Spherical Astronomy Lecture 01 Cristiano Guidorzi Physics Department University of Ferrara Astrophysical Measurements Cristiano Guidorzi (Università di Ferrara) 12/04/2010 1 / 15 Outline 1 Spherical Trigonometry Basics Spherical Triangle: angles and sides relations 2 The Earth Coordinates Geodetic/geocentric latitude Cristiano Guidorzi (Università di Ferrara) 12/04/2010 2 / 15 Spherical Trigonometry Basics Outline 1 Spherical Trigonometry Basics Spherical Triangle: angles and sides relations 2 The Earth Coordinates Geodetic/geocentric latitude Cristiano Guidorzi (Università di Ferrara) 12/04/2010 3 / 15 Spherical Trigonometry Basics Great Circles and Triangles Small and Great Circles. Spherical triangle: arcs of great circles Cristiano Guidorzi (Università di Ferrara) 12/04/2010 4 / 15 Spherical Trigonometry Basics Spherical angles and sides Angles & sides Unique correspondence a ↔ BC, b ↔ AC, c ↔ AB Spherical Excess E = A + B + C − 180◦ it’s not constant Cristiano Guidorzi (Università di Ferrara) 12/04/2010 5 / 15 Spherical Trigonometry Basics Spherical Excess and triangle area Spherical Excess A (∆) : triangle ∆ area S(A)/4 π r2 = A/π A+B+C π 4 π r2 = 4 π r2 + 4 A (∆) A (∆) = (A + B + C − π ) r2 = E r2 Cristiano Guidorzi (Università di Ferrara) 12/04/2010 6 / 15 Spherical Trigonometry Spherical Triangle: angles and sides relations Outline 1 Spherical Trigonometry Basics Spherical Triangle: angles and sides relations 2 The Earth Coordinates Geodetic/geocentric latitude Cristiano Guidorzi (Università di Ferrara) 12/04/2010 7 / 15 Spherical Trigonometry Spherical Triangle: angles and sides relations Spherical Triangle: angles and sides relations Rotation around x axis by angle χ x = cos ψ cos θ y = sin ψ cos θ z = sin θ x′ = cos ψ ′ cos θ ′ y′ = sin ψ ′ cos θ ′ ′ z = sin θ ′ x′ = x y′ = y cos χ + z sin χ z′ = −y sin χ + z cos χ Cristiano Guidorzi (Università di Ferrara) 12/04/2010 8 / 15 Spherical Trigonometry Spherical Triangle: angles and sides relations Spherical Triangle: angles and sides relations Replacing, we get: cos ψ ′ cos θ ′ = cos ψ cos θ sin ψ ′ cos θ ′ = sin ψ cos θ cos χ + sin θ sin χ sin θ ′ = − sin ψ cos θ sin χ + sin θ cos χ with spherical angles (P is now C) ψ = A − 90◦ , ψ ′ = 90◦ − B, Cristiano Guidorzi (Università di Ferrara) θ = 90◦ − b θ ′ = 90◦ − a, χ =c 12/04/2010 9 / 15 Spherical Trigonometry Spherical Triangle: angles and sides relations Spherical Triangle: angles and sides relations Replacing, we get: cos (90◦ − B)cos (90◦ − a) = cos (A − 90◦ ) cos (90◦ − b) sin (90◦ − B)cos (90◦ − a) = sin (A − 90◦ ) cos (90◦ − b) cos c + sin(90◦ − b)sin c sin (90◦ − a) = − sin (A − 90◦ ) cos (90◦ − b) sin c + sin(90◦ − b) cos c or sin B sin a = sin A sin b cos B sina = − cos A sin b cos c + cosb sin c cos a = cos A sinb sin c + cosb cos c Cristiano Guidorzi (Università di Ferrara) 12/04/2010 10 / 15 Spherical Trigonometry Spherical Triangle: angles and sides relations Sine and Cosine Formulas Sine Formula sin b sin c sin a = = sin A sin B sin C Sine-Cosine Formula sin c cos b = sin b cos c cos A + sina cos B Cosine Formula cos a = cos b cos c + sinb sin c cos A Cristiano Guidorzi (Università di Ferrara) 12/04/2010 11 / 15 The Earth Coordinates Outline 1 Spherical Trigonometry Basics Spherical Triangle: angles and sides relations 2 The Earth Coordinates Geodetic/geocentric latitude Cristiano Guidorzi (Università di Ferrara) 12/04/2010 12 / 15 The Earth Coordinates Coordinates Coordinates Longitude: angle between the meridian and that of Greenwich Latitude geodetic (geographical) φ (plumb line) geocentric φ ′ Cristiano Guidorzi (Università di Ferrara) 12/04/2010 13 / 15 The Earth Geodetic/geocentric latitude Outline 1 Spherical Trigonometry Basics Spherical Triangle: angles and sides relations 2 The Earth Coordinates Geodetic/geocentric latitude Cristiano Guidorzi (Università di Ferrara) 12/04/2010 14 / 15 The Earth Geodetic/geocentric latitude Geodetic/geocentric latitude Geodetic Reference System 1980 (GRS-80) Ellipsoid: equatorial radius: a = 6, 378 Km polar radius: b = 6, 357 Km flattening: f = (a − b)/b = 1/298 x2 a2 y2 + b2 = 1 tan φ ′ = y x tan φ = − dx dy = ⇒ tan φ ′ = b2 a2 a2 y b2 x tan φ = (1 − e2 ) tan φ Max difference ∆φ = φ − φ ′ = 11.5′ (@ 45◦) Cristiano Guidorzi (Università di Ferrara) 12/04/2010 15 / 15
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