University:
Massachusetts Institute of Technology
Course:
16.346 | Astrodynamics
Academic year:

2019
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301

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2
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lirmzinlelfuu5

Lecture 5 Euler Angles Euler Angles Ω, i, ω Page 84 Deﬁne two sets of orbital plane unit vectors: (1) ie ≡ iξ and ip ≡ iη referenced to pericenter, and (2) in and im referenced to the ascending node. with ih ≡ iζ . Each of these can be expressed in terms of the Euler angles as components along the reference axes unit vectors ix , iy , iz using vector algebra. We have: in = cos Ω ix + sin Ω iy ih = sin i in × iz + cos i iz = sin Ω sin i ix − cos Ω sin i iy + cos i iz = 3 ix + m3 iy + n3 iz im = ih × in = − sin Ω cos i ix + cos Ω cos i iy + sin i iz ie = cos ω in + sin ω im = (cos Ω cos ω − sin Ω sin ω cos i) ix + (sin Ω cos ω + cos Ω sin ω cos i) iy + sin ω sin i iz = 1 ix + m1 iy + n1 iz ip = ih × ie = −(cos Ω sin ω + sin Ω cos ω cos i) ix − (sin Ω sin ω − cos Ω cos ω cos i) iy + cos ω sin i iz = 2 ix + m2 iy + n2 iz ⎡ ⎤ ⎡ ie 1 ⎣ ip ⎦ = ⎣ 2 ih 3 so that m1 m2 m3 ⎤⎡ ⎤ n1 ix n2 ⎦ ⎣ i y ⎦ n3 iz Fig. 2.2 from An Introduction to the Mathematics and Methods of Astrodynamics. Courtesy of AIAA. Used with permission. 16.346 Astrodynamics Lecture 5 Position and Velocity Vectors in Reference Coordinates See Problem 3–21 The rotation matrix to transform between orbital plane and reference coordinates is R= cos Ω cos ω − sin Ω sin ω cos i sin Ω cos ω + cos Ω sin ω cos i sin ω sin i rorbital plane sin Ω sin i − cos Ω sin i cos i ⎡ ⎡ with ⎤ cos f = r ⎣ sin f ⎦ 0 − cos Ω sin ω − sin Ω cos ω cos i − sin Ω sin ω + cos Ω cos ω cos i cos ω sin i vorbital plane ⎤ − sin f µ = ⎣ e + cos f ⎦ h 0 The proper expressions for position and velocity in reference coordinates are obtained by premultiplying the position and velocity vectors, expressed in orbital plane coordinates, by the rotation matrix R. We also need the trigonometric identities sin θ = sin(ω + f ) = sin ω cos f + cos ω sin f cos θ = cos(ω + f ) = cos ω cos f − sin ω sin f to obtain the position and velocity vectors in the form: r(cos Ω cos θ − sin Ω sin θ cos i) ix r= + r(sin Ω cos θ + cos Ω sin θ cos i) iy + r sin θ sin i iz and µ [cos Ω(sin θ + e sin ω) + sin Ω(cos θ + e cos ω) cos i] ix h µ − [sin Ω(sin θ + e sin ω) − cos Ω(cos θ + e cos ω) cos i] iy h µ + (cos θ + e cos ω) sin i iz h where θ = ω + f is called the argument of latitude. v=− Terminology #3.4, Page 160–161 Time of pericenter passage τ Argument of latitude θ =ω+f Angle of inclination i True longitude L= Longitude of ascending node Ω Mean anomaly M = n(t − τ ) Argument of pericenter Mean longitude l= Longitude of pericenter ω = Ω + ω Mean longitude at epoch = +f + M = nt + − nτ Then the mean anomaly is determined from: M = nt + − 16.346 Astrodynamics Lecture 5 (4.39)

Lecture 5: Euler Angles

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