Appendix : Rigorous Derivation of Realizable Launch Inclination • Launch Initial Conditions North Pole Greenwich Meridian • Position: λ, Latitude z Ω, Longitude (Inertial) Launch meridian h, Altitude
Equator
y Re
δ
x
Vernal Equinox θg
1 MAE 5540 - Propulsion Systems What Happens at Launch?
2 MAE 5540 - Propulsion Systems What Happens at Launch? • Velocity and Position at Burnout Determine Orbit
Final Stage Burnout
3 MAE 5540 - Propulsion Systems What HappensWhat happens at Launch on Launch? • Example Kennedy Space Center (KSC) Due-East Launch 28.5° Inclination Orbit
28.5°
4 MAE 5540 - Propulsion Systems Launch? …
5 MAE 5540 - Propulsion Systems Launch?
6 MAE 5540 - Propulsion Systems What Happens on Launch? (cont'd) • ExampleLaunch? Kennedy Space Center (KSC) 30° South (from east)Launch Direction 40.44° Orbit Inclination Huh?
40.44° 28.5°
7 MAE 5540 - Propulsion Systems Achieved Orbit Inclinations 0° Launch Angle 28.6° Inclination -90° Launch Angle 90° Inclination = 30° Launch Angle 40.44° Inclination -30° Launch Angle 40.44° Inclination
What?! Doesn't add up ...
8 MAE 5540 - Propulsion Systems "Spherical Geometry"
KennedySpherical Geometry Space Center (KSC) 30° North (from east)Launch Direction • Non-Euclidian Space ... 40.44° Orbit Inclination Addition Doesn't Exist! in this Abstract Space
φ −− launch angle φ λ −− launch latitude i −− inclination angle "Spherical Triangle" λ i
9 MAE 5540 - Propulsion Systems "Spherical Geometry" Spherical … “fixed earth • Non-Euclidian Space ... Addition Doesn't approximation” Exist! in this Abstract Space cos i = cos φ cos λ φ −− launch angle λ −− launch latitude i inclination angle Launch Angle sometimes expressed as −− "azimuth" ... angle from local true north
90° - φ ⇒ degrees Az = π - φ ⇒ radians 2
• Goingcos (toi )have= ctoos push(λ the)⋅ s"faithin(Az) • then a miracle occurs: Meter" pretty hard here 10 MAE 5540 - Propulsion Systems Direct Launch Inclination Angle
180° i = cos-1 cos 30° π × cos 28.6° π = 40.44° π 180° 180° 180° i = cos-1 cos -30° π × cos 28.6° π = 40.44 π 180° 180°
11 MAE 5540 - Propulsion Systems Achievable Direct-Launch Achievable Direct Launch Inclination Angles Inclination Angles
160 Orbit Inclination Angle, °
• No way to get directly 140 to Equatorial orbit from KSC °
120 •28.6° inclination is lower limit
100 Retrograde Orbit 80
60
Orbit Inclination, Inclination, Orbit Polar orbit 40 KSC Launch Latitude, 28.6°
20 -200 -100 0 100 200 Launch Angle (Counterclockwise ° from due east) 12 MAE 5540 - Propulsion Systems Achievable Direct-Launch InclinationAchievable … Angles
160 Orbit Inclination Angle, °
° Almost can Achieve 120 Direct launch to Equatorial orbit
80 Retrograde Orbit
40
Orbit Inclination, Inclination, Orbit French Guyana (Ariane) Polar orbit Launch Latitude, 5.3°
0 -200 -100 0 100 200 Launch Angle (Counterclockwise ° from due east) 13 MAE 5540 - Propulsion Systems Bottom Line
Bottom Line
See following slides for rigorous calculation • Physically Impossible to Launch Directly into an orbit with a Lower inclination Angle than the Launch latitude •Physically Possible to launch directly into any orbit with an inclination angle greater than or equal to launch latitude 14 MAE 5540 - Propulsion Systems • Launch Initial Conditions
North Pole Greenwich Meridian z Launch meridian • Launch Initial Conditions (Inertial Coordinates) Equator y Re
δ
x
Vernal Equinox θg Sidereal hour angle 15 MAE 5540 - Propulsion Systems Sidereal Hour Angle
Ignore for now!
16 MAE 5540 - Propulsion Systems Computing the Hour Angle
17 MAE 5540 - Propulsion Systems • Earth Radius
• Earth radius as Function of Latitude
18 MAE 5540 - Propulsion Systems • Initial Velocity Vector after the rocket “turns the corner”
V0
• Launch Initial Conditions
• Velocity: Vo (earth relative vel.) γ (flight path angle) φ (launch azimuth)
19 MAE 5540 - Propulsion Systems • Initial Velocity Vector after the rocket “turns the corner”
• Launch Initial Conditions
North Pole Greenwich Meridian z Launch meridian
Equator
y Re + rotational velocity of earth
δ
x
Vernal Equinox θg
20 MAE 5540 - Propulsion Systems • Earth “Boost”
Earth Rotates Under the Orbit • Launch Initial Conditions Fixed in (earth) Inertial Space V boost acts due east
V"boost"
21 MAE 5540 - Propulsion Systems • Angular Velocity of Earth
Earth Rotates Under the Orbit • Launch Initial Conditions Fixed in (earth) Inertial Space V boost acts due east
V"boost"
22 MAE 5540 - Propulsion Systems • Initial Velocity Vector after the rocket “turns the corner”
• Launch Initial Conditions
23 MAE 5540 - Propulsion Systems Inertial Velocity (Initial condition) Rotate Through Latitude First 2-rotation
No 1 (X-axis) rotation Needed Why? z` already points Towards center of earth 24 MAE 5540 - Propulsion Systems Inertial Velocity (Initial condition)
Rotate Through Longitude Next
25 MAE 5540 - Propulsion Systems Inertial Velocity (Initial condition)
26 MAE 5540 - Propulsion Systems Initial Conditions In Inertial Coordinates
27 MAE 5540 - Propulsion Systems Out-of-plane orbital elements (cont’d)
l =R x V 28 MAE 5540 - Propulsion Systems 29 MAE 5540 - Propulsion Systems 30 MAE 5540 - Propulsion Systems 31 MAE 5540 - Propulsion Systems 32 MAE 5540 - Propulsion Systems 33 MAE 5540 - Propulsion Systems