ASEN 5050 SPACEFLIGHT DYNAMICS Prox Ops, Lambert

Transcription

1 ASEN 5050 SPACEFLIGHT DYNAMICS Prox Ops, Lambert Prof. Jeffrey S. Parker University of Colorado Boulder Lecture 15: ProxOps, Lambert 1

2 Announcements Homework #5 is due next Friday 10/10 CAETE by Friday 10/17 Homework #6 will be due Friday 10/17 CAETE by Friday 10/24 Solutions will be available in class on 10/17 and online by 10/24. If you turn in HW6 early, me and I ll send you the solutions to check your work. Mid-term Exam will be handed out Friday, 10/17 and will be due Wed 10/22. (CAETE 10/29) Take-home. Open book, open notes. Once you start the exam you have to be finished within 24 hours. It should take 2-3 hours. Reading: Chapter 6, 7.6 Lecture 15: ProxOps, Lambert 2

3 Space News Comet Siding Spring will pass by Mars on October 19 th, making things really exciting for all of the Mars vehicles. Lecture 15: ProxOps, Lambert 3

4 Space News The Rosetta Comet or 67P/Churyumov- Gerasimenko fired its jets last week just published. Lecture 15: ProxOps, Lambert 4

5 ASEN 5050 SPACEFLIGHT DYNAMICS Prox Ops Prof. Jeffrey S. Parker University of Colorado Boulder Lecture 15: ProxOps, Lambert 5

6 CW / Hill Equations Enter Clohessy/Wiltshire (1960) And Hill (1878) (notice the timeline here when did Sputnik launch? Gemini was in need of this!) Lecture 15: ProxOps, Lambert 6

7 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 7

8 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Use RSW coordinate system (may be different from NASA) Target satellite has two-body motion: rtgt r µ = tgt 3 r The interceptor is allowed to have thrusting r r int int = µ + F 3 rint Then r = r r r = r r So, rel int tgt r r r µ µ = + F + rel r r int 3 int tgt rel 3 tgt tgt int tgt Lecture 15: ProxOps, Lambert 8

9 Clohessy-Wiltshire (CW) Equations If we assume circular motion, (Hill s Equations) µ 3 r tgt 2 = ω, ω = 0, r rel R = ω 2 { x ˆR + yŝ + z W ˆ 3x ˆR }+ F + 2ω y ˆR 2ω xŝ +ω 2 x ˆR +ω 2 yŝ Thus, 2 x 2ωy 3ω x = y + 2ω x = f z + ω 2 z = f f x y z CW or Hill s Equations Assume F = 0 (good for impulsive ΔV maneuvers, not for continuous thrust targeting). Lecture 15: ProxOps, Lambert 9

10 Assumptions Please Take Note: We ve assumed a lot of things We ve assumed that the relative distance is very small We ve assumed circular orbits These equations do fall off as you break these assumptions! Lecture 15: ProxOps, Lambert 10

11 Clohessy-Wiltshire (CW) Equations (Hill s Equations) The above equations can be solved (see book: Algorithm 48) leaving: x( t) = x 0 y t ω sinωt " 3x y 0 % " $ 'cosωt + 4x y 0 % $ ' # ω & # ω & " % $ 'sinωt + 2 x 0 # ω & ω cosωt 6ωx y 0 ( ) = 6x y 0 z( t) = z 0 cosωt + z 0 ω sinωt ( ) t + y 0 2 x 0 $ " # ω % ' & x t Lecture 15: ProxOps, Lambert ( ) = x 0 cosωt + 3ωx y 0 ( ) = 6ωx y 0 ( ) = z 0 ω sinωt + z 0 cosωt y t z t ( )sinωt ( )cosωt 2 x 0 sinωt 6ωx y 0 ( ) So, given x0, y0, z0, x 0, y 0, z 0 of interceptor, can compute x, y, z of interceptor at future time. x, y, z, 11

12 Applications What can we do with these equations? Estimate where the satellite will go after executing a small maneuver. Rendezvous and prox ops! Examples. First from the book. Lecture 15: ProxOps, Lambert 12

13 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 13

14 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 14

15 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 15

16 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 16

17 Hubble s Drift from Shuttle RSW Coordinate Frame Lecture 15: ProxOps, Lambert 17

18 Clohessy-Wiltshire (CW) Equations (Hill s Equations) x( t) = x 0 y t ω sinωt " 3x y 0 % " $ 'cosωt + 4x y 0 % $ ' # ω & # ω & " % $ 'sinωt + 2 x 0 # ω & ω cosωt 6ωx y 0 ( ) = 6x y 0 z( t) = z 0 cosωt + z 0 ω sinωt ( ) t + y 0 2 x 0 $ " # ω % ' & Lecture 15: ProxOps, Lambert 18

19 Clohessy-Wiltshire (CW) Equations (Hill s Equations) x( t) = x 0 y t ω sinωt " 3x y 0 % " $ 'cosωt + 4x y 0 % $ ' # ω & # ω & " % $ 'sinωt + 2 x 0 # ω & ω cosωt 6ωx y 0 ( ) = 6x y 0 z( t) = z 0 cosωt + z 0 ω sinωt ( ) t + y 0 2 x 0 $ " # ω % ' & Lecture 15: ProxOps, Lambert 19

20 Clohessy-Wiltshire (CW) Equations (Hill s Equations) We can also determine ΔV needed for rendezvous. Given x 0, y 0, z 0, we want to determine x, y 0, z 0 0 necessary to make x=y=z=0. Set first 3 equations to zero, and solve for x, y,. x y z ωx = = z ( 4 3cosωt) + 2( 1 cosωt) ( 6 x ( ωt sinωt) y ) ω sinωt 2ωx ( 4 3cosωt)( 1 cosωt) ( 4 sinωt 3ωt) sinωt + 4( 1 cosωt) = z ω cotωt 0 0 sinωt Assumptions: 1. Satellites only a few km apart 2. Target in circular orbit 3. No external forces (drag, etc.) y 0 Lecture 15: ProxOps, Lambert 20

21 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 21

22 Clohessy-Wiltshire (CW) Equations (Hill s Equations) Lecture 15: ProxOps, Lambert 22

23 Clohessy-Wiltshire (CW) Equations (Hill s Equations) NOTE: This is not the Delta-V, this is the new required relative velocity! Lecture 15: ProxOps, Lambert 23

24 CW/H Examples Scenario 1: Use eqs to compute x(t), y(t), and z(t) RSW coordinates, relative to deployer. Note: in the CW/H equations, the reference state doesn t move it is the origin! Deployment: x 0, y 0, z 0 = 0, Delta-V = non-zero Scenario 1b: Use eqs to compute x(t), y(t), and z(t) RSW coordinates, relative to reference. Initial state: non-zero Lecture 15: ProxOps, Lambert 24

25 CW/H Equations Rendezvous For rendezvous we usually specify the coordinates relative to the target vehicle and set x, y, and z to zero Though if there s a docking port, then that will be offset from the center of mass of the vehicle. Define RSW targets: x, y, z (often zero) Initial state in the RSW frame Lecture 15: ProxOps, Lambert Target: some constant 25 values in the RSW frame

26 CW/H Equations Rendezvous For rendezvous we usually specify the coordinates relative to the target vehicle and set x, y, and z to zero Though if there s a docking port, then that will be offset from the center of mass of the vehicle. Define RSW targets: x, y, z (often zero) This is easy if the targets are zero (Eq. 6-66) This is harder if they re not! Initial state in the RSW frame Velocity needed to get onto transfer Lecture 15: ProxOps, Lambert Target: some constant 26 values in the RSW frame

27 CW/H Equations Rendezvous For rendezvous we usually specify the coordinates relative to the target vehicle and set x, y, and z to zero Though if there s a docking port, then that will be offset from the center of mass of the vehicle. Define RSW targets: x, y, z (often zero) This is easy if the targets are zero (Eq. 6-66) This is harder if they re not! Initial state in the RSW frame Velocity needed to get onto transfer The Delta-V is the difference of these velocities Target: some constant values in the RSW frame Lecture 15: ProxOps, Lambert 27

28 CW / H Scenario 3: A jetpack-wielding astronaut leaves the shuttle and then returns. Lecture 15: ProxOps, Lambert Shuttle: reference frame 28

29 CW / H Scenario 3: A jetpack-wielding astronaut leaves the shuttle and then returns. Result of deployment Lecture 15: ProxOps, Lambert Shuttle: reference frame 29

30 CW / H Scenario 3: A jetpack-wielding astronaut leaves the shuttle and then returns. Rendezvous trajectory (Eq 6-66) Result of deployment Lecture 15: ProxOps, Lambert Shuttle: reference frame 30

31 CW / H Scenario 3: A jetpack-wielding astronaut leaves the shuttle and then returns. Delta-V is the difference of these velocities. Rendezvous trajectory (Eq 6-66) Result of deployment Lecture 15: ProxOps, Lambert Shuttle: reference frame 31

32 Comparing Hill to Keplerian Propagation Deviation: 100 meters 1 cm/s Position Error: ~1.4 meters/day Lecture 15: ProxOps, Lambert 32

33 Comparing Hill to High-Fidelity, e=0.15 Generally returns to 0.0 near one apsis Position Error: ~9.5 km/day Lecture 15: ProxOps, Lambert 33

34 Comparing Hill to High-Fidelity, e=0.73 Generally returns to 0.0 Position Error: ~4.2 km/day Lecture 15: ProxOps, Lambert 34

35 Analyzing Prox Ops Lecture 15: ProxOps, Lambert 35

36 Analyzing Prox Ops What happens if you just change x0? Radial displacement? Lecture 15: ProxOps, Lambert 36

37 Analyzing Prox Ops Lecture 15: ProxOps, Lambert 37

38 Analyzing Prox Ops What happens if you just change vx0? Radial velocity change? Lecture 15: ProxOps, Lambert 38

39 Analyzing Prox Ops Lecture 15: ProxOps, Lambert 39

40 Analyzing Prox Ops What happens if you just change vy0? Tangential velocity change? Lecture 15: ProxOps, Lambert 40

41 Analyzing Prox Ops Lecture 15: ProxOps, Lambert 41

42 Analyzing Prox Ops What happens if you just change vz0? Out-of-plane velocity change? Lecture 15: ProxOps, Lambert 42

43 Analyzing Prox Ops Lecture 15: ProxOps, Lambert 43

44 ASEN 5050 SPACEFLIGHT DYNAMICS Lambert s Problem Prof. Jeffrey S. Parker University of Colorado Boulder Lecture 15: ProxOps, Lambert 44

45 Lambert s Problem Lambert s Problem has been formulated for several applications: Orbit determination. Given two observations of a satellite/ asteroid/comet at two different times, what is the orbit of the object? Passive object and all observations are in the same orbit. Satellite transfer. How do you construct a transfer orbit that connects one position vector to another position vector at different times? Transfers between any two orbits about the Earth, Sun, or other body. Lecture 15: ProxOps, Lambert 45

46 Lambert s Problem Given two positions and the time-of-flight between them, determine the orbit between the two positions. Lecture 15: ProxOps, Lambert 46

47 Orbit Transfer We ll consider orbit transfers in general, though the OD problem is always another application. Lecture 15: ProxOps, Lambert 47

48 Orbit Transfer Note: there s no need to perform the transfer in < 1 revolution. Multi-rev solutions also exist. Lecture 15: ProxOps, Lambert 48

49 Orbit Transfer Consider a transfer from Earth orbit to Mars orbit about the Sun: Lecture 15: ProxOps, Lambert 49

50 Orbit Transfer Orbit Transfer True Anomaly Change Short Way Δν < 180 Long Way Δν > 180 Hohmann Transfer (assuming coplanar) Δν = 180 Type I 0 < Δν < 180 Type II 180 < Δν < 360 Type III 360 < Δν < 540 Type IV 540 < Δν < 720 Lecture 15: ProxOps, Lambert 50

51 Lambert s Problem Given: Find: ~R 0 ~ Rf t 0 t f ~V 0 ~ Vf Numerous solutions available. Some are robust, some are fast, a few are both Some handle parabolic and hyperbolic solutions as well as elliptical solutions All solutions require some sort of iteration or expansion to build a transfer, typically finding the semi-major axis that achieves an orbit with the desired Δt. Lecture 15: ProxOps, Lambert 51

52 Ellipse d 1 + d 2 =2a d 1 d 2 Lecture 15: ProxOps, Lambert 52

53 Ellipse d 1 + d 2 =2a d 1 d 2 d 3 d 4 d 3 + d 4 =2a Lecture 15: ProxOps, Lambert 53

54 Ellipse A transfer from r 1 to r 2 will be on an ellipse, with the central body occupying one focus. Where s the 2 nd focus? r 1 r 2 Lecture 15: ProxOps, Lambert 54

55 Ellipse A transfer from r 1 to r 2 will be on an ellipse, with the central body occupying one focus. Where s the 2 nd focus? 2a r 1 r 1 r 2 Lecture 15: ProxOps, Lambert 55

56 Ellipse A transfer from r 1 to r 2 will be on an ellipse, with the central body occupying one focus. Where s the 2 nd focus? 2a r 1 r 1 r 2 2a r 2 Lecture 15: ProxOps, Lambert 56

57 Ellipse A transfer from r 1 to r 2 will be on an ellipse, with the central body occupying one focus. Where s the 2 nd focus? Focus is one of these Try different a values until you hit your TOF 2a r 1 r 1 r 2 2a r 2 Lecture 15: ProxOps, Lambert 57

58 Lambert s Problem Lecture 15: ProxOps, Lambert 58

59 Lambert s Problem Find the chord, c, using the law of cosines and cos Δν = c 0 r r 0 r r 2 2 = r0 + r 2r r cos Δν sin(δν) = t m 1 cos 2 (Δν) 0 Define the semiperimeter, s, as half the sum of the sides of the triangle created by the position vectors and the chord r0 + r + s = 2 c We know the sum of the distances from the foci to any point on the ellipse equals twice the semi-major axis, thus Lecture 15: ProxOps, Lambert 2a = r + ( 2a r ) 59

60 Lambert s Problem Lecture 15: ProxOps, Lambert 60

61 Lambert s Problem The minimum-energy solution: where the chord length equals the sum of the two radii (a single secondary focus) Thus, 2a r + 2a r0 = s 2 r 0 a = min + r 4 c + c = (anything less doesn t have enough energy) Lecture 15: ProxOps, Lambert 61

62 If If Δν > 180, then t > t min, then α Lambert s Problem e β e = β = 2π α e e Elliptic Orbits Hyperbolic Orbits ' α e sin% & 2 ' βe sin% & 2 $ " # $ " # = = r r r + c 4a + r c 4a = = s 2a s c 2a sinh ' α h % & 2 sinh ' β h % & 2 $ " # $ " # = = r r r + c 4a + r c 4a = = s 2a s c 2a t = = 3 a µ a µ Lecture 15: ProxOps, Lambert [ α sin( α ) ( β sin( β )] 3 e [ sinh( α ) α ( sinh( β ) β )] h e h e e h h Lambert s Solution 62

63 Lambert s Problem For minimum energy t v min 0 = = a µ 3 min e = π 3 βe sin / = [ π β + sin( β )] µ p & min % r r0 r sin Δν $ elliptic orbit α e, * 1 + e r p s c s min { 1 cos Δν} ) ' r0 ( # "! Lecture 15: ProxOps, Lambert 63

64 Universal Variables A very clear, robust, and straightforward solution. There are a few faster solutions, but this one is pretty clean. Begin with the general form of Kepler s equation: Lecture 15: ProxOps, Lambert 64

65 Universal Variables Simplify Lecture 15: ProxOps, Lambert 65

66 Universal Variables Define Universal Variables: Lecture 15: ProxOps, Lambert 66

67 Universal Variables Lecture 15: ProxOps, Lambert 67

68 Universal Variables Use the trigonometric identity Lecture 15: ProxOps, Lambert 68

69 Universal Variables Now we need somewhere to go Let s work on converting this to true anomaly, via: Lecture 15: ProxOps, Lambert 69

70 Universal Variables Multiply by a convenient factoring expression: Lecture 15: ProxOps, Lambert 70

71 Universal Variables Collect into pieces that can be replaced by true anomaly Lecture 15: ProxOps, Lambert 71

72 Universal Variables Substitute in true anomaly: Lecture 15: ProxOps, Lambert 72

73 Trig identity again: Universal Variables Lecture 15: ProxOps, Lambert 73

74 Universal Variables Note: Lecture 15: ProxOps, Lambert 74

75 Use some substitutions: Universal Variables Lecture 15: ProxOps, Lambert 75

76 Universal Variables Summary: Lecture 15: ProxOps, Lambert 76

77 Universal Variables It is useful to convert to f and g series (remember those!?) Lecture 15: ProxOps, Lambert 77

78 UV Algorithm Lecture 15: ProxOps, Lambert 78

79 A few details on the Universal Variables algorithm Lecture 15: ProxOps, Lambert 79

80 Universal Variables Let s first consider our Universal Variables Lambert Solver. Given: R 0, R f, ΔT Find the value of ψ that yields a minimum-energy transfer with the proper transfer duration. Applied to building a Type I transfer Lecture 15: ProxOps, Lambert 80

81 Single-Rev Earth-Venus Type I -4π 4π 2 Lecture 15: ProxOps, Lambert 81

82 Single-Rev Earth-Venus Type I -4π 4π 2 Lecture 15: ProxOps, Lambert 82

83 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 83

84 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 84

85 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 85

86 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 86

87 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 87

88 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 88

89 Note: Bisection method -4π 4π 2 Lecture 15: ProxOps, Lambert 89

90 Time history of bisection method: Note: Bisection method Requires 42 steps to hit a tolerance of 10-5 seconds! Lecture 15: ProxOps, Lambert 90

91 Note: Newton Raphson method Lecture 15: ProxOps, Lambert 91

92 Note: Newton Raphson method Lecture 15: ProxOps, Lambert 92

93 Note: Newton Raphson method Lecture 15: ProxOps, Lambert 93

94 Note: Newton Raphson method Lecture 15: ProxOps, Lambert 94

95 Note: Newton Raphson method Time history of Newton Raphson method: Requires 6 steps to hit a tolerance of 10-5 seconds! Note: This CAN break in certain circumstances. With current computers, this isn t a HUGE speedup, so robustness may be preferable. Lecture 15: ProxOps, Lambert 95

96 Note: Newton Raphson Log Method Note log scale -4π 4π 2 Lecture 15: ProxOps, Lambert 96

97 Single-Rev Earth-Venus Type I -4π 4π 2 Lecture 15: ProxOps, Lambert 97

98 Single-Rev Earth-Venus Type II -4π 4π 2 Lecture 15: ProxOps, Lambert 98

99 Interesting: 10-day transfer -4π 4π 2 Lecture 15: ProxOps, Lambert 99

100 Interesting: 950-day transfer -4π 4π 2 Lecture 15: ProxOps, Lambert 100

101 UV Algorithm Lecture 15: ProxOps, Lambert 101

102 Multi-Rev Seems like it would be better to perform a multi-rev solution over 950 days than a Type II transfer! Lecture 15: ProxOps, Lambert 102

103 A few details The universal variables construct ψ represents the following transfer types: 8 >< Type of Transfer >: < 0, Hyperbolic =0, Parabolic 0 < < 4 2, 0 revolutions elliptical 4n 2 2 < < 4(n + 1) 2 2, n revolutions elliptical Lecture 15: ProxOps, Lambert 103

104 Multi-Rev ψ ψ ψ ψ Lecture 15: ProxOps, Lambert 104

105 Earth-Venus in 850 days Type IV Lecture 15: ProxOps, Lambert 105

106 Earth-Venus in 850 days Heliocentric View Distance to Sun Lecture 15: ProxOps, Lambert 106

107 Earth-Venus in 850 days Type VI Lecture 15: ProxOps, Lambert 107

108 Earth-Venus in 850 days Heliocentric View Distance to Sun Lecture 15: ProxOps, Lambert 108

109 What about Type III and V? Type III Type IV Type V Type VI Lecture 15: ProxOps, Lambert 109

110 Earth-Venus in 850 days Type III Lecture 15: ProxOps, Lambert 110

111 Earth-Venus in 850 days Heliocentric View Distance to Sun Lecture 15: ProxOps, Lambert 111

112 Earth-Venus in 850 days Type V Lecture 15: ProxOps, Lambert 112

113 Earth-Venus in 850 days Heliocentric View Distance to Sun Lecture 15: ProxOps, Lambert 113

114 Summary The bisection method requires modifications for multi-rev. Also requires modifications for odd- and even-type transfers. Newton Raphson is very fast, but not as robust. If you re interested in surveying numerous revolution combinations then it may be just as well to use the bisection method to improve robustness Lecture 15: ProxOps, Lambert 114

115 Types II - VI Lecture 15: ProxOps, Lambert 115