Talk:Earth radius

I believe that in the comment where A is the authalic surface area of Earth, the word authalic is not needed. As the term means equal area, the "authalic surface area" of the Earth is identical to the surface area of the Earth! Making the distinction between the two may confuse people into thinking that they do not really understand what authalic means. If anything, make it clear A_r is the authalic radius.

Question: Why is this comment

Note: Earth radius is sometimes used as a unit of distance, especially in astronomy and geology. It is usually denoted by RE.

part of the "Volumetric radius" section?

Should it be perhaps "Earth volumetric radius is sometimes used...", or does this comment refers to the general case, situation in which it should be moved somewhere in the main text?

Radius vs. Sea Level

"The radius of Earth (or any other planet) is the distance from its center to a point on its surface at mean sea level." This doesn't seem right. Why is the mean radius of a planet exactly the sea level? The sea level of the earth is rising, does this mean the radius of the earth is increasing too? Why would the melting of the ice caps cause the land to rise? Anyone have a cite? -anon

Radius of curvature

I have changed the formula for Rc near the end of this section to reverse M and N, since when traversing along a meridian (with alpha = 0, i.e. facing North) one is moving along "M" the meridianal radius and when moving (e.g at the equator) eastward (for alpha = + 90 degrees) one is moving along "N" the radius of curvature in the prime vertical. JimC (C&B) 17:57, 7 January 2007 (UTC)

You just described it right——facing north-south the RoC = M, while facing east-west = N——so cos(0)M = M and sin(90°)N = N, which is how it was (your change went and inverted it!

Whoops, okay, I see the problem: Presenting it with MN as the numerator cancels and reverses the placement of M and N in the denominator. I've changed it to make it clearer. P=) ~Kaimbridge~ 15:21, 8 January 2007 (UTC)

"Neridional" radius of curvature???

The heading Neridonal appeared in a 1 April 2007 edit. Surely this should be Meridional. I don't think neridional is even a real word. Certainly it's not in any online dictionary. Also a Google search for it only gives 149 hits, and almost all of those hits are either this article or people who have copied this article verbatim (like answers.com). I'm going to edit this. If someone feels there is some really burning reason why it should be 'neridional' and wants to revert, please add a note here why such an obscure word that doesn't appear in dictionaries should be used. I'm looking at you, User:Kaimbridge, since you made that 1st April edit ... surely you wouldn't be playing an April Fool's joke on Wikipedia ;). Dr algorythm 03:19, 18 June 2007 (UTC)

LOL!!! Sorry about that——I'm blind in one eye and the other eye's no bargain, with an acutely advancing cataract further complicating things. So, needless to say, a few typos may escape me! P=) ...Hmmm, how come it took someone two and a half months to catch?!? P=/ ~Kaimbridge~14:27, 18 June 2007 (UTC)

ERROR?

On this page the polar radius is derived as being larger than the equatorial radius. Please check this derivation. —The preceding unsigned comment was added by 216.59.226.224 (talk) 04:48, 25 April 2007 (UTC).

What do you mean, a (equatorial radius) = 6378.135 and b (polar) = 6356.75? ~Kaimbridge~10:49, 25 April 2007 (UTC)

Regarding the above given values a and b, I think they should be corrected and precised as follows: a=6378.137 km (according to GRS80 and GRS84) and b~= 6356.752 km (as a derived value). Gil

Latitude dependent radius error?

When using the formula *RADII WITH LOCATION DEPENDENCE* I've had a not logical result, for a 43ºS. calculation. My carefully result was no logical because it was 5417.374347 Km., not between the equatorial and polar radius, can spmebody explain this abnormal fact, I will be very happy if somebody explains me this anomaly, Tks. Carlos J.J.Vial <carlosjjv@terra.cl>Carlos J. Jiménez Vial 21:02, 3 August 2007 (UTC)

Well, if its the actual radius at 43º, it should be about 6368 km (depending on the actual values of a and b); but if it involves cos(43º), 5417 is actually too big——it should be somewheres between 4633 and 4681 km (based on the arcradii extremes of 6335 and 6399 km)! What values does it give for 0 and 90°? ~Kaimbridge~17:37, 4 August 2007 (UTC)

I am interested on the equation of estimating earth radius depend on latitude, could you give me information the reference of this equation (from where this equation come from?) Marksteven2 04:45, 3 December 2008

Please sign your talk-page posts with four tildes.

These may help:

• http://www.mathworks.com/access/helpdesk/help/toolbox/aeroblks/geocentrictogeodeticlatitude.html
• http://lists.maptools.org/pipermail/proj/2004-April/001212.html
• http://mathforum.org/library/drmath/view/61528.html

-Ac44ck (talk) 07:47, 3 December 2008 (UTC)

SI Units

Why are we not using SI units here (radii in meters)? MeddlerOnTheRoof (talk) 02:21, 9 May 2008 (UTC)

Changes to section on authalic radius

I removed this text:

This number is derived by square rooting the average (latitudinally cosine corrected) geometric mean of the meridional and transverse equatorial, or "normal" (i.e., perpendicular), arcradii of all surface points on the spheroid

It may be that the number approximates the root of the average. I don't find evidence that this average is used to derive the exact expression. One derivation is here.

I also modified this text

the radius of a hypothetical perfect sphere which has the same, geometric mean oriented surface area as the spheroid.

The "mean oriented surface area" of a perfect sphere is zero, as each elemental vector area has an equal antipodal counterpart. The vector sum of the oriented area is zero. -Ac44ck (talk) 02:43, 1 November 2008 (UTC)

Approximation

The original Meridional definition of a meter was one ten-millionth the distance from the North Pole to the Equator, the circumference of the earth is then (about) 40 million meters. Treating the earth as a perfect sphere then gives a nice, and easy to remember approximation for the average radius of the earth: (2/pi)*10^7 meters. This works out to 6366.2km, which is within 0.1% of all of the various figures for the average radius. Pulu (talk) 06:14, 12 November 2008 (UTC)

Reference or monograph

A reference for radius of curvature section was recently expanded. It seems lengthy for a reference.

If the subject in this reference is to be expounded upon, I suggest moving the longer treatment to the great circle distance article, which doesn't seem to include any approximate formulae at the moment.

I found the reference written with:

A related application of M and N: if two nearby points have the difference in latitude of $d\phi \,\!$ and longitude of $d\lambda \,\!$ (in radians) with mean latitude $\phi \,\!$, then the distance D between them is

$D\approx {\sqrt {(Md\phi )^{2}+(N\cos \phi d\lambda )^{2}}}.\,\!$

I tweaked it because two-glyph symbols like $d\lambda \,\!$ may be difficult for some readers to parse and $\phi \,\!$ seemed to be used in different ways. So I changed it to:

A related application of M and N: if two nearby points have the difference in latitude of $d\phi \,\!$ and longitude of $d\lambda \,\!$ (in radians) with $M\,\!$ and $N\,\!$ calculated at mean latitude $\phi _{m}\,\!$, then the distance D between them is

$D\approx {\sqrt {(M\cdot d\phi )^{2}+(N\cdot \cos \phi _{m}\cdot d\lambda )^{2}}}.\,\!$

The current version contains undefined symbols, such as $dH\,\!$. Why the new symbol is multiplied by a trigonometric identity to obtain $dH={\sqrt {(dH\cos(\alpha ))^{2}+(dH\sin(\alpha ))^{2}}}\,\!$ is lost on me. And current indications are that the results are exact. The previous version gave a formula that was an approximation. Perhaps the current formula is exact for a differential distance $dD.\,\!$ But the introducing paragraph advertises that a way to calculate $D\,\!$ is to be presented, and the now-more-elaborate treatment doesn't seem to deliver it.

I wondered whether the formula for $D\approx ...\,\!$ belonged in a reference when I tweaked it. But it is similar to the procedure prescribed by the FCC for moderate distances in 47 CFR 73.208, an allusion to which was recently omitted from the external references in this or a related article, and I didn't want to find a place for it in the article. That it is an application of the Pythagorean theorem over short distances on a curved surface seemed more easily discernable when the formula was by itself. - Ac44ck (talk) 22:54, 3 December 2008 (UTC)

I'm the guilty party! P=)

Everything you said appears right (I, too, questioned whether it should be presented as a footnote reference, but no one moved it, and now you tweaked it, so I attempted to clarify it further). I added H ("hypotenuse") to isolate the angular distance.

First of all, the FCC formula (a better presentation of it is here) is the same formula——"KPD_lat" is M and "KPD_lon" is cos(lat)N, in binomial series expansion form, based on the Clarke 1866 spheroid (a = 6378.2064, b = 6356.5838). As for "(w)hy the new symbol is multiplied by a trigonometric identity to obtain $\scriptstyle {dH={\sqrt {(dH\cos(\alpha ))^{2}+(dH\sin(\alpha ))^{2}}}}\,\!$", that is important because $\scriptstyle {M\cos(\alpha )H=M\Delta \phi }\,\!$ and $\scriptstyle {N\sin(\alpha )H=N\cos(\phi _{m})\Delta \lambda }\,\!$, so if you can isolate and separate H, the left over $\scriptstyle {\sqrt {(M\cos(\alpha ))^{2}+(N\sin(\alpha ))^{2}}}\,\!$ is the transverse meridional radius of curvature, T, or arcradius, $\scriptstyle {\overset {{}_{\smile }}{R}}\,\!$, that can be applied to orthodromic calculations (by using the orthodromic azimuth), for any distance (as for infinitesimal distances, it equals the geodetic distance), though it is technically the "elliptical great-circle distance", as it follows the geographical delineation, rather than the geodetic!

Something that needs to be corrected, though, is a distinction between spherical/geographical and geodetic azimuth ("whole" and "local"):

${\tilde {\alpha }}({\widehat {\sigma }})=\lim _{H\to 0}{\tilde {\alpha }}=\arctan \left({\frac {N}{M}}\tan({\widehat {\alpha }})\right);\,\!$

$R_{c}={\frac {{}_{1}}{{\frac {\cos ^{2}({\tilde {\alpha }})}{M}}+{\frac {\sin ^{2}({\tilde {\alpha }})}{N}}}};\,\!$

${\begin{aligned}{\color {white}{\frac {\big |}{}}}{\overset {{}_{\smile }}{R}}=T&={\sqrt {(M\cos({\widehat {\alpha }}))^{2}+(N\sin({\widehat {\alpha }}))^{2}}},\\&={\frac {{}_{1}}{\sqrt {\left({\frac {\cos({\tilde {\alpha }}({\widehat {\sigma }}))}{M}}\right)^{2}+\left({\frac {\sin({\tilde {\alpha }}({\widehat {\sigma }}))}{N}}\right)^{2}}}};\end{aligned}}\,\!$

I give a more in-depth analysis here, though some of the notation has evolved since then (e.g., "O" is now "T", and the loxodromic azimuth notation has simplified).

As for adding it to the great circle distance article, I would suggest changing the section to "Approximate elliptical great-circle distance formula" (and after giving the FCC form, let $\scriptstyle {SA_{m}={\frac {\cos(LAT_{1})\cos(LAT_{2})\sin(\Delta \lambda )}{\cos(LAT_{m})\sin(\Delta {\widehat {\sigma }})}}}\,\!$, $\scriptstyle {{\color {white}{\frac {\big |}{}}}KPD_{m}={\sqrt {M^{2}(LAT_{m}){\big (}1-SA_{m}^{2}{\big )}+{\big (}N(LAT_{m})SA_{m}{\big )}^{2}}}}{\color {white}.}\,\!$, then $\scriptstyle {DIST=KPD_{m}\times \Delta {\widehat {\sigma }}}\,\!$) and moving it to after the "A worked example" section.

If you are really interested in geodetic formulation, in general, I would STRONGLY recommend Richard Rapp's Geometric Geodesy (both parts), now downloadable in PDF, free, from OSU, here: This is the bible of geodetic formulation! P=)  ~Kaimbridge~ (talk) 18:07, 4 December 2008 (UTC)

Thanks for this and for your reply at Talk:Great-circle_distance. I moved the section in the great circle distance article, as you suggested. I would hesitate to bring up "elliptical" in an article on "great-_circle_ distance". At the moment, the character string "ellip" appears only in two words near the bottom of the page.

I still don't understand the relationship between $\scriptstyle {dD}\,\!$ and $\scriptstyle {D}\,\!$ in the current note in this article. As this Earth radius article recognizes the earth is not spherical, it seems that the accuracy of any constant-radius formula, such as $\scriptstyle {dD={\overset {{}_{\smile }}{R}}\cdot {dH}}$ would decrease with range. It looks like a formula for a differential distance as opposed to one which is applicable for any distance.

Thanks for the pointer to Rapp's documents. Part 1 is 189 pages; Part 2 is 225 pages! I have only scanned them so far, but they look very interesting.

You have a much better handle on this topic than I do. I wouldn't feel comfortable copying things from the note here to the great-circle distance article. And on reflection, things that talk about the elliptical shape of the earth seem out of place there. Which makes the FCC formula out of place there, too. A short-distance approximation using the Pythagorean theorem would not be so complicated on a _sphere_.

There seems to be good stuff in the note here, but it makes for a longish note compared with its neighbors in the "Notes and references" section. I don't know how standard the term "great ellipse" is, but a section to address distances along such a curve seems to be missing. The article on Earth _Radius_ may not be the place for it. The article on geodesy doesn't seem to be so much about application. The Vincenty's formulae article seems to be the only treatment of non-spherical distance. Maybe a separate article is needed to discuss exact and approximate distances along the surface of a spheroid? One name for the new article might be "Geodetic formulae".-Ac44ck (talk) 21:03, 4 December 2008 (UTC)

Right, $\scriptstyle {dD={\overset {{}_{\smile }}{R}}\cdot {dH}}\,\!$ is the formula for a differential distance, just like $\scriptstyle {dD=M\cdot {d\phi }}\,\!$ is the formula for a differential meridional distance: So how do you find a distance of any length along a meridian? By finding the average value of M between the latitudes and multiply by the difference: $\scriptstyle {D=M_{avg}\Delta \phi }\,\!$. Likewise, to find the elliptical great-circle (technically, due to the way it is differentiated, I think this would be the proper term, rather than "great-ellipse") distance, you would find the average value of T (or $\scriptstyle {\overset {\smile }{R}}\,\!$) along the great circle between the points and multiply by the angular distance: $\scriptstyle {D=T_{avg}\Delta {\widehat {\sigma }}}\,\!$. Andoyer's Approximationapproximates this. I have a more direct/simpler form of his equation, but I'm shutting down now, so I'll give it here tomorrow! P=) ~Kaimbridge~ (talk) 20:39, 5 December 2008 (UTC)

Will (eventually) add to Geographic distance, instead.  ~Kaimbridge~ (talk) 02:14, 7 December 2008 (UTC)

Contrary to what the first paragraph in the note says about showing how "D between them can be found", the current version doesn't seem to present a way to find D. Instead, it provides something that needs to be averaged via a forthcoming mechanism, which I gather will be quite more complicated than the formula which existed in the note last week. I don't think we need to make an either-or choice here. Both versions could be helpful to have in an article.

Interesting stuff, but finding distances along the surface doesn't seem on-topic in the Earth _radius_ article.

Agreed, though another possibility would be to change it from a reference for R_c, to its own subsection underneath ("Transverse meridional or arcradius", $\scriptstyle {T{\mbox{ or }}{\overset {\smile }{R}}}\,\!$), or even its own section, "Radius of arc", $\scriptstyle {\overset {\smile }{R}}{\mbox{ or }}R_{a}\,\!$ to highlight

${\tilde {\alpha }}({\widehat {\sigma }})=\lim _{H\to 0}{\tilde {\alpha }}=\arctan \left({\frac {N}{M}}\tan({\widehat {\alpha }})\right);\,\!$

${\begin{aligned}{\color {white}{\frac {\big |}{}}}{\overset {{}_{\smile }}{R}}=T&={\sqrt {(M\cos({\widehat {\alpha }}))^{2}+(N\sin({\widehat {\alpha }}))^{2}}},\\&={\frac {{}_{1}}{\sqrt {\left({\frac {\cos({\tilde {\alpha }}({\widehat {\sigma }}))}{M}}\right)^{2}+\left({\frac {\sin({\tilde {\alpha }}({\widehat {\sigma }}))}{N}}\right)^{2}}}};\end{aligned}}\,\!$

and its relationship to

$R_{c}={\frac {{}_{1}}{{\frac {\cos ^{2}({\tilde {\alpha }})}{M}}+{\frac {\sin ^{2}({\tilde {\alpha }})}{N}}}};\,\!$

if you didn't think it would violate NOR!  ~Kaimbridge~ (talk) 02:14, 7 December 2008 (UTC)

What do you think of having a separate article on "Geodetic formulae" with its own content (including the FCC formula and the growing contents of the note under discussion here) plus links to the Vincenty's formulae and great-circle distance articles?

The term "elliptical great-circle" seems awkward to me. A circle is a special kind of ellipse. The term reads something like an "elliptical special ellipse". The term "elliptical great-perimeter" comes to mind as an alternative. -Ac44ck (talk) 21:25, 5 December 2008 (UTC)

The thing is, I think, where a great circle or great ellipse would be a straight arc ("----------"), the way this would slightly deviate from the whole geodetic delineation at each increment (as you are delineating it spherically, then squashing it down to an ellipse) would result in a broken arc, something like "-’¯‘-’¯‘-", thus it wouldn't be a smooth geodetic elliptic arc, but increments of spherical arc elliptically adjusted, independently (i.e., a polygonal arc?).  ~Kaimbridge~ (talk) 02:14, 7 December 2008 (UTC)

First note

The first note is not clear to me:

The center of the Earth is somewhat model dependent. Exceptions to the cited range occur near the South Pole and along the equator. Also, differences due to variation of mass density within the planet and tidal forces require data for the entire surface of the Earth and are not included here.

• How is the "center of the earth" model dependent? Wouldn't the center of the ellipsoid be at the same location as that of a spherical model? The center of curvature varies for the ellipsoid, but the center of the ellipsoid is a unique point. The geometric "center of the earth" may not be at the center of mass. I don't see a need to distinguish between "geometric center" and "center of mass" here. The "center of gravity" for an observer on the surface may be elsewhere, but that is a different concept.
• How are the limits "exceptions to what cited range"? Why only the South Pole and not the North Pole?
• How is an anomaly near the South Pole related to the location of the center of the earth or the stated value of the polar radius?
• How do tidal forces affect the location of the center of the earth? Or is this referring to deformation of the surface by tidal forces? The discussion so far in the article seems to assume that the surface is a static shape of some form (spherical, ellipsoidal) as opposed to a moving surface.
• Two of the three wikilinks are already in the "See also" section.

Do we need to keep this footnote? -Ac44ck (talk) 20:48, 26 December 2008 (UTC)

I rewrote the first paragraph. It had become incoherent. I think you still have questions, but let's start over, referring to the new first paragraph. Strebe (talk) 02:24, 27 December 2008 (UTC)

Every shape (solid or not) has a radius at every point. A straight line has an infinite radius at each point. I thought the phrase "a radius that is the same by all measures" was fairly elegant. The article covers both "radius to a geometric center" and "radius of curvature" (which generally varies with both latitude and azimuth). Those are different ways to measure the radius. They are the same for all points (in every direction) on a sphere.

The mention that the radius "varies from place to place" might say something about the role of topography in that variation. The radius is constant at a given latitude for the models discussed in the article.

The part about "Exceptions to this range occur near the South Pole" is confusing to me. The polar radius is one of the limits in the "6,356.750 km to 6,378.135 km" range, so it isn't an exception. The polar radius is equal for both poles, so I don't see a need to single out the South pole. Likewise, the equator is one of the limits, not an exception.

Other prior questions remain. - Ac44ck (talk) 03:28, 27 December 2008 (UTC)

I am not aware of any definition of radius that applies to anything but (hyper)spheres. If you know otherwise, please cite it? I reverted your changes to the lead paragraph for several reasons. For one, the problem with definition of radius. For another, it's too chatty. For yet another, the restriction of the discussion to a geometric model of the sphere is not made until later in the paragraph; hence it makes no sense to discuss variation by "latitude" when the discussion is about the geoid at that point. It is true that the south pole disclaimer is bogus; local topography there never exceeds the equatorial radius. But the equatorial disclaimer holds; local topography exceeds the model ellipsoid radius. Still, it is poorly written as is because the notion of a model ellipsoid has not even been introduced and yet it is comparing measurements against that. I'll make another attempt. It sure was a mess before! I think we're making progress. Strebe (talk) 09:50, 27 December 2008 (UTC)