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16.346 Astrodynamics
Fall 2008
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Lecture 17
The Battin —Vaughan Algorithm for the BVP
Developing the Algorithm
The right hand side of the cubic equation
E − sin E
y
3 − y
2 = m
4 tan3 12 E
can be expressed as the derivative of a hypergeometric function
Utilizing series manipulations, we find
1
2
sin E =
1
2E
tan
12 E
= tan
12 E(1 − x + x
2 − · · ·)
1 + x
= tan
12 E(1 −
13 x +
15 x
2 − · · ·)
d
E − sin E
=
13 −
25 x +
37 x
2 −
49 x
3 + · · · = − F ( 12 , 1;
32 ; −x)
3 1
dx
4 tan 2 E
where the hypergeometric function and its derivative are
√
arctan x
1
3
= (1 −
13 x +
51 x
2 −
71 x
3 + 91 x
4 − · · ·)
F (
2 , 1;
2 ; −x) = √
x
√
√
dF
d � arctan x �
1 � arctan x
1 �
√
√
=
=
−
= −(
13 −
25 x +
73 x
2 −
94 x
3 + · · ·)
dx
dx
2x
1 + x
x
x
The function F (
12 , 1;
23 ; −x) also has a continued fraction representation
√
1
arctan x
1
1
3
F (
2 , 1;
2 ; −x) = √
=
=
1 + xG
x
x
1+
4x
3+
9x
5+
7 + ...
1
G=
3+
4x
9x
5+
16x
7+
9 + ..
16.346 Astrodynamics
=
1
3+
4x
ξ
where
ξ =5+
.
Lecture 17
9x
16x
7+
25x
9+
36x
11 +
13 + . .
.
1 �
1 �
(1 − G)F
(2x + ξ)F G
2x + ξ
dF
=
F−
=−
=−
=−
dx
2x
1+x
2(1 + x)
ξ(1 + x)
(1 + x)[4x + ξ(3 + x)]
Finally, a successive substitution algorithm, paralleling the Gauss algorithm, results using:
��
x=
1−�
2
�2
+
m
1+�
−
2
y
2
and
y 3 − y 2
=
m(2x + ξ)
(1 + x)[4x + ξ(3 + x)]
Graphics of a Successive Substitution Algorithm
Fig. 7.5 from An Introduction to the Mathematics and Methods
of Astrodynamics. Courtesy of AIAA. Used with permission.
16.346 Astrodynamics
Lecture 17
Page 333
Improving the Convergence of the New Algorithm
1. Time equation with free parameter β
�
�
r
r0 �
1 µ
(ψ − sin ψ) + 0 [sin ψ − β(ψ − sin ψ)]
(t2 − t1 ) = 1 + β
3
2 a
a
a
√
arctan x
dF
√
2. With
F =
and
h1 = 2βx(1 + x)
dx
x
the time equation can be written as
�
�
h1 (x)
dF
3
2
2
+
where
h2 (x) = −m
y − y − h1 (x)y − h2(x) = 0
dx
(� + x)(1 + x)
3. Differentiate and obtain
�
�
� 2
�
� dy
� 2
m
d
dh1
d F
1
2
+m
+ h1 (x)
=0
− y −
3y −2(1+h1 )y
dx
(� + x)(1 + x) dx
dx2
dx (� + x)(1 + x)
At the solution point, the coefficient of dh1 /dx is, of course, zero.
Then, for dy/dx = 0 at solution point, we must have
d
d2 F
1
=0
+ h1 (x)
2
dx
dx (� + x)(1 + x)
from which we calculate h1 (x) [or β(x) which we don’t really need] and finally h2 (x).
16.346 Astrodynamics
Lecture 17
Battin–Vaughan Algorithm
√
Given r1 , r2 , θ , µ (t2 − t1 ) ≡ k(t2 − t1 )
1. Calculate
A = 12 (r1 + r2 )
C =A+B
√
B = r1 r2 cos 12 θ
�
0 parabola, hyperbola
2. Initialize x =
� ellipse
�=
A−B
C
m=
[k(t2 − t1 )]2
C3
3. Calculate
ξ(x) = 5 +
9x
16x
7+
25x
9+
36x
11 +
13 + . .
.
Note: Instead of the continued fraction for ξ(x) , we can use
√
√
4x( x − arctan x )
√
√
ξ(x) =
(3 + x) arctan x − 3 x
for elliptic orbits which are not close to parabolic.
4. Calculate
H = (1 + 2x + �)[4x + (3 + x)ξ(x)]
h1 (x) =
1
(� + x)2 [1 + 3x + ξ(x)]
H
and
h2 (x) =
m
[x − � + ξ(x)]
H
5. Solve the cubic y 3 − y 2 − h1 y 2 − h2 = 0
Note:
�
�b
2
+1
converts the cubic to
z 3 − 3z = 2b
y = (1 + h1 )
3
z
�
�
2 cosh( 13 arccosh b) b ≥ 1
27h2
where
b =
+1
and
z=
4(1 + h1 )3
2 cos( 13 arccos b)
b<1
�
m
A
B2
+
−
6. Determine new x =
C2
y2
C
7. Repeat until x no longer changes.
8. Calculate the orbital elements:
4xy2
1
=
a
Cm
16.346 Astrodynamics
cy 2 (1 + x)2
p
=
pm
2Cm
or
Lecture 17
p =
r1 r2 y 2 (1 + x)2 sin2
Cm
1
2
θ