FW: Re: calculating the azimuth between two lat/lon's
- To: mathgroup at smc.vnet.net
- Subject: [mg34946] FW: [mg34790] Re: calculating the azimuth between two lat/lon's
- From: "DrBob" <majort at cox-internet.com>
- Date: Fri, 14 Jun 2002 02:38:50 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
I think azimuth is far more complicated than rainer indicated. Copy the following notebook expression into an empty notebook, for a discussion of calculation of azimuth, plus examples using the CityData package. If I've missed the boat somewhere, I'm sure someone will let me know!! Bobby Treat (* copy starting here --- with or without this comment *) Notebook[{ Cell[CellGroupData[{ Cell["Derivation", "Subsubtitle"], Cell["\<\ Using the Law of Sines and Law of Cosines for spherical triangles, we \ can solve for the azimuth from ptA to ptB. First we define the a, b, c, A, B, and C as follows:\ \>", "Text"], Cell[BoxData[ \(info = TableForm[{{a, "\<arc angle, ptB to north \ (90\[Degree]-latitude)\>"}, {A, "\<angle at ptA (Azimuth)\>"}, {b, \ "\<arc angle, ptA to north (90\[Degree]-latitude)\>"}, {B, "\<angle \ at ptB\>"}, {c, "\<arc angle, ptA to ptB\>"}, {C, "\<angle at \ north\>"}}]\)], "Input"], Cell["\<\ The Cosine of an arc angle is the dot product of vectors terminating \ the arc, with each angle is in the interval {0, \[Pi]}:\ \>", "Text"], Cell[BoxData[ \(c \[Equal] ArcCos[ptA . ptB]\)], "Input", Evaluatable->False], Cell["Angle C is the difference in longitudes at ptA and ptB:", \ "Text"], Cell[BoxData[ \(C \[Equal] ArcCos[Cos[p[a] - p[b]]]\)], "Input", Evaluatable->False], Cell["The Law of Sines is:", "Text"], Cell[BoxData[ \(Sin[a]\/Sin[A] \[Equal] Sin[b]\/Sin[B] \[Equal] Sin[c]\/Sin[C]\)], "Input", Evaluatable->False], Cell["Hence the azimuth A satisfies", "Text"], Cell[BoxData[ \(\(Sin[A] \[Equal] \(Sin[a] Sin[C]\)\/Sin[c];\)\)], "Input", Evaluatable->False], Cell["The Law of Cosines at ptA tells us:", "Text"], Cell[BoxData[ \(Cos[a] \[Equal] Cos[b] Cos[c] + Sin[b] Sin[c] Cos[A]\)], "Input", Evaluatable->False], Cell["from which we can derive", "Text"], Cell[BoxData[ \(Cos[ A] \[Equal] \(Cos[a] - Cos[b] Cos[c]\)\/\(Sin[b] Sin[c]\)\)], \ "Input", Evaluatable->False], Cell["Hence A is determined by", "Text"], Cell[BoxData[{ \(A \[Equal] ArcTan[Cos[A], Sin[A]]\), "\[IndentingNewLine]", \(\ \ \ \ \(\(\[Equal]\)\(ArcTan[\(Cos[a] - Cos[b] \ Cos[c]\)\/\(Sin[b] Sin[c]\), \(Sin[a] Sin[C]\)\/Sin[c]]\)\)\), "\ \[IndentingNewLine]", \(\ \ \ \ \(\(\[Equal]\)\(ArcTan[Cos[a] - Cos[b] Cos[c], Sin[a] Sin[b], Sin[C]]\)\)\)}], "Input", Evaluatable->False] }, Open ]], Cell[CellGroupData[{ Cell["Coordinate Functions", "Subsubtitle", InitializationCell->True], Cell[BoxData[ \(<< Miscellaneous`CityData`\)], "Input", InitializationCell->True], Cell["\<\ The following computes a vector in three dimensions from its \ spherical coordinates. t[a] is latitude and p[a] is longitude. All \ input angles are in radians.\ \>", "Text", InitializationCell->True], Cell[BoxData[ \(v[a_] := {Cos[t[a]]\ Cos[p[a]], Cos[t[a]]\ Sin[p[a]], Sin[t[a]]}\)], "Input", InitializationCell->True], Cell["\<\ The functions p and t can be computed from latitude and longitude \ data in the CityData format as follows:\ \>", "Text", InitializationCell->True], Cell[BoxData[{ \(t[{{d_, m_, s_: 0}, {___}}] := \((d + m/60 + s/360)\) Pi/180\), "\[IndentingNewLine]", \(t[{{d_, m_: 0}, {___}}] := \((d + m/60)\) Pi/180\), "\[IndentingNewLine]", \(t[{{d_: 0}, {___}}] := d\ Pi/180\), "\[IndentingNewLine]", \(t[name_String] := t[CityData[name, CityPosition]]\), "\[IndentingNewLine]", \(t[name_List] := t[CityData[name, CityPosition]]\), "\[IndentingNewLine]", \(p[{{___}, {d_, m_, s_: 0}}] := \((d + m/60 + s/360)\) Pi/180\), "\[IndentingNewLine]", \(p[{{___}, {d_, m_: 0}}] := \((d + m/60)\) Pi/180\), "\[IndentingNewLine]", \(p[{{___}, {d_: 0}}] := d\ Pi/180\), "\[IndentingNewLine]", \(p[name_String] := p[CityData[name, CityPosition]]\), "\[IndentingNewLine]", \(p[name_List] := p[CityData[name, CityPosition]]\), "\[IndentingNewLine]", \(\(t["\<north\>"] = Pi/2;\)\), "\[IndentingNewLine]", \(\(p["\<north\>"] = 0;\)\), "\[IndentingNewLine]", \(\(north = v["\<north\>"];\)\)}], "Input", InitializationCell->True], Cell[BoxData[{ \(\(Unprotect[ArcTan];\)\), "\[IndentingNewLine]", \(\(ArcTan[0. , 0. ] = 0;\)\), "\[IndentingNewLine]", \(\(Protect[ArcTan];\)\)}], "Input"], Cell[BoxData[{ \(ClearAll[azimuth]\), "\n", \(azimuth[A_, B_] := Module[{ptA, ptB, a, b, c, CC}, \[IndentingNewLine]ptA = v[A]; \[IndentingNewLine]ptB = v[B]; \[IndentingNewLine]a = Pi/2 - t[B]; \[IndentingNewLine]b = Pi/2 - t[A]; \[IndentingNewLine]c = ArcCos[ptA . ptB]; \[IndentingNewLine]CC = Mod[p[B] - p[A], 2 Pi, \(-Pi\)]; \[IndentingNewLine]Mod[ ArcTan[\((north . ptB)\) - \((north . ptA)\) \((ptA . ptB)\), Sin[a] Sin[b] Sin[CC]], 2. Pi]/ N[Degree]\[IndentingNewLine]]\)}], "Input", InitializationCell->True] }, Open ]], Cell[CellGroupData[{ Cell["Examples from City Data", "Subsubtitle", InitializationCell->True], Cell["\<\ There are 258 cities in the CityData database. To save time on \ slower computers, we'll take a sample of them:\ \>", "Text"], Cell[BoxData[{ \(\(origin = "\<Dallas\>";\)\), "\n", \(count = Length[CityData[CityPosition]]\), "\[IndentingNewLine]", \(indices = Union[\(Random[Integer, {1, 258}] &\) /@ Range[40]]\)}], "Input", InitializationCell->True], Cell[BoxData[{ \(\(pts = Sort[\(CityData[CityPosition]\)[\([indices]\)], OrderedQ[{azimuth[origin, #1], azimuth[origin, #2]}] &];\)\), "\n", \(TableForm[\({p[#]/N[Degree], t[#]/N[Degree], azimuth[origin, #]} &\) /@ pts, TableHeadings \[Rule] {pts, {"\<Longitude\>", "\<Latitude\>", "\ \<Azimuth\>"}}]\)}], "Input", InitializationCell->True] }, Open ]] }, FrontEndVersion->"4.1 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 711}}, AutoGeneratedPackage->Automatic, WindowSize->{1016, 673}, WindowMargins->{{0, Automatic}, {Automatic, 0}}, Magnification->1.25 ] (* copy ending here --- with or without this comment *)