Lifting Entry 2. Basic planar dynamics of motion, again Yet another equilibrium glide Hypersonic phugoid motion MARYLAND U N I V E R S I T Y O F

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "Lifting Entry 2. Basic planar dynamics of motion, again Yet another equilibrium glide Hypersonic phugoid motion MARYLAND U N I V E R S I T Y O F"

Ολυμπία Γούσιος
5 χρόνια πριν
Προβολές:

glide Hypersonic phugoid motion MARYAN 1

1 ifting Entry Basic planar dynamics of motion, again Yet another equilibrium glide Hypersonic phugoid motion MARYAN avid. Akin - All rights reserved ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

2 Basic Equations of Motion From last time, mv v, s r v dγ γ dt = v m + r g cos γ horizontal dv v mg dt = r g sin γ m Assume (at entry velocities close to orbital) g = v r v dγ dt = m = ρv c A m = ρv (1) MARYAN dv dt = m = ρv () ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

3 Solving for Velocity ivide () by (1) dv dt v dγ dt = ρv ρv = dv v = dγ / (3) v v e dv v = 1 / γ γ e dγ ln v v e = (γ γ e) / = v = e γ γ e / v e (4) MARYAN 3 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

4 ifferential Elements As before, ifferentiating (6), dh dt = v sin γ ρ = ρ o e h dρ dt = ρ o h s e h dh dt = ρ dh h s dt dρ dt = ρ h s v sin γ (5) (6) (7) MARYAN 4 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

5 ifferential Elements () Solve (7) for v v = h s sin γ 1 ρ dρ dt Substitute (8) into (1) and rewrite as (8) dγ dt = ρv = ρ sin γ 1 ρ dρ dt dγ dt = h s sin γ dρ (9) MARYAN 5 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

6 Solving for Flight Path Angle γ MARYAN γ e sin γdγ = h s cos γ cos γ e = h s cos γ = h s 6 ρ 0 ρ ρ + cos γ e (11) γ = cos 1 ρ + cos γ e (1) Note that the negative sign was inserted because cos -1 is ambiguous as to direction, and the flight path angle on entry should be >0. dρ (9) (10) ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

7 Flight Path Angle and Velocity Equations γ = cos 1 ρ oe h Rewrite (4) as v = v e exp γ γ e = v e exp / and substitute into (13) + cos γ e γe γ / (13) v = v e exp 1 / γ e + cos 1 ρ oe h + cos γ e (14) MARYAN 7 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

8 eceleration Along the flight path, m = ρv Perpendicular to the flight path, m = ρv Total deceleration (not in g s) n = + m MARYAN = 1 m m n = ρ ov 8 + = ρv e h (15) ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

9 Fiddling with Algebra n = ρ o Substitute in (14) n = ρ ov e X 1+ MARYAN exp / 9 1+ γ e + cos 1 n = ρ ov e e h v γ e + cos 1 1+ ρ oe h ρ oe h e X / + cos γ e + cos γ e (16) (17) ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

10 More Algebra Set dn dh =0 0= ρ ov e 1+ e h e X / / dx dh + 1 h s e h e X / (18) Factoring out common terms, 1 = h s / dx m dh (19) MARYAN 10 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

11 Even More Algebra From (13), γ = cos 1 ρ oe h + cos γ e X = γ e γ = cos 1 (cos γ)+γ e Y cos γ X = γ e cos 1 Y (0) MARYAN 11 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

12 Trigonometry, for a Change Trig identity - d(cos 1 u) dx = 1 1 u du dx (1) dx dh = Y ) d(cos 1 dh = 1 1 Y dy dh dx dh = 1 1 cos γ d(cos γ) dh MARYAN 1 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

13 Back to the Algebra From (13), cos γ = h s dx dh = 1 1 cos γ dx dh = 1 sin γ d dh ρ oe h ρ o ρ o + cos γ e e h 1 h s + cos γ e e h () MARYAN dx dh = 13 ρ o sin γ e h (3) ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

14 Maximum eceleration Case If we go back to the n max case, (19) gives 1 = dx m h s dh 1 h s = / MARYAN 14 / ρ o hm sin γ m e h s h m, γ m are values at n max 1 = ρ o hm h s β sin γ m e h s sin γ m = ρ oh s β hm e h s (4) (5) ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

15 et s Go Back to Algebra et Φ ρ oh s cos γ m =Φe h m h s sin γ m =Φe h m h s + cos γ e Φe h m h s et H e h m h s + cos γ e + Φe h m h s =1 (ΦH + cos γ e ) +(ΦH) =1 MARYAN 15 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

16 Algebra is Fun, on t You Think? Φ H + ΦH cos γ e + cos γ e 1=0 Φ H + ΦH cos γ e sin γ e =0 H = Φ cos γ e ± 4Φ cos γ e + 4(Φ )sin γ e 4Φ cos γ e ± cos γ e +sin γ e H = 1 Φ H = 1 Φ MARYAN cos γ e ± 16 1+sin γ e ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

17 Maximum eceleration Equations Skipping some painful algebra and trig, ρ o h s h m = h s ln sin γ e 4+ csc γ e cot γ e cos γ m = cos γ e (/)sinγ e 4+(/) csc γ e (/) cot γ e MARYAN v m = v e e γ m γe / n max = ρ ov e vm ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

18 Phugoid Oscillations Assume a shallow, linear entry trajectory: γ 1 0 ḣ = v sin γ = vγ v γ g = 1 mg v v c γ o 0 Small perturbations = h = h 1 + h γ = γ 1 + γ γ MARYAN 18 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

19 Perturbation Analysis ḣ = vγ ḣ = ḣ1 + ḣ = γ 1v 1 + ḣ = ḣ ḣ =(v 1 + v 1 )(γ 1 + γ) =v 1 γ + v 1 γ v 1 g ḣ = v 1 γ γ = mg + 1 mg 1 v 1 v c MARYAN v 1 g γ = mg = ḧ g 19 ifting Atmospheric Entry II ENAE aunch and Entry Vehicle esign

Σχετικά έγγραφα

Lifting Entry (continued)

Lifting Entry (continued) ifting Entry (continued) Basic planar dynamics of motion, again Yet another equilibrium glide Hypersonic phugoid motion Planar state equations MARYAN 1 01 avid. Akin - All rights reserved http://spacecraft.ssl.umd.edu