EE 495 Modern Navigation Systems
Navigation Mathematics
Linear Velocity
Lecture 9
EE 495 Modern Navigation Systems
Slide 1 of 15
Navigation Mathematics:
Linear Velocity - A Fixed point in a Rotating Frame
• CASE 1: A Fixed point in a Rotating Frame
Lecture 9
EE 495 Modern Navigation Systems
Slide 2 of 15
Navigation Mathematics :
Linear Velocity - A Fixed point in a Rotating Frame
• Consider the motion of a fixed point (origin of frame
{2}) in a rotating frame (frame {1}) as seen from an
inertial (frame {0}):
A Fixed Point in a Rotating Frame
Z
 Frame {1} rotates (about 𝑘)
wrt frame {0}
 The frame {2} (origin) is
fixed wrt frame {1}
• Position:
1
Z0
0.5
k
0
Y1
{1}
{0}
Z1
-0.5
r12 r02
X0
-1
r020 (t )  r010  C10 (t ) r121
 C (t ) r
0
1
1
12
{2}
Y0
Z2
-1
Y2
X1
X2
-0.5
0
-1
-0.5
0.5
Lecture 9
EE 495 Modern Navigation Systems
0
1
0.5
1
Slide 3 of 15
Navigation Mathematics :
Linear Velocity - A Fixed point in a Rotating Frame
• Velocity: r 0 (t )  d C 0 (t ) r 1
 1 12 
02
dt
C10  001 C10  C10 101
 C10 (t ) r112
  001 (t ) C10 (t ) r121
   
0
 01
(t )   C10 (t ) r112  r120  C10 (t ) r121
 010 (t )  r120 (t )
Angular velocity of
frame {1} wrt frame {0}
Lecture 9
r
Radius of rotation
EE 495 Modern Navigation Systems
Slide 4 of 15
Navigation Mathematics :
Linear Velocity - An Example
• What is our velocity in the classroom (wrt the ECI frame)?
0
r002 (t )  01
(t )  C10 (t )r121
 rici (t )  iei (t )  Cei (t )reec
 Frame {0} is the ECI frame
 Frame {1} is the ECEF frame
 Orig of frame {2} is our classroom.
r  Rz , LON Ry , LAT
e
ec
1 
0  Earth's Radius

 
0 
 2.0 
  4.8   106 (meters)
 3.6 
iei (t )  Cei (t )rece
Let , Cei (t )  I 33
 0 
iei (t )   0  106 (rad/s)
72.9 
 353 
 146  (m/s)


 0 
 790 
 327  mph


 0 
Speed ~855 mph
Lecture 9
LAT = 34 32’ 24” (N)
LON = 112 28’ 4” (W)
Assume that the e & i-frames are aligned
EE 495 Modern Navigation Systems
Slide 5 of 15
Navigation Mathematics :
Linear Velocity - A Fixed point in a Rotating Frame
• Acceleration:

d
r (t ) 
010 (t )   C10 (t ) r121 
dt
0
02


 C 
(t )   C (t ) r    (t )    (t )C (t ) r 
(t )  r (t )   (t )   (t )  r (t ) 
0
 010 (t )   C10 (t ) r121   01
(t )  C10 (t ) r121
 010
 010
0
1
1
12
0
12
0
01
Transverse accel (𝜔𝑟)
Lecture 9
0
01
0
01
0
01
0
1
0
1
0
01
C10
1
12
0
12
Centripetal accel (𝜔2 𝑟)
EE 495 Modern Navigation Systems
Slide 6 of 15
Navigation Mathematics :
Linear Velocity - An Example
• What is our acceleration (wrt the ECI frame) due only to motion?
 Not including our gravitational mass attraction!!
0
0
r020  01
 r120  01
 010  r120   rici  iei  reci  iei  iie  reci 
 2.0 
reci (t )   4.8  106 (meters)
 3.6 
 0 
iei (t )   0  106 (rad/s)
72.9 
0  iie  iei  reci 
Lecture 9
Let , Cei (t )  I 33
0 
iei (t )  0  (rad/s 2 )
0 
 0.01 
0.026  (m/s 2 )


 0 
~3 mg
Assume that the e & i-frames are aligned
EE 495 Modern Navigation Systems
Slide 7 of 15
Navigation Mathematics :
Linear Velocity - A Fixed point in a Rotating Frame
• A simulation (case1.m):
0.5
r  0.8
 0.1
1
12
 1/ 3
ˆ0 
k01   2 / 3 
 2 / 3 
A Fixed Point in a Rotating Frame
  26t  t 2
 1/ 3
ˆ0
0
01  k01   2 / 3   26  2t 
 2 / 3 
0.8
x
y
z
r121
0.7
0.6
0.4
0.2
0.1
0
1
2
3
4
5
6
Time in sec
7
8
9
10
0
01
0
12
0
01
0
01
0
12

Z2
-1
X2
-0.5
-1
0.6
0.8
0
0.5
x
y
z
0
02
0.1
Velocity in m/s
r
0.2
x"
y"
z"
r
0.2
0
-0.2
-0.4
Accel in m/s 2
0.4
r002
1
The X, Y, & Z Acceleration of 2 Origin in frame 0: a002(t)
X
0
02
0.3
0.4
0.6
Y
0.4
-0.5
Acceleration
0.5
The X, Y, & Z Velocity of 2 Origin in frame 0: v002(t)
1
Position in m
Y2
X1
1
Motion in Frame {0}
{2}
Y0
-1
Velocity
The X, Y, & Z Position of 2 Origin in frame 0: r002(t)
0
-0.2
x'
y'
z'
-0.4
0
-0.1
-0.2
-0.3
-0.6
-0.6
-0.4
-0.8
-1
r12 r02
X0
0
Position
0.2
Y1
{1}
{0}
Z1
-0.5
r    r      r
0
02
k
0
0
r020 (t )  01
(t )  r120 (t )
0.3
Z0
0.5
r020 (t )  r010  C10 (t ) r121
0.5
0
1
Z
0.9
Position in m
Motion in Frame {1}
The X, Y, & Z Position of 2 Origin in frame 1: r112(t)
1
0
1
Lecture 9
2
3
4
5
6
Time in sec
7
8
9
10
-0.8
0
1
2
3
4
5
6
Time in sec
7
8
EE 495 Modern Navigation Systems
9
10
-0.5
0
1
2
3
4
5
6
Time in sec
7
8
Slide 8 of 15
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Navigation Mathematics:
Linear Velocity - A Moving point in a Rotating Frame
• CASE 2: A Moving point in a Rotating Frame
Lecture 9
EE 495 Modern Navigation Systems
Slide 9 of 15
Navigation Mathematics :
Linear Velocity - A Moving point in a Rotating Frame
• Consider the motion of a moving point (origin of
frame {2}) in a rotating frame (frame {1}) as seen from
an inertial (frame {0}):
A Moving Point in a Rotating Frame
 Frame 1 rotates (about k)
wrt frame 0
 The origin of frame 2 is
moving wrt frame 1
1
Z
• Position:
2
Z0
0
k
-1
X0
Now a function of Time!!
r (t )  r  C (t ) r (t )
0
02
0
01
0
1
 C (t ) r (t )
0
1
Lecture 9
1
12
1
12
{1}
{0}
Z1
Y0
-2
-2
Y1
r02 (t )
{2}
r12 (t )
Z2
X1
Y2
X2
-3
0
-2
-1
EE 495 Modern Navigation Systems
2
Y
0
1
X
Slide 10 of 15
Navigation Mathematics :
Kinematics – Linear Velocity
• Velocity:
r020 (t ) 
d
C10 (t ) r121 (t ) 

dt
 C (t ) r (t )  C (t ) r (t )
0
1
1
12
0
1
1
12
C10  001 C10
 001 (t )C10 (t ) r121 (t )  C10 (t ) r112 (t ) NOTE : r120 (t )  C10 (t ) r121 (t )
 010 (t )   C10 (t ) r112 (t )   C10 (t ) r121 (t )
 001 (t )  r120 (t )  C10 (t ) r112 (t )
0
NOTE: Be careful about differentiating 𝑟12
directly
Can differentiate 𝑟𝑖𝑗𝑖 directly but NOT 𝑟𝑖𝑗𝑘
Lecture 9
EE 495 Modern Navigation Systems
Slide 11 of 15
Navigation Mathematics :
Kinematics – Linear Velocity
• Acceleration: (Omitting explicit ?(t) for clarity)

d
r 
010   C10 r121   C10 r112
dt
0
02





0
0
 01
  C10 r121   01
 C10 r121  001  C10 r121  C10 r121  C10 r112

C10 r121  001  C10 r121
0
C10  01
C10



0
0
0
0
 01
 r120  01
 01
 r120   201
 C10 r121  C10 r121
Centripetal accel (2r)
Transverse accel
Lecture 9
Coriolis accel (2v)
EE 495 Modern Navigation Systems
The accel of {2} org as seen
in {1}, but resolved in {0}.
Slide 12 of 15
Navigation Mathematics :
Kinematics – Linear Velocity
• Why do toilets drain clockwise in the Northern hemisphere and
counter-clockwise in the Southern hemisphere?


0
0
0
0
r121  C01  r020  01
 r120  01
 01
 r120   201
 C10 r121 


ie

In truth, the effect is too
small to be relevant!
v
v

v
v

v

v
v
v
Lecture 9
EE 495 Modern Navigation Systems
Slide 13 of 15
Navigation Mathematics :
Kinematics – Linear Velocity
• Position:
0.5sin(t ) 
r121 (t )   t  0.1t 2 
 0.1t 
r002 (t )  C10 (t ) r121 (t )
 1/ 3
ˆ
k010   2 / 3 
 2 / 3 
C10 (t )  R ˆ 0
k
 (t )  26 t  t 2
01 , ( t )

 1/ 3
ˆ
010  k010    2 / 3   26  2t 
 2 / 3 
0.5cos(t ) 
r121 (t )   1  0.2t 
 0.1 
• Velocity:
0
r020 (t )  01
(t )  r120 (t )  C10 (t ) r112 (t )
• Acceleration


0
0
0
0
r020  01
 r120  01
 01
 r120   201
 C10 r121  C10 r121
Lecture 9
 1/ 3
010   2 / 3   2 
 2 / 3 
 0.5sin(t ) 
r121 (t )   0.2 


0
EE 495 Modern Navigation Systems
Slide 14 of 15
Navigation Mathematics :
Kinematics – Linear Velocity
• A simulation (case2.m):
x
y
z
0.8
Velocity in m/s
r
1
0.5
0
1
2
3
4
5
6
Time in sec
7
8
9
0.4
0.2
0.2
0
1
12
r
-0.6
-0.6
-0.8
-0.8
1
2
3
Position
The X, Y, & Z Position of 2 Origin in frame 0: r002(t)
5
6
Time in sec
7
8
9
-1
10
1
The X, Y, & Z Velocity of 2 Origin in frame 0: v002(t)
2
x
y
z
r
Velocity in m/s
0
-0.5
x'
y'
z'
0
02
r
1
0.5
4
5
6
Time in sec
7
8
9
10
The X, Y, & Z Acceleration of 2 Origin in frame 0: a002(t)
1.5
0.5
0
-0.5
x"
y"
z"
r020
1
Accel in m/s 2
0
02
3
2
1.5
1
0
Acceleration
2
2
Position in m
Motion in Frame {0}
4
Velocity
2.5
1.5
0
-0.4
0
r121
-0.2
-0.4
-1
10
0.6
0.4
-0.2
x"
y"
z"
0.8
x'
y'
z'
0.6
1
12
0
1
Accel in m/s 2
2
-0.5
The X, Y, & Z Acceleration of 2 Origin in frame 1: a112(t)
1
1.5
  26t  t 2
The X, Y, & Z Velocity of 2 Origin in frame 1: v112(t)
2.5
Position in m
Motion in Frame {1}
The X, Y, & Z Position of 2 Origin in frame 1: r112(t)
 1/ 3
ˆ0 
k01   2 / 3 
 2 / 3 
0.5sin(t ) 
r   t  0.1t 2 
 0.1t 
1
12
0.5
0
-0.5
-1
-1
-1
-1.5
-1.5
-1.5
-2
-2.5
0
1
Lecture 9
2
3
4
5
6
Time in sec
7
8
9
10
-2
0
1
2
3
4
5
6
Time in sec
7
8
EE 495 Modern Navigation Systems
9
10
-2
0
1
2
3
4
5
6
Time in sec
7
8
Slide 15 of 15
9
10