Orbit Determination
3.5 Transforming Orbital Elements to Cartesian Coordinates
3.5.1 Coordinate Transformations
Let us begin with an example of a simple transformation where the two coordinate frames differ by a single rotation about a primary axis. Figure 3.8 shows the rotation of orthogonal axes XYZ about its +Z axis through angleχ to produce the orthogonal
“primed” axes X Y Z . The angular rotation χ is positive as defined by the “right-hand rule”; that is, curling the fingers of the right hand about the Z axis so that the thumb points“up” in Figure 3.8. Therefore, the new Z is aligned with the original Z axis and the new X and Y axes are in the same plane as the original X–Y plane. Now, suppose we have the position vectorrXYZ= 5I + 5K as shown in Figure 3.8 where IJK are unit vec-tors along the original XYZ coordinate system. The same position vector expressed in the X Y Z coordinate system is
rX Y Z =
cosχ sinχ 0
−sinχ cosχ 0
0 0 1
rXYZ (3.26)
Using the position vectorrXYZ= 5I + 5K shown in Figure 3.8, we have
rX Y Z =
cosχ sinχ 0
−sinχ cosχ 0
0 0 1
5 0 5
Carrying out the matrix-vector multiplication, we obtainrX Y Z = 5cosχI − 5sinχJ + 5K where I J K are unit vectors along the new “primed” coordinate system X Y Z . The Z-components ofrX Y Z andrXYZare identical because the rotation to establish X Y Z was about the +Z axis [note that the elements of the third row and third column of the rotation matrix in Eq. (3.26) are zero except for the“1” in the lower right corner that preserves the Z component]. The reader should note that the vector norms of rXYZ
and rX Y Z are both 50 which shows that the magnitude of the position vector did not change after the coordinate transformation.
Let us rewrite the coordinate transformation denoted by Eq. (3.26) as
rX Y Z =C χ rXYZ (3.27)
whereC(χ) is the rotation matrix associated with a positive rotation about the +Z axis through angleχ
C χ =
cosχ sinχ 0
−sinχ cosχ 0
0 0 1
(3.28)
Rotation matrixC(χ) is an orthogonal matrix; that is, its rows and columns are orthog-onal unit vectors [the rows (or columns) are orthogorthog-onal because the dot product of any two rows (or columns) is zero]. Because matrixC(χ) is orthogonal, it has the properties
X Y
Z Z ′
X ′
Y ′ χ
χ χ
rXYZ= 5I + 5K
5 5 0
Figure 3.8 Coordinate transformation: positive rotation about the Z axis.
C χ CT χ = CT χ C χ = I (3.29)
C−1 χ = CT χ (3.30)
where superscript T indicates the transpose of matrixC(χ) and I is the 3 × 3 identity matrix. Multiplying both sides of Eq. (3.27) byC−1 χ = CT χ yields
rXYZ=CT χ rX Y Z (3.31)
which is the transformation from the X Y Z frame to the XYZ frame.
A coordinate transformation may involve a positive rotation about the +X axis.
Figure 3.9 shows the rotation of Cartesian frame XYZ about its +X axis through angle α to produce the frame X Y Z . A position vector rXYZexpressed in the XYZ coordinate system may be transformed using the rotation matrix
rX Y Z =
1 0 0
0 cosα sinα 0 −sinα cosα
rXYZ
=A α rXYZ
(3.32)
The rotation matrixA(α) is orthogonal and the X component of rXYZis unaltered by the transformation.
Figure 3.10 shows the rotation of Cartesian frame XYZ about its +Y axis through angle β to produce the frame X Y Z . Transformation of vector rXYZfrom the XYZ coordinate system to X Y Z is
rX Y Z =
cosβ 0 −sinβ
0 1 0
sinβ 0 cosβ rXYZ
=B β rXYZ
(3.33)
Equations (3.32), (3.33), and (3.27) are the coordinate transformations from the XYZ frame to a new frame X Y Z created after a single positive rotation about a principle axis.
X Y
Z ′ Z
X ′
α α Y ′
α
0
Figure 3.9 Coordinate transformation: positive rotation about the X axis.
Remember that our goal is to develop a transformation between the ECI (orIJK) coor-dinate system and the perifocalPQW system. In general, the IJK system can be aligned with thePQW system after three successive rotations, as Figure 3.11 illustrates. The first rotation is about theK (or +Z) axis through the ascending node angle Ω (Figure 3.11a).
After this first rotation, theI (or +X) axis becomes the X axis (i.e., the ascending node vectorn) and the X Y axes remain in the equatorial plane. The second rotation is about the intermediate X axis through the inclination i (Figure 3.11b). This rotation establishes the angular momentum vectorh (note that the Z axis in Figure 3.11b is aligned with the W axis of the perifocal system). The third and final rotation is about the Z (or W) axis through the argument of perigeeω (Figure 3.11c) in order to establish the P axis (perigee direction) and the perifocalPQW coordinate system. Next, let us apply the appropriate rotation matrices to the position vectorrECIthat is expressed inIJK coordinates:
First rotation about +Z axis rX Y Z =C Ω rECI (3.34) Second rotation about +X axis rX Y Z =A i rX Y Z (3.35) Third rotation about +Z axis rPQW=C ω rX Y Z (3.36) Note that there is no Y-axis rotation in our sequence of rotations fromIJK to PQW.
Finally, we can combine Eqs. (3.34)–(3.36) to yield the direct computation of rPQWfrom IJK coordinates rECI:
rPQW=C ω A i C Ω rECI (3.37)
Equation (3.37) transforms the position vectorrECI(expressed inIJK coordinates) to position vector rPQW (expressed in PQW coordinates). The order of multiplying the rotation matrices is important and must be preserved: the first rotationC(Ω) is the last matrix in the left-to-right matrix multiplication presented in Eq. (3.37).
Remember that our overall goal is to derive the state vector (r,v) in the ECI (or IJK) coordinate system from the orbital elements. Our first step involved computing the posi-tion and velocity vectorsrPQWandvPQW in perifocal coordinates [see Eqs. (3.18) and (3.23)]. Therefore, we know the left-hand side of Eq. (3.37). We can solve Eq. (3.37) forrECIby multiplying both sides by the matrix inverse ofC(ω)A(i)C(Ω)
X Y
Z Z ′
X ′ β Y ′
β β
0
Figure 3.10 Coordinate transformation: positive rotation about the Y axis.
I J X ′
Y ′
Ω
Ω Ω
0
Ascending node, n (a)
i i Angular
momentum, h Z ′
X ′
Y ′
0
Ascending node,n
Y ′′
Z ′′
i
X ′′
(b)
ω ω
Angular momentum,h
X ′′
Y ′
0
Ascending node,n
Y ′′
Z ′′
ω
P Q
W
(c)
Perigee direction, e
Figure 3.11 Three rotations of IJK to establish PQW: (a) positive rotation about K through angle Ω to establish the ascending node vector; (b) positive rotation about X’ through angle i to establish the angular momentum vector; and (c) positive rotation about Z’’ (W) through angle ω to establish thePQW frame.
rECI= C ω A i C Ω −1rPQW (3.38) Comparing Eq. (3.38) with Eq. (3.24) we see that the inverse of matrixC(ω)A(i)C(Ω) is the overall rotation matrixR that transforms any vector expressed in the PQW frame to the ECI (IJK) frame. Because the product of three orthogonal matrices is also orthogonal, the inverse of matrixC(ω)A(i)C(Ω) is its transpose. Hence, the overall rotation matrix is
R =
cosω sinω 0
−sinω cosω 0
0 0 1
1 0 0
0 cosi sini 0 −sini cosi
cosΩ sinΩ 0
−sinΩ cosΩ 0
0 0 1
T
Carrying out the matrix multiplications and transposing the result, we obtain the overall rotation matrix fromPQW to IJK:
R =
cΩcω−sΩsωci −cΩsω−sΩcωci sΩsi
sΩcω+ cΩsωci −sΩsω+ cΩcωci −cΩsi
sωsi cωsi ci
(3.39)
We have used the short-hand notation cα= cosα and sα= sinα for the cosine and sine of the three rotation angles. Multiplying rotation matrixR and any vector expressed in the PQW frame will transform it to IJK coordinates.
Let us summarize the steps for transforming the classical orbital elements to the state vector (r,v) in the ECI (or IJK) frame:
1) Using Eqs. (3.18) and (3.23), determine the satellite’s position and velocity vectors in thePQW frame, rPQWandvPQW.
2) Use Eq. (3.39) to determine the overall rotation matrix fromPQW to IJK and then useR in Eqs. (3.24) and (3.25) to compute the state vector (rECI,vECI) in the ECI coor-dinate system.
This process is a bit cumbersome for calculations with a hand-held calculator. However, it is easy to see that MATLAB is well-suited to carry out the matrix-vector multiplica-tions. We encourage the reader to develop a general-purpose M-file that will compute (rECI,vECI) given an arbitrary set of six classical orbital elements.
Example 3.2 Let us consider again the Molniya orbit presented in Example 3.1.
A satellite in a Molniya orbit has the following orbital elements at epoch time t0 a= 26,564 km
e= 0 7411 i= 63 4 Ω = 200 ω = −90 θ0= 30
Determine the satellite’s state vector at this epoch in the Earth-centered inertial (ECI) coordinate system.
First, we use semimajor axis a and eccentricity e to compute parameter p and angular momentum h:
p= a 1−e2 = 11, 974 3 km, h = pμ = 69,086 5 km2/s Orbital radius atθ0= 30 is determined by the trajectory equation
r= p
1 + e cosθ= 7, 293 3 km
The satellite’s position vector in the PQW frame is computed using Eq. (3.18)
rPQW=
rcosθ0
rsinθ0
0
=
6, 316 21 3, 646 67
0
km = 6, 316 21P + 3,646 67Q km
Equation (3.23) allows us to calculate the satellite’s velocity vector in PQW
vPQW=
−μ h sinθ0
μ
h e+ cosθ0
0
=
−2 8848 9 2724
0
km/s =−2 8848P + 9 2724Q km/s
The first step is complete. Next, use Eq. (3.39) to determine the overall rotation matrix fromPQW to IJK
R =
cΩcω−sΩsωci −cΩsω−sΩcωci sΩsi
sΩcω+ cΩsωci −sΩsω+ cΩcωci −cΩsi sωsi cωsi ci
=
−0 1531 −0 9397 −0 3058 0 4208 −0 3420 0 8402
−0 8942 0 0 4478
where we have used cΩ= cosΩ = –0.9397, sω= sinω = −1, ci= cosi = 0.4478, and so on.
Finally, the position and velocity vectors in ECI coordinates are
rECI=RrPQW=
−0 1531 −0 9397 −0 3058 0 4208 −0 3420 0 8402
−0 8942 0 0 4478
6, 316 21 3, 646 67
0
=
−4,394 0 1, 410 3
−5,647 7 km
Or,rECI= – 4,394 0I + 1,410 3J – 5,647 7K km
vECI=RvPQW=
−0 1531 −0 9397 −0 3058 0 4208 −0 3420 0 8402
−0 8942 0 0 4478
−2 8848 9 2724
0
=
−8 2715
−4 3852 2 5794
km/s
Or,vECI= – 8 2715I – 4 3852J + 2 5794K km/s
As a check, the reader can compare the vector norms ofrECIandrPQW(both should equal r = 7,293.3 km) and the vector norms of vECI and vPQW (both should equal 9.7108 km/s).