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16.346 Astrodynamics
Fall 2008
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Lecture 31 The Calculus of Variations & Lunar Landing Guidance
The Brachistochrone Problem
In a vertical xy -plane a smooth curve y = f (x) connects the origin with a point P (x1 , y1 )
in such a way that the time taken by a particle sliding without friction from O to P along
the curve propelled by gravity is as short as possible. What is the curve?
Assume the positive y -axis is vertically downward. Then the equation of motion is
dy
d2 s
with ds2 = dx2 + dy 2
m 2 = mg sin γ = mg
dt
ds
dy ds
d2 s ds
=g
dt2 dt
ds dt
�
�
2
dy
ds �
d ds
= 2g
=⇒
= 2gy
dt dt
dt
dt
Then
�
�
T
x1
dt = T =
0
0
ds
1
√
=√
2gy
2g
Deriving Euler’s Equation
To minimize the integral
�
x1
I=
�
0
x1
�
1
1 + y 2
dx = √
√
y
2g
F (x, y, y ) dx
let
x0
Then
dI
=
dα
�
x1 �
x0
�
x1
F (y, y ) dx
0
y(x, α) = ym (x) + α�(x)
�
�
� x1 �
∂F
∂F d�
∂F
d ∂F
�(x) + dx =
−
�(x) dx
∂y
∂y dx
∂y
dx ∂y x0
Therefore, from the Fundamental Lemma of the Calculus of Variations
d ∂F
∂F
−
=0
∂y
dx ∂y is a Necessary Condition which F must satisfy if the integral I is to be a minimum.
Special Case of Euler’s Equation
�
�
�
�
∂F ∂F
d ∂F
∂F
∂F
d
F − y =
+
−
y =
dx
∂y
∂x
∂y
dx ∂y
∂x
�
��
�
=0
which will be zero if F is not a function of x . Therefore
Also
F−
∂F y = constant
∂y Prob. 11–33
which establishes the necessary condition used to solve the Brachistochrone Problem.
16.346 Astrodynamics
Lecture 31
Solution of the Brachistochrone Problem
If T is to be a minimum, then, using Euler’s Special Case of the Necessary Condition, we
have
�
� �
y
2
y(1 + y ) = 2c
dy
or
dx = x =
2c − y
Now let
y = 2c sin2 θ = c(1 − cos 2θ)
�
so that
x = 2c
(1 − cos 2θ) dθ = c(2θ − sin 2θ)
Therefore, the equation of the curve in parametric form is
x = c(φ − sin φ)
with
y = c(1 − cos φ)
φ = 2θ
and represents a cycloid—the path of a point on a circle of radius c as it rolls along the
underside of the x axis.
Terminal State Vector Control
Find the acceleration vector a(t) to minimize
� t1
�
2
J=
a(t) dt =
t0
t1
aT(t)a(t) dt
t0
subject to
dr
=v
dt
dv
=a
dt
Define the Admissible Functions:
r(t0 ) = r0
v(t0 ) = v0
δ(t0 ) = δ(t1 ) = 0
r(t, α) = rm (t) + αδ(t)
v(t, α) = vm (t) + αδ (t)
a(t, α) = am (t) + αδ (t)
Then
�
where
�
t1
t1
T
J(α) =
t0
am(t)am (t) dt + 2α
16.346 Astrodynamics
r(t1 ) = r1
v(t1 ) = v1
T
t0
δ (t0 ) = δ (t1 ) = 0
δ (t0 ) = δ (t1 ) = 0
am(t)δ (t) dt + α
Lecture 31
�
2
t1
t0
δ (t) δ (t) dt
T
A Necessary Condition for
�
t1
�
t1
�
t1
T
T
T
2
δ (t) δ (t) dt
J(α) =
am(t)am (t) dt + 2α am(t)δ (t) dt + α
t0
t0
to be a minimum is that
�
�
t1
dJ
��
T
= 0 = 2 am
(t)δ (t) dt
dα
�
α=0
t0
Use integration by parts
�t1 �
� t1
�
T
T
am(t)δ dt = am(t)δ (t)��
−
t0
t0
�t1
�
dam(t)
�
δ (t)
�
= −
dt
t0
T
�
dJ ��
=0
dα
�
α=0
Hence
t0
� t1 T
daTm(t) dδ(t)
dam(t) dδ(t)
dt = 0 −
dt
dt
dt
dt
dt
t0
t0
�
t1 2 T
�
t1 2 T
d am(t)
d am(t)
δ(t) dt = 0 + δ(t) dt
+
2
dt
dt2
t0
t0
t1
�
=⇒
t1
t0
T
d 2 am
(t)
δ (t) dt = 0
2
dt
Again using the Fundamental Lemma of the Calculus of Variations it follows that
Therefore, with tgo
T
(t)
d 2 am
= 0T
2
dt
= t1 − t , we have
am (t) = c1 t + c2 =
=⇒
am (t) = c1 t + c2
4
6
[v1 − v(t)] + 2 {r1 − [r(t) + v1 tgo ]
tgo
tgo
Lunar-Landing Guidance for Apollo Missions
To include the effects of gravity
a(t) = aT (t) + g(r)
we could use
aT (t) =
4
6
[v1 − v(t)] + 2 {r1 − [r(t) + v1 tgo ]} − g[r(t)]
tgo
tgo
for the thrust acceleration which would be an exact solution if g were constant.
16.346 Astrodynamics
Lecture 31