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Archive 1

Section "Radius for spherical Earth" needs shortening

I removed some on "Remove redundant stuff about average radii" on 2013-05-02. It was restored by 88.8C.80.7A. Ac44ck took the material out again. Then it was put back by 93.115.83.247. I'd like to take it out again! However, before I do, I would like to give one or both of the contributors who think this material should stay the opportunity to explain their position. (I will also note that the material includes 3 dubious links.) cffk (talk) 18:33, 7 June 2013 (UTC)

To follow up on an earlier question I asked you... What is your mean radius for a triaxial ellipsoid with (approximately equal) semi-axes ? You gave me an answer before (June 19); but when I probed you about it, you appeared to hedge. cffk (talk) 01:17, 29 June 2013 (UTC)

Yes, I understand what you mean with and do agree with your results as you define them (just to make sure its valuation is clear, is just the approximation of “the mean radius of these great ellipses”, not the actual mean/average radius, right?), but I think part of the problem in our descriptions is in the recognition and meaning of the vertex. Look again at the great-circle segments:
This is the transverse graticule perspective, with and four great circles shown crossing its equatorial vertex: The meridional, equatorial and two delineated Â(rc path)s. But now each of these four great circles that intersect that equatorial vertex extend out to the transverse equator, with each crossing the vertex latitude unique to their angle: The meridional’s is at φ_v = 90°, the equatorial’s at φ_v = 0°, and the Âs’ at φ_v = 90° - Â. So for each great circle, we are talking about two vertex latitudes, φ' in and φ_v at the great circle’s crossing of its transverse equator. What I was referring to by “So each has an infinite number of great circles crossing that vertex ...” is the facing (in this case, equatorial) vertex: In the great-circle segment image, only four great circles are shown crossing the facing vertex, but, in fact, there are an infinite number——one for each Â.
Hmmm, let’s focus a minute on the extreme points of a great circle/ellipse: The point crossing the equator is its node, while its highest point, which is at its transverse equator, is its vertex. So a single set of “differently angled great circle/ellipse” originate from a single node, which is . So why would you consider any other , which doesn’t have a full set of “differently angled great circle/ellipse” (in the extreme case, only contains one great circle/ellipse angle——north-south!)?
Try this: Imagine attaching a marker to and having it draw each Â. Wouldn’t the whole globe be covered? ~Kaimbridge~ (talk) 05:49, 30 June 2013 (UTC)

You're hopelessly stuck on the case of an ellipsoid of revolution. Whatever scheme you're using to sample great ellipses needs to begin covering all the great ellipses on an arbitrary ellipsoid. Once you've got that sampling scheme settled, you can simplify the resulting integrals in the case of an ellipsoid of revolution. You, on the other hand, start with an ellipsoid of revolution and because it looks like the great ellipses emanating from a single point on the equator cover the ellipsoid assume that this is a sensible way to sample. It isn't. If you really were confident in your approach you would say, "Go ahead and average over all possible starting points; but I think you're wasting your time and you'll recover the result I obtained more simply." (But of course, you don't recover the same result and that points out the basic problem with your sampling scheme.)

So to break you have your bad habits, let's, for a while, only discuss the case of a triaxial ellipsoid semi-axes ? In the limit where the semi-axes are nearly equal I get for the mean radius. What's your result? cffk (talk) 11:07, 30 June 2013 (UTC)

Okay, fine, only let’s make the semi-axes and——for this stage of the triaxial discussion——let there be only one Cartesian set, , where
Applying these to an Earthy scalene ellipsoid,

The same type averaging can be applied to your (2a+b)/3 (or, in this case, (a_x+a_y+b)/3) modeling.
The scalene modeling is just one more step in “auxiliaryization”:
  • Sphere: ;
  • Spheroid integrand: Create an auxilary sphere; ;
  • Scalene Ellipsoid integrand: Create an auxiliary sphere from an auxiliary spheroid; ;
The actual calculation and approximations of the mean/average (arc)radii appear to utilize a_m, as the secondary, surface/arc oriented Cartesian set uses a_m as a constant, with the axial variation transferred to a variable form of b, b(λ), which I will address next (I’m out of time now! P=)...though I did touch upon it back up at Ellipsoid axes, radii, Cartesian coordinates.  ~Kaimbridge~ (talk) 18:53, 30 June 2013 (UTC)

So it looks like you're saying that the mean is . Is this right? (I skipped over all the sections where you substituted numerical values above, since this just obscures what's going on.) However, I'm worried about the weasel words you added near the beginning: "let there be only one Cartesian set". Surely the result shouldn't depend on the coordinate system I choose? And certainly any sensible "average radius" you define for a nearly spherical ellipsoid should be independent of coordinate system. Can you clarify? cffk (talk) 19:17, 30 June 2013 (UTC)

Basically, yes, .
There are two Cartesian sets, one for the geocentric radius, R, and one for surface/arc (the parametric arcradius), S. In the simpler case of the spheroid:
Maybe I’m just wording it wrong as a second “set”. How would you define R and S?  ~Kaimbridge~ (talk) 19:40, 30 June 2013 (UTC)

In an effort to keep you focused on the general case and away from treating one axis specially, I'm going to stick with for the semi-axes. Why is your result not invariant on interchanging the semi-axes (e.g., swapping and )? cffk (talk) 19:52, 30 June 2013 (UTC)

I’m not sure what you mean by “interchanging the semi-axes”, but I base the general ellipsoid on the parametric integrand for surface area:
 
Isolating and restructuring the radii into equivalent formations with a common a factor,
thus

From this, four prime possibilities for the general form for Qr become apparent:
Using the Previously defined scalene ellipsoid and converting to a,b,c denotation,
If you compare integrations of variable Qr’s, one for a(λ),b(λ) and a third for both, and their averagings,
any of the four compounding models (Qr_m1, Qr_m2, Qr_m3, Qr_m4) seem plausible, with all of them simplifying to the ((3a^2+b^2)/4)^.5 spheroidal form: While the latter three appear closer to the intergration averagings, up to now I have favored Qr_m1, as that seemed most appropriate for the great-ellipse averaging, since it appeared to consist of the most basic averages, (a^2_m+b^2)/2 for the meridians and (a^2_x+a^2_y)/2 for the equator, but Qr_m3 may be the actual proper model, as it is the same form as Qr_m1, except that it uses b^2(45°) instead of b^2, and that mid-b value equals the average of b_x and b_y. Would that address/reflect your “interchanging the semi-axes”?  ~Kaimbridge~ (talk) 05:48, 4 July 2013 (UTC)

Triaxial example calculation

I think cffk's question is clear.

Let's bring his question back into focus: What is the mean radius for a triaxial ellipsoid with approximately equal semi-axes?

Let's use numbers: a = 1.01, b = 1.02, c = 1.03

The average according to cffk is (a + b + c) / 3 = (1.01 + 1.02 + 1.03) / 3 = 1.02

The average according to Kaimbridge is (3a + 3b + 2c) / 8 = (3 * 1.01 + 3 * 1.02 + 2 * 1.03) / 8 = 1.01875

Now let's swap b and c:

a = 1.01, b = 1.03, c = 1.02

The average according to cffk is (a + b + c) / 3 = (1.01 + 1.03 + 1.02) / 3 = 1.02

The average according to Kaimbridge is (3a + 3b + 2c) / 8 = (3 * 1.01 + 3 * 1.03 + 2 * 1.02) / 8 = 1.02

The average according to Kaimbridge was affected by changing the orientation of the ellipsoid.

The average radius is to be used for calculating the distance between random points on the ellipsoid. The orientation of the ellipsoid should not matter when calculating the average radius for this purpose. - Ac44ck (talk) 07:50, 4 July 2013 (UTC)

Thanks, Ac44ck. Yes, this is what I meant by interchanging the axes and your last paragraph nicely summarizes why this is important. cffk (talk) 09:07, 4 July 2013 (UTC)
Well, by doing that, you are changing it from oblate to prolate, where everything changes.
What I think it comes down to is, we are just talking about different degrees of averaging: I’m talking about the principal circumferential average, which is the average of all the different circumferences (circles or ellipses) emanating from a single equatorial node, while you are talking about the compound circumferential average, which is the average of all different circumferences emanating from all of the different vertex latitudes, which approximates the volume/surface area radius.  ~Kaimbridge~ (talk) 20:34, 4 July 2013 (UTC)

Well I'm not sure that prolate and oblate are meaningful for a triaxial ellipsoid. But never mind... I think we can close off the discussion now. By computing "the average of all the different circumferences emanating from a single equatorial node", you are not sampling great circles in a manner that makes any sense for computing distances between arbitrary points on the earth (which was the reason that I started this thread). Finally, a better way to describe the averaging I am doing is "rotationally invariant". That's clearer (and shorter) than "compound circumferential average, which is the average of all different circumferences emanating from all of the different vertex latitudes, which approximates the volume/surface area radius". Thanks for the interesting discussion! cffk (talk) 22:17, 4 July 2013 (UTC)

Yes, I can get a little wordy at times (I suppose you can blame it on my Aspergic wiring!). P=)
Probably an even more succinct identification of what I (and others who) consider the two principal——meridional and equatorial——circumferences as the boundaries for calculating the average great-circle/ellipse and its (arc)radius, would be to classify it as the mid-range oriented, average circumference.
Also, a follow-up point regarding the interchanging either a_x or a_y with b: I think the ellipsoid’s orientation certainly is important, in terms of dealing with circumferences. Consider an extreme, yes, spheroid case: Let’s use semi-major and minor axes of 10,000 and 1000, respectively. Are you really saying orientation of the axes isn’t relevant, in terms of its shape? Of course it is! The oblatum is like a pancake, while the prolatum is like an upright pipe! Would tilting them 90° sideways make one look like the other? Of course not, the curvature——and, therefore, circumferences——is completely different. Thus, the corresponding average radius should be different, too!
Yes, I realize trying to define arc paths and distances on such extremes is probably difficult to outright impossible (get out the string and globe!), but the example does highlight the different curvature dynamics. Now consider the two extreme equivalent scalene examples:
  1. a_x = 10,000, a_y = 9000, b =   1000;
  2. a_x =   1000, a_y =  900, b = 10,000;
While a bit slanted or lopsided, the same general curvature dynamics are produced. Right?  ~Kaimbridge~ (talk) 06:19, 7 July 2013 (UTC)
Ecclesiastes 3:1-3:
There is an appointed time for everything.
...
a time to give up as lost.
Orientation does not matter for the distance between randomly distributed points. Shape is another issue. The disagreement was over the type of sampling being done. If going from the general triaxial case to the specific spheroidal case requires a change in sampling method, something is wrong with with the sampling method. And there is something wrong with your sampling method that averages only from the equator. I came to a similar though less rigorous conclusion in a discussion with you here:
Talk:Spherical_Earth/Archive_1#Formula_for_ellipsoidal_quadratic_mean
In this present discussion, Cffk identified the issues much more clearly. With more competence and rigor, he inescapably pointed out the error in your sampling method.
Besides those who parrot the misinformation inserted into Wikipedia, who are the "others" that "consider the two principal——meridional and equatorial——circumferences as the boundaries for calculating the average great-circle/ellipse and its (arc)radius"?
Contrary to the parroting by others elsewhere, your so-called "ellipsoidal quadratic mean" is not the best radius to use for calculations of great circle distances between random points on the Earth. -Ac44ck (talk) 09:37, 7 July 2013 (UTC)
If you go to near the beginning of this discussion, it was requoted from one of the sourcesgiven:
/* The circumference of the earth at the equator is 24,901.55 mi (40,075.16 km),
   but the circumference through the poles is 24,859.82 mi (40,008 km).
   Geometric mean radius from this info is 6372.8km, or 3959.88mi
This concept/approach wasn’t just pulled out of mid-air: If you go to page 5 of Richard Rapp’s Geometric Geodesy (Vol II),you will see that the “latitude”s and “longitude”s of a distance calculation are first converted to “arc from E(quator) to P(oint)” (i.e., transverse latitudes), measured from “azimuth of specific geodesic at the equator” (i.e., transverse longitude/meridian, or “arc path”)——none of the formularies state it as transverse latitudes/longitude, but that is what they are (again, I didn’t make up these terms, I got them from the USIGS Glossary,which got them from the DoD’s MIL-HDBK-850——found after numerous, unsuccessful online inquiries, such as this one,back 10 years ago!).
So, yes, based on the fact that distances are found exclusively on great circles/ellipses crossing the equator (and, as the above source demonstrates, the meridional/polar and equatorial circumferences are the principal circumferences cited when describing a planet’s size and shape), orientation certainly does matter when defining a graticule used in calculations, and any such great-circle radius average should likewise be based on an equatorial orientation. In terms of my particular ellipsoidal quadratic mean,that is a different issue, which I would agree crosses into original research. But just a general “6372.8” type value, with the simple formulaic structure given (such as the most basic mid-range average and/or the above source’s geometric mean) certainly wouldn’t be out of place to present as a possibility——I’m not saying this article should endorse it as “the right choice”, just a possibility that is out there and that that is what the “6372” value in any particular given source is based on (perhaps a section in Earth radius could be created, where a more detailed analysis of the pros and cons of 6372.8 vs. 6371.0 could be presented).  ~Kaimbridge~ (talk) 17:33, 7 July 2013 (UTC)

Your citation of Rapp is completely bogus. He discusses the solution of the geodesic problem in the conventional terms of spherical geometry. He does not use the terms "transverse latitude" and "transverse longitude". Your imposition of a "transverse graticule" is an inappropriate use of techniques from map projection. I and everyone else I know of in this business are able to solve geodesic problem without using the transverse graticule. Indeed your use of the graticule locks you into a coordinate dependent way of thinking that prevents you from seeing your errors. If you insist on tying yourself to the equator, then you need to sample azimuth using sin(alpha_0) = 2*rand - 1. Incidentally, the circles which are equidistant from a given point are called "geodesic circles". Gauss (1826) showed that, for an arbitrary surface, the geodesics from a single point and the corresponding circles are mutually orthogonal.

However, to get to the point, you have never enunciated any property of your ellipsoidal quadratic mean that might be useful to someone measuring distances (despite my numerous requests). All you have is a bunch of hand-waving arguments when what is needed is a table similar to the one I showed on 2013:06:21 20:04. Against we have the fact that use of (2a+b)/3 minimizes the variance in the errors in distance measurements, a concrete result that a user can take to the bank. You do yourself and the community a disservice by recommending (3a+b)/4.

Please do not add a section to Earth Radius discussing this. cffk (talk) 18:14, 7 July 2013 (UTC)

What do you mean citation is bogus? I said “none of the formularies state it as transverse latitudes/longitude, but that is what they are”. I know he (and others) don’t define it as such (but should). As for “sample azimuth using sin(alpha_0) = 2*rand - 1”, it is actually “sigma” that is the one that would need the adjustment (we disagree on that), but even if it did, the averaging comes out to about “6374.58”, which, if you average with “6371.0”, you get “6372.79”! But again, the skewing is a whole different issue (and don’t worry, I have no intention of adding it to Earth radius! P=).  ~Kaimbridge~ (talk) 20:51, 7 July 2013 (UTC)
"What do you mean citation is bogus?" I mean that transverse latitudes and longitudes are terms of art from the world of map projections. Rapp would not be viewing the geodesic problem in these terms. In any case, the use of transverse latitude and longitude is restricted to spherical map projections. They aren't, for example, particularly useful in constructing the transverse Mercator projection. So the relevance to the problem at hand (geodesics on an ellipsoid) is questionable. Is "sigma" the arc length from the equator? How can that be adjusted if you're averaging the full circumference of the great ellipses? cffk (talk) 21:14, 7 July 2013 (UTC)
What “terms of art”? These are just the conceptual identities.
The middle, transverse graticule, is the left, conjugate graticule, “pulled down” from the pole to the equator: Why are you so adverse to giving their definitions a corresponding name, something along the lines of,
  • Transverse Latitude—Angular distance along a great arc from its equatorial node to P (in contrast to a “regular”, conjugate latitude’s equivalent definition of “angular distance along P’s meridian, from the equator to P”);
  • Transverse Longitude—Azimuth of great arc at the equator (in contrast to a conjugate longitude’s definition of “polar azimuth of P, measured from the prime meridian”); ???
Regarding “sigma”: Yes, it is the arc length from the equator and, no, I don’t think it should be adjusted. You say random latitudes need to be skewed (φ_random = arccos(2*rand-1)) to compensate for the longitude merging at the poles (i.e., “cos(φ)Δλ”), so in the transverse case, the merging is with A, at the equatorial node, (i.e., “sin(σ)ΔA”, since σ is the transverse colatitude), which means the skewing should be σ_random = arcsin(2*rand-1), not A_random.  ~Kaimbridge~ (talk) 19:45, 8 July 2013 (UTC)
Kaimbridge, you say "orientation certainly does matter when defining a graticule used in calculations". That's as may be. However, if the results of your calculation depend on the orientation, there's a fundamental error with the way you posed the problem. cffk (talk) 11:13, 8 July 2013 (UTC)
This may be just misinterpreted semantics (in terms of “orientation”), but I’m just stating what should be obvious: If you take either a spheroid or scalene ellipsoid, tilt it——say 30°——define the now most northernest point as “the new north pole” and redefine the conjugate φ and λ graticule accordingly and attempt to calculate distances, the results are going to be bogus.  ~Kaimbridge~ (talk) 19:45, 8 July 2013 (UTC)
Kaimbridge, you say "orientation certainly does matter when defining a graticule used in calculations". That's as may be. However, if the results of your calculation depend on the orientation, there's a fundamental error with the way you posed the problem. cffk (talk) 11:13, 8 July 2013 (UTC)

Kaimbridge: You (and I and the other participants in this discusson) should be worried that this discussion has gone on so long (apparently without any definite resolution as far as you're concerned). And it seems that this isn't the first time that this has happened. Why is this?

The problem is that your page about the ellipsoidal quadratic mean radius makes statements that cannot be tested. The concluding sentence is a good example "Given all of these considerations, it would seem that the ellipsoidal quadratic mean provides the ideal (approximative) great-circle radius for an ellipsoid." There are a number of red flags here: "would seem", "ideal", "approximative". However, the big one is that you don't establish a concrete test for "idealness"; in the jargon, your work isn't falsifiable. This moves your work out of the realm of science; because of this, we should probably have cut the disussion off long before now. cffk (talk) 17:34, 8 July 2013 (UTC)

I think we have the two different averagings identified (I’m talking about a principal averaging, where all different angle possibilities are only counted once, while you are referring to a compounded average, where all different total perspective averages are combined).
Again, could it be just bad semantics (“average”), on my part? E.g., Is there a difference between “average” and “rectified” (i.e., could my “principle average” be the “rectified great-circle radius”?)? Or is it the same thing?  ~Kaimbridge~ (talk) 19:45, 8 July 2013 (UTC)

Kaimbridge, Following the advice in my previous comment, I'm going to hold off replying until you give me something concrete I can test. Otherwise we're just going around in circles. I looking for something along the lines of:

Select random pairs of points and determine the true geodesic distance and the great circle distance, using a radius of r = (3a+b)/4; generate a table like this:
 lat1      lon1     lat2     lon2     geod      g-c     %err
54.1220  -26.6448 -36.8476  -29.9051 10083964 10123039  0.387
11.8042  -46.6654 -49.3498  -77.1150  7401984  7425986  0.324
54.6973   86.8521  -2.3602  119.0341  6981592  6999420  0.255
52.6389    8.8880  10.1733 -114.3053 11215289 11206605 -0.077
51.6027 -123.4414  -0.0507   -9.1026 11666729 11664592 -0.018
...
The radius r = (3a+b)/4 is the "ideal" radius because ...

Please fill in the "..." using statistics from the table. See my comment of 2013-06-21 20:04 for the equivalent result for r = (2a+b)/3. cffk (talk) 20:12, 8 July 2013 (UTC)

Okay. P=)  ~Kaimbridge~ (talk) 20:26, 8 July 2013 (UTC)
Archive 1