Robotic Mobility Above the Surface

Transcription

1 Free Space Relative Orbital Motion Airless Major Bodies (moons) Gaseous Environments (Mars, Venus, Titan) Lighter-than- air (balloons, dirigibles) Heavier-than- air (aircraft, helicopters) David L. Akin - All rights reserved

2 Propulsive Motion in Free Space Basic motion governed by Newton s Law ẍ (actually, F = m ) F = ma Over a distance d and time t, assuming the motion is predominately coasting, V = 2 d t (required to accelerate and decelerate) The rocket equation (relates propellant to ΔV) m final m o = e V V exhaust 2

3 Cost of Propulsive Maneuvering Assuming V V exhaust Use the Taylor s Series expansion of e m final m o 1 V V exhaust Since m o =m initial =m prop +m final, m prop m o 2 d V exhaust t or m prop m o 2 V travel V exhaust 3

4 Hill s Equations (Proximity Operations) Linearized equations of motion relative to a target in circular orbit in a rotating Cartesian reference frame ẍ = 3n 2 x + 2nẏ + a dx ÿ = 2nẋ + a dy z = n 2 z + a dz Ref: J. E. Prussing and B. A. Conway, Orbital Mechanics Oxford University Press, 1993 n = µ a 3 adx, ady, adz are disturbing accelerations (e.g., thrust, solar pressure) 4

5 Clohessy-Wiltshire ( CW ) Equations Force-free solutions to Hill s Equations x(t) = [4 3 cos (nt)]x o + y(t) = 6[sin (nt) nt]x o + y o 2 n sin (nt) n x o + 2 [1 cos (nt)] n y o [1 cos (nt)] x o + 4 sin (nt) 3nt n y o ẋ(t) =3n sin (nt)x o + cos(nt)ẋ o +2sin(nt)ẏ o ẏ(t) = 6n [1 cos (nt)] x o 2sin(nt)ẋ o + [4 cos (nt) 3] ẏ o z(t) = z o cos (nt) + z o sin (nt) n ż(t) = z o n sin (nt)+ż o cos (nt) 5

6 V-Bar Approach 0.01 m/sec m/sec m/sec %$ #$!"&#$!"##$!'#$!(#$!)#$!&#$!%$ #$!"#$ 6

7 R-Bar Approach Approach from along the radius vector ( Rbar ) Gravity gradients decelerate spacecraft approach velocity - low contamination approach Used for Mir, ISS docking approaches Ref: Collins, Meissinger, and Bell, Small Orbit Transfer Vehicle (OTV) for On-Orbit Satellite Servicing and Resupply, 15th USU Small Satellite Conference,

8 Hopping (Airless Flat Planet) V v, h γ V V h, d Use F=ma for vertical motion V v = g t flt =2V v /g h = V v t 1 2 gt2 Constant velocity in horizontal direction produces d = V h t flt =2 V hv v g V h = V cos γ; V v = V sin γ d =2 V 2 sin γ cos γ g 8 = V 2 g sin (2γ)

9 Hopping (Airless Flat Planet) V v, h V Horizontal distance is maximized when sin (2γ) = 1 γ V h, d γ opt = π 2 = 45o d max = V 2 g V = gd V total =2V =2 gd V v h max = V v g 1 2 g h max = V 2 2 gd 4g = 4g Vv g = d V v = V 2

10 An Example of Propulsive Gliding 10

11 Propulsive Gliding (Airless Flat Planet) T = mg V g Assume horizontal velocity is V V h =2V t flt = d/v Total ΔV becomes (includes acceleration and deceleration) V v = gt flt = gd V V total = V v + V h =2V + gd V 11

12 Propulsive Gliding (Airless Flat Planet) V 2V + gd V Want to choose V to minimize =0 2 gd V 2 =0 V opt = gd 2 gd 2 V total =2 2 + gd gd =2 2 gd 12

13 Delta-V for Hopping and Gliding Delta-V (m/sec) Distance (m) Ballistic Hop 13 Propulsive Glide

14 Hopping (Spherical Planet) a = r v =2v o v = µ 2 r 1 a p r = 1+ecos ν = a(1 e2 ) 1 e cos θ 1 e cos θ 2 1 e 2 v = µ r 1 e 2 r(1 e cos θ v e =0 r(1 e cos θ)( 2e) + (1 e2 )r( cos θ) r 2 (1 e cos θ) 2 =0 14

15 Hopping (Spherical Planet) 2er 2e 2 r cos θ r cos θ + re 2 cos θ =0 cos θe 2 2e + cos θ =0 e opt = 2 ± cos 2 θ 2 cos θ = 1 ± sin θ cos θ + produces e>1 (hyperbolic orbit); gives elliptical orbit e opt = 1 sin θ cos θ a opt = r 1 eopt cos θ 1 e 2 opt 15

16 Propulsive Gliding (Airless Round Planet) ω 2 r = V 2 r V g Assume horizontal velocity is V V h =2V t flt = d/v Total ΔV becomes (includes acceleration and deceleration) V v = g V 2 r t flt = gd V dv r V total = V v + V h =2V + gd V dv r 16

17 Propulsive Gliding (Airless Round Planet) Want to choose V to minimize 2V + gd V V dv =0 2 gd r V 2 d r =0 gd V opt = 2 d r gd 2 d r V total =2 2 d + gd gd d gd r r 2 d r V total =2 2 d r gd 17

18 Hopping on Flat and Round Bodies Delta-V (m/sec) Distance (m) Ballistic Hop Propulsive Glide Hop on Sphere Glide on Sphere 18

19 Nondimensional Forms Define ν V dg ρ d r η h max d ν flat glide =2 2 ν flat hop =2 η = 1 4 ν spherical glide =2 2 ρ (0 ρ 1) 19

20 Multiple Hops Assume n hops between origin and destination At each intermediate touchdown, v v has to be reversed V total =2V + 2(n 1)V v t peak = V v g t total =2nt peak =2n V v g d = V h t total = 2n g V hv v V v = 2gh max 2η ν v = η h max ν V d/n V h = dg n dg ν 2nV h = 1 1 v 2 2nη 20

21 Multiple Hop Analysis ν =2ν + 2(n 1)ν v ν =2 νv 2 + νh 2 + 2(n 1)ν v 2η ν =2 n + 1 2η + 2(n 1) 8nη n ν η = 1 2 2η n + 1 n 1 2 8nη 2 +(n 1) nη =0 8nη nalytically messy, but note that for n =1 η opt = 1 4 (In general, solve numerically) 21

22 Optimal Solutions for Multiple Hops η opt 0.3 ν Number of Hops (n) Number of Hops (n) 22

23 Hopping Between Different Altitudes Relative to starting point, landing elevation h 2 v 1 =(v h,v v1 ) v 2 =(v h,v v2 ) v v1 = v v2 t peak = v v1 h = v v1 t 1 2 gt2 h g peak = 1 2 v 2 v1 g v v1 = 2gh peak From peak, v v = gt fall ; h = h peak 1 2 gt2 fall h 2 = h peak 1 vv2 2 2 t fall = 2 g g (h peak h 2 ) v v2 = 2g(h peak h 2 ) 23

24 Optimal Hop with Altitude Change vv1 2 d = v h (t peak + t fall )=v h g + g (h peak h 2 ) 2h peak 2 d = v h + g g (h peak h 2 ) d 2hpeak g = v h + 2(h peak h 2 ) d g v h = 2hpeak + 2(h peak h 2 ) v = v 2h + v2v1 + vh 2 + v2 v2 24

25 Nondimensional Form of Equations Remember that ν v ; η h peak dg d ; λ h 2 d ν = ν 2h + ν2v1 + νh 2 + ν2 v2 ν v1 = 2η ν v2 = 2(η λ) 1 ν h = 2η + 2(η λ) ν = η + 2η + 2(η λ) 1 2η + 2(η λ) 2 + 2(η λ) 25

26 Optimization of Height-Changing Hop This is not going to be one where you can take the derivative and set equal to zero, so use the equation to find a numerical optimization Set λ =0 to check for plain hop solution 1 ν =2 8η +2η η opt =

27 Trajectory Design for Height Change ν 2 η opt Height Change λ

28 Apollo Concept of Lunar Flying Vehicle from Study of One-Man Lunar Flying Vehicle - Final Report Volume 1: Summary North American Rockwell, NASA CR , August

29 Apollo 15 Revisited: LFV Sortie Basic assumptions Vehicle inert mass=300 kg Crew mass=150 kg Science package=100 kg Total propellant=130 kg 29

30 Apollo 15 Revisited: Leg 1 Base camp to bottom of rille Distance 3 km Altitude change -150 m ΔV=139 m/sec Propellant used=22 kg Collect 20 kg of samples at landing site 30

31 Apollo 15 Revisited: Leg 2 Propulsive glide along bottom of rille Distance 2 km No net altitude change ΔV=160 m/sec Propellant used=25 kg Collect 20 kg of samples at landing site; leave 25 kg science package 31

32 Apollo 15 Revisited: Leg 3 Hop to top of mountain Distance 15 km Altitude change 1600 m ΔV=310 m/sec Propellant used=46 kg Collect 30 kg of samples at landing site; leave 50 kg science package 32

33 Apollo 15 Revisited: Leg 4 Return to base Distance 12 km Altitude change m ΔV=278 m/sec Propellant used=37 kg Return with 25 kg of science equipment and 70 kg of samples 33

34 Apollo 15 Revisited: Discussion Current minimum estimates are for 400 kg of residual propellants in Altair at landing - would support three equivalent sorties Presence of water ice or ISRU propellant production at outpost would easily support moderate flier mission requirements Challenges in routine refueling of cryogenic propellants on the lunar surface, reliable flight and landing control system 34

35 Landing Impact Attenuation Cannot rely on achieving perfect zero velocity at touchdown Specifications for landing conditions Vertical velocity 3 m/sec Horizontal velocity 1 m/sec Kinetic Energy = 1 2 mv2 = 1 2 m(v2 h + v 2 v) Max case 500 kg vehicle = E = 2500Nm 35

36 Mars Phoenix Lander 36

37 Apollo Lunar Module UNIVERSITY OF 37

38 Landing Deceleration Look at 3 m/sec vertical velocity Constant force deceleration 1 2 mv2 = Fd t decel = Spring deceleration F = kx k = mv2 v a desired d v2 = F m d = a desiredd d = 1 2 Fdx = 1 2 mv2 a peak = kd m 38 v 2 a desired a desired dcm t d sec 1/6 g /2 g g g g

39 Effect of Lateral Velocity at Touchdown Resolve torques around landing gear footpad h mg w F h F v θ = τ tot I tot θ = F hh F v w mgw I cg + m 2 Worst cases - hit obstacle (high force), landing downhill Issue: rotational velocity induced is counteracted by vehicle weight Will vehicle rotation stop before overturn limit? 39

40 Atmospheric Neutral Buoyancy Given an enclosed volume V of gas with density ρ Lift force is V(ρ atm -ρ) - must be mg on Earth ~1 kg lift/cubic meter of He on Mars ~10 gms lift/cubic meter of He Horizontal velocity at equilibrium is identical to wind speed Interior pressure generally identical to ambient (except for superpressure balloons) Can generate low density through choice of gas, heating 40

41 Dynamic Atmospheric Lift Drag Lift Thrust D = 1 2 ρv2 Sc D Weight L = 1 2 ρv2 Sc L For steady, level flight: T = D L = W W = L = D L D = T L D 41 L = 1 2 ρv2 Sc D L D

42 Mars Atmosphere ρ =0.020 kg m 3 T = 210 K g =3.71 R = m sec 2 γ = J kg K Speed of sound a = γrt = m sec 42

43 Aircraft Flight Performance v stall = U-2 high-altitude spy plane Cruises at 70,000+ feet m=18,000 kg b=32 m S~64 m 2 mg S U-2 v stall(mars) = m sec 2 2 ρc L(max) 43

44 Looking for Equation for Range Efficiency = propulsive power fuel power = Tv e ṁ f h η overall = Tv e ṁ f h dw dt = ṁ f g = W L D T ṁ f g dw dt = Wv e h g L D Tv e ṁ f h = Wv e h g L D η overall 44

45 More Aerial Range Rewrite and integrate dw W = v edt h L g D η overall = ln W = C h g v e t L D η overall Initial conditions - Range = h g Range = at t =0W = W init C = ln W init L D η overall ln W init V L D gsfc ln W init W final W final -->Breguet Range Equation 45