2a^{2}+b^{2}}{3}}}$

instead of

$Q_{r}={\sqrt {\frac {3a^{2}+b^{2}}{4}}}?$

"Quadratic mean" is another name for "root mean square" (RMS).

Euler's approximation for the perimeter of an ellipse in Section 5 of this page

http://www.ebyte.it/library/docs/math05a/EllipsePerimeterApprox05.html

uses the RMS of the two half-axes as the "radius" in the formula for the perimeter of a circle.

An extension to three dimensions would use an analogous RMS value in calculating the surface area of an ellipsoid.

An ellipse is a two-dimensional curve. The term "elliptical quadratic mean", which is used in this Spherical Earth article, is the RMS of the two distances along orthogonal axes which define a two-dimensional shape:

$Q_{m}={\sqrt {a^{2}/2+b^{2}/2}}.$

An ellipsoid is a three-dimensional surface. Doesn't the term "ellipsoidal quadratic mean" suggest the RMS values of the three distances along orthogonal axes which define a three-dimensional shape? In the case of an ellipsoid, wouldn't that value be

$Q_{m}={\sqrt {a^{2}/3+b^{2}/3+c^{2}/3}}?$

In the special case of a spheroid, where a = c, that simplifies to:

$Q_{m}={\sqrt {a^{2}/3+b^{2}/3+a^{2}/3}}={\sqrt {2a^{2}/3+b^{2}/3}}.$

But the formula given in the article for "ellipsoidal quadratic mean" is equivalent to:

$Q_{r}={\sqrt {a^{2}/4+a^{2}/4+a^{2}/4+b^{2}/4}}.$

This appears to be a weighted RMS, double-counting an equatorial radius in some direction.

Shouldn't the "ellipsoidal quadratic mean" radius of the earth be

$Q_{r}={\sqrt {a^{2}/3+a^{2}/3+b^{2}/3}}={\sqrt {\frac {2a^{2}+b^{2}}{3}}}?$ -Ac44ck (talk) 16:56, 26 October 2008 (UTC)

No, because the equator is double weighted, as you have both X and Y based "a" values for the equator (a_x and a_y), and the meridional (a_m):

${\color {white}{\frac {\bigg |}{}}}{\sqrt {\frac {{\frac {a_{m}^{2}+b^{2}}{2}}+{\frac {a_{x}^{2}+a_{y}^{2}}{2}}}{2}}}={\sqrt {\frac {3a^{2}+b^{2}}{4}}};\,\!$

I give a more "verbose" explanation here! P=) ~Kaimbridge~ (talk) 19:06, 26 October 2008 (UTC)

Thanks for your reply and the link to the detailed discussion. My current thoughts are these:

1. I don't understand why graticule projections are used instead of standard polar coordinates.
2. The stated goal is to find the radius which preserves the average circumference. If one succeeds in doing that, wouldn't the use of that value in the formula for the surface area of a sphere yield the actual surface area of the spheroid of interest? That is, wouldn't the radius which preserves the average circumference _be_ the authalic radius?
3. As the "elliptical quadratic mean" approximates the radius of an equal-perimeter circle, wouldn't the simple RMS of the half-axes of the spheroid model be an approximation of the authalic radius?
4. Mention of a scalene ellipsoid (where the equatorial radius varies) doesn't seem relevant for the oblate spheroid model of the earth, where the equatorial radius is constant.

For the moment, I am still leaning toward the notion that the "ellipsoidal quadratic mean" should count the equatorial radius only twice, not three times.

It appears that the "ellipsoidal quadratic mean" is an approximation of the authalic radius. If so, shouldn't the preferred radius in the article on great-circle distance be the authalic radius?-Ac44ck (talk) 04:48, 27 October 2008 (UTC)

Removed the hand-wringing in my two intermediate edits and changed back to authalic radius because "authalic" means "equal-area" according to http://en.wiktionary.org/wiki/authalic. The definition of "authalic radius" seems a bit murky to me at the moment — equal to _what_?

• The Earth radius article says it is related to the "geometric mean oriented surface area" of a spheroid.
• This spherical Earth says it is related to the "surface area of the real Earth". But what area is that? As the "real" length of a coastline cannot be known, the "real" area of the earth cannot be known. Which of the following do we call the "real" area of the earth?

• The area of the WGS84 reference ellipsoid?
• The area of an oblate spheroid? Defined by what values for radii?
• Something else?

-Ac44ck (talk) 17:10, 27 October 2008 (UTC)

The verbiage in Spherical Earth was wrong. I corrected it. Obviously no one is using analytic formulæ to model the surface area of the "real Earth". All the spherical models are computed against the oblate ellipsoid chosen to model the earth.

I do not understand your inquiry about "graticule projections" versus "polar coordinates". What is a "graticule projection"? In general I do not understand your usage of the term "graticule" throughout your comments. "Graticule" means the latitude/longitude network shown on a map.

Strebe (talk) 19:21, 27 October 2008 (UTC)

Thanks for clarifying the text in Spherical Earth.

I was referring to the "graticule perspectives" in the archived talk page.

I have reconsidered some of my earlier comments modified them in this edit.

And I have a new comment:

The approximation $Tr\approx {\sqrt {\frac {Mr^{2}+{\overline {a}}^{2}}{2}}}$ appears to be the elliptical quadratic mean of two orthogonal mean radii, one of which is evaluated for a unique circumference: the equator. Tr may be a useful value for one who is standing on the equator and asks, "What is the average circumference for all directions available to me from here?" I don't believe that it answers the same question for a random location on the earth. This reinforces my opinion that the simple RMS of the half-axes would accurately be called the "ellipsoidal quadratic mean", as it would not give undue weight to the equatorial radius when considering the earth as a whole. It seems to me that Tr yields a slightly higher value than one should use for calculating a circumference in a random direction at a random location on the earth. -Ac44ck (talk) 20:02, 27 October 2008 (UTC)

I removed my formula which made the same mistake of giving preference to a location on the equator. For someone standing at a pole, the question "What is the average circumference for all directions available to me from here?" is answered using 'Mr' because all available directions are only variations in longitude. For someone standing at the equator, the question is answered using 'Tr'. At all intermediate latitudes, a value somewhere between 'Mr' and 'Tr' would be appropriate. The approximation for 'Mr' counts the equatorial radius once. The approximation for 'Tr' counts the equatorial radius three times. What seems to me like the proper interpretation of the "elliptical quadratic mean" counts it twice. Curiously, that is exactly half way "between" the extreme values and may be suitable for a random location on Earth. -Ac44ck (talk) 22:37, 27 October 2008 (UTC)

Ahh, you just about have it figured out! P=)

As you said,

For someone standing at a pole, the question "What is the average circumference for all directions available to me from here?" is answered using 'Mr' because all available directions are only variations in longitude.

If you calculate random distances along a meridian and divide each distance by its respective latitude difference, you will have the mean arcradius of that distance: If you calculate the length of the meridian in increments of infinitesimal distances and average all of the mean arcradii together, you will get Mr. Then, as you said,

For someone standing at the equator, the question is answered using 'Tr'.

Right, exactly!

The transverse value, Tr, includes all of the different mean arcradii for every differently angled great circle/circumference, while Mr only defines the north-south ellipse (and an averaging at an oblique perspective——each of the circumferences of which cross the equator, meaning they are circumferences of different transverse meridians and, therefore, included in Tr——is only partially inclusive), thus for random distances at random angles, Tr is the all inclusive average arcradius, while the surface area's authalic radius is based on latitude cosine adjusted, (north-south×east-west)^.5 paths.

As for the weighing of the ellipsoidal quadratic mean using only one equatorial a,

${\color {white}{\frac {\Big |}{}}}Tr\approx {\sqrt {\frac {Mr^{2}+a^{2}}{2}}}\approx {\sqrt {\frac {{\frac {a^{2}+b^{2}}{2}}+a^{2}}{2}}}={\sqrt {\frac {3a^{2}+b^{2}}{4}}}=Qr;{\color {white}.}\,\!$

so, again, a is triple weighted! P=) ~Kaimbridge~ (talk) 23:05, 28 October 2008 (UTC)

Wouldn't it be necessary to evaluate your Circ'(B,A,D) function for all values of B to consider "every differently angled great circle"? The archived talk page indicates that Tr is evaluated at B = 90°. Mr is calculated at B = 0°. What happens at B = 45°?

The statement that the triple-weighted equatorial radius is applicable at all locations other than a pole suggests that:

• Mr is the correct value to use if one is standing on a pole
• Tr is the correct value to use if they move one inch away from the pole.

That seems like too much of a step function to me.

I haven't deciphered everything that is in the archived talk page, but I note that:

• There is no mention of the Circ' function after presenting the formula for the triple-weighted equatorial radius.
• It says "let's look at differentiation by second latitude: Where the first latitude is set to 0 and the longitude difference to 90°". This seems to consider only directions which radiate from a point on the equator. It doesn't seem to cover the general case where someone may be standing anywhere on Earth and asking about the "average circumference".

It doesn't seem to me that Tr is applicable for "every differently angled great circle".

Is the triple-weighted equatorial radius endorsed by others, or is this original research? - Ac44ck (talk) 23:58, 28 October 2008 (UTC)

OHHHHHHHHHH, you are soooo close, but you are still missing the critical point: The north-south meridian (Mr) and the equator (a) are just the boundaries——all lengths/distances are either along or between them.

Take two places with any latitudes but different longitudes and pull a string between them: The string will not follow a north-south meridian, since the longitudes are different, BUT if you extend the string past the equator (if it isn't already) you will see it DOES follow a transverse meridian. It has nothing to do with where one is standing: Yes, if one is standing on the pole and measuring only north-south distances then Mr would be the mean arcradius...but the path/distance between NYC and LA is not north-south, or Boston to London, or Tampa to Bejing or Paris to Tokyo. So what single, mean/average, elliptic great-circle arcradius would apply to all of them: a? Mr? No! Tr, since that covers ALL transverse meridians. ~Kaimbridge~ (talk) 01:00, 29 October 2008 (UTC)

I think it quite important where one is standing. The assertion is that Tr gives the "average circumference". That means that there should be someplace on Earth where I can observe that average. I think that I probably would observe that average while standing on the equator. I am sure that I would observe a value lower than that average while standing on a pole. It seems reasonable to me that intermediate values — all lower than Tr — would be observed at intermediate latitudes. I believe that Tr is a maximum, not an average.

I can convince myself that the approximation for $T_{r}\,\!$may work for a point on the equator. For other points, I would have to see the Circ' function spelled out using functions that I recognize. And then I would probably need help getting from what I imagine to be a messy integral to what is presented as a clean, universally applicable approximation. Again: is this generally-accepted math that I can look up somewhere on the internet, or is $T_{r}\,\!$ your own creation? - Ac44ck (talk) 03:47, 29 October 2008 (UTC)

Pt.1: Tr and Circ are just my conceptual terms, used here in the talk pages (not in the articles themselves).

In terms of standing at a point of the average value, I think you are looking at it wrong.

The average value of something is near the middle, not at the endpoints, so the approximate value that Mr represents is found by standing at the mid-latitude (45°). At the other extreme we have the equator, which is constant——a.

Therefore, let's let Mr = Tr(0) and a = Tr(90°): The average arcradius of the mid circumference would be Tr(45°), which can be approximated by finding the mean arcradius of the distance from φ_s = 0 to φ_f = 45°, where Δλ = 90° (about 6372.810). So, just as φ = 45° is the point of approximation for Tr(0)/Mr, Tr(45°) is the path of approximation between Tr(0) and Tr(90°). Next, the midpoint of Tr(45°) is found halfway along the 45° arc path, $\scriptstyle {\widehat {\mathrm {A} }}\,\!$, at the 45° transverse co-latitude, $\scriptstyle {\widehat {\sigma }}\,\!$, with a great-circle azimuth, $\scriptstyle {{\widehat {\alpha }}({\widehat {\sigma }})}\,\!$, at that point, $\scriptstyle {{\widehat {\alpha }}({\widehat {\sigma }}_{m})}\,\!$, approximately equaling 54.73561° (i.e., $\scriptstyle {\cos ^{2}({\widehat {\alpha }}({\widehat {\sigma }}_{m}))={\frac {1}{3}}}\,\!$). Keeping in mind $\scriptstyle {\sin(\phi )=\cos({\widehat {\mathrm {A} }})\sin({\widehat {\sigma }})}\,\!$, the mid-latitude of $\scriptstyle {\widehat {\mathrm {A} }}\,\!$ = 45°, $\scriptstyle {\phi _{m}}\,\!$, equals 30°:

$\sin(30^{\circ })=\cos(45^{\circ })\sin(45^{\circ });\,\!$

The conjugate form of the arcradius integrand is

${\color {white}{\Bigg |}}T(\phi )={\sqrt {(M(\phi )\cos({\widehat {\alpha }}({\widehat {\sigma }})))^{2}+(N(\phi )\sin({\widehat {\alpha }}({\widehat {\sigma }})))^{2}}};\,\!$

So what does M(30°) and N(30°) equal?

Where a = 6378.137 and b = 6356.752,

${\color {white}{\frac {\Big |}{}}}\sin(30^{\circ })={\sqrt {\frac {1}{4}}}={\frac {1}{2}};\quad \cos(30^{\circ })={\frac {\sqrt {3}}{2}}={\sqrt {\frac {3}{4}}}\approx .8660254;\,\!$

${\begin{aligned}M(30^{\circ })&={\frac {(ab)^{2}}{((a\cos(30^{\circ }))^{2}+(b\sin(30^{\circ }))^{2})^{1.5}}},\\&={\frac {(ab)^{2}}{\left({\frac {3}{4}}a^{2}+{\frac {1}{4}}b^{2}\right)^{1.5}}}={\frac {(ab)^{2}}{\left({\frac {3a^{2}+b^{2}}{4}}\right)^{1.5}}}={\frac {(ab)^{2}}{Qr^{3}}}\;!\,!\,!{\color {white}.}\\&\approx 6351.37671;\end{aligned}}\,\!$

${\begin{aligned}N(30^{\circ })&={\frac {a^{2}}{\sqrt {(a\cos(30^{\circ }))^{2}+(b\sin(30^{\circ }))^{2}}}},\\&={\frac {a^{2}}{\sqrt {{\frac {3}{4}}a^{2}+{\frac {1}{4}}b^{2}}}}={\frac {a^{2}}{\sqrt {\frac {3a^{2}+b^{2}}{4}}}}={\frac {a^{2}}{Qr}}\;!\,!\,!{\color {white}.}\\&\approx 6383.48100;\end{aligned}}\,\!$

${\begin{aligned}{\color {white}{\Big |}}T(30^{\circ })&={\sqrt {(M(30^{\circ })\cos({\widehat {\alpha }}({\widehat {\sigma }}_{m})))^{2}+(N(30^{\circ })\sin({\widehat {\alpha }}({\widehat {\sigma }}_{m})))^{2}}},\\&={\sqrt {N^{2}(30^{\circ }){\frac {2}{3}}+M^{2}(30^{\circ }){\frac {1}{3}}}}={\sqrt {{\frac {1}{3}}\left({\frac {2a^{4}}{Qr^{2}}}+{\frac {(ab)^{4}}{Qr^{6}}}\right)}},\\&=Qr{\sqrt {{\frac {1}{3Qr^{4}}}\left(2a^{4}+({\frac {ab}{Qr}})^{4}\right)}}\approx {6372.79754},\\&{\color {white}=}\qquad \qquad \qquad \qquad \quad \;\;\approx {Qr}\approx {6372.79748};{\color {white}{}_{\big |}}\end{aligned}};\,\!$

Here is the kicker:

${\color {white}{\frac {\Big |}{}}}{\sqrt {\frac {M^{2}(30^{\circ })+N^{2}(30^{\circ })}{2}}}\approx {6367.44909}\;!\,!\,!{\color {white}.}\,\!$

So, using $\scriptstyle {M(30^{\circ })}\,\!$ and $\scriptstyle {N(30^{\circ })}\,\!$ instead of a and b, your prescribed weights are utilized!

${\color {white}{\frac {\Big |}{}}}Mr\approx {\sqrt {\frac {N^{2}(30^{\circ })+M^{2}(30^{\circ })}{2}}};\quad \,Tr\approx {\sqrt {\frac {2N^{2}(30^{\circ })+M^{2}(30^{\circ })}{3}}};\,\!$

Thus, the approximate value of Tr is found standing at latitude 30°, in the direction of 54.73561°: If you look at the quadrant of a globe, the midpoint——in terms of area——appears (though it could be just an illusion) to be at lat = 30°, not 45°.

(Pt.2, below) ~Kaimbridge~ (talk) 01:00, 31 October 2008 (UTC)

I am now sure that the formula $T_{r}\approx {\sqrt {\frac {3a^{2}+b^{2}}{4}}}\,\!$ is applicable only at the equator.

First, the form of the $Circ'(B,A,D)\,\!$ function used to derive it is overly complex for the task at hand. It accommodates a scalene ellipsoid. We don't need such generality for this task. All we need is a function to evaluate the perimeter of the great ellipse that passes through three points:

1. 0° latitude, 90° east longitude
2. 0° latitude, 90° west longitude
3. $\delta ^{\circ }\,\!$ latitude, 0° longitude

where $\delta \,\!$ = any latitude of interest

Let's use the "oblique graticule perspective" and use the (latitude, longitude) coordinates $(\delta ,0)\,\!$ to define the point which appears in the center of our view. Because we have an oblate spheroid, we can use symmetry to obtain a worldwide average by considering points found only on the Prime Meridian at 0° longitude (or its continuation at 180° longitude, but I'll use the one name to refer to the great ellipse which contains them both).

• The Prime Meridian (0° longitude) will always appear as the vertical axis at the center of our view.
• We will always see the edge of a great ellipse as the horizontal axis at the center of our view. That ellipse passes through the points $(0,-90^{\circ }),(\delta ,0),(0,90^{\circ }).\,\!$

Let's say we have some function $P(\delta )\,\!$ to calculate the perimeter of the ellipse which appears as a horizontal line at the center of our view.

The average radius of our great ellipse is:

$R_{GE}(\delta )={\frac {1}{2\pi }}P(\delta )$

We can now find values for $M_{r}\,\!$ and ${\overline {a}}:\,\!$

• $M_{r}={\frac {1}{2\pi }}P(90^{\circ })\,\!$

The meridian at 90° longitude (which has exactly the same perimeter as the Prime Meridian) appears as a horizontal line in this view (as seen from a point far above a pole).

• ${\overline {a}}={\frac {1}{2\pi }}P(0^{\circ })\,\!$

The equator appears as a horizontal line in this view (as seen from a point far above the equator).

Now we can approximate an answer to the question: "What is the average circumference for all directions available to one who is standing at the center of our view?" Let's consider only two of those directions:

• If they travel along the vertical axis, they are moving along the Prime Meridian, so the perimeter in that direction is always $P(90^{\circ })\,\!$.
• If they travel along the horizontal axis, they are moving along a great ellipse whose perimeter is $P(\delta )\,\!$.

We might justify saying that the average circumference for all directions available from the center of our view may be approximated by the RMS (root mean square) of those two perimeters:

${\overline {P}}(\delta )\approx {\sqrt {(P(90^{\circ }))^{2}/2+(P(\delta ))^{2}/2}}.\,\!$

Then:

${\overline {R}}(\delta )\approx {\sqrt {({\frac {1}{2\pi }}P(90^{\circ }))^{2}/2+({\frac {1}{2\pi }}P(\delta ))^{2}/2}}.\,\!$

Using the approximation above for $M_{r}:\,\!$

${\overline {R}}(\delta )\approx {\sqrt {(M_{r})^{2}/2+({\frac {1}{2\pi }}P(\delta ))^{2}/2}}.\,\!$

The extreme values are:

• ${\overline {R}}(0^{\circ })\approx {\sqrt {(M_{r})^{2}/2+({\overline {a}})^{2}/2}}\approx T_{r}\,\!$
• ${\overline {R}}(90^{\circ })\approx {\sqrt {(M_{r})^{2}/2+(M_{r})^{2}/2}}\approx M_{r}.\,\!$

The average would be:

$\mathbb {R} _{avg}=\left({\frac {2}{\pi }}\right)\int _{0}^{\pi /2}{\overline {R}}(\delta )d\delta$

I am not prepared to evaluate this integral, but I am confident that the result is somewhere between $T_{r}\,\!$ and $M_{r}.\,\!$ The RMS seems be an approximate average elsewhere in this discussion. My first guess is that this may not be far off:

$\mathbb {R} _{avg}\approx {\sqrt {(M_{r})^{2}/2+(T_{r})^{2}/2}}\approx {\sqrt {5a^{2}/8+3b^{2}/8}}.\,\!$

The assertion that $T_{r}\,\!$ is applicable worldwide seems to be provably false. Unless there are citations to the contrary, I suspect that it was original research. That doesn't make it wrong; all research was original at some point. But in this case, it appears that it was wrong and it seems unfortunate that other websites are parroting the conclusion found in Wikipedia:

http://www.google.com/search?q=%22ellipsoidal+quadratic+mean%22

I believe that $Q_{m}={\sqrt {\frac {a^{2}+b^{2}+c^{2}}{3}}}$ is the proper interpretation of the term "ellipsoidal quadratic mean", and that the name suggests a particular kind of unweighted RMS value — where a, b, and c are measured along orthogonal axes.

And I don't believe that $\mathbb {R} _{avg}\,\!$ is the best value to use in calculating great-circle distances. It would be an average radius for distances between random (latitude, longitude) coordinates. But that isn't the same thing as the distance between random locations on Earth. The distinction arises from the discussions here:

• http://mathworld.wolfram.com/SpherePointPicking.html
• http://groups.google.com/group/sci.math/msg/c6e5c86af2f5664e?hl=en

Using an unweighted RMS, I would expect $Q_{r}={\sqrt {\frac {2a^{2}+b^{2}}{3}}}$. The approximation above for $\mathbb {R} _{avg}\,\!$ seems to give too much weight to the polar radius. The multiplier in the polar radius term is 0.333... for the unweighted RMS; it is 0.375 for $\mathbb {R} _{avg}\,\!$. That is consistent with what is said in the links above about picking points via random latitude and longitude: the polar regions are over-represented.

I don't see a reason to use anything other than the authalic radius for random calculations of great-circle distances. -Ac44ck (talk) 23:39, 29 October 2008 (UTC)

Pt.2: First of all, other than the north-south meridian, do you think the proper term is "great ellipse" or, due to the nature of the elliptic curvature with non-north-south arcs, the proper description is "elliptic great circle"?

As for point of perspective, the equator is the universal perspective. All great circle and geodetic distance calculations are based on distances from, and angles at, the equator. Look at Vincenty's benchmark formula(PDF) (my notation equivalent in [brackets]):

${\alpha \;\;[{\tilde {\mathrm {A} }}]}\,\!$ azimuth of the geodesic at the equator.

$\sigma \;\;[\Delta {\tilde {\sigma }}]\,\!$ angular distance P₁ P₂ $[{\tilde {\sigma }}_{f}-{\tilde {\sigma }}_{s}]\,\!$ on the sphere.

$\sigma _{1}[{\tilde {\sigma }}_{s}]\,\!$ angular distance on the sphere from the equator to P₁.

$\sigma _{m}[{\tilde {\sigma }}_{m}]\,\!$ angular distance on the sphere from the equator to the midpoint of the line.

You find what arc path (which crosses the equator) the two points occupy; find the ellipsoidal/parametric/auxiliary angular distance from the equator to the points (which equals the transverse co-latitude) and subtract the two; find the mean (auxiliary) arcradius between the two points and multiply it by the subtracted angular distance.

When you stand at an "oblique graticule perspective", that point you are standing at can also be considered the forepoint latitude on the transverse equator, where the standpoint is on the (conjugate) equator at a longitude difference of 90°, and the arc path is 90° minus the forepoint latitude.

Regarding randomness, go back and look at the UBasic section of my verbose explanation at Talk:Earth: The key is to base the randomization on the transverse coordinates rather than the common conjugate ("Lat"/"Long") coordinates.

In terms of the authalic valuation, that is more a loxodromic than orthodromic quantity, as the integration is (north-south)×(east-west): cos(φ)M(φ)N(φ).

 ~Kaimbridge~ (talk) 01:00, 31 October 2008 (UTC)

I am replying to both parts of Kaimbridge's replies in this block. This section is already long. Scanning it for in-line replies isn't something I want to start.

>Tr and Circ are just my conceptual terms

Is this original research? I asked that twice before.

K: The variables ("Tr" and "Circ") are but not the actual transverseand obliquegraticule concept.

>used here in the talk pages (not in the articles themselves).

But the conclusion is used in this and other articles, at least one claiming that Tr is "the best approximation of Earth's average transverse meridional radius". I believe that I have shown the claim to be false.

K: No you are talking about the average oblique meridional arcradius (which includes the conjugate and transverse).

>the approximate value that Mr represents is found by standing at the mid-latitude (45°).

Isn't Mr a constant at all latitudes?

K: Mr is the average of the whole (conjugate) meridional circumference——its approximate value is found at the midpoint, here at lat = 45° (if, say, you wanted the average meridional arcradius between lats 40° and 80°, Mr(40°,80°), the average's approximate value would be found at lat = 60°)

>Therefore, let's let Mr = Tr(0) and a = Tr(90°)

Why is Tr now a function of angle? Wasn't it claimed earlier that the one value (triple-weighting the equatorial radius) "includes all of the different mean arcradii for every differently angled great circle/circumference" and that "for random distances at random angles, Tr is the all inclusive average arcradius"?

K: Sorry about any confusion (I'm trying to keep notation as concise as possible): Mr = Mr(0,90°)——the average value of M from the equator (0) to the pole (90°)——and $\scriptstyle {Tr({\widehat {\mathrm {A} }})=Tr({\widehat {\mathrm {A} }};0,90^{\circ })}\,\!$——the average arcradius from the conjugate equator out to the transverse equator, along $\scriptstyle {\widehat {\mathrm {A} }}\,\!$——and Tr is the average of all $\scriptstyle {Tr({\widehat {\mathrm {A} }})}\,\!$s, from $\scriptstyle {\widehat {\mathrm {A} }}\,\!$ = 0 to 90°.

>So what does M(30°) and N(30°) equal?

Where are M and N defined? What is the list of equations which follow intended to show?

K: M and N are the principal radii of curvature.

>Here is the kicker: ... = 6367.44909!!!

I don't understand the significance of this result.

K: The average of M(30°) and N(30°) (not 45°) approximately equals Mr!!

>Thus, the approximate value of Tr is found standing at latitude 30°, in the direction of 54.73561°

The claim was that Tr is related to the average circumference. I would not look in only one direction to find the value which applies "for random distances at random angles".

K: Right, but, except at the poles, the arcradius (unlike the radius) is different for all directions at a given point: Even along the meridian, you are facing north-south, thus using the average value of M——the average east-west arcradius along a meridian would be found using N, in which case you would be finding the mean arcradius of the transverse equator.

I want to be able to go to a place where the average may be observed, measure the circumference in every possible direction, and average them. The only place where this average is Tr (with the triple-weighted 'a') is on the equator.

K: RIGHT! And, orthodromically (i.e., great circle-wise), ALL random distances at random angles exist along transverse meridians (which includes the conjugate meridian and equator), which means any and all distances at any and all angles are included, ONCE, in Tr——just like a section of a meridian is only used once in the calculation of Mr——whereas some may be included several times in your oblique valuation, which, then, DOES introduce problems of randomness and over-representation.

>As for point of perspective, the equator is the universal perspective.

The equator is a reference plane, from which angles are measured in polar coordinates. Perspectives may be from anywhere on the globe. There is no "universal perspective".

K: See above response.

>Look at Vincenty's benchmark formula

I didn't spend a lot of time looking for it. The pdf file contains scanned images; the text search feature in Adobe reader didn't help. Where is Vincenty finding an "average circumference"?

K: His formula calculates geodetic distance, not average circumference. At the top of pg.89 he gives the definitions I quoted.

>You find what arc path (which crosses the equator) the two points occupy ...

I don't understand how this relates to finding the worldwide average circumference.

K: Because this is how you calculate the distance between two points. Again, any and all distances at any and all angles exist along arc paths, which means the average circumference of all of the different arc paths would include all of the different possible combinations of distances, which means the quadrant distance (i.e., from the conjugate to transverse equators), at different angles.

>Regarding randomness, go back and look at the UBasic section of my verbose explanation

Leading mathematicians describe, in the links I provided, how to find random locations on Earth.

K: Again, see my response above.

I don't understand the full form of your Circ' function. I gather that it can look a different graticule perspectives. And it can look at any point within one of those perspectives. Where does it do a spin through local azimuths to evaluate the average circumference?

My P(delta) function is pretty simple -- it returns the perimeter of a well-defined ellipse at a given latitude.

My P_bar(delta) function does a spin through local azimuths and returns the average circumference at a given latitude.

I think I gave a convincing demonstration that the radius you seek to calculate a worldwide "average circumference" is in the neighborhood of R_avg = sqrt(5a^2/8 + 3b^2/8). Do you disagree with this conclusion? If so, where do you think I went wrong in my derivation?

K: I don't think either of us are wrong, we may be just butting heads over different concepts.

You are taking an infinite number of "worldwide average circumferences" (one for each oblique position/vertex) and averaging them together for a grand average, which I would agree ${\color {white}{\frac {\Big |}{}}}\approx {\sqrt {\frac {Mr^{2}+Tr^{2}}{2}}}\approx {\sqrt {\frac {5a^{2}+3b^{2}}{8}}}\,\!$.

What I am talking about is a single worldwide average circumference, the averaging of which only involves touching a given point once, which is achieved by averaging all of the transverse meridian values together.

Rather than "elliptic" and "ellipsoidal", would "conjugate" and "transverse" be better?

• Conjugate Quadratic Mean: ${\color {white}{\frac {\Big |}{}}}{\sqrt {\frac {a^{2}+b^{2}}{2}}};\,\!$
• Transverse Quadratic Mean: ${\color {white}{\frac {\Big |}{}}}{\sqrt {\frac {3a^{2}+b^{2}}{4}}};\,\!$

>In terms of the authalic valuation, that is more a loxodromic than orthodromic quantity

I am out of my element here. Is that a bad thing?

Well, yes. Great circles are arc paths, meaning orthodomic, so an arcradius average used as a spherical great-circle radius should be orthodromically found. Plus, the authalic radius is integrated in squared form and then rooted at the end: If you calculated a narrow sliver of surface area from the equator to the pole, its radius would not look anything like Mr.

I am about done with this discussion.

Are there independent sources which favor the use of sqrt(3a^2/4 + b^2/4) in calculating random great-circle distances? If not, why should Wikipedia include this formula or the value that it produces? - Ac44ck (talk) 03:34, 31 October 2008 (UTC)

K: Besides the ones that may have gotten it from one of our articles, there are others that use variations of this value, including "Test Report on the November 2005 NATO RTG-40 Active Imager Land Field Trials" Pg.15,"Comparison of AVHRR, MODIS and VEGETATION for land cover mapping and drought monitoring at 1 km spatial resolution" Pg.s 35, 37, 51, "BULLETIN DE L’INSTITUT ROYAL DES SCIENCES NATURELLES DE BELGIQUE" Pg.43, "Overview", even several book citingsusing 6373, most of them long before Wikipedia was even a gleam in someone's eye! P=) ~Kaimbridge~ (talk) 18:39, 31 October 2008 (UTC)

Kaimbridge, please add those citations to the two articles, then—though variations of this value sounds suspicious and bears clarification. I was about to remove the average circumference formulæ because none of my sources list any such thing. We need to be citing the rest of the material as well, and I will get to some of it eventually, but the other metrics have not met with any controversy and references are readily available for them, so they do not seem quite so urgent. Strebe (talk) 19:58, 31 October 2008 (UTC)

I looked at the given links:

http://www.arl.army.mil/arlreports/2006/ARL-TR-4010.pdf

Page 15 does use the value 6372.8 km for the mean radius. I didn't find mention of why they chose this value.

This may be self-referential:

https://dspace.lib.cranfield.ac.uk/bitstream/1826/2903/1/Toukiloglou,%20Pericles_PhD.pdf

Pages 35, 37 and 51 do use the value 6372.795 km; but page 310 gives the following as a reference:

Wikipedia, 2006a. United Kingdom. http://en.wikipedia.org/wiki/Uk (9 January 2006).

Curiously, the "Spherical models" section (containing the value in question) was added to the "Spherical Earth" article the day before: [(8 January 2006)].

http://striweb.si.edu/basset/PdFs/Copy%20of%20BassetEtAl2007-IBISCA-PanamaSmaller.pdf

Page 43 does use 6372.8 km as the "mean earth radius". I didn't find mention of why they chose this value.

I didn't find the sqrt(3a^2/4 + b^2/4) formula in any of the links above. Do any of the book citings give an independent derivation of 6373 km via sqrt(3a^2/4 + b^2/4)?

I want to know the "best" radius to use for calculating a great-circle distance between two random locations on Earth. The following values (measured in kilometers) seem to be easily obtainable:

• Volumetric radius = ((a^2)*b)^(1/3) = 6371.0007
• Authalic radius = a manageable logarithmic expression = 6371.0071
• Simple average of orthogonal radii = 2a/3 + b/3 = 6371.0087
• RMS of orthogonal radii = sqrt(2a^2/3 + b^2/3) = 6371.0166
• R_avg = sqrt(5a^2/8 + 3b^2/8) = 6370.126
• Tr = sqrt(3a^2/4 + b^2/4) = 6372.7975

The first four values above are quite close together. I am not persuaded that the largest of the six values is the "best" one to use in calculating a great-circle distance between two random locations on Earth.

The last four values above are all approximations of something.

The authalic radius is an exact value for the reference ellipsoid. The authalic radius is related to things only on the surface of the reference ellipsoid. The authalic radius describes a sphere that is

1. Related to the reference ellipsoid in an easily-understandable way, while
2. Considering points found only on the surface of the reference ellipsoid.

"Great-circle" distances are measured along the surface of a sphere.

I am persuaded that the authalic radius is the "best" one to use in calculating a great-circle distance between two random locations on Earth. - Ac44ck (talk) 21:04, 31 October 2008 (UTC)

I replaced the controversial formula with the IUGG "official" definition. Strebe (talk) 22:01, 31 October 2008 (UTC)

I don't have an IUGG document. Do they relate R_1 to the "average circumference"? I don't find a suggestion of it via Google searches like this. I calculate (2*6378.137 + 6356.752)/3 = 6371.00866667. I would round to three decimals as 6371.009 km. - Ac44ck (talk) 23:01, 31 October 2008 (UTC)

The IUGG calls it the "mean" radius, not "average". The main article calls the IUGG value specifically a "mean", though the discussion is about "average". I see no harm in that, especially since there is no sharp distinction between the two terms. There is a mention of IUGG "mean circumference" here: http://www.jqjacobs.net/astro/xls/aegeo.xls. It is not clear where this number comes from — certainly it is not $2\pi R_{1}$ as one would expect, and I do not find anything official from the IUGG about it. Thanks for catching the typo; I see that I was also off by three orders of magnitude! Strebe (talk) 23:22, 31 October 2008 (UTC)