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16.346 Astrodynamics
Fall 2008
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Lecture 20
Powered Flight Guidance
#11.7
Powered Flight Maneuver Equations
vr (t, r)
Required velocity:
Velocity-to-be-gained:
vg (t)
Vehicle velocity:
v(t)
v + vg = vr
dvr
∂vr
∂vr dr
∂vr
∂vr
∂vr
∂vr
∂vr
dv dvg
+
=
=
+
=
+
v=
+
vr −
v
dt
dt
dt
∂t
∂r dt
∂t
∂r
∂t
∂r
∂r g
dvg
∂vr
g(r) + aT +
= g(r) −
v
dt
∂r g
dvg
∂v
dvg
∂vr
= − r
vg − aT =⇒
= −C vg − aT where C =
dt
∂r
∂r
dt
This is the fundamental equation for the velocity-to-be-gained.
Constant Gravity Field Example
dr
=v
dt
dv
=g
dt
r(t) = r0 + (t − t0 )v0 + 12 (t − t0 )2 g
r(t0 ) = r0
v(t0 ) = v0
=⇒
or, with change of notation,
r(t1 ) = r(t) + (t1 − t)vr (t) + 12 (t1 − t)2 g
vr (t) =
Hence
C (t) =
1
[r1 − r(t) − 12 (t1 − t)2 g]
t1 − t
1
∂ vr
=−
I
∂r
t1 − t
=⇒
dvg
1
=
v − aT
dt
t1 − t g
To optimize the control of the thrust direction for the purpose of minimizing the fuel
consumption, we convert the equation for the velocity-to-be-gained vector to scalar form
by multiplying both sides sides by vg
dvg
d
2
d
= (vg · vg ) = vg2 =
v 2 − 2aT · vg
dt
dt
dt
t1 − t g
d
(t1 − t) vg2 = 2vg2 − 2(t1 − t)aT · vg
dt
Next, integrate by parts from the current time t to the time of engine cut-off tco
�tco �
tco
�
tco
�
2
2
�
(t1 − t)vg (t)
�
+
v
g (t) dt =
[2v
g2 (t) − 2(t1 − t)aT · vg ] dt
2vg ·
�
tco
t
t
t
t
[2(t1 − t)aT · vg − vg2 ] dt = (t1 − t)vg2 (t) = constant at the time t
dvg
× v
g = 0
dt
to maximize the integrand—the latter is referred to as Cross-Product Steering
To minimize the time interval tco − t, choose aT vg or equivalently,
16.346 Astrodynamics
Lecture 20
Vehicle Orientation Prior to Cross-Product Steering
dvg
dvr
− g(r) so that
= p(t) − aT
dt
dt
Then, to determine the direction of the thrust acceleration vector aT
p(t) = −C (t)vg (t) =
Define a vector
dvg
× vg = 0
dt
=⇒
aT × vg = p × vg
(aT × vg ) × vg = (p × vg ) × vg
(aT · vg )vg − vg2 aT = (p · vg )vg − vg2 p
aT = p + (q − ivg · p) ivg
To obtain q :
so that
where q = aT · ivg
aT · aT = a2T = [p + (q − ivg · p) i vg ] · [p + (q − ivg · p) i vg ]
q =
a
2T − p
2 + (ivg · p)2
Clearly, cross-product steering is not possible unless a
2T > p2 .
Hyperbolic Injection Guidance
vr =
12 v∞ [(D + 1)i∞ + (D − 1)ir ]
T
2
v
∞
(r + i∞
r)(D2 − 1) = 4µ
where
To obtain the C matrix, first calculate:
∂D 1
∂i
∂v
r
=
12 v∞ (i∞ + ir )
+
2 v
∞ (D − 1)
r
∂r
∂r
∂r
∂ r
1 ∂r
1 ∂r
1
∂ir
=
=
−
2 r
= (I − ir iT
r)
∂r
∂r
r
r
∂r r
∂r
r
2
∂D
2 (D − 1)
= −v
∞
(i∞ + ir )T
∂r
8µD
Then
C =
v∞ (D − 1)
v 3 (D2 − 1)2
(I − ir irT ) − ∞
(i∞ + ir )(i∞ + ir ) T
2r
16µD
Circular Orbit Insertion Guidance
µ
µ
vr =
in × ir = Sn r
3
r
r
where in and Sn are
 

nx
0
−nz



in = n y
nz
0
and
Sn =
nz
−ny nx
Then
C =
16.346 Astrodynamics
µ
S (I −
r3 n
T
3
2 ir ir )
Lecture 20

ny
−nx 
0