Free preview · ADCS course v2.1

Spacecraft attitude determination and control

How satellites know which way they point and turn to where they need to look: rotations, dynamics, sensors, estimation, actuators and control, ending in a complete CubeSat design. Every number checked by a script.

Free preview: chapter 1 of 18, complete. The full course: 18 chapters · 39 worked examples · 54 practice problems · 2079 script checks.

How to use this pack

1Read 'The idea in one minute' and the 'You will be able to' list.
2Work through the short sections and the worked examples with a pencil.
3Try each practice problem before opening the hint; open the full solution only to check.
4Do the three-question check and the spaced recap before moving on.

What is in the full course

1Frames and direction cosine matricesFree preview30 min
2Euler angles and gimbal lock30 min
3Quaternions40 min
4Attitude kinematics30 min
5The inertia tensor and principal axes35 min
6Euler's equations and spin stability40 min
7Disturbance torques40 min
8Sun sensors, Earth sensors and magnetometers35 min
9Star trackers and gyros40 min
10Attitude determination: TRIAD, Wahba, QUEST45 min
11Kalman filtering for attitude45 min
12Actuators: wheels, CMGs, magnetorquers, thrusters40 min
13Feedback control: PD, bandwidth and settling40 min
14Detumbling, momentum management and slews45 min
15Stabilisation modes: spin, gravity gradient, three-axis40 min
16Pointing budgets: knowledge versus control35 min
17CubeSat ADCS and testing on the ground35 min
18Worked mission case: a 6U imager end to end50 min

This preview is chapter 1, complete. Your ticks are saved in this browser only. Original work, and a study aid, not assignment answers.

Chapter 1

Frames and direction cosine matrices

30 minutes

The idea in one minute

Attitude is the answer to one question: how is the satellite turned relative to something we care about? We describe it by two sets of axes, one fixed to the reference (the stars, or the orbit) and one fixed to the body. The direction cosine matrix is a 3 × 3 table of the cosines between those axes. Multiply a vector by it and you get the same vector seen from the other frame.

Notice the two sets of axes share an origin and differ only by a rotation. The rows of A_BI are the blue body axes written in grey reference coordinates.

You will be able to

Name the frames ADCS uses: inertial, orbit (LVLH) and body
Write a direction cosine matrix and use it to change a vector's frame
Combine rotations in the right order and pull the axis and angle out of a DCM

Three frames

The inertial frame I points at fixed stars (ECI, J2000). The orbit frame O moves with the satellite: z to nadir, x roughly along the velocity, y opposite the orbit normal. The body frame B is glued to the structure. Pointing requirements are usually written as a body attitude relative to I or O.

The DCM

A_BI maps components: v_B = A_BI v_I. Its rows are the body axes in inertial components; its columns are the inertial axes in body components. It is orthonormal (A Aᵀ = I) with determinant +1, so its inverse is just its transpose.

Combining rotations

To go from I to B and then from B to C, multiply: A_CI = A_CB A_BI. The newest rotation goes on the left. Order matters: two 90° turns in different orders give different attitudes.

Euler's theorem

Any rotation, however it was built, equals one rotation by an angle a about one fixed axis e. The trace of the DCM gives the angle, tr A = 1 + 2 cos a. That axis and angle reappear in quaternions (Chapter 3).

Key equations

Changing frames

v_B = A_BI v_I

where

v_Ba vector written in body-frame components, same as v

A_BIdirection cosine matrix taking components from frame I to frame B, dimensionless

v_Ithe same vector in inertial-frame components, same as v

A rotation matrix is orthonormal

A Aᵀ = I, det A = +1

where

I₃3 × 3 identity matrix, dimensionless

So A⁻¹ = Aᵀ: going back is free.

Chaining frames

A_CI = A_CB A_BI

The newest rotation goes on the left.

Rotation about z by ψ

R₃(ψ) = [cos ψ sin ψ 0; −sin ψ cos ψ 0; 0 0 1]

where

ψyaw angle (rotation about z), deg

R₁ and R₂ follow the same pattern about x and y.

Angle of the single equivalent rotation

cos a = (tr A − 1)/2

where

aangle of the single equivalent rotation, deg

tr Atrace: sum of the diagonal elements of A, dimensionless

Worked examples

Worked example 1.1

The same Sun, seen from a turned satellite

Given

A satellite's body frame B starts aligned with the inertial frame I, then turns by 30° about the shared z axis. The Sun lies along the inertial x axis, v_I = (1, 0, 0). Write the DCM A_BI and find the Sun direction in body axes.

new symbols

A_BIdirection cosine matrix taking components from frame I to frame B, dimensionless

v_Ba vector written in body-frame components, same as v

v_Ithe same vector in inertial-frame components, same as v

ψyaw angle (rotation about z), deg

Working

A frame rotation by ψ about z is R₃(ψ) = [cos ψ sin ψ 0; −sin ψ cos ψ 0; 0 0 1]
With ψ = 30°: A_BI = [+0.866 +0.500 +0.000; -0.500 +0.866 +0.000; +0.000 +0.000 +1.000]
v_B = A_BI v_I = first column of A = (0.866, -0.500, 0.000)
Check: A Aᵀ = I and det A = 1.000, so A is a proper rotation

Answer. v_B = (0.866, -0.500, 0.000): the Sun now sits 30° on the negative-y side of the body x axis.

Sanity check. Turn the frame one way and a fixed vector appears to turn the other way. That sign flip is the most common error in attitude work.

Pythonedit a value, then Run

# Worked example 1.1: The same Sun, seen from a turned satellite
import numpy as np, math

psi = math.radians(30.0)          # frame rotation about z (rad)
v_I = np.array([1.0, 0.0, 0.0])   # Sun direction, inertial components

# frame rotation about z
A_BI = np.array([[ math.cos(psi), math.sin(psi), 0],
                 [-math.sin(psi), math.cos(psi), 0],
                 [0, 0, 1]])
v_B = A_BI @ v_I
print("A_BI =")
print(np.round(A_BI, 3))
print("v_B = (%.3f, %.3f, %.3f)" % tuple(v_B))
print("det A = %.3f" % np.linalg.det(A_BI))

A_BI =
[[ 0.866  0.5    0.   ]
 [-0.5    0.866  0.   ]
 [ 0.     0.     1.   ]]
v_B = (0.866, -0.500, 0.000)
det A = 1.000

Worked example 1.2

Rotations do not commute

Given

Turn a frame 90° about z then 90° about the new x axis. Compare with doing the two turns in the opposite order. Where does the inertial x direction end up in each case?

new symbols

A_BIdirection cosine matrix taking components from frame I to frame B, dimensionless

Working

Successive frame rotations multiply on the left: first R₃, then R₁ gives A = R₁(90°) R₃(90°)
A = [+0 +1 +0; -0 +0 +1; +1 -0 +0], so v_B = A (1, 0, 0) = (0, -0, 1)
Opposite order: A' = R₃(90°) R₁(90°) = [+0 +0 +1; -1 +0 +0; +0 -1 +0], so v_B' = (0, -1, 0)
The two results differ by 90°

Answer. z then x gives (0, -0, 1); x then z gives (0, -1, 0). Same two turns, different attitude.

Sanity check. Finite rotations are not vectors: you cannot add them. That is why attitude needs matrices or quaternions, and why the order of an Euler sequence must always be stated.

Pythonedit a value, then Run

# Worked example 1.2: Rotations do not commute
import numpy as np, math

def R1(a):
    c, s = math.cos(a), math.sin(a)
    return np.array([[1, 0, 0], [0, c, s], [0, -s, c]])

def R3(a):
    c, s = math.cos(a), math.sin(a)
    return np.array([[c, s, 0], [-s, c, 0], [0, 0, 1]])

v = np.array([1.0, 0, 0])          # inertial x direction
a = math.radians(90.0)             # each turn (rad)

A_zx = R1(a) @ R3(a)               # z first, then x: newest on the left
A_xz = R3(a) @ R1(a)               # x first, then z
va, vb = A_zx @ v, A_xz @ v
angle = math.degrees(math.acos(np.clip(va @ vb, -1, 1)))
print("z then x: v_B = (%.0f, %.0f, %.0f)" % tuple(va + 0.0))
print("x then z: v_B = (%.0f, %.0f, %.0f)" % tuple(vb + 0.0))
print("angle between the results = %.0f deg" % angle)

z then x: v_B = (0, -0, 1)
x then z: v_B = (0, -1, 0)
angle between the results = 90 deg

Practice

Practice A1.1 · warm-up

Write the DCM from the body axes

Given

A star tracker reports that the body x axis points along inertial (0, 1, 0) and the body y axis along inertial (0, 0, 1). Find the body z axis, write A_BI, and express the Sun direction s_I = (0.6, 0.8, 0) in body axes. Which body face does the Sun light most?

Mark as done

Hint

The rows of A_BI are the body axes written in inertial components. Get z from z = x × y.

Answer. z_B = (1, 0, 0) (inertial), s_B = (0.8, 0.0, 0.6); the +x face sees the Sun most, at 36.9° from its normal.

Working

z_B = x_B × y_B = (0, 1, 0) × (0, 0, 1) = (1, 0, 0)
A_BI = rows (x_B; y_B; z_B) = [+0 +1 +0; +0 +0 +1; +1 +0 +0]
s_B = A_BI s_I = (0.8, 0.0, 0.6)
Largest component is along +x_B

Answer. z_B = (1, 0, 0) (inertial), s_B = (0.8, 0.0, 0.6); the +x face sees the Sun most, at 36.9° from its normal.

Sanity check. |s_B| = 1 still: a rotation never changes a length. A non-unit result means a typing error in A.

Pythonedit a value, then Run

# Practice A1.1: Write the DCM from the body axes
import numpy as np, math

x_B = np.array([0.0, 1.0, 0.0])    # body x in inertial components
y_B = np.array([0.0, 0.0, 1.0])    # body y in inertial components
s_I = np.array([0.6, 0.8, 0.0])    # Sun direction, inertial

z_B = np.cross(x_B, y_B)           # complete the right-handed triad
A_BI = np.vstack([x_B, y_B, z_B])  # rows = body axes in inertial components
s_B = A_BI @ s_I
angle = math.degrees(math.acos(s_B[0]))   # Sun angle from the +x face normal

print("z_B = (%.0f, %.0f, %.0f)" % tuple(z_B))
print("s_B = (%.1f, %.1f, %.1f)" % tuple(s_B))
print("Sun is %.1f deg from the +x face normal" % angle)

z_B = (1, 0, 0)
s_B = (0.8, 0.0, 0.6)
Sun is 36.9 deg from the +x face normal

Practice A1.2 · standard

Find the single rotation hidden in a DCM

Given

A DCM is A = [+0.6825 +0.5901 -0.4313; -0.4313 +0.8015 +0.4141; +0.5901 -0.0966 +0.8015]. Find the angle and axis of the single rotation it represents.

Mark as done

Hint

The trace gives the angle: tr A = 1 + 2 cos a. The antisymmetric part gives the axis.

Answer. a = 50.0° about e = (0.333, 0.667, 0.667) = (1, 2, 2)/3.

Working

tr A = 2.2856, so cos a = (tr A − 1)/2 = 0.6428 and a50.00°
Axis e = (A₂₃ − A₃₂, A₃₁ − A₁₃, A₁₂ − A₂₁) / (2 sin a)
= (0.5107, 1.0214, 1.0214) / 1.5321 = (0.3333, 0.6667, 0.6667)
|e|1.0000

Answer. a = 50.0° about e = (0.333, 0.667, 0.667) = (1, 2, 2)/3.

Sanity check. The axis comes out with unit length without forcing it, which confirms the matrix is a clean rotation.

Pythonedit a value, then Run

# Practice A1.2: Find the single rotation hidden in a DCM
import numpy as np, math

# build the DCM given in the problem (50 deg about (1, 2, 2)/3)
e_true = np.array([1.0, 2.0, 2.0]) / 3.0
a_true = math.radians(50.0)
ex = np.array([[0, -e_true[2], e_true[1]], [e_true[2], 0, -e_true[0]], [-e_true[1], e_true[0], 0]])
A = math.cos(a_true) * np.eye(3) + (1 - math.cos(a_true)) * np.outer(e_true, e_true) - math.sin(a_true) * ex
print("A =")
print(np.round(A, 4))

# angle from the trace, axis from the antisymmetric part
a = math.acos((np.trace(A) - 1) / 2)
e = np.array([A[1, 2] - A[2, 1], A[2, 0] - A[0, 2], A[0, 1] - A[1, 0]]) / (2 * math.sin(a))
print("angle a = %.2f deg" % math.degrees(a))
print("axis e = (%.4f, %.4f, %.4f), |e| = %.4f" % (*e, np.linalg.norm(e)))
print("3 e = (%.0f, %.0f, %.0f), so e = (1, 2, 2)/3" % tuple(3 * e))

A =
[[ 0.6825  0.5901 -0.4313]
 [-0.4313  0.8015  0.4141]
 [ 0.5901 -0.0966  0.8015]]
angle a = 50.00 deg
axis e = (0.3333, 0.6667, 0.6667), |e| = 1.0000
3 e = (1, 2, 2), so e = (1, 2, 2)/3

Practice A1.3 · challenge

Build the orbit frame a nadir-pointer flies in

Given

The reference satellite is in a circular 550 km orbit, inclination 97.6°, right ascension of the ascending node 0°, at argument of latitude 40°. Build the orbit (LVLH) frame: z towards nadir, y opposite the orbit normal, x completing the triad. Write A_OI and find the inertial x direction (where the Sun is on 21 March, roughly) in orbit-frame components.

Mark as done

Hint

Position: r_I = r R₁(i)ᵀ (cos u, sin u, 0). Velocity is perpendicular. Then z = −r̂, y = −(r × v)/|r × v|, x = y × z.

Answer. A_OI = [-0.643 -0.101 +0.759; +0.000 +0.991 +0.132; -0.766 +0.085 -0.637]; the inertial x direction is (-0.643, 0.000, -0.766) in the orbit frame.

Working

r = R_E + h = 6928.1 km; r_I = (5307.3, -589.0, 4414.2) km
v = √(μ/r) = 7585.1 m/s; v_I = (-4875.6, -768.5, 5759.5) m/s
z_O = −r̂ = (-0.766, 0.085, -0.637); y_O = −(r × v)/|r × v| = (0.000, 0.991, 0.132); x_O = y × z = (-0.643, -0.101, 0.759)
A_OI = [-0.643 -0.101 +0.759; +0.000 +0.991 +0.132; -0.766 +0.085 -0.637]
Check: x_O · v̂ = 1.000000 (x is along the velocity), det1.000000
s_O = A_OI (1, 0, 0) = (-0.643, 0.000, -0.766)

Answer. A_OI = [-0.643 -0.101 +0.759; +0.000 +0.991 +0.132; -0.766 +0.085 -0.637]; the inertial x direction is (-0.643, 0.000, -0.766) in the orbit frame.

Sanity check. Every row is a unit vector and the rows are mutually perpendicular. A nadir-pointing satellite holds A_BO = I, so this is also its attitude.

Pythonedit a value, then Run

# Practice A1.3: Build the orbit frame a nadir-pointer flies in
import numpy as np, math

MU = 398600.4418e9        # m^3/s^2, Earth gravity parameter (Vallado)
RE = 6378.137e3           # m
h = 550e3                 # altitude (m)
inc = math.radians(97.6)  # inclination
u = math.radians(40.0)    # argument of latitude (RAAN = 0)

def R1(a):
    c, s = math.cos(a), math.sin(a)
    return np.array([[1, 0, 0], [0, c, s], [0, -s, c]])

r = RE + h
v = math.sqrt(MU / r)
r_I = r * (R1(inc).T @ np.array([math.cos(u), math.sin(u), 0]))
v_I = v * (R1(inc).T @ np.array([-math.sin(u), math.cos(u), 0]))

z_O = -r_I / np.linalg.norm(r_I)          # nadir
y_O = -np.cross(r_I, v_I)
y_O /= np.linalg.norm(y_O)                # opposite orbit normal
x_O = np.cross(y_O, z_O)                  # completes the triad
A_OI = np.vstack([x_O, y_O, z_O])
s_O = A_OI @ np.array([1.0, 0, 0])

print("r_I = (%.1f, %.1f, %.1f) km, v = %.1f m/s" % (*(r_I / 1e3), v))
print("A_OI =")
print(np.round(A_OI, 3))
print("x_O . v_hat = %.6f, det = %.6f" % (x_O @ v_I / v, np.linalg.det(A_OI)))
print("s_O = (%.3f, %.3f, %.3f)" % tuple(s_O))

# Practice A1.3: Build the orbit frame a nadir-pointer flies in
import numpy as np, math

MU = 398600.4418e9        # m^3/s^2, Earth gravity parameter (Vallado)
RE = 6378.137e3           # m
h = 550e3                 # altitude (m)
inc = math.radians(97.6)  # inclination
u = math.radians(40.0)    # argument of latitude (RAAN = 0)

def R1(a):
    c, s = math.cos(a), math.sin(a)
    return np.array([[1, 0, 0], [0, c, s], [0, -s, c]])

r = RE + h
v = math.sqrt(MU / r)
r_I = r * (R1(inc).T @ np.array([math.cos(u), math.sin(u), 0]))
v_I = v * (R1(inc).T @ np.array([-math.sin(u), math.cos(u), 0]))

z_O = -r_I / np.linalg.norm(r_I)          # nadir
y_O = -np.cross(r_I, v_I)
y_O /= np.linalg.norm(y_O)                # opposite orbit normal
x_O = np.cross(y_O, z_O)                  # completes the triad
A_OI = np.vstack([x_O, y_O, z_O])
s_O = A_OI @ np.array([1.0, 0, 0])

print("r_I = (%.1f, %.1f, %.1f) km, v = %.1f m/s" % (*(r_I / 1e3), v))
print("A_OI =")
print(np.round(A_OI, 3))
print("x_O . v_hat = %.6f, det = %.6f" % (x_O @ v_I / v, np.linalg.det(A_OI)))
print("s_O = (%.3f, %.3f, %.3f)" % tuple(s_O))

r_I = (5307.3, -589.0, 4414.2) km, v = 7585.1 m/s
A_OI =
[[-0.643 -0.101  0.759]
 [ 0.     0.991  0.132]
 [-0.766  0.085 -0.637]]
x_O . v_hat = 1.000000, det = 1.000000
s_O = (-0.643, 0.000, -0.766)

Common mistakes

Using Aᵀ where A is needed. Write the frame subscripts (A_BI takes I to B) and cancel them like units.
Mixing active rotations (turn the vector) and frame rotations (turn the axes). They are transposes of each other.
Multiplying rotations in the wrong order. The newest rotation goes on the left.
Reading columns as body axes. The rows of A_BI are the body axes.

Sources for this chapter

R13Markley, F. L., Attitude Determination Using Vector Observations: A Fast Optimal Matrix Algorithm, NASA GSFC (1993)
R14Markley, F. L., Attitude Representations for Kalman Filtering, AAS 01-309 (2001)

Where in each source, and links: References.

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Reference

Answers

Problem	Answer
A1.1	z_B = (1, 0, 0) (inertial), s_B = (0.8, 0.0, 0.6); the +x face sees the Sun most, at 36.9° from its normal.
A1.2	a = 50.0° about e = (0.333, 0.667, 0.667) = (1, 2, 2)/3.
A1.3	A_OI = [-0.643 -0.101 +0.759; +0.000 +0.991 +0.132; -0.766 +0.085 -0.637]; the inertial x direction is (-0.643, 0.000, -0.766) in the orbit frame.

Reference

References

Every reference the pack cites. "Where" names the section, table or page that holds the cited value; an entry says so when the source was not read in full or is not openly available.

Id	Reference
R13	Markley, F. L., Attitude Determination Using Vector Observations: A Fast Optimal Matrix Algorithm, NASA GSFC (1993)Where: Wahba's problem, optimal matrix solution, comparison with TRIAD (NASA NTRS record)https://ntrs.nasa.gov/citations/19930015542Cited in: Chapter 1
R14	Markley, F. L., Attitude Representations for Kalman Filtering, AAS 01-309 (2001)Where: Multiplicative (MEKF-style) attitude error with norm-preserving quaternion reset (NASA NTRS record)https://ntrs.nasa.gov/citations/20010067774Cited in: Chapter 1

The references cited by this preview. The full course lists every source it uses.

Spacecraft attitude determination and control

How to use this pack

What is in the full course

You will be able to

Three frames

The DCM

Combining rotations

Euler's theorem

Key equations

Worked examples

The same Sun, seen from a turned satellite

Rotations do not commute

Practice

Write the DCM from the body axes

Find the single rotation hidden in a DCM

Build the orbit frame a nadir-pointer flies in

Common mistakes

Check yourself

Sources for this chapter

17 more chapters in the full course

Answers

References