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MathGroup Archive 2002

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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 *)




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