#### Transcript Multiobjective Optimization of Multiple-Impulse Transfer

```Multiobjective Optimization of
Multiple-Impulse Transfer
between Two Coplanar Orbits
using Genetic Algorithm
Abbas Kafaee Razavi‡,Vahid Hamed Azimi¶
Department of Aerospace Engineering, Sharif University of Technology, Tehran, Iran
¶ Mathematical group, Department of Science, K. N. Toosi University of Technology, Tehran,
*†‡
2
Introduction
• Optimal transfer between two coplanar elliptical
orbits
• Multiobjective optimization
▫ Transfer time
▫ Total impulse
• Multiple-impulse maneuvers
• Two type of parameterization:
▫ Based on true anomaly, normal and tangential
velocities of impulse points
▫ Position vector & true anomaly of impulse points
3
Multi-impulsive transfer model
4
First method
• First we are at the initial orbit then with choosing
optional true anomaly ,parallel velocity and normal
velocity of impulse point we enter the first transfer
orbit and with repeating this process for next transfers
finally we arrive at the target orbit.
5
First method

(cont.)
First we apply primary impulse so we have,
(1  e )
r0  a1
1  e1cos (1 )
2
1
h1  a1 (1  e ) 
2
1
6
First method
(cont.)
• In the moment of (i+1)th impulse applied at the
random position of Δ𝜓 𝑖 we are at the reference axis,
(1  ei )
ri  ai
1  ei cos (i  i )
2
7
First method

(cont.)
The values of ΔV‖ and ΔV⊥ are chosen from
genetic algorithm in random way so we obtain the
value of total impulse for each (i),
Vi  V i 2  Vi 2
• And for the velocity before (i+1)th impulse we have,
V 0i 

2
(
)(
)
ai
ri
8
First method

(cont.)
Also V𝑖 is the value of velocity after applying
(i+1)th impulse:
V1i  (V0i  V i )2  Vi 2
• The main half-axis for (i+1)th orbit and also the
flight path angle are calculated as following:
ai 1 
2(

V 1i 2
2

ri
hi
 i =cos (
)
rV
i i
1
)
9
First method

(cont.)
For the angular momentum and eccentricity
we have
hi 1 =hi +ri V i cos( 0i )-ri Vi sin( 0i )
e i 1 
1  hi 12
ai 1
10
First method
(cont.)
• The condition of answer being, arriving to the
final orbit, is equivalent to S ≥ 1
e n 2
e n 1
2
S  (D
cos(- )+
cos(k))
1 D
1 D
e n 2
e n 1
2
(D
sin(- )+
sin(k))
1 D
1 D
11
First method
(cont.)
• With letting out the values of D & K in the last
equation we have,
1  e n 12
D  an 1
an  2 (1  e n  2 2 )
k=n1 -n1;
12
First method

(cont.)
The total value of (n+1) impulse,
n
Vt  V0 n1 2  V1 n1 2  2V0 n1V1 n1 cos( 1 n1  0 n1 )   Vi
i 1
13
First method
(cont.)
• In this optimization we should calculate the sum
of transfer times from initial to final orbit,
- 1
E i + 1 = cos (
- 1
E 1i + 1 = cos (
ai + 1 -ri
ai + 1ei + 1
)
ai + 1 -ri + 1
ai + 1ei + 1
)
14
First method
(cont.)
t1  0
t i 1  (E 1i 1  E i 1  e i 1 (sin (E 1i 1 )  sin (E i 1 )))
n
T   t i 1
i 1
ai 11.5
 0.5
15
Second method
• In the following subject the other method for
optimal transfer between two arbitrary orbits is
presented, With this difference in this method 𝑅𝑖
(position vector)
and
θ𝑖 (true anomaly of
impulse points) are entered into the calculations
16
Second method
(cont.)
• The values of 𝑅𝑖 and 𝑅𝑖+1 are chosen random
• from genetic algorithm method which have two
components 𝑅𝑥 &𝑅𝑦 .
17
Second method
(cont.)
• The angle between these two vector is,
- 1
D j i = cos (
v v
R i ×R i + 1
Ri Ri + 1
)
18
Second method

(cont.)
Also the eccentricity value ,the unit eccentricity vector
and the angular momentum are,
ei + 1 =
Ri
- 1
Ri+ 1
Ri
cos(qi + 1 + D j i ) - cos(qi + 1 )
Ri+ 1
19
Second method
v
ei + 1
ei + 1
=
v
hi + 1
hi + 1
v v
R i ´ hi + 1
R ihi + 1
=
(cont.)
sin(qi + 1 ) +
Ri ´
v
Ri
Ri
cos(qi + 1 )
Rj
RiRi + 1
20
Second method
(cont.)
• The main half-axis of transfer orbits is,
ai + 1 =

R i (1 + ei + 1 cos(qi + 1 ))
(1 - ei2+ 1 )
The value of angular momentum for transfer orbits
is,
hi =
mai (1 - ei 2 )
21
Second method

(cont.)
The calculation method of velocity vector and the
impulses values is
é-sin( q + D j )
ù
ê
i+ 1
i
ú
m ê
ú
vb,i + 1 =
êei + 1 + cos( qi + 1 + D j i )ú
hi + 1 ê
ú
0
ê
ú
û
ë
é-sin ( q
ù
)
ê
i+ 1
ú
m ê
ú
v a ,i =
e
+
cos(
q
)
ê
i+ 1 ú
hi + 1 ê i + 1
ú
ê0
ú
û
ë
22
Second method
(cont.)
• 𝛚𝑖+1 is the perigee angle for the orbit transfer point to
the reference axis,
- 1
wi + 1 = cos (
v
ei + 1 ×iˆ
ei + 1
)
23
Second method
(cont.)
écos( w ) sin( w ) 0ù
i
i
ê
ú
ê
ú
Qi = ê-sin( wi ) cos( wi ) 0ú
ê
ú
0
1ú
êë 0
û
v
- 1
V b,i = Q ´ vb,i
24
Second method
(cont.)
v
Va ,i = Q - 1 ´ va ,i

The sum of impulses of this multi-impulse transfer
is,
n+1
D Vt =
å
i= 1
n+1
DVi =
å
i= 1
v
v v
v
(Va ,i - V b,i ) ×(Va ,i - V b,i )
25
Second method
(cont.)
• Time is calculated like the previous method,
a i + 1 -R i
- 1
E i = cos (
- 1
(a i + 1ei + 1 )
E1i = cos (
a i + 1 -R i + 1
a i + 1ei + 1
)
)
26
Second method
ti =
a
1.5
i+ 1
m
( E1i - E i - ei + 1(sin ( E 1i ) - sin ( E i ) ) )
n
T =
å
i= 1
(cont.)
ti
27
Initial and Tangent orbit parameters
a1 = 7000
a 2 = 130000
e1 = 0.4
e2 = 0.1
 = 22.92 (deg)
28
Dual-impulse transfer (Second method)
5
1.5
x 10
1
0.5
Initial orbit
Y (km)
Target orbit
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
X (km)
0.5
1
1.5
5
x 10
29
Pareto-optimal solution of dual-impulse transfer
(Second method)
28
Vmin =3.37820 (km/s)
26
24
T =27.81627 (h)
Transfer time (s)
22
20
18
16
14
12
10
8
2
4
6
8
10
 V (km/s)
12
14
16
30
Tri-impulse transfer (Second method)
5
1.5
x 10
1
0.5
0
Y (km)
-0.5
-1
-1.5
-2
-2.5
-3
-3
-2.5
-2
-1.5
-1
-0.5
X (km)
0
0.5
1
1.5
5
x 10
31
Pareto-optimal solution of tri-impulse transfer
(Second method)
5
10
x 10
Vmin =3.16955 (km/s)
9
8
T = 251.396 (h)
Transfer time (s)
7
6
5
4
3
2
3
3.5
4
4.5
 V (km/s)
5
5.5
6
32
Dual-impulse transfer (First method)
5
1.5
x 10
1
Y (km)
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
X (km)
0.5
1
1.5
5
x 10
33
Pareto-optimal solution of dual-impulse
transfer (First method)
4
10
x 10
Vmin = 3.265168 (km/s)
9
T =26.984 (h)
Transfer time (s)
8
7
6
5
4
3
3
3.5
4
4.5
 V (km/s)
5
5.5
34
Tri-impulse transfer (First method)
5
1.5
x 10
1
Y (km/s)
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
X (km/s)
0.5
1
1.5
5
x 10
35
Pareto-optimal solution of tri-impulse
transfer (First method)
4
10
x 10
Vmin = 3.27381 (km/s)
9
T =26.557 (h)
Transfer time (s)
8
7
6
5
4
3.2
3.4
3.6
3.8
4
4.2
 V (km/s)
4.4
4.6
4.8
5
5.2
36
5
1.5
x 10
1
Y (km/s)
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
X (km/s)
0.5
1
1.5
5
x 10
37
transfer (First method)
4
10
x 10
Vmin = 3.26815 (km/s)
Transfer time (s)
9
T = 26.162 (h)
8
7
6
5
4
3
3
3.5
4
4.5
 V (km/s)
5
5.5
```