Talk:Great-circle distance/Archive 1

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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)

Extended discussion about averaging ellipses and arcradius

I came upon it when looking at a factsheet on Earth's shape and saw the given average radius as "6372" . I thought that number seemed a little odd so I did a little poking around and found similar values given in various papers, mostly "6372.8"s pg.2 pg.15 pg.8 pg.43, and a couple of "6372.790"s and a 6373 pg.265 pg.45 pg.1969, as well as many different program source code papers, such as this one (about half way down): /* 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 But most just give the values, the only one that talks about it in any depth and gives any real derivation is the article about " Ellipsoidal Quadratic Mean" (which I agree is a little over the top for just this article). And since "6372.8" appears the actual approximation of Earth's great circle radius equivalent (the radius of the mean meridian and equator circumferences), it would seem appropriate to at least identify it here in the Great-Circle Distance article, even if only in a brief, comparative relationship mention, wrt meridional vs. equatorial vs. volume/surface area. 88.8C.80.7A (talk) 03:41, 9 June 2013 (UTC) Thank you, 88.8C.80.7A, for responding! I see two problems with the extra material. 1. It's silly trying to make an intrinsically approximate result very precise. Tim Zukas in the beginning of the section makes the point pretty succintly. You can minimize the maximum relative error by taking r = (a + b)/2 and then the error is 0.5% (= 3*f/2). If you take r = a or r = b the error is modestly bigger 0.67% (= 2*f). With r = (2*a + b)/3 or (3*a + b)/4 the errors are 0.56% (= 5*f/3) and 0.58% (= 7*f/4) respectively. The bottom line is that as long as you can live with a 1% relative error, any of these radii are suitable. To minimize the maximum error, use (a + b)/2; for consistency with other applications, use (2*a + b)/3 (the IUGG recommendation). If you need better than 1% accuracy, then use the ellipsoidal formulas. 2.The link for mean quadratic radius takes me to a site which is short on hard facts. "This particular average/mean arcradius is the natural choice as a general purpose approximation of an ellipsoid's great-circle/spherical radius..." is merely an assertion. "Remarkable coproximity", "inherent elliptic quantity", and "conjugate equator" are ill-defined. I can't make any sense of the "Spheroidal Isopathic Median". And to what end? To increase the relative error in spherical distances from 0.56% to 0.58%! cffk (talk) 18:15, 9 June 2013 (UTC) Hang on a second now. When you say .5% error, .5% error to what? This article and particular approximation is specific to a "great-circle radius", meaning the radius of all the different circumferences, not just the meridian ("6367.4") and equator ("6378.1"). The "quadratic mean" article says that there are different ways to differentiate circumferences, with the results ranging from about 6372.7904 to 6372.7994 geodeticly, to a definitive 6372.80165 for an average "great ellipse" circumference (all using a reference spheroid where a and b are about .002 km smaller than the WGS84 equivalents). Meanwhile the mid-circumference of the above, 6372.7994 version is found at ${\displaystyle \scriptstyle \phi _{1}=0,\phi _{2}=45^{\circ },\Delta \lambda =90^{\circ }}$, which for WGS84 gives a radius of about 6372.8014 km ${\displaystyle \scriptstyle \left({\frac {2}{\pi }}10010.373\,km\right)}$. Anyway you look at it, the theoretically precise "average great-circle radius" (or more correctly, "average great-ellipse radius"?) is somewhere between 6372.7904 and 6372.8017 km, giving a difference of only ${\displaystyle \scriptstyle \Delta .0113\,km}$! So the approximation range and error should be based on these parameters, not "a", "b", "6367" or "6371", and the given "(3a+b)/4" and "(a3b)1/4" provide about the same great-circle radius approximation error as "(a+b)/2" and "(ab)1/2" do for the meridional radius. I do agree that the explanation shouldn't get too involved, maybe something like the above quote ("the circumference of the earth...") would be adequate (maybe expound more in Earth radius?). Again, for the reason you first deleted it, the article isn't about radius, it is about great circle distance, which "6372.8" seems the axiomatic choice, just like "6367.4" is for meridional distance. 88.8C.80.7A (talk) 19:11, 10 June 2013 (UTC) The problem that this section is addressing is the following: you're considering geodesics on an ellipsoid (equatorial radius = a, polar semi-axis = b); however you want instead to do the simpler calculation of great circle distances on a sphere (radius = r). The natural question is "what is a good (or best) value to use for r?" To make this precise it's necessary to define a metric for goodness. Here's mine: (1) consider all possible end points (lat1,lon1) and (lat2,lon2); (2) for each pair, compute the ellipsoidal distance s12 and spherical distance a12; (3) define the relative error as err = abs(a12-s12)/s12; (4) my metric is then max(err) where the maximum is taken over all the pairs. Note that this definition includes geodesics going in all possible directions. For r = (a + b)/2, max(err) = 0.5%. For r = (3*a + b)/4, max(err) = 0.58%. It's possible to verify these assertions experimentally by trying out many pairs of randomly chosen points. However, you can quickly determine which geodesics are giving you the biggest errors and this lets you write down analytic expressions (thus 0.5% = 3*f/2 for f small). The key here is that I've got a well defined metric and a easily verified result. The "quadratic mean" link never precisely defines "average great-circle radius". And even if it did, the connection would need be made to the problem at hand, namely what is the error involved in using a sphere instead of an ellipsoid to compute distances. I also don't really know what the term "differently angled circumference" means. I suppose it refers to a great ellipse? But how is this relevant to a discussion of geodesics which don't lie on great ellipses. If you would like to propose some concrete alternative measure of error that the quadratic mean radius minimizes, then at least I would have something to check. cffk (talk) 20:31, 10 June 2013 (UTC) ADDENDUM: The "quadratic mean" link talks about means. So here's another way of determining an optimal r. Compute s12 as above, but compute a12 on a sphere of radius 1. The expression R = s12/a12 gives a mean radius for each geodesic. Define r = average of R over all possible pairs (uniformly distributed on the sphere). This results in r = (2*a + b) / 3. Normally, it's preferable to minimize the maximum error (as I did before). However this definition in terms of the mean is a reasonable approach and it's nice that this ends up with the IUGG recommended radius. For consistency across different applications, you should standardize on the IUGG value. cffk (talk) 21:05, 10 June 2013 (UTC) I don't really understand the nature of your particular "metric for goodness". My determination of the "true" average great-circle radius for a given geodesic is ${\displaystyle r_{avg}=\scriptstyle {\frac {geodesic}{\Delta {\widehat {\sigma }}}}}$. The way to approximate the global average is to calculate random distances of different lengths at different places and average all of the ${\displaystyle r_{avg}}$s together (a particular example of this is approximating the meridional radius by dividing random north-south distances by their latitude distances and averaging them together). Of course you would need 100s, 1000s, maybe even millions of examples to get a good, stable average. 88.8C.80.7A (talk) 20:34, 11 June 2013 (UTC) I've done precisely this experiment! Here's the matlab/octave code function meanr = meanr(a, f), % estimate a mean radius spherical radius to use so that the great-circle % distance approximated the ellipsoidal geodesic distance. % Requires: % http://www.mathworks.com/matlabcentral/fileexchange/39108 num = 1000000; lat1 = 180*asin(2*rand(num,1)-1)/pi; lon1 = 360*rand(num,1)-180; lat2 = 180*asin(2*rand(num,1)-1)/pi; lon2 = 360*rand(num,1)-180; ell = [a, flat2ecc(f)]; s12 = geoddistance(lat1, lon1, lat2, lon2, ell); a12 = geoddistance(lat1, lon1, lat2, lon2, [1, 0]); meanr = [mean(s12)/mean(a12), mean(s12./a12), mean(s12.^2./a12)/mean(s12)]; end The last line calculates the average 3 different ways. They all end up being close to (2*a+b)/3 and not (3*a+b)/4. This experiment is really quick (it takes just 10 secs with octave). If you don't have the ability to do such an experiment yourself and you want me to try a different average, feel free to suggest one. If you want to put your finger on where the quadratic mean derivation goes astray it's in averaging one of the meridional circumferences with the equatorial circumference. Better would be to average the perimeters of all the principal ellipses. However I would always want to back up such a seat-of-the-pants derivation with some hard numbers. cffk (talk) 00:33, 12 June 2013 (UTC) P.S. Incidentally the quantity (min(s12./a12)+max(s12./a12))/2 in the code above gives the radius which minimizes the maximum relative error. This is close to (a + b)/2. cffk (talk) 01:17, 12 June 2013 (UTC) P.P.S. I reread the "ellipsoidal quadratic mean radius" page to pinpoint where the analysis goes awry. It turns out that the premise of the page "if one calculated random distances of varying lengths and extracted the average/mean arcradius of each distance, the cumulative average will be significantly greater than Ar" is wrong. Too bad that the author of this page didn't do a simple numerical experiment to verify his hypothesis (and doing the average analytically is simple enough). Then he wouldn't have caused us so much grief. cffk (talk) 11:27, 12 June 2013 (UTC) Well, I guess it’s time I come out of the shadows (this page is on my WatchList), as I’m the infamous author of the “quadratic mean” article! P=) First off, yes, I’ve done much empirical sampling, all of it coming to the ≈“6372.8” conclusion——for starters, IIRC, there was something about using a -90°->+90° range for latitude and -180°->+180° for longitude that skewed the average too low: Try reducing the range to a quadrant (lat = 0->90°, ∆long = 0->90°). Better yet, as you say, “average the perimeters of all the principal ellipses”: Change your above source code to lat1 = long1 = 0, lat2 = 90*rand(num,1) (or, 180*rand(num,1)-90), and long2 = 90°. Each result is the quarter-circumference for a given arc path (which, in this case, is 90°-lat2, the circumference’s vertex latitude). The average arcradius for each of these distances = 2/π*s12. Based on this, “vertex latitude flavored differentiation”, the actual average arcradius (i.e., “great-circle radius”) is about 6372.8014 km (based on a=6378.137, b=6356.752), which is what your code above, with my modifications, should now reflect (btw, my program of choice——and, with which, all my experimentation is done on——is UBasic). Does it? Does “num = 1000000” mean the sample size? If so, and that only takes about 10 sec.s, try letting num = 10,000,000, with a printout of the running average at every 1,000,000. Backing up, I suppose the first thing to clarify is the graticule perspective: The left, conjugate graticule is the lat/long webbing that everyone is familiar with. However, it is the middle, transverse graticule (here, “transverse” meaning the polar vertex is “pulled down” to the conjugate equator and the transverse equator is the conjugate one pulled back, so that it is the facing perimeter, 90° away from the central meridian) that distances are calculated, along those “differently angled circumferences” (i.e., “arc paths”, “azimuths at the equator”, “vertex latitudes”——whatever you want to call them). In the case of geodetic distances, an auxiliary sphere is created, with elliptical/parametric/auxiliary coordinates (including arc path). I have much more to say, including breaking down the individual sample entries to infinitesimal small lengths and presenting the actual arcradius at a given point, but this should hopefully turn you in the right direction! P=)  ~Kaimbridge~ (talk) 16:46, 12 June 2013 (UTC) Yes, 1000000 is my sample size. When I used the term "principal ellipses" I meant the three orthogonal ellipses with semi-axes (a1,a2), (a1,b), (a2,b), for a triaxial ellipsoid with semi-axes (a1,a2,b). Performing this average for an oblate ellipsoid, a1 = a2, gives (2*a+b)/3 as the average radius. I notice a contradiction in your first sentence. "all [my experimental sampling] coming to the 6372.8 conclusion" vs "using [-90,90] for lat and [-180,180] for lon skewed the average too low". This seems to indicate that at least some of your experiments came up with a result different from (3*a+b)/4! The experiment that you suggest, i.e., taking lat1 = 0, lon1 = 0, lat2 = 90*rand(num,1), lon2 = 90, does indeed yield an average of (3*a+b)/4. I understand your reasoning here; you're trying to sample a quarter circumference at all the different vertex latitudes. But you've made a blunder! The distribution of vertex latitudes is not uniformly distributed in [0,90]. It's skewed towards the pole. To account for that you need to sample lat2 = acosd(cos(asin(rand(num,1))).*sin(pi*rand(num,1))); This is the distribution of vertex latitudes for geodesics thru a random latitude = asin(rand(num,1)) with a random azimuth = pi*rand(num,1). This is equivalent to lat2 = acosd(rand(num,1)) If you do this that the average is (2*a+b)/3. cffk (talk) 19:31, 12 June 2013 (UTC) Ahh, okay, NOW we are getting somewheres! Using the three boundaries of meridian_x, meridian_y and the equator, then yes, average arcradius ≈ (2*6367.449+6378.137)/3 ≈ (2*6378.137+6356.752 ≈ 6371.01! But that is for approximation of width x height area, not differentiated arc, which the single meridian and equator mean approximate. If you want to talk total 360°, true geodetic (rather than great-ellipse) circumferences, then there is only the equator and an infinite number of meridional ellipses, since once you move the slightest bit off the equator, all arc paths shift north-south (since that is the shortest distance——i.e., the “geodesic”), in which case you would have average arcradius = (6378.137 + (∞-1)*6367.449)/∞ ≈ 6367.449 (realizing, of course, that “∞” is just a theoretical quantification of everything! P=). But let’s put aside the average geodetic circumference for the moment and focus on the average great-ellipse circumference. Do you have the same issue/approach with that (particularly the vertex latitude distribution)? Again, I have much more to say and present, yet. ~Kaimbridge~ (talk) 20:38, 12 June 2013 (UTC) My remarks about the distribution of vertex latitudes apply also to the circumferences of great ellipses. The distribution is not flat in [0,90] but skewed to the pole according to acosd(rand(num,1)). You can easily verify this by looking at random great circles on a sphere. If you correct the distribution of vertex latitudes, then your mean radius will become (2*a+b)/3 and everyone will be happy. cffk (talk) 21:16, 12 June 2013 (UTC) That doesn't make sense! How can it be a defined latitude yet random. Say for example, vertex latitude 37. 37 is 37, not a random number. 88.8C.80.7A (talk) 14:41, 13 June 2013 (UTC) I'm not sure I understand your question. We are carrying out the exercise of averaging the "effective radius" of geodesic segments to obtain a mean radius for the sphere. We can do this by using integrals. However it's sometimes quicker and easier to do the average by generating lots of random samples. In both cases it's crucial that the integral or random sampling be correctly weighted to correspond to uniformly distributed great circles. cffk (talk) 17:44, 13 June 2013 (UTC) Yes it does, but it doesn’t apply in this case. A vertex latitude, lat_v, is a “pole point”: Look at the graticule perspectives image, above. The oblique perspective on the right appears to have a lat_v of 60°, with an infinite number of meridians intersecting it, at all different angles——one for each azimuth. However, only one of them, the one that is east-west at the vertex, equals 90° - lat_v when it crosses the (conjugate) equator, 90° away from lat_v (in contrast, the north-south meridian that lat_v occupies, equals the length of lat_v from lat_v down to the equator, with an arc path value of 0). Thus there is only one oblique meridian intersecting lat_v that is considered lat_v, in terms of co-arc path identity——the one intersecting Lat_v at 90°. Now let me make the case (I think) you are trying to make...and why I think it is flawed. Let’s create a generic “circumference integrand”, ${\displaystyle {\mbox{Circ}}'(\phi _{v},{\widehat {\alpha }}_{v},{\widehat {\sigma }})\,\!}$ For a single lat_v (here, φ_v), along a single meridian, α_v (which, for a transverse meridian, equals Â, and for a “regular”, conjugate meridian, λ), the average arcradius is, ${\displaystyle {\overset {\,_{\smile }}{\!R}}_{avg}={\frac {1}{\pi }}\int _{{\widehat {\sigma }}_{s}}^{{\widehat {\sigma }}_{f}}{\mbox{Circ}}'(\phi _{v},{\widehat {\alpha }}_{v},{\widehat {\sigma }})d{\widehat {\sigma }};\,\!}$ ${\displaystyle ({\widehat {\sigma }}_{s}=\phi _{v},\;{\widehat {\sigma }}_{f}=\phi _{v}+180^{\circ })\,\!}$ which, for an arcpath from the conjugate equator to the transverse equator, can be reduced to ${\displaystyle {\overset {\,_{\smile }}{\!R}}_{avg}={\frac {2}{\pi }}\int _{0}^{90^{\circ }}{\mbox{Circ}}'(0,{\widehat {\mathrm {A} }},{\widehat {\sigma }})d{\widehat {\sigma }};\,\!}$ Now, for the average arcradius of all circumferences, based on a transverse vertex, ${\displaystyle {\overset {\,_{\smile }}{\!R}}_{avg}=\left({\frac {2}{\pi }}\right)^{2}\int _{0}^{90^{\circ }}\int _{0}^{90^{\circ }}{\mbox{Circ}}'(0,{\widehat {\mathrm {A} }},{\widehat {\sigma }})d{\widehat {\mathrm {A} }}d{\widehat {\sigma }}\;\approx {\frac {3a+b}{4}};\,\!}$ For the regular, polar vertex, we have ${\displaystyle {\overset {\,_{\smile }}{\!R}}_{avg}=\left({\frac {2}{\pi }}\right)^{2}\int _{0}^{90^{\circ }}\int _{0}^{90^{\circ }}{\mbox{Circ}}'(90^{\circ },{\widehat {\lambda }},{\widehat {\sigma }})d{\widehat {\lambda }}d{\widehat {\sigma }},\,\!}$ which can be reduced to ${\displaystyle {\overset {\,_{\smile }}{\!R}}_{avg}={\frac {2}{\pi }}\int _{0}^{90^{\circ }}{\mbox{Circ}}'(0,0,\phi )d\;\phi \approx {\frac {a+b}{2}};\,\!}$ what you are asserting it should be is ${\displaystyle {\overset {\,_{\smile }}{\!R}}_{avg}={\frac {4}{\pi ^{3}}}\int _{0}^{90^{\circ }}\int _{0}^{90^{\circ }}\int _{{\widehat {\sigma }}_{s}}^{{\widehat {\sigma }}_{f}}{\mbox{Circ}}'(\phi _{v},{\widehat {\alpha }}_{v},{\widehat {\sigma }})d\phi _{v}d{\widehat {\alpha }}_{v}d{\widehat {\sigma }}\approx {\frac {2a+b}{3}};\,\!}$ Here is the problem. Look at the conjugate graticule: Can you connect, say, Los Angeles and New York along a single meridian? No, only points that are north-south to one another can. Now try it with the given oblique perspective. While the meridians are a little more angled, you still can’t! It is only with the transverse graticle that you can. It is only the transverse graticule that you can connect any two points along a single meridian/arc path. If you try creating a total average arcradius by averaging all of the different average arcradii for the different vertex latitudes, as you suggest, you are distorting the average because, again, only the transverse graticule allows universal connection between two points. But, you know what, I think I remember the reason your random ±90° for lat and ±180° for long, is skewed. Distances are calculated using transverse coordinates, where ${\displaystyle \phi =\arcsin(\cos({\widehat {\mathrm {A} }})\sin({\widehat {\sigma }}))\,\!}$ and ${\displaystyle \lambda =\arctan(\sin({\widehat {\mathrm {A} }})\tan({\widehat {\sigma }}))+C\,\!}$, so let TL1 = 360*rand(num,1)-180,TL2 = TL1 + 180*rand(num,1) and AP = 90*rand(num,1). Then let lat1 = acosd(cos(AP)*sin(TL1)), lat2 = acosd(cos(AP)*sin(TL2)), lon1 = atand(sin(AP)*tan(TL1)) and lon2 = atand(sin(AP)*tan(TL2)) (the “C”s cancel). Now try your sampling with this. I kind of rushed towards the end here, so if something seems out of place, it is probably just a typo.  ~Kaimbridge~ (talk) 20:13, 13 June 2013 (UTC) Kaimbridge, I'm sorry but I couldn't really follow your arguments. Your terminology is very nonstandard. I manage to deal with great circles without the need to introduce "transverse graticules", "conjugate equators", and the like. I dare say these concepts help you, but it would really help if you could translate these terms into more conventional ones. I can, however, understand the experiment you suggest in the second to last paragraph. Translated into words, this says sample a random piece (TL1..TL2) of a great circle which intersects the equation at a random azimuth AP. The error you make is to assume that AP should be uniformly sampled, AP = 90*rand(num,1); instead AP should have a lambertian distribution AP = asind(rand(1)-1). This is precisely the same issue that I had with your distribution of vertex latitudes. Can I ask either or both of you to state the problems you see in the original sampling scheme I proposed, namely to average over the great-circle segments joining two randomly chosen points on the sphere? This is by far the simplest experiment to formulate. It samples great circle segments uniformly (if you don't agree, then please say why); and the average clearly results in (2*a+b)/3. This is the motivation of the ellipsoidal quadratic mean calculation; it's just that it corrects a flaw (which I have pinpointed) in that analysis. cffk (talk) 22:33, 13 June 2013 (UTC) Another way of understanding how to sample great circles uniformly is to picture the great circle as the rim of a gyroscope (ignore the precessing in this picture). The great circle is then fully specified by the latitude and longitude of its shaft, [lat0,lon0]. To sample great circles uniformly, we need to sample the shaft direction uniformly, i.e., take lat0 = asin(2*U-1) lon0 = 2*pi*U (U is uniform in [0,1]). Since the latitude of the vertex is given by latv = lat0 > 0 ? pi/2 - lat0 : pi/2 + la0 The distribution of latv is given by latv = acos(U) cffk (talk) 14:08, 14 June 2013 (UTC) Are you referring to this idea: 10000 Pointsin Monte Carlo Techniques: Introduction? 88.8C.80.7A (talk) 17:43, 14 June 2013 (UTC) Yes. The average in the ellipsoidal quadratic mean link essentially uses the left hand figure for the distribution of "shaft positions". This means that the great circles which lie closest to the equator are unfairly favored and this is what leads to an overestimate of the mean radius. cffk (talk) 17:55, 14 June 2013 (UTC) WHAT! That is what you are talking about? So all you are adjusting for is the shrinking latitude value as the pole is approached: ${\displaystyle \cos(\phi )\Delta \lambda \,\!}$?? Of course I am aware of that concept——it’s just not applicable here. You still aren’t appreciating the concept of the transverse graticule. Look at the graticule perspectives image, again: All three perspectives show the exact same geography——the north pole is at the top of all three globes——only the graticule webbing is shifted. The distance between arc paths are the same out at the perimeter, i.e., transverse equator: It is at the “normal”, conjugate equator that area is shrunken/lost. So, if anything, it is TL that needs cosine adjusting (though, since TL is actually transverse colatitude), it would need sine adjusting. But that still doesn’t apply here (I think I did try calculating a “transverse authalic radius” and the result was either close to a or about 6374). Here is proof why: Let’s say I wanted to calculate the average (regular, conjugate) meridian by calculating all of them and average them together. According to your reasoning, the averaging would be skewed at the poles, but they are not, because you are averaging them all together and you are only going to get one answer, since all meridians have the same value: ≈ 6367.449 km! Well that same result applies to the average transverse meridian, i.e, arc path, only each meridian/path has a different value! P=)  ~Kaimbridge~ (talk) 20:50, 14 June 2013 (UTC) Before I go down that road (which would entail my understanding more of your notation and terminology than I do now), can I repeat my request that you comment on the straightforward sampling scheme that I proposed: pick two random points on a sphere use this to define a sample great circle? Asked another way, if I were to draw great circles at random on the sphere, what do you think the distribution of vertex latitudes should be (or equivalently the distribution of equatorial azimuths)? (Incidentally, when I look at your pictures it seems that you mean to say "oblique graticule" instead of "transverse graticule". However, I could easily have misunderstood you.) If I can get the discussion back to what should go on the Great Circle Distance page. This discussion was helpful in showing three things (a) using a spherical radius of (2*a+b)/3 minimizes the RMS relative error in distance calculations; (b) this same radius is the IUGG recommended mean radius (which gets things like the volume and area of the ellipsoid correct to first order) and therefore, for consistency, is the recommended radius to use for great circle distance calculations; (c) that there's an error in the ellipsoidal quadratic mean link (at least as it applies to this discussion) which means that there's no good reason to use (3*a+b)/4 as the spherical radius for distance calculations. cffk (talk) 21:41, 14 June 2013 (UTC) I'm sorry Cffk, I admit I don't understand everything Kaimbridge is saying, but if I take a globe and tilt it down so that the north pole is facing me at the equator level, all of the meridians (his "arc paths") are equal spaced up at the pole and all the way down, along what was the equator that is now vertical, and the shrinking "spaceage" between meridians is right in front of me down here at the vertex where the equator should be, just like the middle picture of the "graticule perspectives" shows.88.8C.80.7A (talk) 22:47, 15 June 2013 (UTC) The article says "There are different approximations used as a radius for a non-spherical body such as Earth" The reader will ask: used for what? If the writer claims one "average" is better than another he means it's better for one purpose, and he needs to spell out what that purpose is, in the article. Cffk seems to be saying when calculating distance on the Earth, using his "average radius" will minimize the average or rms error for all possible great-circle tracks. Does anyone think he's wrong about that? If he's right then the article might as well mention that "average radius", spelling out what it's for. If someone has some other use for "average radius", let's hear it. No harm putting more than one in the article-- just tell the readers what each one is good for. If it isn't good for anything, why mention it? Tim Zukas (talk) 23:10, 15 June 2013 (UTC) Right, that is what I thought. Just identify what each of them specifically approximates ("a" for the equator, "(a+b)/2" the meridians and geocentric radius, "(2*a+b)/3", volume/surface area and "(3*a+b)/4" for the average circumference, though I think it was right before as average "great-circle radius" rather than "great-ellipse radius"). Do you understand what he means about the different graticule perspectives? 88.8C.80.7A (talk) 19:20, 16 June 2013 (UTC) Just listing these lengths with no extra information is not very useful for most readers. Why should a user prefer (a+b)/2 to a (specifically in the context of measuring distances)? In any case, there are errors in your characterizations of these lengths: the mean geocentric radius is not (a+b)/2 (it is (2*a+b)/3); the avarage circumference is not (3*a+b)/4 (it also is (2*a+b)/3). At a minimum these errors need to be corrected. (I believe that Kaimbridge and, maybe, 88.8C.80.7A are still contesting point 2. Are they?) From my perspective the important points are: For consistency with IUGG recommendations and other uses of the spherical model of the earth, use R_1 = (2*a+b)/3. This particular value minimizes (to first order in f) the RMS relative error in the distance. The maximum relative error for the WGS84 ellipsoid is 0.56%. If you need better accuracy calculate the distance with an ellipsoidal geodesic. Any other recommendations for a spherical radius should be subsidiary to this choice. (I note the (a+b)/2 gives a smaller maximum relative error of 0.51%; however in light of point 1, (2*a+b)/3 is the prefered choice.) cffk (talk) 21:32, 16 June 2013 (UTC) Isn't the geocentric radius (which equals a at the equator and b at the poles) the same in all directions at a given point, so it is just an integration of a and b, not the geometric mean of M and N (since it is not "surface radius" but "center to surface radius")? 88.8C.80.7A (talk) 00:06, 17 June 2013 (UTC) You're right about the values of the geocentric radius at the equator and the poles. However the "integration of a and b" has to be over the surface of the ellipsoid and there are more points near the equator than near the poles. Hence the average is closer to a than to b. Doing the math, you get (2*a+b)/3. cffk (talk) 01:26, 17 June 2013 (UTC) There's a good reason why the averages over the whole ellipsoid result in (2*a+b)/3. Consider any such surface or volume average for an ellipoid with semi-axes a = r+ea, b = r+eb, c = r+ec. This may be expressed as f(r,ea,eb,ec) which is symmetric under interchange of its last three arguments. Taylor expand this in the limit of small ea, eb, ec. The coefficents of the ea, eb, ec must be the same (by symmetry). Thus f(r,ea,eb,ec) = A(r) + B(r)*(ea+eb+ec) + second order terms. Chose r = (a + b + c)/3 so that ea + eb + ec = 0 and we get f(r,ea,eb,ec) = A((a+b+c)/3) to first order. For an oblate ellipsoid we have c = a and f = A((2*a+b)/3). cffk (talk) 03:08, 17 June 2013 (UTC) The article says "McCaw[8] recommends just using the volume/surface area based approximation." This makes it sound like that radius is somehow the best for calculating distance, which it presumably isn't. Anyone know what he actually said? In any case, a section on other-radius-approximations should make clear that there's no silver bullet. One radius minimizes the maximum error, another radius presumably minimizes the average error; none of them does better than 0.5%. (No one seems to know what the other ones are good for.) Tim Zukas (talk) 16:10, 17 June 2013 (UTC) The McCaw quote is (halfway down on p260): Distances on the Sphere—It is not in general the arc σ on the sphere but the length s of the arc σ on the earth's surface which is demanded. In purely spherical working s is derived from σ on the assumption that the earth is a sphere of known radius r. A good mean value for r is simply the arithmetic mean of the three axes of the spheroid of reference; in other words, if a and b are the semi-axes major and minor, r = (2a + b)/3. This value of r is less than 2 metres greater than the radius of the sphere whose surface has the same area as the surface of the earth. If f is the compression, then r = a - fa/3; so that if f = 1/296 and a = 6378356 meters, r = 6371173 meters. The radius r may therefore be taken as 6371170 metres in round numbers. I think this is sound advice (only to be updated to reflect the values for modern reference ellipsoids). cffk (talk) 17:21, 17 June 2013 (UTC) Cffk: Okay, I think I know how to force your acknowledgment of my approach and result, even though not conventionally recognized, is a valid one. Let’s say Earth was scalene and I wanted to know the average meridional, strictly north-south circumference radius: Where a_x = 6383.1004, a_y = 6373.1707 and b = 6356.752, meridional_x’s, mer_x’s, average radius = 6369.933 and mer_y’s = 6364.964. For this exercise (I’ve experimented with scalene surface area and I realize it may actually be more complicated than this, but this is good enough for this demonstration), for a given meridian/longitude, let ${\displaystyle \scriptstyle a=a(\lambda )={\sqrt {(a_{x}\cos(\lambda ))^{2}+(a_{y}\sin(\lambda ))^{2}}}{}_{\color {white}.}.\,\!}$ I get (mer_x + mer_y)/2 = 6367.4485, ((mer^2_x + mer^2_y)/2)^.5 ≈ 6367.44898471 and (mer_x*mer_y)^.5 ≈ 6367.44801529. If a(45°) (≈ 6378.137482362) is used, then mer_45 ≈ 6367.449230215. So I would say the approximate, average meridional radius is ≈ 6367.45——would you agree? If not and you insist that surface area skewing needs to be adjusted for, what would you suggest it should be, and then set a back to a constant 6378.137 and redo that exact same computation: Do you get a similar, different value than about 6367.449?  ~Kaimbridge~ (talk) 16:49, 17 June 2013 (UTC) Are you trying to say 6367449 meters is the best radius to use when calculating the length of a north-south line? Or are you saying it's the best for all lines? Tim Zukas (talk) 18:07, 17 June 2013 (UTC) In the above example, yes. I was defining Earth as scalene, meaning the equator is elliptical and, therefore, all of the meridians have different values: So what would you use for a meridian value? You could average all of the different meridians (from a(0) to a(90°)) or just average the two boundary values (6369.933 and 6364.964) together and get about 6367.449, which is the approximate spheroidal value, where all of the meridians have the same value: ≈ 6367.449! If he is consistent, Cffk is saying that, because the length of longitude difference shortens as you move from the equator to pole, the weight of the radius being averaged needs to take this into account, and I’m saying, no, a degree of great-circle arc is the same at the pole as it is at the equator, and that the same holds true for all of the “differently angled great-circles/ellipses”, that the middle, transverse graticule perspective shows (not the facing small circles...those are the transverse co-latitudes) and that “(3a + b)/4” approximates the average of. Incidentally, I didn’t make up the terms, they are in the USIGS Glossary.  ~Kaimbridge~ (talk) 20:06, 17 June 2013 (UTC) The term "average meridional, strictly north-south circumference radius" is not precisely defined. If you're asking for the mean circumference of a meridian, then it's close to pi*(a+b) for an ellipsoid of revolution. If you're asking the appropriate radius to use for portions of great circles which happen run north-south, I would need to understand exactly what the application is before knowing how to take the average. However I'm none the wiser after this exercise! For calculating a mean radius for use in distance calculations, the end points should be uniformly distributed over the sphere. That is sin(lat1) and sin(lat2) should be uniformly sampled in [-1,1] and lon1 and lon2 should be uniformly sampled in [-180d,180d]. If you sample in this way (and you've never committed yourself as to whether you think this is right), you get a mean of (2*a+b)/3. Incidently, Kaimbridge, does your result Qr = (3*a+b)/4 still hold for a prolate ellipsoid (where a is the equatorial radius and b is the polar semi-axis)? If not, what's your formula in that case. Also, what's the generalization to a triaxial ellipsoid (semiaxes ax, ay, b)? cffk (talk) 20:39, 17 June 2013 (UTC) Kaimbridge, you look at your transverse graticule picture and are convinced that you are sampling great circles fairly. I look at it and say "I'm not sure". Why? Because the picture won't appear the same looked from different angles. Why is this important? Because that's the only way to be sure that the sampling is rotationally invariant. It might be that your sampling is fair; but it's not obvious and it needs to be checked. I did that and discovered (what I should have remembered) that angular dependence of an isotropic flux through a surface is not uniform but lambertian. However the simple numerical experiment that I did to ascertain this behavior was to take great circles between randomly and uniformly chosen pairs of points. Such sampling is clearly rotationally invariant and if you plot the distribution of equatorial azimuths for the resulting great circles you recover the lambertian distribution. I hope this helps. cffk (talk) 01:06, 18 June 2013 (UTC) Yes, by “average meridional, strictly north-south circumference radius” I mean π*(a+b). But for the scalene case, each meridian has its own value, π*(a(λ)+b) (or, more properly, π*(a_m+b(λ)), as I illustrate below), meaning there is a different value for each meridian! As for “the appropriate radius to use for portions of great circles which happen (to) run north-south”, it is just the average radius to use for random north-south distances/lengths (e.g., for someone at the North Pole wanting to calculate distances from their location....You wouldn’t want to use 6371.0, would you? No, you would use 6367.449——right?): Adjust your above sampling program to let Δλ = 0. Do you think there needs to be a sampling (cosine) adjustment in that case? This is the exact same approach used in the averaging for any arc path: ${\displaystyle \lambda \rightarrow {\widehat {A}};\quad \Delta {\widehat {A}}=0;\quad \Delta \phi \rightarrow \Delta {\widehat {\sigma }};{\color {white}{\Big .}}\,\!}$ Now do that to all arc paths (i.e., transverse meridians, which as the above USIGS glossary linkstates, means “perpendicular to a transverse equator”, not perpendicular to its corresponding, conjugate state, as all “perpendicular meridians” just equal the (conjugate) equator!) and average them all together, noting that there is no cosine reduction (or, in this case, sine reduction), just as there is no cosine reduction in averaging the different (conjugate) meridional values in the scalene case, since each meridian has equal weight in the scalene averaging. A given arc path only exists on a graticule perspective which has its vertex latitude lesser than or equal to the arc path’s (I think I worded that right). Consider these two examples:   On the left are two great-circle arc segments on different arc paths (transverse meridians). Since the arc paths aren’t north-south, they can’t exist on the left, conjugate graticule (since their vertex latitudes are less than the conjugate perspective’s vertex latitude, 90°), nor can the right arc path on the right, oblique graticule, for the same reason. You can horizontally push a given perspective along its particular vertex latitude, all the way around and back, but only the transverse graticule is able to always connect any two points on a single path for a given perspective (think Los Angeles and New York). In the scalene case, just like you would have to average all of the different (conjugate) meridian values (one for each λ) to find the global scalene average, you would have to average all of the different arc path values, not just for a single transverse, equatorial vertex, but for all the different equatorial vertices, one for each λ (0->90°). But, as has been pointed out, such a complicated, precise calculation is unnecessary——just a simple averaging of the two principal (different) circumferences, the meridional and equatorial. In terms of “sampling great circles fairly”, the approach is to get one (and only one) for each angle, which means ${\displaystyle \scriptstyle {\widehat {A}}=0\rightarrow 90^{\circ }\,\!}$, except in the scalene case, where (as just previously stated) ${\displaystyle \scriptstyle {\widehat {A}}=0\rightarrow 90^{\circ }\,\!}$ is found and averaged for each λ, 0->90°, with all of these averages averaged into a final, scalene average. If you were to average the transverse meridional average (6372.8) with the conjugate meridional average (6367.449), you would have one set of “differently angled” great circles and a second set of north-south great circles, the averaging of the two sets equaling about 6370.125, which approximates the volume/surface area, as you assert. Again, though, distance between two points is always along a single path, it never “jumps paths”, meaning it doesn’t matter how much area there is between paths. At φ = 60°, Δλ is half as wide as it is on the equator: Does that have any bearing on calculating distance or just arcradius along one of the λs within the Δλ? As for the general formula for the ellipsoidal quadratic mean (or, perhaps more technically, triaxial quadratic mean), it becomes apparent when the two Cartesian coordinate sets are defined. This would seem to apply to oblate, prolate and scalene ellipsoids. Ellipsoid axes, radii, Cartesian coordinates There are two Cartesian coordinate sets used on an ellipsoid, the geocentric and a second, auxiliary set (I’m not sure of its formal name), used for surface calculations, such as distance (geodetic and great-elliptic) and surface area. Geocentric, auxiliary radii Cartesian sets ${\displaystyle {\begin{aligned}{\color {white}.}&\qquad \mathbf {\overline {\underline {|\;Geocentric\;|}}} ^{\color {white}|}\\a(\lambda )&={\sqrt {{\big (}a_{x}\cos(\lambda ){\big )}^{2}+{\big (}a_{y}\sin(\lambda ){\big )}^{2}}};^{\color {white}{\big |}}\\&\qquad ^{(a_{x}=a(0);\quad \;a_{y}=a(90^{\circ }))}\\X&=a_{x}\cos(\beta )\cos(\lambda );\\Y&=a_{y}\cos(\beta )\sin(\lambda );\\{\color {white}.}{\sqrt {X^{2}+Y^{2}}}&=x(\lambda )=a(\lambda )\cos(\beta );\\Z&=\quad \;y\,=b\sin(\beta );_{\color {white}{\big |}}\\R&=R(\beta )={\sqrt {x^{2}(\lambda )+y^{2}}}={\sqrt {X^{2}+Y^{2}+Z^{2}}},{\color {white}.}\\&={\sqrt {{\big (}a(\lambda )\cos(\beta ){\big )}^{2}+{\big (}b\sin(\beta ){\big )}^{2}}};{\color {white}{\Big |}}\end{aligned}}\,\!}$ ${\displaystyle {\begin{aligned}&\mathbf {\overline {\underline {|\;Complementary/Auxiliary\;Surface\;|}}} ^{\color {white}|}\\{\acute {a}}_{x}=b_{x}&=b{\frac {a_{y}}{a_{m}}}=b{\sqrt {\frac {a_{y}}{a_{x}}}};\qquad {\acute {a}}_{y}=b_{y}=b{\frac {a_{x}}{a_{m}}}\;\,=b{\sqrt {\frac {a_{x}}{a_{y}}}};{\color {white}.}\\&\qquad ^{(b_{x}=b(0);\quad \;b_{y}=b(90^{\circ });\quad \;b_{x}b_{y}=b^{2})}\\{\acute {a}}(\lambda )=b(\lambda )&=b{\frac {a(90^{\circ }\pm \lambda )}{a_{m}}}=b{\sqrt {{\frac {a_{x}}{a_{y}}}\sin ^{2}(\lambda )+{\frac {a_{y}}{a_{x}}}\cos ^{2}(\lambda )}};\\\qquad {\acute {b}}=a_{m}&={\sqrt {a_{x}a_{y}}};_{\color {white}{\big |}}\\{\acute {X}}&=b_{x}\cos(\beta )\cos(\lambda )=b{\frac {a_{y}}{\sqrt {a_{x}a_{y}}}}\cos(\beta )\cos(\lambda );\\{\acute {Y}}&=b_{y}\cos(\beta )\sin(\lambda )=b{\frac {a_{x}}{\sqrt {a_{x}a_{y}}}}\cos(\beta )\sin(\lambda );\\{\sqrt {{\acute {X}}^{2}+{\acute {Y}}^{2}}}&={\acute {x}}(\lambda )\,=b(\lambda )\cos(\beta );\\{\acute {Z}}&=\quad \;{\acute {y}}\,=a_{m}\sin(\beta );\\S&=S(\beta )={\sqrt {{\acute {x}}^{2}(\lambda )+{\acute {y}}^{2}}}={\sqrt {{\acute {X}}^{2}+{\acute {Y}}^{2}+{\acute {Z}}^{2}}},\\&={\sqrt {{\big (}b(\lambda )\cos(\beta ){\big )}^{2}+{\big (}a_{m}\sin(\beta ){\big )}^{2}}};{\color {white}{\Big |}}\end{aligned}}\,\!}$ These surface radii compositions are based on the recognized parametric area integrand, ${\displaystyle \scriptstyle {\cos(\beta ){\sqrt {{\Big (}{\big (}a_{y}b\cos(\lambda ){\big )}^{2}+{\big (}a_{x}b\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}a_{x}a_{y}\sin(\beta ){\big )}^{2}}}}^{\color {white}|}\,\!}$  ${\displaystyle {\begin{aligned}\mathbf {\overline {\underline {|\;Pa}}} &\mathbf {\overline {\underline {rametric\;Surface\;Area\;Integrand\;|}}} ^{\color {white}|}\\a_{m}\cos(\beta )S(\beta )&=a_{m}\cos(\beta ){\sqrt {{\acute {X}}^{2}+{\acute {Y}}^{2}+{\acute {Z}}^{2}}}^{\color {white}|},\\&=\scriptstyle {a_{m}\cos(\beta ){\sqrt {{\big (}{\frac {a(90^{\circ }\pm \lambda )b}{a_{m}}}\cos(\beta ){\big )}^{2}+{\big (}a_{m}\sin(\beta ){\big )}^{2}}}},\\&=\scriptstyle {\cos(\beta ){\sqrt {{\big (}a(90^{\circ }\pm \lambda )b\cos(\beta ){\big )}^{2}+{\big (}a_{x}a_{y}\sin(\beta ){\big )}^{2}}}},\\&=\scriptstyle {\cos(\beta ){\sqrt {{\Big (}{\big (}a_{y}b\cos(\lambda ){\big )}^{2}+{\big (}a_{x}b\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}a_{x}a_{y}\sin(\beta ){\big )}^{2}}}},{\color {white}.}\\\scriptstyle {\mathrm {(simplifying} }\;&\scriptstyle {\mathrm {to\;the\;spheroidal\;case)} }\\&=\scriptstyle {\cos(\beta ){\sqrt {{\Big (}{\big (}ab\cos(\lambda ){\big )}^{2}+{\big (}ab\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}aa\sin(\beta ){\big )}^{2}}}},\\&=\scriptstyle {a\cos(\beta ){\sqrt {{\big (}b\cos(\beta )\cos(\lambda ){\big )}^{2}+{\big (}b\cos(\beta )\sin(\lambda ){\big )}^{2}+{\big (}a\sin(\beta ){\big )}^{2}}}},\\&=a\cos(\beta ){\sqrt {{\big (}b\cos(\beta ){\big )}^{2}+{\big (}a\sin(\beta ){\big )}^{2}}};{\color {white}{\Big |}}\end{aligned}}\,\!}$ As can be seen, for surface calculations such as distance (and, specific to this discussion, arcradii), a is a constant geometric mean, a_m, with the elliptical variation and dynamics transferred to an auxiliary, variable b, b(λ). Given that, the ellipsoidal quadratic mean, Qr, has the general ellipsoidal form ${\displaystyle {\begin{aligned}Qr&={\sqrt {\frac {{\frac {a_{x}a_{y}+b_{x}b_{y}}{2}}+{\frac {a_{x}^{2}+a_{y}^{2}}{2}}}{2}}}^{\color {white}|}={\sqrt {\frac {a_{x}^{2}+a_{x}a_{y}+a_{y}^{2}+b^{2}}{4}}},{\color {white}.}\\&\approx {\frac {{\frac {a_{m}+b}{2}}+{\frac {a_{x}+a_{y}}{2}}}{2}}^{\color {white}|}={\sqrt {\frac {a_{x}+a_{m}+a_{y}+b}{4}}}\approx {\sqrt[{4}]{a_{x}^{1.5}a_{y}^{1.5}b}};{\color {white}.}\end{aligned}}\,\!}$ which, reduced to the spheroidal case, simplifies to ${\displaystyle {\begin{aligned}Qr&={\sqrt {\frac {{\frac {a^{2}+b^{2}}{2}}+{\frac {a^{2}+a^{2}}{2}}}{2}}}^{\color {white}|}={\sqrt {\frac {3a^{2}+b^{2}}{4}}},\\&\approx {\frac {{\frac {a+b}{2}}+{\frac {a+a}{2}}}{2}}={\sqrt {\frac {3a+b}{4}}}\approx {\sqrt[{4}]{a^{3}b}}.{\color {white}.}\end{aligned}}\,\!}$ And, just to clarify, “Qr” is meant to specifically denote the quadratic mean, not the simple or geometric means.  ~Kaimbridge~ (talk) 18:54, 19 June 2013 (UTC) Kaimbridge, if I substitute ${\displaystyle a_{x}=a}$ and ${\displaystyle a_{y}=b}$, I get ${\displaystyle Q_{r}=(a^{3}b^{5})^{1/8}}$ as the ellipsoidal quadratic mean for a prolate ellipsoid with equatorial radius ${\displaystyle b}$ and polar semi-axis ${\displaystyle a}$. But if I go through your original reasoning (which didn't specifically depend on the oblateness of the ellipsoid), I would get ${\displaystyle Q_{r}=(ab^{3})^{1/4}}$ (switching ${\displaystyle a}$ and ${\displaystyle b}$ in your oblate formula). Where's my mistake? cffk (talk) 19:44, 19 June 2013 (UTC) No, a_x = a_y = a. b_x = b_y = b. Look at them in the Cartesian coordinate subsection. As for prolate, a and b stay in their respective positions, just like the Cartesians.  ~Kaimbridge~ (talk) 20:42, 19 June 2013 (UTC) I'm confused. So ${\displaystyle Q_{r}=(a_{x}^{3}a_{y}^{3}b^{2})^{1/8}}$ isn't a general formula that I can apply to a triaxial ellipsoid with semi-axes ${\displaystyle a_{x}}$, ${\displaystyle a_{y}}$, and ${\displaystyle b}$? (I guess I failed to notice ${\displaystyle b_{x}}$ and ${\displaystyle b_{y}}$. How come you need 4 independent axes to define an ellipsoid?) cffk (talk) 21:06, 19 June 2013 (UTC) Kaimbridge, can you post your code where you verify that the distribution of equatorial azimuths is flat when you consider the great circles joining two random points (your "Los Angeles" and "New York")? Here's my code and this results in a lambertian distribution: % Requires: % http://www.mathworks.com/matlabcentral/fileexchange/39108 num=1000000; % number of samples % random points 1 and 2 lat1 = asind(2*rand(num,1)-1); lon1 = 180*(2*rand(num,1)-1); lat2 = asind(2*rand(num,1)-1); lon2 = 180*(2*rand(num,1)-1); % solve for great circle (sphere = [1,0]) [s12, azi1, azi2] = geoddistance(lat1, lon1, lat2, lon2, [1,0]); % compute northerly equatorial azimuth (Clairaut relation) azi0 = asind(sind(azi1) .* cosd(lat1)); hist(azi0, 100); What's the error in my approach? cffk (talk) 21:34, 19 June 2013 (UTC) I haven’t got time to fully answer now, but do you see the collapsible “Geocentric, auxiliary radii Cartesian sets” section with these, images? E.g., b_x=b*(a_y/a_x)^.5, thus the a's cancel in the spheroidal cases, and b_x = b. As for Los Angeles-New York, I’m just saying that they are not on the same north-south meridian!  ~Kaimbridge~ (talk) 00:31, 20 June 2013 (UTC) Yes, I saw "Geocentric, auxiliary radii Cartesian sets" section. I followed this first part and got lost with the second and third parts. I'm just asking for the general formula for ${\displaystyle Q_{r}}$ for the triaxial ellipsoid? Then surely I can make substitutions on ${\displaystyle a_{x}}$, ${\displaystyle a_{y}}$, and ${\displaystyle b}$ to recover special cases? On LA-NY, I still can't understand why you think uniformly sampling the equatorial azimuth is remotely the right thing to do. At this point, I believe we've aired the issues enough. ${\displaystyle R_{1}=(2a+b)/3}$ is clearly a good choice in that it minimizes the RMS relative error in distance measurements. ${\displaystyle (a+b)/2}$ might also be useful since it minimizes the maximum relative error. I haven't yet heard an understandable explanation of why ${\displaystyle Q_{r}=(3a+b)/4}$ might be a suitable choice; by "understandable" I mean that it can be explained purely in terms of the accuracy of distance calculations (spherical vs geodesic), without the extra "baggage" of graticules, conjugate equators, and the like. Kaimbridge or 88.8C.80.7A, do you want to take a final crack that this? What is a required is a statement along the lines of "if ${\displaystyle s_{12}}$ is the geodesic between points 1 and 2 on an ellipsoid and ${\displaystyle s'_{12}}$ is the great circle distance between the same points on a sphere of radius ${\displaystyle r=Q_{r}}$, then the follow statement can be made about the distribution of relative errors ${\displaystyle e=(s'_{12}-s_{12})/s_{12}}$ for randomly chosen points 1 and 2: fill in here please!". cffk (talk) 01:12, 20 June 2013 (UTC) I read in some articles that the "6367" is refered to as "the radius on an auxiliary sphere". Isn't that what Kaimbridge is really doing? That is, each "arc path" represents an "auxiliary sphere", so that the "average circumference" is the averaging of all the different "auxiliary spheres" and deriving the average radius from that? 88.8C.80.7A (talk) 20:35, 20 June 2013 (UTC) By "6367", do you mean (a+b)/2 (which is roughly what it is for the WGS84)? Or did you mean to refer to (3a+b)/4 which is Kaimbridge's quadratic mean? Anyway, I'm not really sure what Kaimbridge is doing. He has a particular way of doing averaging which doesn't sample great circles uniformly; this means that it's difficult to see how it applies to distance calculations. In any case, speaking in terms of "auxiliary spheres" is not going to be very useful to someone trying to figure out which radius to use for distance calculations (I don't think that "auxiliary sphere" has any precise definition). He just wants a handle on the error involved in using a particular spherical radius. cffk (talk) 20:55, 20 June 2013 (UTC) I mean "(a+b)/2" for the meridian mean. I think I understand what is going on. Kaimbridge is basing the average on a definite integral where the meridian and equator are the boundaries, while you are using the x and y meridians and the equator to find the surface area (like for a triangle) and using the radius from that. I just tend to agree with Kaimbridge that averaging the two circumference extremes (meridian and equator) seems to be the obvious choice for a simple great-circle radius average approximation. 88.8C.80.7A (talk) 19:06, 21 June 2013 (UTC) A more accurate characterization of my approach is that it averages over all possible great ellipses (uniformly sampling the orientation of the ellipses). In this way, the definition is independent of how the ellipsoid is oriented and there's no necessity to label one of the semi-axes as the polar one and the other (possibly unequal) ones as equatorial. After doing the average it turns out that the average is equivalent to averaging over the 3 principal ellipses (in the case where the 3 semi-axes are nearly equal). But that was not my starting point. You and Kaimbridge have decided on different way of doing the averaging and of course you are free to do this and thereby define your own average radius. Neither of you has yet articulated why this should be used by someone who wants to estimate geodesic distances via a great circle calculation. Here's how I would explain why R_1 = (2*a+b)/3 is a good choice. Select random pairs of points and determine the true geodesic distance and the great circle distance (using R_1). Make a table: lat1 lon1 lat2 lon2 geod g-c %err 54.1220 -26.6448 -36.8476 -29.9051 10083964 10120209 0.359 11.8042 -46.6654 -49.3498 -77.1150 7401985 7423910 0.296 54.6973 86.8521 -2.3602 119.0341 6981592 6997463 0.227 52.6389 8.8880 10.1733 -114.3053 11215289 11203472 -0.105 51.6027 -123.4414 -0.0507 -9.1026 11666730 11661331 -0.046 .... With this choice of radius, we have (to first order in the flattening): mean(%err) = 0 mean(geod) = mean(g-c) The calculation is easy to follow and the resulting properties are clearly "good". Can you make a similar case for the quadratic mean? cffk (talk) 20:04, 21 June 2013 (UTC) Cffk: I’ve been focused on whole arc paths and you with sampling and equal surface area. Try this: What would you consider the midpoint of a quadrant, taking into account direction (not just north-south)? Hint: Try using a string to physically trace it. (I won’t be able to respond until tomorrow)  ~Kaimbridge~ (talk) 20:29, 21 June 2013 (UTC) Kaimbridge: Please define "quadrant". Is it a line, an area, or what? I think you use quadrant to refer to an eighth of the earth, e.g., lat in [0d,90d], lon in [0d,90d]. Is this right? The midpoint of this particular quadrant is at lon = 45d, lat = 35.26d (tan(lat) = 1/sqrt(2)). (I'm confused where direction comes into this; so I probably misinterpreted your question.) cffk (talk) 20:52, 21 June 2013 (UTC) Quick answer: Yes, quadrant equals φ = 0->90° and Δλ = 90°. As for your preliminary answer, not quite. You are on the right arc path (45°), but wrong mid-lat: Since it is at the midpoint of the midpath, ${\displaystyle \scriptstyle {\widehat {\sigma }}_{mp}={\frac {\Delta {\widehat {\sigma }}}{2}}=45^{\circ }\,\!}$, your answer gives a common value of ≈ 54.73561° (which is what the——spherical?, geographic(al)?——azimuth of the line at the midpoint should be), meaning your mid-lat is a little too high (hint: Remember, ${\displaystyle \scriptstyle \sin(\phi _{p})=\cos({\widehat {\mathrm {A} }})\sin({\widehat {\sigma }}_{p})\,\!}$, thus ${\displaystyle \scriptstyle \sin(\phi _{mp})=\cos(45^{\circ })\sin(45^{\circ })\,\!}$).  ~Kaimbridge~ (talk) 16:43, 22 June 2013 (UTC) Kaimbridge, I wish you didn't talk in riddles. I think you're telling me that the midpoint is φ = 30°, λ = 45°. How do you define midpoint to arrive at this answer? (It's not equally distant from the three corners of the equilateral triangle.) Where is this going anyway? To return to another point you made, that you've "been focused on whole arc paths". I reiterate: the mean circumference of a great ellipse on an ellipsoid with (approximately equal) semi-axes a, b, c is ${\displaystyle 2\pi (a+b+c)/3}$. So the resulting "mean radius" for an oblate ellipsoid (${\displaystyle c=a}$) is ${\displaystyle (2a+b)/3}$. cffk (talk) 19:11, 22 June 2013 (UTC) No, I am saying φ_mp = 30° and λ_mp ≈ 35.26439° (i.e, arctan(.5^.5)), which gives ${\displaystyle \scriptstyle {\widehat {\mathrm {A} }},\,{\widehat {\sigma }}_{mp}\,\!}$ both values of 45°. This is based on the simple division of a great circle: A great circle has an angular distance of 360°, ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}=360^{\circ }\,\!}$; a quarter of its circumference, therefore equals 90°, ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}=90^{\circ }\,\!}$; the midpoint of that quarter of a great circle (forget about φ and λ at the moment, we are only dealing with a single great circle) is at the halfway point of that quarter-circumference, i.e., ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}=45^{\circ }\,\!}$. Are we in agreement so far? Now let’s assume that particular great circle has been upright, facing us. Now turn it horizontally 90° so that it looks like just a vertical straight line (i.e., the perimeter, not face, is facing you). Next, tilt it sideways so that it is now at a 45°angle and slice it through the center of a sphere of equal diameter: The top of that 45° tilted great circle is at the sphere’s mid-vertex latitude, also 45°, keeping in mind that that whole “vertex great circle”, perpendicular to the sphere’s equator——hence, “transverse equator”——is also 360° in angular distance, or 90° per quadrant (1° of spherical arc always being the same length throughout, from the poles to the equator). As the distance from the equator out to the vertex, along the great circle, is always 90°, then the midpoint, ${\displaystyle \scriptstyle {\widehat {\sigma }}_{mp}\,\!}$ of that quarter-great circle is always 45°, no matter what the angle of that great circle (i.e., arc path, ${\displaystyle \scriptstyle {\widehat {\mathrm {A} }}\,\!}$) is. You are focused on looking at it with respect to surface area and I’m focused on looking at it in terms of arc. An ellipsoid is regarded as a “three dimensional analogue of an ellipse”. You are looking at it as being composed of north-south ellipses from left to right, ellipse_xz to ellipse_yz and going from there, adjusting for longitude difference shrinkage as the poles are approached. In contrast, in terms of calculating distances along great arcs, I see the ellipsoid as being composed of “differently angled ellipses” from vertical to horizontal, from ellipse_xz to circle/ellipse_xy (incidently, I don’t know if you just miswrote, but in the traditional notation, the polar radius is c, not b, I suspect the logic being that the equator is considered the initial ellipse, with a-b radii). Just consider how geodetic distances are calculated: The φs and λs are first converted to βs and ωs, then to their transverse counterparts, σs and Α/α_eq, to calculate along a specially created (or, as noted earlier, “auxiliary”) great circle (surface area, equal or skewed, is irrelevant). Well let’s try applying the concept physically. Take a piece of string and hold one end on the equator of a globe, and the other end at its pole (${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}={\frac {\pi }{2}}=90^{\circ }\,\!}$): This length represents the north-south quarter-circumference. Now, still holding one end on the equator, slowly pull the other end down along the transverse equator (i.e., ${\displaystyle \scriptstyle \Delta \lambda =\Delta {\widehat {\sigma }}=90^{\circ }\,\!}$), until it sets east-west on the conjugate equator: The final string length represents the equatorial, east-west quarter-circumference, the radius of which equals a. If the string recorded or “memorized” all of the infinitesimally differentiated arc path lengths——the fact that the string swept across the (quarter) face of the globe from north-south down to east-west means that it covered the whole surface area, once (and only once)——averaged them all together and divided by half-π, you would have the average arcradius of all the “differently angled circumferences”. This average would represent the (arc)radius for any and all ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}\,\!}$ possibilities between any two points along any arc path (“circumference”), once (and only once). That would be my definition of an ellipsoid’s average great-circle radius. Perhaps the simplest and clearest illustration would be to take a pie with a diameter of 180 mm and divide it into twelve even slices, each quadrant having three slices, the perimeter segment of each slice being exactly 30°. Each piece has a unique but constant depth, throughout: The upper slice has a depth of 108, the middle 120, and the lower slice, 138. What is the average depth? Since each slice is exactly the same size in terms of width (30°) and length (radius equals 90 mm), it would seem that the average slice depth would be (108+120+138)/3 = 122 ≈ 123 = (108+138)/2, both of which are close to the mid-slice depth. Now increase the number of equi-width/90 mm radius slices for each quadrant from 3 with a angular width of 30°, to infinity, with an angular width of infinitesimality: How would you find the average depth in this case? The same way: Add them all up and divide by infinity, an approximation of which equals the mean of the upper and lower slices. Now change the 90 mm radius to a 90° arc and the infinite number of plane depth slices become quarter-circumfernces of different sized great circles (or different sized great-ellipses with the different arcradii of each averaged together to create a unique, spherical radius for each). Now let’s narrow down the midpoint with the string and globe. The midpath would be ${\displaystyle \scriptstyle {\widehat {\mathrm {A} }}=45^{\circ }\,\!}$: If you take the string, hold one end on the equator and the other end 90° away (Δλ) at φ = 45° (the mid vertex latitude), you will see that the midpoint of that path, ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}_{mp}\,\!}$, equals 45° (the same as all quarter-circumferences measured from the conjugate to transverse equators). Is the north-south distance from the pole to φ_mp and from the (conjugate) equator to φ_mp 45° each way? No, it is two thirds towards the equator where——bringing your favorite element into consideration——there is more surface area! I believe it would be in the middle subtriangle of this (imagine it being a spherical quadrant) triangle(one end of the string would be at the lower right corner of the big triangle, and the other at the upper right corner of the middle subtriangle). Or, in this image, where the “h”s intersect (the big difference between this plane triangle and the spherical equivalent being that AB, AC, BC, Aa, Bb and Cc on the spherical equivalent all equal 90° in length, and that the h intersection is the 45° midpoint of Aa and Bb, but not Cc, where it is two thirds down!). So, breaking down the midpath of Earth, from the whole quadrant to the midpoint, itself, where “DxS” is the geodetic distance (traditionally just “s”) we have Midpoint arcradius ${\displaystyle a=6378.137;\quad \;b=6356.752;\quad \;Qr=6372.797477595906624842;{\color {white}|}\,\!}$ ${\displaystyle {\underline {{\color {white}{\tfrac {D}{\Delta }}}_{_{\Delta {\widehat {\sigma }}}}\;}}\qquad {\underline {{\color {white}\qquad }_{\phi _{s}}\qquad \qquad \quad _{\phi _{f}}\qquad \quad \qquad _{\Delta \lambda }{\color {white}\quad }{\qquad {\tfrac {DxS}{\Delta {\widehat {\sigma }}}}}\qquad \qquad \quad }}_{\color {white}|}\,\!}$ 90° 0°, 45°, 90°, 6372.809808 km 10 27.03402084, 32.79775133, 9.43868403, 6372.797830 1 29.71049001, 30.28783051, .94281968, 6372.797541 .15 29.999997113249, 30.000002886751, .000009428090, 6372.797537721364 .110 29.999999999971, 30.000000000029, .000000000094, 6372.797537721364 Thus ${\displaystyle \scriptstyle {\widehat {\mathrm {A} }}={\widehat {\sigma }}_{mp}=45^{\circ }\,\!}$, and ${\displaystyle \scriptstyle \phi =30^{\circ };\quad {\widehat {\alpha }}_{mp}=54.73561031725^{\circ };\quad \;{\tilde {\alpha }}({\widehat {\alpha }}_{mp})={\tilde {\alpha }}_{mp}=54.87167633668^{\circ };{\color {white}|}\,\!}$ ${\displaystyle {\begin{aligned}M&=M\{45,45\}=\;\,{\frac {(ab)^{2}}{Qr^{3}}}\,\,=6351.3767100568876;\\N&=N\{45,45\}=\;\;\,{\frac {a^{2}}{Qr}}\;\;\;=6383.4809961849728;\\{\frac {M+N}{2}}&={\frac {a^{2}}{2Qr}}{\Bigg (}{\bigg (}{\frac {b}{Qr}}{\bigg )}^{2}+1{\Bigg )}=6367.4288531209302;\\{\color {white}_{\Bigg |}}{\sqrt {\frac {M^{2}+N^{2}}{2}}}&={\frac {a^{2}}{Qr^{3}}}{\sqrt {{\frac {1}{2}}{\Big (}b^{4}+Qr^{4}{\Big )}}}=6367.4490866322499;\\\quad \;G\{45,45\}&={\frac {dxS}{d{\widehat {\sigma }}}}={\sqrt {(\cos({\widehat {\alpha }}_{mp})M)^{2}+(\sin({\widehat {\alpha }}_{mp})N)^{2}}}={\sqrt {{\frac {1}{3}}M^{2}+{\frac {2}{3}}N^{2}}},{\color {white}.}\\&={\sqrt {{\frac {1}{3Qr^{2}}}{\Bigg (}{\bigg (}{\frac {ab}{Qr}}{\bigg )}^{4}+2a^{4}{\Bigg )}}}={\frac {a^{2}}{Qr}}{\sqrt {{\frac {1}{3}}{\Bigg (}{\bigg (}{\frac {b}{Qr}}{\bigg )}^{4}+2{\Bigg )}}},\\&=6372.7975377213639884;{\color {white}|}\end{aligned}}\,\!}$ Therefore, Qr actually plays a role in the composition of the midpoint arcradius and even in approximating the average meridional (arc)radius (6367.449)! This appears to be, in fact, the midpoint value of THE general radius of curvature, not “Euler's radius of curvature”, which is usually quoted as such! Radii of curvature ${\displaystyle {}_{\color {white}.}{\widehat {\alpha }}_{sf}=\scriptstyle \arcsin \left(\cos(\phi _{f}){\tfrac {\sin(\Delta \lambda )}{\sin(\Delta {\widehat {\sigma }})}}\right);{}_{\color {white}.}\,\!}$ ("spherical" (? geographical?) azimuth from P_s to P_f) ${\displaystyle {}_{\color {white}.}{\tilde {\alpha }}_{sf}=\scriptstyle \arcsin \left(\cos(\beta _{f}){\tfrac {\sin(\Delta \omega )}{\sin(\Delta {\tilde {\sigma }})}}\right);{}_{\color {white}.}\,\!}$ (geodetic azimuth from P_s to P_f) ${\displaystyle {\begin{aligned}{}_{\color {white}x}{\tilde {\alpha }}({\widehat {\alpha }}_{p})&=\scriptstyle \arctan \left({\tfrac {\tan({\widehat {\alpha }}_{p})}{s'(\phi _{p})}}\right),{\color {white}.}\\&=\lim _{\Delta {\tilde {\sigma }}\to 0}{\tilde {\alpha }}_{p};{}_{\color {white}{\big |}}\\\end{aligned}}\,\!}$ (great-elliptic azimuth at P_p) ${\displaystyle {\color {white}.}\sin(\phi _{p})=\sin(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}_{p}\})=\cos({\widehat {\mathrm {A} }})\sin({\widehat {\sigma }}_{p});\;\;\tan(\lambda _{p})=\sin({\widehat {\mathrm {A} }})\tan({\widehat {\sigma }}_{p})+{\mathcal {C}};{}_{\color {white}.}\,\!}$ ${\displaystyle {\color {white}.}\tan(\beta _{p})=\tan(\beta \{{\tilde {\mathrm {A} }},{\tilde {\sigma }}_{p}\})={\frac {b}{a}}_{\color {white}|}\tan(\phi _{p});\qquad \tan(\omega _{p})=\sin({\tilde {\mathrm {A} }})\tan({\tilde {\sigma }}_{p})+{\mathcal {C}};{}_{\color {white}.}\,\!}$ (where ${\displaystyle {\color {white}_{.}}{\mathcal {C}}^{\color {white}i}\,\!}$ is the Constant of integration)   ${\displaystyle {}_{\color {white}.}\sin({\widehat {\sigma }}_{p})=\sec({\widehat {\mathrm {A} }})\sin(\phi _{p});\quad \sin({\widehat {\mathrm {A} }})=\cos(\phi _{v})=\cos(\phi _{p}){\big |}\sin({\widehat {\alpha }}_{p}){\big |};{}_{\color {white}.}\,\!}$   ${\displaystyle {}_{\color {white}.}\sin({\tilde {\sigma }}_{p})={\sec({\tilde {\mathrm {A} }})\sin(\beta _{p})};\quad \sin({\tilde {\mathrm {A} }})=\cos(\beta _{v})=\cos(\beta _{p}){\big |}\sin({\tilde {\alpha }}_{p}){\big |};{}_{\color {white}.}\,\!}$ ${\displaystyle {}_{\color {white}.}\cos(\Delta {\widehat {\sigma }})=\sin(\phi _{s})\sin(\phi _{f})+\cos(\phi _{s})\cos(\phi _{f})\cos(\Delta \lambda );{}_{\color {white}.}\,\!}$ ${\displaystyle {}_{\color {white}.}\cos(\Delta {\tilde {\sigma }})=\sin(\beta _{s})\sin(\beta _{f})+\cos(\beta _{s})\cos(\beta _{f})\cos(\Delta \omega );{}_{\color {white}.}\,\!}$ ${\displaystyle {\begin{aligned}&\mathbf {\overline {\underline {|\;Principal\;Radii\;Integrands\;|}}} ^{\color {white}|}\\{\color {white}.}M=M(\phi )=M\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}&={\frac {(ab)^{2}}{{\Big (}{\big (}a\cos(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}){\big )}^{2}+{\big (}b\sin(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}){\big )}^{2}{\Big )}^{1.5}}};{}_{\color {white}.}\\N=N(\phi )=N\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}&={\frac {a^{2}}{\sqrt {{\big (}a\cos(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}){\big )}^{2}+{\big (}b\sin(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}){\big )}^{2}}}};\\m'={\frac {M^{\color {white}{\big |}}}{a_{\color {white}{\big |}}}};\quad n'&={\frac {N}{a}}={\frac {b}{a^{2}}}S;\quad {\mbox{(unit integrands)}};\\S=S(\beta )=S\{{\tilde {\mathrm {A} }},{\tilde {\sigma }}\}&={\sqrt {{\big (}a\sin(\beta \{{\tilde {\mathrm {A} }},{\tilde {\sigma }}\}){\big )}^{2}+{\big (}b\cos(\beta \{{\tilde {\mathrm {A} }},{\tilde {\sigma }}\}){\big )}^{2}}};\\&={\frac {ab}{\sqrt {{\big (}a\cos(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}){\big )}^{2}+{\big (}b\sin(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}){\big )}^{2}}}};{}_{\color {white}.}\\&=a\cdot C'\{{\tilde {\mathrm {A} }},{\tilde {\sigma }}\}=a\cdot s'\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\};_{\color {white}{\big |}}\\\end{aligned}}\,\!}$ ${\displaystyle {\begin{aligned}&\qquad \mathbf {\overline {\underline {|\;Relationship\;of\;M\;and\;S\;|}}} ^{\color {white}|}\\\beta '(\phi )&={\frac {1}{\phi '(\beta )}}_{\color {white}X}=\arctan '\left({\frac {b}{a}}\tan(\phi )\right)^{\color {white}|}=n'(\phi )s'(\phi )={\sqrt {m'(\phi )n'(\phi )}};{\color {white}.}\\{\color {white}X}m'(\phi )&=C'(\beta (\phi ))\beta '(\phi )=s'(\phi )n'(\phi )s'(\phi )=n'(\phi )s'^{2}(\phi );\\s'(\phi )&=m'(\phi (\beta ))\phi '(\beta );\\&DxM=\int _{\phi _{s}}^{\phi _{f}}M(\phi )d\phi =\int _{\beta _{s}}^{\beta _{f}}S(\beta )d\beta$ ;\end{aligned}}\,\!} By changing ${\displaystyle \scriptstyle {S(\beta (\phi ))\beta '(\phi )}\,\!}$ to ${\displaystyle \scriptstyle {S\{{\tilde {\mathrm {A} }}({\widehat {\sigma }}),{\tilde {\sigma }}({\widehat {\sigma }})\}{\tilde {\sigma }}'({\widehat {\sigma }})}\,\!}$, the arcradius (“geographic radius of curvature”, “geodetic radius of curvature”, “great-elliptic radius of curvature” or “great arc radius of curvature”——I’m not sure what the formal “radius of curvature” designation would be) can be isolated. ${\displaystyle {\begin{aligned}&\qquad \qquad \qquad \qquad \qquad \qquad \mathbf {\overline {\underline {|\;Arcradius\;|}}} ^{\color {white}|}\\&^{\color {white}{\big |}}\qquad \;g'\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}=C'\{{\tilde {\mathrm {A} }}({\widehat {\sigma }}),{\tilde {\sigma }}({\widehat {\sigma }})\}{\tilde {\sigma }}'({\widehat {\sigma }})=s'\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}{\tilde {\sigma }}'({\widehat {\sigma }}),\\&\qquad \qquad \qquad \,={\sqrt {{\big (}m'(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\})\cos({\widehat {\alpha }}){\big )}^{2}+{\big (}n'(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\})\sin({\widehat {\alpha }}){\big )}^{2}}},\\&\qquad \qquad \qquad ={\sqrt {\frac {{\big (}m'\{{\widehat {A}},{\widehat {\sigma }}\}\cos({\widehat {A}})\cos({\widehat {\sigma }}){\big )}^{2}+{\big (}n'\{{\widehat {A}},{\widehat {\sigma }}\}\sin({\widehat {A}}){\big )}^{2}}{{\big (}\cos({\widehat {A}})\cos({\widehat {\sigma }}){\big )}^{2}+\sin ^{2}({\widehat {A}})}}};\\{\color {white}.}G&=\;\;G\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}\;\;={\sqrt {\frac {{\big (}M\{{\widehat {A}},{\widehat {\sigma }}\}\cos({\widehat {A}})\cos({\widehat {\sigma }}){\big )}^{2}+{\big (}N\{{\widehat {A}},{\widehat {\sigma }}\}\sin({\widehat {A}}){\big )}^{2}}{{\big (}\cos({\widehat {A}})\cos({\widehat {\sigma }}){\big )}^{2}+\sin ^{2}({\widehat {A}})}}}^{\color {white}{\Big |}}\!\!,\\&\qquad \qquad \qquad ={\sqrt {M^{2}\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}+{\Big (}{\tfrac {\sin({\widehat {\mathrm {A} }})}{\cos(\phi \{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\})}}{\Big )}^{2}{\Big (}N^{2}\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}-M^{2}\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}{\Big )}}},{\color {white}.}\\&=\;\;G({\widehat {\alpha }},\phi )\;\;\;\,\!={\sqrt {{\big (}M(\phi )\cos({\widehat {\alpha }}){\big )}^{2}+{\big (}N(\phi )\sin({\widehat {\alpha }}){\big )}^{2}}},\\&=G({\tilde {\alpha }}({\widehat {\alpha }}),\phi )={\frac {1}{\sqrt {\left({\frac {\cos({\tilde {\alpha }}({\widehat {\alpha }}))}{M(\phi )}}\right)^{2}+\left({\frac {\sin({\tilde {\alpha }}({\widehat {\alpha }}))}{N(\phi )}}\right)^{2}}}};\\&{\color {white}.}\qquad \qquad \qquad \qquad \quad \scriptstyle {\neq }\;{\tfrac {1}{{\tfrac {\cos ^{2}({\tilde {\alpha }})}{M(\phi )}}+{\tfrac {\sin ^{2}({\tilde {\alpha }})}{N(\phi )}}}};\\&\qquad \qquad \qquad \qquad \quad ^{\scriptstyle {\mathrm {(Euler's\;radius\;of\;curvature)} }}\\&\quad DxG=\int _{{\widehat {\sigma }}_{s}}^{{\widehat {\sigma }}_{f}}G\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}d{\widehat {\sigma }};\quad DxS=\int _{{\tilde {\sigma }}_{s}}^{{\tilde {\sigma }}_{f}}S\{{\tilde {\mathrm {A} }},{\tilde {\sigma }}\}d{\tilde {\sigma }};\\&\quad _{\color {white}{\big |}}\lim _{\Delta {\widehat {\sigma }}\to 0}DxG=\lim _{\Delta {\tilde {\sigma }}\to 0}DxS\!\!:\;\;G=\lim _{\Delta {\widehat {\sigma }}\to 0}{\frac {DxG}{\Delta {\widehat {\sigma }}}}=\lim _{\Delta {\widehat {\sigma }}\to 0}{\frac {DxS}{\Delta {\widehat {\sigma }}}};\end{aligned}}\,\!}$ It is with this arcradius, G, that the great-ellipse distance, DxG, can be found. Try calculating miniscule distances (e.g., about ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}\leq .1^{10}\,\!}$) and dividing by ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}\,\!}$: You should see G equaling the result. Now, rather than calculating random distances and dividing by its ${\displaystyle \scriptstyle \Delta {\widehat {\sigma }}\,\!}$ to find the average arcradius, you can find the actual arcradius for any single point along an arc path and, thereby, systematically find the actual average great-elliptic arcradius: ${\displaystyle {\begin{aligned}{\overset {\,_{\smile }}{\!R}}({\widehat {\mathrm {A} }})&=\int _{0}^{90^{\circ ^{\color {white}|}}}\!\!G\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}d{\widehat {\sigma }};\\Gr&=\int _{0_{\color {white}|}}^{90^{\circ }}{\overset {\,_{\smile }}{\!R}}({\widehat {\mathrm {A} }})d{\widehat {\mathrm {A} }}=\int _{0}^{90^{\circ }}\int _{0}^{90^{\circ }}G\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}d{\widehat {\mathrm {A} }}d{\widehat {\sigma }};{\color {white}.}\end{aligned}}\,\!}$  ~Kaimbridge~ (talk) 19:47, 24 June 2013 (UTC) (1) Your construction for the midpoint treats one of the vertices specially. You start from (lat,lon) = (0,0) and get (30,35.26) for the midpoint. However if I follow your reasoning and start from (0,90), I get (30,54.74); similarly starting from (90,0) gets me (45,45). For this reason, I don't find your definition of midpoint useful. The specific problem is your taking sigma_mp = 45; there's no reason "why" the midpoint should be half way between the vertex and the opposite edge. (2) I followed your construction for your definition of the average great-circle radius. It is also not a useful one, chiefly because it doesn't sample great-circles fairly. However, let me illustrate the problems in your terms by using the triaxial ellipsoid (semi-axes a, b, c parallel to the x, y, z axes as in your figure). If I carried out exactly your construction pivoting about the point (a,0,0), I would end up with (2a+b+c)/4 (to first order). But why should I favor (a,0,0) as the starting point instead of (0,b,0) (which gives me (2b+a+c)/4) or (0,0,c) (which gives me (2c+a+b)/4)? You can't appeal to your assertion that "only the transverse graticule is able to connect any two points", since I haven't yet said whether the equator is the (a,b) ellipse, the (a,c) ellipse, or the (b,c) ellipse. So clearly your definition of average great-circle radius has the problem that it doesn't generalize to a triaxial ellipsoid cleanly. (3) Regarding your last point about averaging using the perimeter of full great ellipses, I've already made the point that this yields (a+b+c)/3 as the mean radius. I think we've exhausted the discussion for now. If all you're interested in is measuring distances, there no reason to use (3a+b)/4 as the spherical radius; so I will update the main page along the lines I indicated earlier. cffk (talk) 21:10, 24 June 2013 (UTC) (1) For what reason don’t you find my definition of midpoint useful? It highlights it perfectly: φs,λs; φf,λf ${\displaystyle \scriptstyle {{\widehat {\mathrm {A} }};\;\;{\widehat {\sigma }}_{s},{\widehat {\sigma }}_{f}}\;\;(\Delta {\widehat {\sigma }})\,\!}$ DxS ${\displaystyle \scriptstyle {DxS/\Delta {\widehat {\sigma }}}\,\!}$ 0, 0; 30,35.264389... 45; 0,45 (45) 4997.19274 6362.6237094 30,35.264389...; 45,90 45;45,90 (45) 5013.19904 6383.0049024 or, from the other direction, 0,90; 30,54.735610... 45; 0,45 (45) 4997.19274 6362.6237094 30,54.735610...; 45,0 45;45,90 (45) 5013.19904 6383.0049024 ∑=10010.39178 .5∑=6372.8143059 0, 0; 0,90 45; 0,90 (90) 10010.38624 6372.8143059 half-π×6372.8143059 = 10010.39330 half-π×6372.8 = 10010.37083 half-π×6371.0 = 10007.54340 half-π×6367.449 = 10001.96550 half-π×6378.137 = 10018.75417 half-π×(6367.449+6378.137)/2 = 10010.35984 Yes, the midpoint should be halfway between the great circle’s equatorial vertex (its node) and transverse equator (vertex latitude). See down in (3). (2) No, you are doing it wrong: You are saying the approximation could be either “(2a+b+c)/4” or just as likely “(a+2b+c)/4” or “(a+b+2c)/4”. First of all, yes, the latter one doesn’t sample great circles evenly, as it just averages the different north-south conjugate meridians, “((a+c)/2 + (b+c)/2)/2”, so that one is out. As for your two, (a,0,0) and (0,b,0), possibilities, it is not either-or, but the average of the two: ((2a+b+c)/4 + (a+2b+c)/4)/2, or (3a+3b+2c)/8 for the approximation. However, as I presented earlier, I believe there is a third, geometric mean of a and b for surface/arc calculations, which is used for the meridional calculations, so the actual above approximation should be ((a+b) + ((ab)^.5+c)))/4, both ways reducing spheroidally to (3a+b)/4. (3) No, averaging the two principal meridians and equator approximates an average surface area radius, which is the more appropriate choice for loxodromical calculations, not for an average of orthodromical circumferences. Okay, let’s look at it that way. Take the string and globe and try stretching it along different latitudes: Does the string stay on the latitudes? No, it arcs towards the poles, as the string traces the orthodromic, not loxodromic, path. As the string length (i.e., ∆λ) increases from 0 to 180°, the arc moves closer to the pole, until it is a north-south arc. Throughout this stretching, the string midpoint (at .5∆λ) is at the transverse equator (vertex latitude) of the circumference arc path that the string tracing occupies at that moment: That midpoint is the midpoint of the whole northern (or southern) facing hemisphere, with the midpoint between the equator (the great circle’s node) and the transverse equator (its vertex latitude) being the midpoint of the quarter-circumference. So, the transverse equator/vertex latitude mid-circumference/great circle of the hemisphere (i.e.,  = φ_v = 45°) is at φ = 45° and, since the left and right halves of the hemisphere are symmetrical (just like the northern and southern hemispheres are to each other), then the midpoint of the unique segment of a hemisphere (i.e., its quadrant) is on  = φ_v = 45°, at φ = 30°. What about the averaging of M and N at my designated midpoint at φ = 30°? Both the simple and quadratic means of M and N approximate the meridional midpoint of (which should be at φ = 45°) and whole average of (6367.4...)! Also, the fact that my Qr is a component of the M and N midpoint valuations (and, therefore, G’s)? Overall, could it be I’m just not being emphatic and specific enough in definition? Namely, that one of the first steps in calculating geodetic distance is determining “azimuth of the geodesic at the equator”(common definition)——i.e., arc path or vertex latitude! This means WHAT? It means that all——I repeat, ALL——distances (and, therefore, all the great circles that the distances are calculated on), of any and all lengths are EXCLUSIVELY based on a transverse vertex, meaning that a “great circle average” should be based on the vertical and horizontal great circle circumferences that the transverse vertex bounds, namely the central meridian and equator.  ~Kaimbridge~ (talk) 05:17, 26 June 2013 (UTC) While I did not analyze it with the rigor demonstrated by Cffk, I also came to the conclusion that the "ellipsoidal quadratic mean" is of questionable utility: Talk:Spherical_Earth/Archive_1#Formula_for_ellipsoidal_quadratic_mean - Ac44ck (talk) 05:48, 26 June 2013 (UTC) Kaimbridge, I think one of the reasons the discussions have gone around in circles is that you don't adequately address the criticisms of your work. Let me give two examples: (1) Your result doesn't satisfy elementary symmetries. For a nearly spherical ellipsoid (axes a,b,c), it shouldn't matter with of the principal ellipses is picked as the equator. With your method it does. You owe us an explanation (preferably a simple one). (2) I understand your assertion about all great circles intersecting the equator. But you're never addressed why you should sample equatorial azimuths uniformly in [0,90] when, for randomly chosen great circles, it's clear that the distribution is not uniform but lambertian. Ac44ck, thanks to the link to the earlier discussion. The fact that these discussions are so voluminous and repetitious is confirmation that Wikipedia shouldn't be used for original research. cffk (talk) 13:26, 26 June 2013 (UTC) Cffk:By now, you seem to understand my goal, but just have a problem with the equator being picked at a single point. Is your objection strictly with the concept of an “X, Y, Z ellipsoid”? In terms of the general concept, what if the mean of the meridian and equator was defined as an approximation of the average circumference of an “x, y spheroid” (of course not as the formal definition)? Would that make the meridian/equator mean, average of “differently angled circumferences” (measured at the equator) and φ_mp = 30° quadrant midpoint properties conceptually valid and acceptable?  ~Kaimbridge~ (talk) 17:36, 26 June 2013 (UTC) I don't necessarily have a problem with starting all your ellipses from a particular point on the equator. I just have a problem with using this as a starting point for the solution. A satisfactory way of solving the problem "what is the mean circumference of great ellipses on an ellipsoid?" would be to average the circumference over all possible mean ellipses. In the special case of an ellipsoid of revolution, you can show this is equivalent to averaging only over the great ellipses which intersect lat,lon = 0,0 with the equatorial azimuth averaged over a lambertian distribution. This doesn't work for a general (nearly spherical) ellipsoid; but there's a simpler average you can do in this case: average over the three principal ellipses (this works for an ellipsoid of revolution too). But in both cases, you have to start with an average over all great ellipses and then show that sampling over a subset of great ellipses yields the same mean. cffk (talk) 18:09, 26 June 2013 (UTC) Okay, let’s pause there. You are saying (in the case of a spheroid) it is legitimate if “averaged over a lambertian distribution”. Do you see the need for Lambertian distribution in the case of the meridional ellipse? If so, then how can it be found via the elliptic integral of the second kind? If you look at the concentric rings in this bullseye, other than right near the middle, the height of the different rings (which are "${\displaystyle \sigma {\color {white}^{.}}\,\!}$"s, in the case of ellipsoids) are the same throughout, from north-south down to east-west. 19:53, 26 June 2013 (UTC) (1) I don't understand your question. My starting point is a straightforward average over all great ellipses. For an ellipsoid of revolution this is equivalent to a lambertian average over ellipses emanating from an equatorial point. I don't see where the meridional ellipse comes into what I said. (2) I'm not sure what your point is with this figure. I hope this doesn't sound impertinent... But have you tried numerically averaging the circumferences of randomly selected great ellipses? Here "randomly" would mean that the normal to the great ellipse is uniformly sampled from a unit sphere. It would help to get a yes or no answer to this question (and if yes, what you got for the average). It would also help to know whether you think this average is relevant to the present discussion (I think it is). cffk (talk) 20:36, 26 June 2013 (UTC) YES!!!!! That is what I’ve been talking about all along!!! I’m referring to an orderly finding of the different ellipses, thus the 'integral 2nd kind finds the north-south ellipse, which ${\displaystyle \scriptstyle {\int _{0}^{90}G\{0,{\widehat {\sigma }}\}d{\widehat {\sigma }}}\,\!}$ would also find. Now find all of the other ${\displaystyle \scriptstyle {\int _{0}^{90}G\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}d{\widehat {\sigma }}}\,\!}$s and average them together. If not this, then what ellipse-perimeter (not surface area) type definite integral from 0->90° would you use?  ~Kaimbridge~ (talk) 20:51, 26 June 2013 (UTC) (1) "Yes" to which question? Did you do the numerical average as I indicated? What was your answer? Can you provide pseudo-code for what you did. (2) You say "orderly" and I say "show me". Your problem is that you've guessed at a way of doing the average without proving that your "orderly" method is equivalent to averaging over all the great ellipses. cffk (talk) 21:02, 26 June 2013 (UTC) Here's the average done analytically. The mean radius of great ellipses is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \left<R\right> = \frac1{4\pi} \int_{-\pi/2}^{\pi/2} \cos\phi'\, d\phi' \int_{-\pi}^{\pi} d\lambda'\, R(\phi',\lambda'),} where ${\displaystyle R(\phi ',\lambda ')}$ is the equivalent radius of a great ellipse lying in the plane normal to ${\displaystyle [\cos \phi '\cos \lambda ',\cos \phi '\sin \lambda ',\sin \phi ']}$. The equatorial azimuth for this great ellipse is ${\displaystyle \alpha _{0}=\left|\phi '\right|}$ and its vertex latitude is ${\displaystyle \phi _{0}={\frac {1}{2}}\pi -\alpha _{0}}$. This great ellipse has semi-axes ${\displaystyle a}$ and ${\displaystyle a\cos ^{2}\phi _{0}+b\sin ^{2}\phi _{0}}$ (to first order in the flattening) and thus its effective radius is ${\displaystyle R(\phi ',\lambda ')={\frac {1}{2}}(a+a\sin ^{2}\phi '+b\cos ^{2}\phi ')}$. Plugging this into the expression for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \left<R\right>} and doing the double integral gives Failed to parse (syntax error): {\displaystyle \left<R\right> \approx \frac13(2a + b)} . For a general ellipsoid with semi-axes ${\displaystyle a_{x},a_{y},b}$, we have ${\displaystyle R(\phi ',\lambda ')={\frac {1}{2}}((a_{x}+a_{y})\sin ^{2}\phi '+((a_{x}+b)\sin ^{2}\lambda '+(a_{y}+b)\cos ^{2}\lambda ')\cos ^{2}\phi '),}$ giving Failed to parse (syntax error): {\displaystyle \left<R\right> \approx \frac13(a_x + a_y + b)} . The averaging that you're doing, Kaimbridge, is ${\displaystyle {\frac {1}{2\pi ^{2}}}\int _{-\pi /2}^{\pi /2}d\phi '\int _{-\pi }^{\pi }d\lambda '\,R(\phi ',\lambda '),}$ which, of course, does not correspond to a uniform average. I hope this helps. cffk (talk) 11:32, 27 June 2013 (UTC) I’m not sure if your ${\displaystyle \scriptstyle {R(\phi ',\lambda ')}\,\!}$ is right: It appears that ${\displaystyle \scriptstyle {\phi '={\widehat {\mathrm {A} }}}\,\!}$, but what about ${\displaystyle \scriptstyle {\lambda '}\,\!}$? Is that supposed to be ${\displaystyle \scriptstyle {\widehat {\sigma }}\,\!}$, and if so (and even if not), where is it used in the spheroidal ${\displaystyle \scriptstyle {R(\phi ',\lambda ')}\,\!}$ integrand? Regardless, addressing point (1) above (“Did you do the numerical average...”): Yes, that was the question I was referring to——that is how I come to my conclusions. The only part of the DxG approximation on the lower right needed for this calculation is “6378.137-cos(Â)^2*10.687854-0.085243*sin(Â)^2”, as all of the “S(ń)C” terms set to 0. Applying it to this ${\displaystyle {\begin{aligned}Gq({\widehat {\mathrm {A} }})&=\int _{0}^{90}G\{{\widehat {\mathrm {A} }},{\widehat {\sigma }}\}{\frac {d{\widehat {\sigma }}}{\Delta {\widehat {\sigma }}}};\\Gr&=\int _{0}^{90}Gq({\widehat {\mathrm {A} }}){\frac {d{\widehat {\mathrm {A} }}}{\Delta {\widehat {\mathrm {A} }}}},\\&\approx \sum _{SS=3}^{UL}{\Bigg (}\sum _{TN=1}^{SS}{\frac {Gq{\Big (}{\tfrac {TN-1}{SS-1}}\cdot 90{\Big )}}{SS-1}}{\Bigg )}-{\frac {1}{2}}{\frac {Gq(0)+Gq(90)}{SS-1}};\end{aligned}}\,\!}$ The maximum error for this approximation formula is at about ±.93 cm @ 100 km, up to ±11.7 cm @ ≥10,000 km (located on  ≈ 62.6° and 27.4°)——keeping in mind that this is measured against the actual great-elliptic distance (“DxG”), not the geodetic distance (“DxS”). The program code is quite simple for this project: 10 Rf=atan(1)/45 20 UL=10:For SS=3 To UL 25 AD = 90:AP_s=0 30 Gr_s=0:V=0:For TN=1 To SS:AP=AP_s+(TN-1)/(SS-1)*AD 40 Gr_s=Gr_s+fnGq_a(AP)/SS:V=V+fnGq_a(AP)/(SS-1):Next TN 45 Gr_t=V-.5*(fnGq_a(AP_s)+fnGq_a(AD))/(SS-1) 50 Print SS,Decimal(,7),Gr_s,Gr_t:Next SS 99 End 100 fnGq_a(AP):Return(6378.137-cos(AP*Rf)^2*(10.687854-0.085243*sin(AP*rf)^2)) This routine uses both the simple (“Gr_s”) and trapezoidal (“Gr_t”) averages for comparison. In my normal calculations, I use (-2,+2) Gaussian Quadrature for integration. If the average arcradii of evenly spaced great ellipses (all sharing the same equatorial vertex) are averaged together, it is clear——immediately with the trapezoidal averaging and eventually with the simple averaging——that the average arcradius of all of the different Gq values is about 6372.8037284. Does it not? SS Â_1, Â_2, ..., Â_ss: Gr_s Gr_t 1 45: 6372.8143838 ---- 2 90, 0: 6372.7930730 ---- 3 90, 45, 0: 6372.8001766 6372.8037284 4 90, 60, 30, 0: 6372.8010645 6372.8037284 5 90, 67.5, 45, 22.5,0: 6372.8015973 6372.8037284 If lines 20 and 50 are changed to 20 For XP=1 To 10:SS=10^XP+1 50 Print SS,Decimal(,7),Gr_s,Gr_t:Next XP you can see that Gr_s doesn’t converge until after 1,000,000! 11 90, 81,..., 9,0: 6372.8027597 6372.8037284 101 90, 89.1,..., .9,0: 6372.8036229 6372.8037284 1001 90, 89.91,..., .09,0: 6372.8037177 6372.8037284 10001 90, 89.991,..., .009,0: 6372.8037273 6372.8037284 100001 90, 89.9991,..., .0009,0: 6372.8037283 6372.8037284 1000001 90,89.99991,...,.00009,0: 6372.8037284 6372.8037284 So what are some other results? Keeping in mind that the least significant digit or two of these results may be off due to rounding, let’s first compare the actual average arcradius (“Gq”, quadrant average of G) for the great ellipses of  = 0°, 30°, 45°, 60° and 90°, with the approximated values (∆Gq = Gq_a(Â) - Gq(Â)):  Gq(Â) ∆Gq Gq_a(Â) 0 6367.4491457009 +.0000002991 6367.4491460000 30 6370.1371253278 -.0000327653 6370.1370925625 45 6372.8143837065 +.0000000435 6372.8143837500 60 6375.4809867969 +.0000327656 6375.4810195625 90 6378.137 .0 6378.137 Now average the Gq values of an  and its quadrant complement, which equals 90°-  = φ_v = Â_c, and you will see that their mean ≈ 6372.8 (as ∆Gq above shows, there is an inverse error symmetry that cancels to below .000001, thus Gq_a equals Gq for this comparison): Â, Â_c .5(Gq(Â)+Gq(Â_c)) 0 90 6372.793073 5 85 6372.793716 10 80 6372.795566 15 75 6372.798401 20 70 6372.801878 22.5 67.5 6372.803728 25 65 6372.805579 30 60 6372.809056 35 55 6372.811891 40 50 6372.813741 45 45 6372.814384 If each of these paired averages are averaged with their symmetrical midpath complement (e.g., (Gq(15)+Gq(30)+(Gq(60)+Gq(75))/4), they all average to 6372.8037285! If you divide the quadrant into three ( = 0->30°, 30°->60°, 60°->90°), you get averages of 6368.3799304, 6372.8125403, 6377.218715, respectively. This can be seen by changing line 25 to 25 AD = 30:AP_s=... 0,30,60 If you find the simple average of the above three trisector averages, you get 6372.8037286, while the trapezoidal average gives 6372.8059315. Likewise, by changing “AD” to 15 and “AP_s” increment to 15° from 0 to 75, the six averages of 6367.6918421, 6369.0680187, 6371.4451757, 6374.1799049, 6376.5394380 and 6377.8979907 give a simple average of 6372.8037283 and a trapezoidal of 6372.8054907. Thus trapezoidal averaging does not work here, as the sectional averages are actual, not approximations, meaning that anything but the simple average will be wrong (though, as these two sectorings show, the error decreases as the number of sectors increases)! This sector averaging can be broken down into smaller and smaller sectors (with AP_s’s increment equaling AD), until AD is infinitesimality and it becomes the Gq integration! This should further reinforce the idea that “6372.803728” is the average great-ellipse radius! If you still don’t think so, then what do you think this “6372.803728” value represents, keeping in mind that we are talking about a spheroid where the nodes of all of the great ellipses comprising this spheroid intersect at a single, equatorial vertex? It certainly is an average of something!  ~Kaimbridge~ (talk) 19:51, 28 June 2013 (UTC) The variables ${\displaystyle \phi ',\lambda '}$ in ${\displaystyle R(\phi ',\lambda ')}$ specify the orientation of the great ellipse; it lies in the plane perpendicular perpendicular to ${\displaystyle [\cos \phi '\cos \lambda ',\cos \phi '\sin \lambda ',\sin \phi ']}$. It happens that, for an ellipsoid of revolution, ${\displaystyle \alpha _{0}=\phi '}$ and thus ${\displaystyle R(\phi ',\lambda ')}$ is independent of ${\displaystyle \lambda '}$ allowing the ${\displaystyle \lambda '}$ integral to be done trivially. Your numerical integration is not over all great ellipses, but only those intersecting the point ${\displaystyle \phi _{0}=0}$, ${\displaystyle \lambda _{0}=0}$, and that is what ${\displaystyle (3a+b)/4}$ represents. What you need to do is to continue your averaging to include all points on the ellipsoid. (Surely this is consistent with your wish to average over all the "differently angled ellipses".) The average radius of great ellipses through a point ${\displaystyle \phi _{0},\lambda _{0}}$ is ${\displaystyle R_{0}(\phi _{0},\lambda _{0})={\frac {1}{2}}(a+b)+{\frac {1}{4}}(a-b)\cos ^{2}\phi _{0}.}$ Check: ${\displaystyle R_{0}(0,\lambda _{0})={\frac {1}{4}}(3a+b)}$ and ${\displaystyle R_{0}({\frac {1}{2}}\pi ,\lambda _{0})={\frac {1}{2}}(a+b)}$. Averaging this over all ${\displaystyle \phi _{0},\lambda _{0}}$ gives ${\displaystyle {\frac {1}{4\pi }}\int _{-\pi /2}^{\pi /2}\cos \phi _{0}\,d\phi _{0}\int _{-\pi }^{\pi }d\lambda _{0}\,R_{0}(\phi _{0},\lambda _{0})={\frac {1}{3}}(2a+b),}$ in agreement with the earlier result. cffk (talk) 20:42, 28 June 2013 (UTC) Ah, so you do know what I am talking about! P=) Okay, so yes, but the flaw with your approach is, each great circle has a vertex, right? So each ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$ has an infinite number of great circles crossing that vertex, but only one that crosses it horizontally, at a 90° azimuth: All of the other great circles intersecting that vertex at different angles have their own ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$ they cross at 90°. So to include each “differently angled ellipse” only once, let ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$ = Gq(90°-φ') = Gq(Â) (thus ${\displaystyle R_{0}(0,\lambda ')\,\!}$ -> Gq(90°), ${\displaystyle R_{0}(45^{\circ },\lambda ')\,\!}$ -> Gq(45°) and ${\displaystyle R_{0}(90^{\circ },\lambda ')\,\!}$ -> Gq(0) for the 3 term average)! More tomorrow.  ~Kaimbridge~ (talk) 00:13, 29 June 2013 (UTC) You may have misunderstood my notation... I define ${\displaystyle R_{0}(\phi _{0},\lambda _{0})}$ as follows: consider all the great ellipses starting at ${\displaystyle \phi _{0},\lambda _{0}}$ and assume that the azimuth at the starting point ${\displaystyle \alpha }$ is uniformly distributed in ${\displaystyle [-\pi ,\pi )}$; then ${\displaystyle R_{0}(\phi _{0},\lambda _{0})}$ is the mean radius of these great ellipses. With this definition, your sentence beginning "So each ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$ has an infinite number of great circles crossing that vertex ..." makes no sense. The vertex of a particular ellipse is, of course, ${\displaystyle \cos ^{-1}(\cos \phi _{0}\sin \alpha )}$. Incidently, ${\displaystyle R_{0}({\frac {1}{4}}\pi ,\lambda _{0})={\frac {1}{8}}(5a+3b)}$; if you're getting a different value, you have misundertood the definition. You seem to think that it's important to include each ellipse only once. But this is not so; in a problem like this, where you're trying to get the averaging right, it might be easier not to make this a requirement. So the averaging I'm talking about with ${\displaystyle R_{0}}$ is over ${\displaystyle \phi _{0}}$, ${\displaystyle \lambda _{0}}$, and ${\displaystyle \alpha }$. Incidently the first average that I discussed, with ${\displaystyle R(\phi ',\lambda ')}$, includes each great ellipse just once. In contrast, your averaging is over only a tiny subset of the great ellipses (which explains why you got the wrong answer).

To follow up on an earlier question I asked you... What is your mean radius for a triaxial ellipsoid with (approximately equal) semi-axes ${\displaystyle a,b,c}$? 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 ${\displaystyle R_{0}\,\!}$ and do agree with your results as you define them (just to make sure its valuation is clear, ${\displaystyle R_{0}\,\!}$ 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 ${\displaystyle R_{0}(0,\lambda ')\,\!}$ 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 ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$ and φ_v at the great circle’s crossing of its transverse equator. What I was referring to by “So each ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$ 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 ${\displaystyle R_{0}(0,\lambda ')\,\!}$. So why would you consider any other ${\displaystyle R_{0}(\phi ',\lambda ')\,\!}$, which doesn’t have a full set of “differently angled great circle/ellipse” (in the extreme case, ${\displaystyle R_{0}(90^{\circ },\lambda ')\,\!}$ only contains one great circle/ellipse angle——north-south!)?

Try this: Imagine attaching a marker to ${\displaystyle R_{0}(0,\lambda ')\,\!}$ 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 ${\displaystyle a,b,c}$? In the limit where the semi-axes are nearly equal I get ${\displaystyle {\frac {1}{3}}(a+b+c)}$ 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 ${\displaystyle a_{x},a_{y},b\,\!}$ and——for this stage of the triaxial discussion——let there be only one Cartesian set, ${\displaystyle X,Y,Z\,\!}$, where

${\displaystyle {\begin{aligned}X&=a_{x}\cos(\phi )\cos(\lambda );\qquad \;a(\lambda )={\sqrt {{\big (}a_{x}\cos(\lambda ){\big )}^{2}+{\big (}a_{y}\sin(\lambda ){\big )}^{2}}};^{\color {white}{\Big |}}\\Y&=a_{y}\cos(\phi )\sin(\lambda );\qquad \;x(\lambda )={\sqrt {X^{2}+Y^{2}}}=a(\lambda )\cos(\phi );\\Z&=b\sin(\phi );\qquad \qquad \qquad \quad \,y=\qquad \;Z\quad \;\;\,=b\sin(\phi );\\&\qquad \quad \;R(\lambda )={\sqrt {X^{2}+Y^{2}+Z^{2}}}={\sqrt {x^{2}(\lambda )+y^{2}}};\end{aligned}}\,\!}$

${\displaystyle {\begin{aligned}R_{0:av1}&={\frac {1}{4}}{\Big (}3a(45^{\circ })+b{\Big )};\\R_{0:av2}&={\frac {1}{4}}{\Big (}a(0)+a(45^{\circ })+a(90^{\circ })+b{\Big )};\\R_{0:av3}&={\frac {2}{\pi }}\int _{0}^{90^{\circ }}R_{(0,\lambda ,\lambda _{0})}d\lambda ={\frac {2}{\pi }}\int _{0}^{90^{\circ }}{\frac {1}{4}}{\Big (}3a(\lambda )+b{\Big )}d\lambda$ ;\\Qr_{av1}&={\sqrt {{\frac {1}{4}}{\Big (}3a^{2}(45^{\circ })+b^{2}{\Big )}}}={\sqrt {{\frac {1}{4}}{\Big (}a^{2}(0)+a^{2}(45^{\circ })+a^{2}(90^{\circ })+b^{2}{\Big )}}};{\color {white}.}\\Qr_{av2}&={\sqrt {{\frac {1}{4}}{\Big (}a_{x}^{2}+a_{x}a_{y}+a_{y}^{2}+b^{2}{\Big )}}};\\Qr_{av3}&={\frac {1}{\pi }}\int _{0}^{90^{\circ }}{\sqrt {{\frac {1}{4}}{\Big (}3a^{2}(\lambda )+b^{2}{\Big )}}}d\lambda ;\end{aligned}}\,\!}

Applying these to an Earthy scalene ellipsoid,

${\displaystyle {\begin{aligned}a_{x}=6383.1004;\quad \;a_{y}=6373.&1707;\quad \;b=6356.752;{\color {white}{\bigg |}}\\a(45^{\circ })={\sqrt {{\frac {a^{2}(0)}{3}}+{\frac {a^{2}(45^{\circ })}{3}}+{\frac {a^{2}(90^{\circ })}{3}}}}&={\sqrt {{\frac {a^{2}(0)}{4}}+{\frac {a^{2}(45^{\circ })}{2}}+{\frac {a^{2}(90^{\circ })}{4}}}},\\={\sqrt {{\frac {2}{\pi }}\int _{0}^{90^{\circ }}a^{2}(\lambda )d\lambda }}&=6378.13748236;\\{\frac {a(0)+a(45^{\circ })+a(90^{\circ })}{3}}&=\mathbf {6378.13619412}$ ;\\{\frac {a(0)}{4}}+{\frac {a(45^{\circ })}{2}}+{\frac {a(90^{\circ })}{4}}&=6378.13651618;\\a_{m}={\sqrt {a_{x}a_{y}}}&=6378.13361764;^{\color {white}{\big |}}\\{\frac {a_{x}+a_{m}+a_{y}}{3}}=6378.13490588;\quad \;&\qquad {\frac {a_{x}}{4}}+{\frac {a_{m}}{2}}+{\frac {a_{y}}{4}}=6378.13458382;{\color {white}{\big .}}\\{\sqrt {\frac {a_{x}^{2}+a_{m}^{2}+a_{y}^{2}}{3}}}=\mathbf {6378.13619412} ;\;&\quad {\sqrt {{\frac {a_{x}^{2}}{4}}+{\frac {a_{m}^{2}}{2}}+{\frac {a_{y}^{2}}{4}}}}=6378.13555000;_{\color {white}{\Big |}}\\{\sqrt[{4}]{a_{x}a_{m}a_{y}b}}={\sqrt[{4}]{a_{x}^{1.5}a_{y}^{1.5}b}}&=6372.78148022;\\R_{0:av1}=6372.79111177;\quad &\;Qr_{av1}=6372.79783967;\\R_{0:av2}=6372.79014559;\quad &\;Qr_{av2}=6372.79687268;\\R_{0:av3}=6372.79038714;\quad &\;Qr_{av3}=6372.79729483;_{\color {white}|}\end{aligned}}\,\!}

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: ${\displaystyle R\Delta \theta \,\!}$;
• Spheroid integrand: Create an auxilary sphere; ${\displaystyle R(\beta )\Delta \beta \,\!}$;
• Scalene Ellipsoid integrand: Create an auxiliary sphere from an auxiliary spheroid; ${\displaystyle R(\beta (\lambda ))\Delta \beta (\lambda )\,\!}$;

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 ${\displaystyle {\frac {1}{8}}(3a+3b+2c)}$. 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, ${\displaystyle {\frac {1}{8}}(3a_{x}+3_{y}+2b)}$.

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:

${\displaystyle {\begin{aligned}R(\beta )&={\sqrt {x^{2}+y^{2}}}={\sqrt {(a\cos(\beta ))^{2}+(b\sin(\beta ))^{2}}};^{\color {white}{\big |}}\\S(\beta )&={\sqrt {{\acute {x}}^{2}+{\acute {y}}^{2}}}={\sqrt {(b\cos(\beta ))^{2}+(a\sin(\beta ))^{2}}};\end{aligned}}\,\!}$

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 ${\displaystyle a,b,c}$ for the semi-axes. Why is your result ${\displaystyle {\frac {1}{8}}(3a+3b+2c)}$ not invariant on interchanging the semi-axes (e.g., swapping ${\displaystyle b}$ and ${\displaystyle c}$)? 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:

 ${\displaystyle {\begin{aligned}{\color {white}.}a\cos(\beta )S(\beta )=&\cos(\beta )\scriptstyle {\sqrt {{\Big (}{\big (}ab\cos(\lambda ){\big )}^{2}+{\big (}ab\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}aa\sin(\beta ){\big )}^{2}}}^{\color {white}|},\\&\qquad \qquad \qquad \qquad \downarrow \qquad \qquad \qquad \quad \downarrow \\&\cos(\beta )\scriptstyle {\sqrt {{\Big (}{\big (}a_{y}b\cos(\lambda ){\big )}^{2}+{\big (}a_{x}b\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}a_{x}a_{y}\sin(\beta ){\big )}^{2}}},{\color {white}x}\\=&\cos(\beta )\scriptstyle {\sqrt {{\Big (}{\big (}{\mathfrak {bc}}\cos(\lambda ){\big )}^{2}+{\big (}{\mathfrak {ac}}\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}{\mathfrak {ab}}\sin(\beta ){\big )}^{2}}};\end{aligned}}\,\!}$

Isolating and restructuring the radii into equivalent formations with a common a factor,

${\displaystyle {\begin{aligned}a_{m}&={\sqrt {a_{x}a_{y}}};{\color {white}^{1}}\\a_{y}b&={\sqrt {a_{x}a_{y}}}\,b{\sqrt {\frac {a_{y}}{a_{x}}}}=a_{m}b{\sqrt {\frac {a_{y}}{a_{x}}}}=a_{m}b_{x};{\color {white}.}\\a_{x}b&={\sqrt {a_{x}a_{y}}}\,b{\sqrt {\frac {a_{x}}{a_{y}}}}=a_{m}b{\sqrt {\frac {a_{x}}{a_{y}}}}=a_{m}b_{y};\\{\color {white}.}b(\lambda )&=b{\sqrt {{\frac {a_{y}}{a_{x}}}\cos ^{2}(\lambda )+{\frac {a_{x}}{a_{y}}}\sin ^{2}(\lambda )}};\\&{\text{or, if you insist,}}\\{\mathfrak {c}}(\lambda )&={\mathfrak {c}}_{\color {white}{\big |}}\!\!{\sqrt {{\frac {\mathfrak {b}}{\mathfrak {a}}}\cos ^{2}(\lambda )+{\frac {\mathfrak {a}}{\mathfrak {b}}}\sin ^{2}(\lambda )}};\end{aligned}}\,\!}$

thus

${\displaystyle {\begin{aligned}{\color {white}.}a_{m}\cos(\beta )S(\beta )=&a_{m}\cos(\beta )\scriptstyle {\sqrt {b^{2}{\big (}{\frac {a_{y}}{a_{x}}}\cos ^{2}(\lambda )+{\frac {a_{x}}{a_{y}}}\sin ^{2}(\lambda ){\big )}\cos ^{2}(\beta )+a_{x}a_{y}\sin ^{2}(\beta )}}^{\color {white}|},\\=&a_{m}\cos(\beta )\scriptstyle {\sqrt {{\Big (}{\big (}b_{x}\cos(\lambda ){\big )}^{2}+{\big (}b_{y}\sin(\lambda ){\big )}^{2}{\Big )}\cos ^{2}(\beta )+{\big (}a_{m}\sin(\beta ){\big )}^{2}}},{\color {white}x}\\=&a_{m}\cos(\beta )\scriptstyle {\sqrt {{\big (}b(\lambda )\cos(\beta ){\big )}^{2}+{\big (}a_{m}\sin(\beta ){\big )}^{2}}},\\=&{\sqrt {\mathfrak {ab}}}\cos(\beta )\scriptstyle {\sqrt {{\big (}{\mathfrak {c}}(\lambda )\cos(\beta ){\big )}^{2}+{\mathfrak {ab}}\sin ^{2}(\beta )}};\end{aligned}}\,\!}$

From this, four prime possibilities for the general form for Qr become apparent:

${\displaystyle {\begin{aligned}{\color {white}.}Qr_{m1}&={\sqrt {\frac {{\frac {a_{x}a_{y}+b_{x}b_{y}}{2}}+{\frac {a_{x}^{2}+a_{y}^{2}}{2}}}{2}}}={\sqrt {\frac {{\frac {a_{m}^{2}+b^{2}}{2}}+{\frac {a_{x}^{2}+a_{y}^{2}}{2}}}{2}}}^{\color {white}|},\\&={\sqrt {\frac {a_{x}^{2}+a_{x}a_{y}+a_{y}^{2}+b^{2}}{4}}};\\Qr_{m2}&={\sqrt {\frac {{\frac {{\tfrac {a_{x}^{2}+b^{2}}{2}}+{\tfrac {a_{y}^{2}+b^{2}}{2}}}{2}}+{\tfrac {a_{x}^{2}+a_{y}^{2}}{2}}}{2}}}={\sqrt {\frac {3a_{x}^{2}+3a_{y}^{2}+2b^{2}}{8}}},\\&={\sqrt {\frac {3a^{2}(45^{\circ })+b^{2}}{4}}};\\Qr_{m3}&={\sqrt {\frac {{\frac {{\tfrac {a_{m}^{2}+b_{x}^{2}}{2}}+{\tfrac {a_{m}^{2}+b_{y}^{2}}{2}}}{2}}+{\tfrac {a_{x}^{2}+a_{y}^{2}}{2}}}{2}}}={\sqrt {\frac {b_{x}^{2}+2a_{x}^{2}+2a_{m}^{2}+2a_{y}^{2}+b_{y}^{2}}{8}}},{\color {white}.}\\&={\sqrt {\frac {b^{2}{\tfrac {a_{y}}{a_{x}}}+2a_{x}^{2}+2a_{x}a_{y}+2a_{y}^{2}+b^{2}{\tfrac {a_{x}}{a_{y}}}}{8}}},\\&={\sqrt {\frac {a_{x}^{2}+a_{x}a_{y}+a_{y}^{2}+b^{2}(45^{\circ })}{4}}};\\Qr_{m4}&={\sqrt {\frac {Qr_{m2}^{2}+Qr_{m3}^{2}}{2}}},\\&={\sqrt {\frac {5a_{x}^{2}+2a_{x}a_{y}+5a_{y}^{2}+b^{2}{\Big (}2+{\tfrac {a_{y}}{a_{x}}}+{\tfrac {a_{x}}{a_{y}}}{\Big )}}{16}}};\end{aligned}}\,\!}$

Using the Previously defined scalene ellipsoid and converting to a,b,c denotation,

${\displaystyle {\begin{aligned}{\color {white}.}{\mathfrak {a}}=a_{x}=6383.1004;\quad \;{\mathfrak {b}}=a_{y}=6373.1707;\quad \;&{\mathfrak {c}}=b=6356.752;{\color {white}{\bigg |}}\\Qr_{m1}={\sqrt {\frac {{\mathfrak {a}}^{2}+{\mathfrak {ab}}+{\mathfrak {b}}^{2}+{\mathfrak {c}}^{2}}{4}}}&=6372.79687268;\\Qr_{m2}={\sqrt {\frac {3{\mathfrak {a}}^{2}+3{\mathfrak {b}}^{2}+2{\mathfrak {c}}^{2}}{8}}}&=6372.79783967;\\Qr_{m3}={\sqrt {\frac {{\mathfrak {c}}^{2}{\tfrac {\mathfrak {b}}{\mathfrak {a}}}+2{\mathfrak {a}}^{2}+2{\mathfrak {ab}}+2{\mathfrak {b}}^{2}+{\mathfrak {c}}^{2}{\tfrac {\mathfrak {a}}{\mathfrak {b}}}}{8}}}&=6372.79783320;\\Qr_{m4}={\sqrt {\frac {5{\mathfrak {a}}^{2}+2{\mathfrak {ab}}+5{\mathfrak {b}}^{2}+{\mathfrak {c}}^{2}{\big (}2+{\frac {\mathfrak {b}}{\mathfrak {a}}}+{\frac {\mathfrak {a}}{\mathfrak {b}}}{\big )}}{16}}}&=6372.79783643;{\color {white}.}\end{aligned}}\,\!}$

If you compare integrations of variable Qr’s, one for a(λ),b(λ) and a third for both, and their averagings,

${\displaystyle {\begin{aligned}Qr_{a}={\frac {1}{\pi }}\int _{0}^{90^{\circ }}{\sqrt {{\frac {1}{4}}{\Big (}3a^{2}(\lambda )+b^{2}{\Big )}}}d\lambda &=6372.7972948266;^{\color {white}{\Big |}}\\Qr_{b}={\frac {1}{\pi }}\int _{0}^{90^{\circ }}{\sqrt {{\frac {1}{4}}{\Big (}3a_{m}^{2}+b^{2}(\lambda ){\Big )}}}d\lambda &=6372.7958394865;\\Qr_{ab}={\frac {1}{\pi }}\int _{0}^{90^{\circ }}{\sqrt {{\frac {1}{4}}{\Big (}3a^{2}(\lambda )+b^{2}(\lambda ){\Big )}}}d\lambda &=6372.7985564120;\\{\frac {Qr_{a}+Qr_{b}}{2}}\approx {\sqrt {\frac {Qr_{a}^{2}+Qr_{b}^{2}}{2}}}&=6372.7965671566;\\{\frac {Qr_{a}+Qr_{ab}+Qr_{b}}{3}}\approx {\sqrt {\frac {Qr_{a}^{2}+Qr_{ab}^{2}+Qr_{b}^{2}}{3}}}&=6372.7978117291;\\{\color {white}.}{\frac {Qr_{a}}{4}}+{\frac {Qr_{ab}}{2}}+{\frac {Qr_{b}}{4}}\approx {\sqrt {{\frac {Qr_{a}^{2}}{4}}+{\frac {Qr_{ab}^{2}}{2}}+{\frac {Qr_{b}^{2}}{4}}}}&=6372.7979979;\end{aligned}}\,\!}$

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)

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Archive 1