Verifying Spherical Coordinate Methods?
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From: W. eWatson on 27 Jul 2010 22:11 I started with the idea of verifying whether I had gotten the correct results. It appears that this is easy to do, although I haven't completed the computation. The idea is to use the Law of Cosines. See for example, <http://www.boeing-727.com/Data/fly%20odds/distance.html>, figures 1 and 2. The central idea is to use co-latitudes. Let me rephrase what I've got. Call it a spherical triangle ABC. A is say at (0,20) [long,lat], B is at [0,30). B is rotated say 45 degrees ccw around A to a position C. The question is what is the lat/long of the point C? After a good bit of Googling on the matter, the calculation doesn't seem to be discussed much at all, except at the more or less "hint" above. A concept somewhat similar is bearing. However, that doesn't seem applicable to finding C, since an end point is not specified.
From: Jon on 27 Jul 2010 23:33 Determine the equation of the circle and its radius resulting from the intersection of a sphere with a plane; if the sphere and plane intersect. |r|^2 = |u x (p-p3|^2 equation of circle where p=(x,y,z) and po=center of sphere and |R|=radius of sphere and |r|=radius of circle and p1,p2,p3 three noncolinear points on plane |u*(p1-po)|=distance from po to plane <= |R| for the sphere to intersect the plane. p3=po+|u*(p1-po)|u u=n/|n| n=(p3-p1)x(p3-p2) |r|={|R|^2-|p3-po|}^(1/2) "W. eWatson" <wolftracks(a)invalid.com> wrote in message news:i2l4ng$op$1(a)news.eternal-september.org... > I'm doing some coordinate transform work on the earths surface using > lat/long using xyz rotation matrices. So far the results look decent but > concocting test cases and results is a bit tricky. Maybe there's a > software tool that would help? > > Here's a simple example I'm working on. x is pointing south, y east and z > through the north pole. I want to pick an arbitrary point and draw a > circle of a radius in degrees around it on the earth. > > Suppose I'm content with one point on the circle to make this easy, and > let's take the set up along 0 longitude. > > Center of circle: (0.0, 20.0) (long,lat) > Point on circle: (0.0, 30.0) > I want the point on the circle to be 90 deg ccw to the west. > > Seemingly, that should be at (-10,20) in the original xyz. But I get > (19.68,-10.63). > > Since I'm drawing a circle (one point here) and it is not (after the > "circle" position rotation) on a great circle, the results seem close > enough. Nevertheless, unless I use another method, I may not have computed > these properly. I could change tests to more favorable spots (center at > (0,0), radius point at (0,90)), but it would be good to have some > definitive result to compare.
From: KBH on 28 Jul 2010 01:53 On Jul 27, 10:11 pm, "W. eWatson" <wolftra...(a)invalid.com> wrote: > I started with the idea of verifying whether I had gotten the correct > results. It appears that this is easy to do, although I haven't > completed the computation. The idea is to use the Law of Cosines. See > for example, <http://www.boeing-727.com/Data/fly%20odds/distance.html>, > figures 1 and 2. The central idea is to use co-latitudes. > > Let me rephrase what I've got. Call it a spherical triangle ABC. A is > say at (0,20) [long,lat], B is at [0,30). B is rotated say 45 degrees > ccw around A to a position C. The question is what is the lat/long of > the point C? > > After a good bit of Googling on the matter, the calculation doesn't seem > to be discussed much at all, except at the more or less "hint" above. A > concept somewhat similar is bearing. However, that doesn't seem > applicable to finding C, since an end point is not specified. Now you're back to the rotation where a new point is set from a held point ? The reason you can't find it on google is because you don't realize that it's called a "forward". Google for Vincenty Inverse/Forward . But for spherical formuals, that don't look any nicer than UTM conversion, google for Aviation Formulary.
From: KBH on 28 Jul 2010 02:32 On Jul 27, 10:11 pm, "W. eWatson" <wolftra...(a)invalid.com> wrote: > I started with the idea of verifying whether I had gotten the correct > results. It appears that this is easy to do, although I haven't > completed the computation. The idea is to use the Law of Cosines. See > for example, > figures 1 and 2. The central idea is to use co-latitudes. > > Let me rephrase what I've got. Call it a spherical triangle ABC. A is > say at (0,20) [long,lat], B is at [0,30). B is rotated say 45 degrees > ccw around A to a position C. The question is what is the lat/long of > the point C? > > After a good bit of Googling on the matter, the calculation doesn't seem > to be discussed much at all, except at the more or less "hint" above. A > concept somewhat similar is bearing. However, that doesn't seem > applicable to finding C, since an end point is not specified. Is the distance from Point A to Point C the same distance as Point A to Point B ? If so then "inverse" A to B to get the direction and distance from A to B. Then rotate the direction 45 degrees and "forward" Point C from Point A using the distance and new direction. Google for Vincenty Inverse and Direct which is actually geodetic. For purely spherical formulas, but different terminologies, google for Aviation Formulary.
From: W. eWatson on 30 Jul 2010 21:51
> Is the distance from Point A to Point C the same distance as Point A > to Point B ? If so then "inverse" A to B to get the direction and > distance from A to B. Then rotate the direction 45 degrees and > "forward" Point C from Point A using the distance and new direction. > > Google for Vincenty Inverse and Direct which is actually geodetic. > > For purely spherical formulas, but different terminologies, google for > Aviation Formulary. > Interesting terminology and methodology. Never heard formulary used in connection with aviation. Here's a link that quite well explains what I am after. <http://www.erikdeman.de/html/sail042e.htm>. Navigation. |