Transcript ASEN 5050 SPACEFLIGHT DYNAMICS

ASEN 5050
SPACEFLIGHT DYNAMICS
Prox Ops, Lambert
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 15: ProxOps, Lambert
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Announcements
• Homework #5 is due next Friday 10/10
– CAETE by Friday 10/17
• Homework #6 will be due Friday 10/17
– CAETE by Friday 10/24
– Solutions will be available in class on 10/17 and online by 10/24. If you
turn in HW6 early, email me and I’ll send you the solutions to check your
work.
• Mid-term Exam will be handed out Friday, 10/17 and will be due
Wed 10/22. (CAETE 10/29)
– Take-home. Open book, open notes.
– Once you start the exam you have to be finished within 24 hours.
– It should take 2-3 hours.
• Reading: Chapter 6, 7.6
Lecture 15: ProxOps, Lambert
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Space News
• Comet Siding Spring will pass by Mars on October 19th,
making things really exciting for all of the Mars vehicles.
Lecture 15: ProxOps, Lambert
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Space News
• The “Rosetta Comet” or 67P/ChuryumovGerasimenko fired its jets last week – just published.
Lecture 15: ProxOps, Lambert
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ASEN 5050
SPACEFLIGHT DYNAMICS
Prox Ops
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 15: ProxOps, Lambert
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CW / Hill Equations
• Enter Clohessy/Wiltshire (1960)
• And Hill (1878)
• (notice the timeline here – when did Sputnik launch?
Gemini was in need of this!)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
Use RSW coordinate system (may be different from NASA)
Target satellite has two-body motion:
m rtgt
rtgt = - 3
rtgt
The interceptor is allowed to have thrusting
mr
rint = - 3int + F
rint
Then rrel = rint - rtgt Þ rrel = rint - rtgt
So, rrel = -
m rint
Lecture 15: ProxOps, Lambert
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int
r
+F+
m rtgt
rtgt3
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
If we assume circular motion,
{
m
3
tgt
r
= w2 ,
w = 0,
}
rrel R = -w 2 xRˆ + ySˆ + zWˆ - 3xRˆ + F + 2w yRˆ - 2w xSˆ + w 2 xRˆ + w 2 ySˆ
Thus,
x - 2wy - 3w 2 x = f x
y + 2w x = f y
CW or Hill’s Equations
z + w2 z = fz
Assume F = 0 (good for impulsive DV maneuvers, not for
continuous thrust targeting).
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Assumptions
• Please Take Note:
–
–
–
–
We’ve assumed a lot of things
We’ve assumed that the relative distance is very small
We’ve assumed circular orbits
These equations do fall off as you break these assumptions!
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
The above equations can be solved (see book: Algorithm 48)
leaving:
æ
æ
2y ö
2y ö
sin w t - ç 3x0 + 0 ÷ cosw t + ç 4x0 + 0 ÷
è
è
w
w ø
w ø
æ
æ
4y ö
2x
2x ö
y ( t ) = ç 6x0 + 0 ÷ sin w t + 0 cosw t - ( 6w x0 + 3y0 ) t + ç y0 - 0 ÷
è
è
w ø
w
w ø
z
z ( t ) = z0 cosw t + 0 sin w t
x (t ) =
x0
w
x ( t ) = x0 cosw t + ( 3w x0 + 2y0 ) sin w t
y ( t ) = ( 6w x0 + 4y0 ) cosw t - 2x0 sin w t - ( 6w x0 + 3y0 )
z ( t ) = -z0w sin w t + z0 cosw t
So, given x0 , y0 , z0 , x0 , y0 , z0 of interceptor, can compute x , y , z ,
x , y , z of interceptor at future time.
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Applications
• What can we do with these equations?
• Estimate where the satellite will go after executing a
small maneuver.
• Rendezvous and prox ops!
• Examples. First from the book.
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Hubble’s Drift from Shuttle
• RSW Coordinate Frame
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
æ
æ
2y ö
2y ö
sin w t - ç 3x0 + 0 ÷ cosw t + ç 4x0 + 0 ÷
è
è
w
w ø
w ø
æ
æ
4y ö
2x
2x ö
y ( t ) = ç 6x0 + 0 ÷ sin w t + 0 cosw t - ( 6w x0 + 3y0 ) t + ç y0 - 0 ÷
è
è
w ø
w
w ø
z
z ( t ) = z0 cosw t + 0 sin w t
x (t ) =
x0
w
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
æ
æ
2y ö
2y ö
sin w t - ç 3x0 + 0 ÷ cosw t + ç 4x0 + 0 ÷
è
è
w
w ø
w ø
æ
æ
4y ö
2x
2x ö
y ( t ) = ç 6x0 + 0 ÷ sin w t + 0 cosw t - ( 6w x0 + 3y0 ) t + ç y0 - 0 ÷
è
è
w ø
w
w ø
z
z ( t ) = z0 cosw t + 0 sin w t
x (t ) =
x0
w
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
We can also determine DV needed for rendezvous. Given x0, y0, z0, we
want to determine x0 , y0 , z0 necessary to make x=y=z=0. Set first 3
equations to zero, and solve for x0 , y0 , z0 .
w x0 (4 - 3 cos w t ) + 2(1 - cos w t )y0
sin w t
(6 x0 (wt - sin wt ) - y0 )w sin wt - 2wx0 (4 - 3 cos wt )(1 - cos wt )
y0 =
(4 sin wt - 3wt )sin wt + 4(1 - cos wt )2
z0 = - z0w cot w t
x0 = -
Assumptions:
1. Satellites only a few km apart
2. Target in circular orbit
3. No external forces (drag, etc.)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
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Clohessy-Wiltshire (CW) Equations
(Hill’s Equations)
NOTE: This is not the Delta-V,
this is the new required relative
velocity!
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CW/H Examples
• Scenario 1:
Use eqs to compute x(t), y(t), and z(t)
RSW coordinates, relative to deployer.
Note: in the CW/H equations,
the reference state doesn’t
move – it is the origin!
Deployment: x0, y0, z0 = 0,
• Scenario 1b:
Lecture 15: ProxOps, Lambert
Delta-V = non-zero
Use eqs to compute x(t), y(t), and z(t)
RSW coordinates, relative to reference.
Initial state: non-zero
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CW/H Equations
• Rendezvous
– For rendezvous we usually specify the coordinates relative to the
target vehicle and set x, y, and z to zero
– Though if there’s a docking port, then that will be offset from
the center of mass of the vehicle.
– Define RSW targets: x, y, z (often zero)
Initial state in the
RSW frame
Lecture 15: ProxOps, Lambert
Target: some constant
values in the RSW frame
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CW/H Equations
• Rendezvous
– For rendezvous we usually specify the coordinates relative to the
target vehicle and set x, y, and z to zero
– Though if there’s a docking port, then that will be offset from
the center of mass of the vehicle.
– Define RSW targets: x, y, z (often zero)
This is easy if the targets are zero (Eq. 6-66)
This is harder if they’re not!
Initial state in the
RSW frame
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Velocity needed to
get onto transfer
Target: some constant
values in the RSW frame
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CW/H Equations
• Rendezvous
– For rendezvous we usually specify the coordinates relative to the
target vehicle and set x, y, and z to zero
– Though if there’s a docking port, then that will be offset from
the center of mass of the vehicle.
– Define RSW targets: x, y, z (often zero)
This is easy if the targets are zero (Eq. 6-66)
This is harder if they’re not!
Initial state in the
RSW frame
Velocity needed to
get onto transfer
The Delta-V is the
difference of these
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velocities
Target: some constant
values in the RSW frame
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CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
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Shuttle: reference frame
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CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
Result of deployment
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Shuttle: reference frame
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CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
Rendezvous trajectory (Eq 6-66)
Result of deployment
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Shuttle: reference frame
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CW / H
• Scenario 3: A jetpack-wielding astronaut leaves the
shuttle and then returns.
Delta-V is the difference
of these velocities.
Rendezvous trajectory (Eq 6-66)
Result of deployment
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Shuttle: reference frame
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Comparing Hill to Keplerian Propagation
Deviation:
100 meters
1 cm/s
Position Error:
~1.4 meters/day
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Comparing Hill to High-Fidelity, e=0.15
Generally returns to
0.0 near one apsis
Position Error:
~9.5 km/day
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Comparing Hill to High-Fidelity, e=0.73
Generally returns to
0.0
Position Error:
~4.2 km/day
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Analyzing Prox Ops
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Analyzing Prox Ops
• What happens if you just change x0?
– Radial displacement?
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Analyzing Prox Ops
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Analyzing Prox Ops
• What happens if you just change vx0?
– Radial velocity change?
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Analyzing Prox Ops
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Analyzing Prox Ops
• What happens if you just change vy0?
– Tangential velocity change?
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Analyzing Prox Ops
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Analyzing Prox Ops
• What happens if you just change vz0?
– Out-of-plane velocity change?
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Analyzing Prox Ops
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ASEN 5050
SPACEFLIGHT DYNAMICS
Lambert’s Problem
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 15: ProxOps, Lambert
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Lambert’s Problem
• Lambert’s Problem has been formulated for several
applications:
– Orbit determination. Given two observations of a
satellite/asteroid/comet at two different times, what is the
orbit of the object?
• Passive object and all observations are in the same orbit.
– Satellite transfer. How do you construct a transfer orbit
that connects one position vector to another position vector
at different times?
• Transfers between any two orbits about the Earth, Sun, or other
body.
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Lambert’s Problem
Given two positions and the time-of-flight between them, determine
the orbit between the two positions.
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Orbit Transfer
• We’ll consider orbit transfers in general, though the OD
problem is always another application.
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Orbit Transfer
• Note: there’s no need to perform the transfer in < 1 revolution.
Multi-rev solutions also exist.
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Orbit Transfer
• Consider a transfer from Earth orbit to Mars orbit about
the Sun:
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Orbit Transfer
Orbit Transfer
True Anomaly Change
“Short Way”
Δν < 180°
“Long Way”
Δν > 180°
Hohmann Transfer (assuming coplanar)
Δν = 180°
Type I
0° < Δν < 180°
Type II
180° < Δν < 360°
Type III
360° < Δν < 540°
Type IV
540° < Δν < 720°
…
…
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Lambert’s Problem
• Given:
• Find:
• Numerous solutions available.
– Some are robust, some are fast, a few are both
– Some handle parabolic and hyperbolic solutions as well as
elliptical solutions
– All solutions require some sort of iteration or expansion to build
a transfer, typically finding the semi-major axis that achieves an
orbit with the desired Δt.
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Ellipse
d1
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d2
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Ellipse
d1
d2
d3
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d4
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Ellipse
A transfer from r1 to r2 will be on an ellipse, with the central body
occupying one focus.
Where’s the 2nd focus?
r1
r2
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Ellipse
A transfer from r1 to r2 will be on an ellipse, with the central body
occupying one focus.
Where’s the 2nd focus?
r1
r2
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Ellipse
A transfer from r1 to r2 will be on an ellipse, with the central body
occupying one focus.
Where’s the 2nd focus?
r1
r2
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Ellipse
A transfer from r1 to r2 will be on an ellipse, with the central body
occupying one focus.
Where’s the 2nd focus?
r1
Focus is one of these
r2
Try different a values
until you hit your TOF
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Lambert’s Problem
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Lambert’s Problem
Find the chord, c, using the law of cosines and cos Dn =
r0 × r
r0 r
c = r02 + r 2 - 2r0 r cos Dn
Define the semiperimeter, s, as half the sum of the sides of the
triangle created by the position vectors and the chord
r0 + r + c
s=
2
We know the sum of the distances from the foci to any point on the
ellipse equals twice the semi-major axis, thus
2a = r + ( 2a - r )
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Lambert’s Problem
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Lambert’s Problem
The minimum-energy solution: where the chord length equals the
sum of the two radii (a single secondary focus)
2a - r + 2a - r0 = c
Thus,
Lecture 15: ProxOps, Lambert
amin =
s r0 + r + c
=
2
4
(anything less doesn’t
have enough energy)
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Lambert’s Problem
If Dn > 180°, then b e = - b e
If t > tmin , then a e = 2p - a e
Elliptic Orbits
Hyperbolic Orbits
r +r +c
s
æa ö
sinç e ÷ = 0
=
4a
2a
è2ø
r0 + r - c
s-c
æ be ö
sinç ÷ =
=
4a
2a
è2ø
t=
=
a3
m
[a
- a3
m
Lecture 15: ProxOps, Lambert
e
r +r +c
s
æa ö
sinhç h ÷ = 0
=
- 4a
- 2a
è 2ø
r0 + r - c
s-c
æ bh ö
sinhç ÷ =
=
- 4a
- 2a
è 2ø
- sin(a e )- (b e - sin(b e ))]
[sinh(a )- a
h
h
- (sinh(b h )- b h )]
Lambert’s
Solution
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Lambert’s Problem
For minimum energy
- - elliptic orbit
- - ae = p
s-c
æ be ö
- - sinç ÷ =
s
è2ø
t min =
3
amin
m
[p - b
m pmin
v0 =
r0 r sin Dn
Lecture 15: ProxOps, Lambert
e
+ sin(b e )]
ì é
ù ü
r
{1 - cos Dn }ú r0 ý
ír - ê1 û þ
î ë pmin
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Universal Variables
• A very clear, robust, and straightforward solution.
– There are a few faster solutions, but this one is pretty clean.
• Begin with the general form of Kepler’s equation:
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Universal Variables
• Simplify
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Universal Variables
• Define Universal Variables:
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Universal Variables
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Universal Variables
• Use the trigonometric identity
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Universal Variables
• Now we need somewhere to go
• Let’s work on converting this to true anomaly, via:
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Universal Variables
• Multiply
by a convenient factoring expression:
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Universal Variables
• Collect into pieces that can be replaced by true
anomaly
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Universal Variables
• Substitute in true anomaly:
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Universal Variables
• Trig identity again:
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Universal Variables
• Note:
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Universal Variables
• Use some substitutions:
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Universal Variables
• Summary:
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Universal Variables
• It is useful to convert to f and g series (remember
those!?)
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UV Algorithm
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A few details on the Universal Variables
algorithm
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Universal Variables
• Let’s first consider our Universal Variables Lambert
Solver.
• Given: R0, Rf, ΔT
• Find the value of ψ that yields a minimum-energy
transfer with the proper transfer duration.
• Applied to building a Type I transfer
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Single-Rev Earth-Venus Type I
-4π
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4π2
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Single-Rev Earth-Venus Type I
-4π
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4π2
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Note: Bisection method
-4π
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4π2
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Note: Bisection method
-4π
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4π2
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Note: Bisection method
-4π
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4π2
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Note: Bisection method
-4π
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4π2
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Note: Bisection method
-4π
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Note: Bisection method
-4π
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4π2
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Note: Bisection method
-4π
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Note: Bisection method
• Time history of
bisection method:
• Requires 42 steps to
hit a tolerance of 10-5
seconds!
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Note: Newton Raphson method
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Note: Newton Raphson method
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Note: Newton Raphson method
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Note: Newton Raphson method
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Note: Newton Raphson method
• Time history of Newton
Raphson method:
• Requires 6 steps to hit a
tolerance of 10-5
seconds!
• Note: This CAN break in
certain circumstances.
• With current computers,
this isn’t a HUGE speedup, so robustness may be
preferable.
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Note: Newton Raphson Log Method
Note log
scale
-4π
Lecture 15: ProxOps, Lambert
4π2
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Single-Rev Earth-Venus Type I
-4π
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4π2
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Single-Rev Earth-Venus Type II
-4π
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4π2
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Interesting: 10-day transfer
-4π
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4π2
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Interesting: 950-day transfer
-4π
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4π2
100
UV Algorithm
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Multi-Rev
• Seems like it would be better to perform a multi-rev
solution over 950 days than a Type II transfer!
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A few details
• The universal variables construct ψ represents the
following transfer types:
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Multi-Rev
ψ
ψ
ψ
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ψ
104
Earth-Venus in 850 days
Type IV
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Earth-Venus in 850 days
Heliocentric View
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Distance to Sun
106
Earth-Venus in 850 days
Type VI
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Earth-Venus in 850 days
Heliocentric View
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Distance to Sun
108
What about Type III and V?
Type III
Type IV
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Type V
Type VI
109
Earth-Venus in 850 days
Type III
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Earth-Venus in 850 days
Heliocentric View
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Distance to Sun
111
Earth-Venus in 850 days
Type V
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Earth-Venus in 850 days
Heliocentric View
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Distance to Sun
113
Summary
• The bisection method requires modifications for
multi-rev.
• Also requires modifications for odd- and even-type
transfers.
• Newton Raphson is very fast, but not as robust.
• If you’re interested in surveying numerous revolution
combinations then it may be just as well to use the
bisection method to improve robustness
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Types II - VI
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