Transcript ARO309_week07-09

ARO309 - Astronautics and
Spacecraft Design
Winter 2014
Try Lam
CalPoly Pomona Aerospace Engineering
Relative Motion
Chapter 7
Relative Motion and Rendezvous
• In this chapter we will look at the relative
dynamics between 2 objects or 2 moving
coordinate frames, especially in close
proximity
• We will also look at the linearized motion,
which leads to the Clohessy-Wiltshire (CW)
equations
Co-Moving LVLH Frame (7.2)
Local Vertical Local
Horizontal (LVLH) Frame
ˆi = rA ,
rA
ˆj = kˆ ´ ˆi,
ˆk = h A
hA
CHASER
(or observer)
TARGET
Co-Moving LVLH Frame
• The target frame is moving at an angular rate of Ω
h A = rA ´ v A = rA vA^kˆ = rA2W kˆ = rA2Ω
where
rA ´ v Aand
Ω=
rA2
dæ1ö
v A × rA
Ω = h A ç 2 ÷ = -2 2 Ω
dt è rA ø
rA
• Chapter 1: Relative motion in the INERTIAL (XYZ)
frame
rrel = rB - rA
v rel = v B - v A - Ω ´ rrel
a rel = a B - a A - Ω ´ rrel - Ω ´ (Ω ´ rrel ) - 2Ω ´ v rel
Co-Moving LVLH Frame
• We need to find the motion in the non-inertial
rotating frame
rrel _ rot = Qiner _ to _ rot rrel _ iner
v rel _ rot = Qiner _ to _ rot v rel _ iner
a rel _ rot = Qiner _ to _ rot a rel _ iner
where Q is the rotating matrix from
Qiner _ to _ rot
é
ê
=ê
ê
êë
ˆi ù é rA / rA
ú ê
ˆj ú = ê kˆ ´ ˆi
ú ê
kˆ úû ê h A / hA
ë
ù
ú
ú
ú
ú
û
Co-Moving LVLH Frame
• Steps to find the relative state given the inertial state
of A and B.
1. Compute the angular momentum of A, hA
2. Compute the unit vectors ˆi, ˆj, and kˆ
3. Compute the rotating matrix Q
4. Compute Ω and Ω
5. Compute the inertial acceleration of A and B
a A = -m rA / rA3, and a B = -m rB / rB3
Co-Moving LVLH Frame
• Steps to find the relative state given the inertial state
of A and B.
6. Compute the relative state in inertial space
7. Compute the relative state in the rotating coordinate
system
rrel _ rot = Qiner _ to _ rot rrel _ iner
v rel _ rot = Qiner _ to _ rot v rel _ iner
a rel _ rot = Qiner _ to _ rot a rel _ iner
Co-Moving LVLH Frame
Rotating Frame
Linearization of the EOM (7.3)
r = R + dr
d r = -R - m
dr / R <<1
R + dr
R + dr
3
neglecting
higher order
terms
ù
mé
3
d r = - 3 êd r - 2 ( R × d r) Rú
û
R ë
R
Linearization of the EOM
Assuming R = R ˆi
é
-2 d x
ê
m
dr = - 3 ê d y
R ê
ë dz
ù
ú
ú = rB - rA = a B - a A
ú
û
Acceleration of B
relative to A in the
inertial frame
darel = dr -Ω ´ dr -Ω ´ (Ω ´ dr) - 2Ω ´ d vrel
h ˆ
Ω= 2 k
R
2 ( V × R) h ˆ
Ω=k
4
R
Linearization of the EOM
After further simplification we get the following EOM
æ 2m h 2 ö
2 ( V × R) h
2h
d x - ç 3 + 4 ÷d x +
dy - 2 dy = 0
4
R
R
èR R ø
æ m h2 ö
2 ( V × R) h
2h
d y + ç 3 - 4 ÷d y dx + 2 dx = 0
4
R
R
èR R ø
dz +
m
R
3
dz = 0
Thus, given some initial state R0 and V0 we can integrate
the above EOM (makes no assumption on the orbit type)
Linearization of the EOM
e = 0.1
e=0
Clohessy-Whiltshire (CW)
Equations (7.4)
( V× R) = 0 and h = m R
Assuming circular orbits:
Then EOM becomes
d x - 3n 2d x - 2n d y = 0
d y + 2n d x = 0
d z + n 2d z = 0
where
n=
m
R
3
=V / R
Clohessy-Whiltshire (CW)
Equations
Where the solution to the CW Equations are:
æ
ö
2
1
2
d x = 4d x0 + d y0 + d x0 sin ( nt ) - ç 3d x0 + d y0 ÷ cos ( nt )
è
ø
n
n
n
æ
ö
2
2
2
d y = d y0 - d x0 - 3 ( 2n d x0 + d y0 ) t + 2 ç 3d x0 + d y0 ÷ sin ( nt ) + d x0 cos ( nt )
è
ø
n
n
n
1
d z = d z0 sin ( nt ) + d z0 cos ( nt )
n
Maneuvers in the CW Frame (7.5)
The position and velocity can be written as
d r ( t ) = Frrd r0 + Frvd v 0
d v ( t ) = Fvrd r0 + Fvvd v 0
where
æ 4 - 3cos nt
0
( ) 0
ç
Frr = ç 6 (sin ( nt ) - nt ) 1
0
ç
ç
0
0 cos ( nt )
è
æ
(1 / n) sin ( nt )
ç
ç
Frv = ç ( 2 / n) ( cos ( nt ) -1)
çç
0
è
( 2 / n) (1- cos ( nt ))
(1 / n) ( 4sin ( nt ) - 3nt )
0
ö
÷
÷
÷
÷
ø
ö
÷
÷
0
÷
(1 / n) sin ( nt ) ÷÷ø
0
Maneuvers in the CW Frame
and
æ
3nsin ( nt )
0
0
ç
Fvr = ç 6n ( cos ( nt ) -1) 0
0
ç
ç
0
0 -nsin ( nt )
è
ö
÷
÷
÷
÷
ø
æ cos nt
2sin ( nt )
0
( )
ç
Fvv = ç -2sin ( nt ) 4cos ( nt ) - 3
0
ç
ç
0
0
cos ( nt )
è
ö
÷
÷
÷
÷
ø
Maneuvers in the CW Frame
Two-Impulse Rendezvous: from
Point B to Point A
Maneuvers in the CW Frame
Two-Impulse Rendezvous: from Point B to Point A
Dv = Dv@B ( t = 0) + Dv@A (t = t f )
where
(
-1
)
Dv@B = -éëFrv ( t f )ùû éëFrr ( t f )ùû d r0 - d v-0
(
-1
)
Dv@A = éëFvr ( t f )ùû - éëFvv ( t f )ùûéëFrv ( t f )ùû éëFrr ( t f )ùû d r0
where d v 0 is the relative velocity in the Rotating frame,
i.e.,
-
dv0 = Qinertial _ to_ rotatingdv = QXx ( v s/c - vtar -Ωtardr)
If the target and s/c are in the same circular orbits then
d v-0 = 0
Maneuvers in the CW Frame
Two-Impulse Rendezvous example:
Rigid Body Dynamics
Attitude Dynamics
Chapter 9-10
Rigid Body Motion
RB = RA + RB/A
Note:
RB/A = constant
dRB/A / dt = w ´ RB/A
vB = vA + w ´ RB/A
a B = aA + a ´ RB/A + w ´ (w ´ RB/A )
Position, Velocity, and Acceleration of points on a rigid body, measure in the
same inertial frame of reference.
Angular Velocity/Acceleration
• When the rigid body is connected to and moving
relative to another rigid body, (example: solar panels
on a rotating s/c) computation of its inertial angular
velocity (ω) and the angular acceleration (α) must be
done with care.
• Let Ω be the inertial angular velocity of the rigid body
dw
a=
+ Ω ´w
dt
Note:
dw
if
a=
dt
Ω =w
Example 9.2
Angular Velocity of Body
Ω = N kˆ
Angular Velocity of Panel
w = -q jˆ + N kˆ
w
rA/O = - sinq iˆ + d
2
æw
ö
v A/O = w ´ rA/O = - ç q cosq + Nd ÷
è2
ø
ˆj + w cosq kˆ
2
ˆi - w N sinq jˆ - w q sinq kˆ
2
2
Example 9.2 (continues)
0
a=
dw
d
+ Ω ´w =
-q jˆ + N kˆ + N kˆ ´ -q jˆ + N kˆ = q N iˆ
dt
dt
(
) ( ) (
)
a A/O = a ´ rA/O + w ´ (w ´ rA/O )
w 2 2
a A/O =
N + q sinq iˆ - N Nd + wq cosq
2
(
)
(
)
ˆj - w q 2 cosq kˆ
2
Example: Gimbal
Ω = q kˆ + N sinq jˆ + N cosq kˆ
wrotor = q iˆ + N sinq jˆ + ( N cosq + wspin ) kˆ
Equations of Motion
• Dynamics are divided to translational and rotational
dynamics
Translational:
Ftrans = m RG
Equations of Motion
• Dynamics are divided to translational and rotational
dynamics
Rotational:
M Pnet =
ò r´R
dm
m
M Pnet =
ò r ´ dF
net
MPnet = HP + v p ´ mvG
If v p = vG then
MGnet = HG
where
HG =
ò r ´ (w ´ r )
dm
Angular Momentum
HG =
ò r ´ (w ´ r )
dm
?
r ´ (w ´ r ) = w ( r × r ) - r (w × r )
æ
2
2
y
+
z
)w x - xyw y - xzw z
ç (
ç
r ´ (w ´ r ) = ç -yxw x + ( x 2 + z 2 ) w y - yzw z
ç
ç -zxw x - zyw y + ( x 2 + y 2 ) w z
è
ö
÷
÷
÷
÷
÷
ø
Angular Momentum
æ H
ç x
HG = ç H y
ç
ç Hz
è
Since:
æ I
ç xx
I = ç I yx
ç
ç I zx
è
I xy
I xz
I yy
I yz
I zy
I zz
Note:
æ
ö ç
÷ ç
÷=ç
÷ ç
÷
ø ç
è
ö
æ w ö
÷
ç x ÷
÷ = [? ] ç w y ÷
÷
ç
÷
÷
ç wz ÷
ø
è
ø
2
2
y
+
z
(
) dm
ò
- ò yx dm
- ò zx dm
HG = Iw
- ò xy dm
ò ( x + z ) dm
- ò zy dm
I yx = I xy , I zx = I xz , and I yz = I zy
2
2
ö
÷
÷
- ò yz dm
÷
÷
2
2
ò ( x + y ) dm ÷ø
- ò xz dm
Angular Momentum
If
I has 2 planes of symmetry then
æ I
ç xx
I =ç 0
ç
ç 0
è
therefore
0
I yy
0
0 ö÷ æ A 0 0 ö
ç
÷
0 ÷=ç 0 B 0 ÷
÷ ç 0 0 C ÷
I zz ÷ø è
ø
H x = A w x, H y = A w y, H z = A wz
Moments of Inertia
Euler’s Equations
• Relating M and w for pure rotation. Assuming
body fixed coordinate is along principal axis of
inertia
• Therefore Mnet = H = Hrelative + W´ H
H=
M net =
( Aw
x
(H
x
Bw y
Hy
Hz
Cw z
)+
T
) = ( Aw
T
x
Bw y
i
j
k
Wx
Wy
Wz
Aw x
Bw y
Cw z
Cw z
)
T
Euler’s Equations
• Assuming that moving frame is the body
frame, then Ω = wthis leads to Euler’s
Equations:
M xnet = Aw x + (C - B) w yw z
M ynet = Bw y + ( A - C ) w zw x
M znet = Cw z + ( B - A) w xw y
Kinetic Energy
1
T=
2
1
ò v dm = 2
m
2
1 2 1
ò v × v dm = 2 mvG + 2 w × HG = Ttrans + Trot
m
1
1 T
Trot = (w x H x + w y H y + w z H z ) = w Iw
2
2
Spinning Top
• Simple axisymmetric top spinning at point 0
Introduces the topic of
1. Precession
2. Nutation
3. Spin
Assumes:
I xx = I yy = A and I zz = C
wp = f
wn = q
Notes:
If A < C (oblate)
If C < A (prolate)
Spinning Top
From the diagram we note 3
rotations:
w = wniˆ + w p Kˆ + ws kˆ
where
Kˆ = sinq jˆ + cosq kˆ
therefore:
wp = f
wn = q
æ w ö æ
wn
ç x ÷ ç
w = ç w y ÷ = ç w p sin q
ç
÷ ç
ç w z ÷ ç w s + w p cosq
è
ø è
ö
÷
÷
÷
÷
ø
Spinning Top
From the diagram we note the
coordinate frame rotation
Ω = wniˆ + w p Kˆ
therefore:
wp = f
\ M 0net
æ Aw
x
ç
= ç Aw y
ç
ç Cw z
è
æ W ö æ
wn
ç x ÷ ç
Ω = ç W y ÷ = ç w p sin q
ç
÷ ç
ç Wz ÷ ç w p cosq
è
ø è
wn = q
ö
iˆ
÷
÷ + Wx
÷
÷
Aw x
ø
jˆ
kˆ
Wy
Wz
Aw y
Cw z
ö
÷
÷
÷
÷
ø
= d kˆ ´ ( -mg) Kˆ = mgd sinq iˆ
Spinning Top
• Some results for a spinning top
– Precession and spin rate are constant w p = ws = 0
– For precession two values exist (in general) for q ¹ 90°
æ
ö
4mgd
A
C
cos
q
C
(
)
çw s ± w s2 ÷
wp =
2
÷
2 ( A - C ) cosq çè
C
ø
– If spin rate is zero then
mgd
w p w =0 = ±
s
(C - A) cosq
if
(C - A) cosq > 0
• If A > C, then top’s axis sweeps a cone below the horizontal plane
• If A < C, then top’s axis sweeps a cone above the horizontal plane
Spinning Top
• Some results for a spinning top
– If
( A - C) cosq = 0
wp =
then
mgd
Cw s
if
( A - C ) cosq = 0
A = C , then precession occurs regardless of title angle
A ¹ C , then precession occurs title angle 90 deg
– If ( A - C) cosq > 0 then a minimum spin rate is required
for steady precession at a constant tilt
• If
• If
2
w s _ MIN =
mgd ( A - C ) cosq
if ( A - C ) cosq > 0
C
– If ( A - C) cosq < 0 then w s ® 0, and w p ® w p w =0
s
Axisymmetric Rotor on Rotating
Platform
q = 90°
w = w p jˆ + ws kˆ
Ω = w p jˆ
iˆ
jˆ
kˆ
MGnet = Ω ´ H = 0
wp
0
0 Aw p
= w p jˆ ´ Cw s kˆ = éëw p ùû ´ Hs
Cw s
Thus, if one applies a torque or moment (x-axis) it will precess, rotating
spin axis toward moment axis
Euler’s Angles (revisited)
• Rotation between body fixed x,y,z to rotation angles
using Euler’s angles (313 rotation)
Qiner _ to_ body = R3 (y ) R1 (q ) R3 (f )
æ
ç
ç
ç
ç
è
wbody = Qiner _ to_ bodywinertal
æ w ö æ w sinq sin y + w cos y ö
n
÷
ç x ÷ ç p
w body = ç w y ÷ = ç w p sin q cos y - w n sin y ÷
÷
ç
÷ ç
ç wz ÷ ç
÷
w s + w p cosq
è
ø è
ø
æ
1
ç
æ
ö
w x sin y + w y cos y )
(
wp ö ç f ÷ ç
sin q
÷
wn ÷ = ç q ÷ = ç
w x cos y - w y sin y
÷ ç
÷ ç
y
ws ÷ ç
÷ ç -1
ø è
ø ç
w x sin y + w y cos y ) + w z
(
è tan q
ö
÷
÷
÷
÷
÷
÷
ø
Euler’s Angles (revisited)
æ w ö æ w sinq sin y + w cos y
n
ç x ÷ ç p
w body = ç w y ÷ = ç w p sin q cos y - w n sin y
ç
÷ ç
ç wz ÷ ç
w s + w p cosq
è
ø è
ö
÷
÷
÷
÷
ø
M xnet = Aw x + (C - B) w yw z
M ynet = Bw y + ( A - C ) w zw x
M znet = Cw z + ( B - A) w xw y
ainertal = QTiner _ to _ bodywbody
Satellite Attitude Dynamics
• Torque Free Motion MG _ net = HG = 0 = HG _ rel + w ´ HG
HG ˆ
cosq =
×k
HG
A - B) w xw y
(
q = wn = HG sin q
Euler’s Equation for Torque Free
Motion
0 = Aw x + (C - A) w yw z
0 = Bw y + ( A - C ) w zw x
Aw x + (C - A) w yw z = 0
A=B
Aw y + ( A - C ) w zw x = 0
0 = Cw z + ( B - A) w xw y
w x - lw y = 0
w y + lw x = 0
Cw z = 0
A-C
l=
w0
A
w z = w 0 = constant
wn = 0
w x + lw x = 0
Euler’s Equation for Torque Free
Motion
C
w xy = w 0 tanq
A
For
Then:
w = w^ + w0 = w^ + w0 kˆ
A-C
ws = y =
w0
A
C ws
wp = f =
A - C cosq
If A > C (prolate), ωp > 0
If A < C (oblate), ωp < 0
Euler’s Equation for Torque Free
Motion
Euler’s Equation for Torque Free
Motion
æ H
ç x
HG = ç H y
ç
ç Hz
è
ö æ Aw
x
÷ ç
÷ = ç Aw y
÷ ç
÷ ç Cw z
ø è
tanq =
ö æ Aw
x
÷ ç
÷ = ç Aw y
÷ ç
÷ ç Cw 0
ø è
A
tan g
C
If A > C (prolate), γ < θ
If A < C (oblate), γ > θ
HG = Aw p
ö
÷
÷
÷
÷
ø
Euler’s Equation for Torque Free
Motion
Stability of Torque-Free S/C
Assumes: w = w 0 kˆ
0 = Aw x + (C - A) w yw z
0 = Bw y + ( A - C ) w zw x
0 = Cw z + ( B - A) w xw y
dw x,y + kdw x,y = 0
A - C) ( B - C) 2
(
k=
w
AB
0
Stability of Torque-Free S/C
dw x,y + kdw x,y = 0
• If k > 0, then dw x,y = c1ei kt + c2 e-i kt solution is
bounded
• A > C and B > C or A < C and B < C
• Therefore, spin is the major axis (oblate) or minor
axis (prolate)
• If k < 0, then dw x,y = c1e + c2 e
solution is
unstable
• A > C > B or A < C < B
• Therefore, spin is the intermediate axis
kt
- kt
Stability of Torque-Free S/C
• With energy dissipation ( Trot < 0 )
1 T
1
1
2
Trot = w Iw = Aw^ + Cw z2
2
2
2
1 æ
1 dw^2 ö
dw ^2
C æ Trot ö
wz =
=2 ç
çTrot - A
÷
÷
Cw z è
2 dt ø
dt
A èC - Aø
dw ^2
<0
dt
dw ^2
>0
dt
if
C > A (oblate) ¬ asymtotically stable
if
C < A ( prolate) ¬ unstable
Stability of Torque-Free S/C
• Kinetic Energy relations
2
1
1
1
H
1æ A-C ö 2
2
2
G
Trot = Aw^ + Cw z =
+ ç
÷ Cw z
2
2
2 A 2è A ø
1 H G2 æ A - C
2 ö
Trot =
cos q ÷
ç1+
ø
2 A è
A
ì
ï
ï
Trot = í
ï
ïî
1 H G2
2 C
1 H G2
2 A
for major axis spinner
for minor axis spinner
Conning Maneuvers
• Maneuver of a purely
spinning S/C with fixed
angular momentum
magnitude
w = w0 kˆ
HG,0 = Cw0 kˆ
DHG = DHG1 + DHG2
DHG =
tf
òM
0
G
dt
Conning Maneuvers
Before the Maneuver
ws = w0
During the Maneuver
æ A-C ö
ws = ç
÷w 0
è A ø
HG = Cw0
ö
C æ w0
÷÷
w P = çç
A è cos (q / 2) ø
Cw 0
H G = Aw p =
cos (q / 2)
Another maneuver is required ΔHG2 after precession 180 deg
Conning Maneuvers
Another maneuver is required ΔHG2 after precession 180 deg.
At the 2nd maneuver we want to stop the precession
(normal to the spin axis):
DHG1 = DHG2
ws = w p
æ C ö
q = 2 cos ç
÷
è A-C ø
-1
Required deflection angle
to precess 180 deg for a
single coning mnvr
t=
p
pA
=
cos (q / 2)
w p Cw 0
DHG = DHG1 + DHG 2
= 2 ( HG 0 tan (q / 2))
» HG 0 q
Gyroscopic Attitude Control
• Momentum
exchange gyros or
reaction wheels can
be used to control
S/C attitude without
thrusters.
• The wheels can be
fixed axis (reaction
wheels) or gimbal 2axis (cmg)
Gyroscopic Attitude Control
(
)
i
HG = IGbus + å IGi w + å IGiwrel
= I w + åI w
s/c
G
i
G
i
rel
MG,net,ext
dHG
=
+ w ´ HG
dt
Example:
H G = ( I p + I w ) w + I wwrel
If external torque free then
HG (t = 0) = HG (t = Dt )
therfore
Dwrel = - (1+ I p / I w ) Dw
Gyroscopic Attitude Control
Example II: S/C with three identical wheels with their axis
along the principal axis of the S/C bus, where the wheels
spin axis moment of inertial is I and other axis are J. Also,
the bus moment of inertia are diagonal elements (A, B, C).
HG = ( I B + I1 + I 2 + I3 ) w + I1w1 + I 2w2 + I3w3