Attitude estimation — triad method (two vector measurements): Exercises

Before we start, one reminder of every symbol we use, in plain words:

Definition Symbol dictionary (nothing used before this)

$\hat{r}_{1}, \hat{r}_{2}$ — the two known directions written in the reference frame (where they should point in space). The little hat $^$ means "length exactly 1."
$\hat{b}_{1}, \hat{b}_{2}$ — the same two directions as measured by the spacecraft, written in its body frame (the frame glued to the vehicle).
Slot 1 / slot 2 — the two "seats" of the recipe. Slot 1 holds the pair you trust the most: $(\hat{r}_{1}, \hat{b}_{1})$ becomes axis 1 exactly. Slot 2 holds the pair $(\hat{r}_{2}, \hat{b}_{2})$ whose angle is discarded — only its direction (which side of axis 1 it lies on) is kept, via the cross product. Rule of thumb: most accurate sensor → slot 1.
$\times$ — the cross product: feed it two arrows, it returns a new arrow perpendicular to both, whose length equals $sin θ$ (the sine of the angle between them).
$∥ v ∥$ — the length of an arrow. Dividing an arrow by its own length gives a unit arrow (points same way, length 1). That is normalizing.
$M_{r}, M_{b}$ — a $3 \times 3$ box holding three unit arrows as its columns (the "triad").
$M^{T}$ — the transpose: flip rows into columns. For a box of orthonormal columns, $M^{T}$ is also its inverse, so $M^{T} M = I$ (the identity, a "do-nothing" box).
$A$ — the attitude matrix: the rotation that turns reference arrows into body arrows, $A \hat{r}_{i} = \hat{b}_{i}$ .

How to read the figures below. The three step-figures walk the geometry visually: first the cross-product construction of one triad (s01), then how a rotation carries the reference triad onto the body triad (s02), and finally the danger zone where the two vectors go parallel (s03). Refer back to them as you solve — the text points to specific arrows.

Figure — Attitude estimation — triad method (two vector measurements)

Look at s01: the magenta arrow $\hat{r}_{1}$ becomes axis 1 unchanged; the violet arrow is $\hat{r}_{1} \times \hat{r}_{2}$ (perpendicular to the plane of the two orange/magenta inputs); the navy arrow closes the right-handed set. Every exercise that "builds a triad" is doing exactly this picture with different numbers.

Level 1 — Recognition

L1.1

State, in one sentence each, why one vector measurement is not enough and why two is enough.

Recall Solution

One is not enough: a single known direction pins the two "aiming" angles but leaves the spacecraft free to spin about that direction — you cannot see that spin, so 1 of 3 rotational freedoms stays unknown. Two is enough: a second non-parallel direction is dragged around by that spin, so its position exposes the last angle — two non-parallel arrows lock all three rotational freedoms.

L1.2

Given $\hat{r}_{1} = (1, 0, 0)$ and $\hat{r}_{2} = (0, 1, 0)$ , compute the first two triad axes $\hat{t}_{1}^{(r)}$ and $\hat{t}_{2}^{(r)}$ .

Recall Solution

$\hat{t}_{1}^{(r)} = \hat{r}_{1} = (1, 0, 0)$ — we always keep the trusted (slot-1) vector as axis 1. Cross product: $(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ . Its length is $1$ , so normalizing changes nothing: $\hat{t}_{2}^{(r)} = (0, 0, 1) .$ This is literally the s01 picture with $\hat{r}_{1} = \overset{x}{^}$ , $\hat{r}_{2} = \overset{y}{^}$ .

L1.3

True or false: the box $M_{r}$ stores the triad axes as its rows.

Recall Solution

False. They are stored as columns: $M_{r} = [\hat{t}_{1}^{(r)} \hat{t}_{2}^{(r)} \hat{t}_{3}^{(r)}]$ . This matters because $A M_{r} = M_{b}$ must act column-by-column ( $A \hat{t}_{i}^{(r)} = \hat{t}_{i}^{(b)}$ ); rows would break that reading.

Level 2 — Application

Refer to s02 while you work these: it shows the finished reference triad and the same triad after the rotation $A$ has carried it into the body frame — that is exactly what $A M_{r} = M_{b}$ does, column by column.

L2.1

Reference: $\hat{r}_{1} = (1, 0, 0)$ , $\hat{r}_{2} = (0, 1, 0)$ . Body (true attitude = $+ 9 0^{\circ}$ about $z$ , right-hand rule): $\hat{b}_{1} = (0, 1, 0)$ , $\hat{b}_{2} = (- 1, 0, 0)$ . Build both triads and find $A = M_{b} M_{r}^{T}$ .

Recall Solution

Reference triad (as in the parent note): $\hat{t}_{1}^{(r)} = (1, 0, 0)$ ; $\hat{r}_{1} \times \hat{r}_{2} = (0, 0, 1)$ so $\hat{t}_{2}^{(r)} = (0, 0, 1)$ ; $\hat{t}_{3}^{(r)} = (1, 0, 0) \times (0, 0, 1) = (0, - 1, 0)$ . $M_{r} = 100001 0 - 1 0$ Body triad: $\hat{t}_{1}^{(b)} = (0, 1, 0)$ ; $(0, 1, 0) \times (- 1, 0, 0) = (0, 0, 1)$ so $\hat{t}_{2}^{(b)} = (0, 0, 1)$ ; $\hat{t}_{3}^{(b)} = (0, 1, 0) \times (0, 0, 1) = (1, 0, 0)$ . $M_{b} = 010001100$ Multiply $A = M_{b} M_{r}^{T}$ : $A = 010 - 1 00 001 = R_{z} (+ 9 0^{\circ}) .$ Sign check with the convention above: $R_{z} (+ 9 0^{\circ})$ sends $\overset{x}{^} = (1, 0, 0)$ to $\overset{y}{^} = (0, 1, 0)$ — exactly $A \hat{r}_{1} = (0, 1, 0) = \hat{b}_{1}$ ✅, and $A \hat{r}_{2} = (- 1, 0, 0) = \hat{b}_{2}$ ✅. Counter-clockwise (looking down $+ z$ ), as promised.

L2.2

Reference: $\hat{r}_{1} = (1, 0, 0)$ , $\hat{r}_{2} = (0, 0, 1)$ . Body: $\hat{b}_{1} = (0, 0, 1)$ , $\hat{b}_{2} = (- 1, 0, 0)$ . Find $A$ and identify the rotation, stating its sign explicitly.

Recall Solution

Reference: $\hat{t}_{1}^{(r)} = (1, 0, 0)$ ; $(1, 0, 0) \times (0, 0, 1) = (0, - 1, 0)$ so $\hat{t}_{2}^{(r)} = (0, - 1, 0)$ ; $\hat{t}_{3}^{(r)} = (1, 0, 0) \times (0, - 1, 0) = (0, 0, - 1)$ . $M_{r} = 100 0 - 1 0 00 - 1$ Body: $\hat{t}_{1}^{(b)} = (0, 0, 1)$ ; $(0, 0, 1) \times (- 1, 0, 0) = (0, - 1, 0)$ so $\hat{t}_{2}^{(b)} = (0, - 1, 0)$ ; $\hat{t}_{3}^{(b)} = (0, 0, 1) \times (0, - 1, 0) = (1, 0, 0)$ . $M_{b} = 001 0 - 1 0 100$ $A = M_{b} M_{r}^{T}$ : $A = 001010 - 1 00 .$ Which sign? Under our right-hand rule, the positive $y$ -rotation is $R_{y} (+ 9 0^{\circ}) = 00 - 1 010100,$ because $R_{y} (+ ϕ)$ sends $\overset{x}{^} = (1, 0, 0)$ toward $\overset{z}{^}$ -and-back with the sign pattern $(cos ϕ, 0, - sin ϕ)$ . Our $A$ has the opposite off-diagonal signs, so it is $A = R_{y} (- 9 0^{\circ}),$ a clockwise $9 0^{\circ}$ turn about $+ y$ (equivalently $+ 9 0^{\circ}$ about $- y$ ). Verify directly: $A \hat{r}_{1} = (0, 0, 1) = \hat{b}_{1}$ ✅, $A \hat{r}_{2} = (- 1, 0, 0) = \hat{b}_{2}$ ✅. Lesson: the numbers alone don't tell you the turn direction — you must fix the convention first, then read the sign off the off-diagonal pattern.

L2.3

For $\hat{r}_{1} = (1, 0, 0)$ , $\hat{r}_{2} = (\frac{1}{2}, \frac{1}{2}, 0)$ (a $4 5^{\circ}$ separation), compute $∥ \hat{r}_{1} \times \hat{r}_{2} ∥$ and confirm it equals $sin 4 5^{\circ}$ .

Recall Solution

$(1, 0, 0) \times (\frac{1}{2}, \frac{1}{2}, 0) = (0, 0, \frac{1}{2})$ . Length $= \frac{1}{2} \approx 0.7071 = sin 4 5^{\circ}$ ✅. This is the geometric meaning of the cross-product length: it measures the separation angle.

Level 3 — Analysis

L3.1

Show that swapping the roles ( $\hat{r}_{1} \leftrightarrow \hat{r}_{2}$ and $\hat{b}_{1} \leftrightarrow \hat{b}_{2}$ ) generally gives a different $A$ when the data are noisy, but the same $A$ when they are noise-free. Explain why.

Recall Solution

Noise-free: if $\hat{b}_{i} = A_{true} \hat{r}_{i}$ exactly for both $i$ , then both triads are exact rotated copies whichever vector we call "1"; TRIAD recovers $A_{true}$ either way, so the answer is identical. Noisy: TRIAD honors slot 1 exactly and throws away slot 2's angle. Put the noisy vector in slot 1 and its full error enters $A$ ; put it in slot 2 and only its plane survives. Different information is discarded, so the two estimates differ. Conclusion: the method is asymmetric — always make the accurate sensor slot 1.

L3.2

The cross product returns length $sin θ$ . As the two reference vectors approach parallel ( $θ \to 0$ ), what happens to $\hat{t}_{2}^{(r)}$ , and why is TRIAD unreliable there?

Recall Solution

$∥ \hat{r}_{1} \times \hat{r}_{2} ∥ = sin θ \to 0$ . We divide by this to normalize, so any tiny measurement wobble in the numerator gets amplified by $1/ sin θ$ , which blows up. Geometrically (see s03): two near-parallel arrows define their common plane very loosely, so the perpendicular $\hat{t}_{2}^{(r)}$ jitters wildly. Best conditioning is near $θ = 9 0^{\circ}$ where $sin θ = 1$ is largest and most stable.

Look at s03: as the orange arrow swings toward the magenta one, the violet cross-product arrow $\hat{r}_{1} \times \hat{r}_{2}$ shrinks toward zero length — that vanishing arrow is the quantity you must divide by, which is why the danger zone amplifies noise so badly.

L3.3

Prove that $A = M_{b} M_{r}^{T}$ satisfies $A A^{T} = I$ and $det A = + 1$ (i.e. it is a genuine rotation, a member of Rotation Matrices and SO(3)).

Recall Solution

Both $M_{r}, M_{b}$ have orthonormal columns, so $M_{r}^{T} M_{r} = I$ and $M_{b}^{T} M_{b} = I$ . Then

=M_bM_r^{\mathsf T}M_rM_b^{\mathsf T}=M_b\,I\,M_b^{\mathsf T}=I.$$ For the determinant: each triad was closed with a right-handed cross product ($\hat{\mathbf t}_3=\hat{\mathbf t}_1\times\hat{\mathbf t}_2$), forcing $\det M_r=\det M_b=+1$. Since determinants multiply, $\det A=\det M_b\cdot\det M_r^{\mathsf T}=(+1)(+1)=+1$. ✅ A proper rotation.

Level 4 — Synthesis

L4.1

A sun sensor (Sun Sensors and Magnetometers) gives the accurate direction; a magnetometer gives a noisier one. Reference: sun $\hat{r}_{S} = (0, 0, 1)$ , field $\hat{r}_{M} = (0, 1, 0)$ . Measured (noise-free for this drill): $\hat{b}_{S} = (1, 0, 0)$ , $\hat{b}_{M} = (0, 0, 1)$ . Which vector goes in slot 1, and what is $A$ ?

Recall Solution

Put the accurate sun vector in slot 1 (most-trusted → slot 1): $\hat{r}_{1} = (0, 0, 1)$ , $\hat{b}_{1} = (1, 0, 0)$ ; $\hat{r}_{2} = (0, 1, 0)$ , $\hat{b}_{2} = (0, 0, 1)$ . Reference: $\hat{t}_{1}^{(r)} = (0, 0, 1)$ ; $(0, 0, 1) \times (0, 1, 0) = (- 1, 0, 0)$ so $\hat{t}_{2}^{(r)} = (- 1, 0, 0)$ ; $\hat{t}_{3}^{(r)} = (0, 0, 1) \times (- 1, 0, 0) = (0, - 1, 0)$ . $M_{r} = 001 - 1 00 0 - 1 0$ Body: $\hat{t}_{1}^{(b)} = (1, 0, 0)$ ; $(1, 0, 0) \times (0, 0, 1) = (0, - 1, 0)$ so $\hat{t}_{2}^{(b)} = (0, - 1, 0)$ ; $\hat{t}_{3}^{(b)} = (1, 0, 0) \times (0, - 1, 0) = (0, 0, - 1)$ . $M_{b} = 100 0 - 1 0 00 - 1$ $A = M_{b} M_{r}^{T}$ : $A = 010001100 .$ Check: $A \hat{r}_{1} = (1, 0, 0) = \hat{b}_{1}$ ✅, $A \hat{r}_{2} = (0, 0, 1) = \hat{b}_{2}$ ✅. This $A$ is a valid rotation ( $det = + 1$ ), cyclically permuting the axes.

L4.2

TRIAD only uses two vectors. Wahba's Problem (solved by the QUEST algorithm or the Davenport q-method) optimally blends many weighted vectors. In one paragraph, state what TRIAD sacrifices and what it gains versus these methods.

Recall Solution

Sacrifices: TRIAD is suboptimal — it trusts one vector fully and discards slot 2's angle, so it does not minimize a weighted least-squares error over both measurements the way Wahba's cost function does, and it cannot fuse three or more sensors or apply per-sensor confidence weights. Gains: it is closed-form and cheap — just cross products, one normalization each, and a transpose-multiply, with no eigenvalue solve (which QUEST algorithm and the Davenport q-method both require). That makes TRIAD ideal for coarse acquisition, fast on-board sanity checks, or initializing an iterative estimator such as the Kalman Filter — Attitude. In short: TRIAD trades the statistically-optimal, multi-sensor answer of Wahba/QUEST/Davenport for speed and simplicity — perfect when you have exactly two vectors and one is clearly more trusted.

L4.3

From the $A$ of L4.1, is the corresponding quaternion (Quaternion Attitude Kinematics) unique? State the scalar part magnitude $∣ q_{0} ∣$ using $cos \frac{ϕ}{2}$ with $1 + 2 cos ϕ = tr (A)$ .

Recall Solution

$tr (A) = 0 + 0 + 0 = 0$ . The rotation angle satisfies $1 + 2 cos ϕ = tr (A) = 0$ , so $cos ϕ = - \frac{1}{2}$ , giving $ϕ = 12 0^{\circ}$ . Then $∣ q_{0} ∣ = cos \frac{ϕ}{2} = cos 6 0^{\circ} = \frac{1}{2}$ . The quaternion is not unique: $q$ and $- q$ describe the same rotation (double cover), so $q_{0} = + \frac{1}{2}$ and $q_{0} = - \frac{1}{2}$ are both valid.

Level 5 — Mastery

Keep s03 in view for L5.1 — it shows the two reference arrows sliding toward parallel and the cross-product arrow collapsing to zero.

L5.1

Reference vectors are exactly collinear: $\hat{r}_{1} = \hat{r}_{2} = (1, 0, 0)$ . Show precisely where TRIAD breaks and describe which piece of the attitude becomes undetermined.

Recall Solution

$\hat{r}_{1} \times \hat{r}_{2} = (1, 0, 0) \times (1, 0, 0) = (0, 0, 0)$ , length $0$ . Normalizing requires dividing by $0$ — undefined, so $\hat{t}_{2}^{(r)}$ cannot be formed and TRIAD fails outright. Physically: a single direction fixes only 2 of 3 freedoms; the spin about that common axis is completely unobservable. No amount of algebra recovers it — you must add a genuinely independent second direction.

L5.2

Second body measurement is noisy: true separation is $9 0^{\circ}$ but $\hat{b}_{2}$ arrives at $8 8^{\circ}$ from $\hat{b}_{1}$ . With slot 1 exact, argue that the estimated first body axis carries zero contribution from this $2^{\circ}$ error, and only the plane tilts.

Recall Solution

By construction $\hat{t}_{1}^{(b)} = \hat{b}_{1}$ , which does not involve $\hat{b}_{2}$ at all — so its estimate is untouched by the $2^{\circ}$ noise (zero contribution). The noisy $\hat{b}_{2}$ enters only through the cross product $\hat{b}_{1} \times \hat{b}_{2}$ , which is then normalized to length 1. Normalizing deletes the magnitude information; what remains is the plane's orientation, which the $2^{\circ}$ shift tilts slightly. Therefore first-order error appears only in the second and third axes $\hat{t}_{2}^{(b)}, \hat{t}_{3}^{(b)}$ (the plane), while axis 1 stays exact. That is the whole reason for assigning the accurate sensor to slot 1: its full precision passes straight through into the estimate, and the weaker sensor can only tilt the remaining plane — a small, bounded effect.

L5.3

Given the noise-free data $\hat{r}_{1} = (1, 0, 0), \hat{r}_{2} = (0, 1, 0)$ , $\hat{b}_{1} = \frac{1}{2} (1, 1, 0), \hat{b}_{2} = \frac{1}{2} (- 1, 1, 0)$ , find $A$ and its rotation angle $ϕ$ from $tr (A) = 1 + 2 cos ϕ$ (state its sign).

Recall Solution

Reference (same as L2.1): $M_{r} = 100001 0 - 1 0$ . Body: $\hat{t}_{1}^{(b)} = \frac{1}{2} (1, 1, 0)$ ; $\hat{b}_{1} \times \hat{b}_{2} = \frac{1}{2} (1, 1, 0) \times (- 1, 1, 0) = \frac{1}{2} (0, 0, 2) = (0, 0, 1)$ , already unit, so $\hat{t}_{2}^{(b)} = (0, 0, 1)$ ; $\hat{t}_{3}^{(b)} = \frac{1}{2} (1, 1, 0) \times (0, 0, 1) = \frac{1}{2} (1, - 1, 0)$ . $M_{b} = \frac{1}{2} \frac{1}{2} 0 001 \frac{1}{2} - \frac{1}{2} 0$ $A = M_{b} M_{r}^{T} = \frac{1}{2} \frac{1}{2} 0 - \frac{1}{2} \frac{1}{2} 0 001$ — this is $R_{z} (+ 4 5^{\circ})$ (positive off-diagonal $- sin ϕ$ in the top row → counter-clockwise about $+ z$ ). $tr (A) = \frac{1}{2} + \frac{1}{2} + 1 = 1 + 2$ . Then $1 + 2 cos ϕ = 1 + 2$ gives $cos ϕ = \frac{2}{2}$ , so $ϕ = + 4 5^{\circ}$ . ✅ Consistent — the body pair is the reference pair rotated $+ 4 5^{\circ}$ about $z$ .

Recall One-line self-test before you close the page

Why $M_{b} M_{r}^{T}$ and never $M_{b} M_{r}$ ? ::: Because the rotation maps reference columns to body columns ( $A M_{r} = M_{b}$ ), so $A = M_{b} M_{r}^{- 1} = M_{b} M_{r}^{T}$ using orthonormality — $M_{b} M_{r}$ has no such meaning. Where does TRIAD literally divide by zero? ::: When $\hat{r}_{1} ∥ \hat{r}_{2}$ (or the body pair), the cross product has length $sin θ = 0$ . Which sensor belongs in slot 1? ::: The more accurate one — it is honored exactly and its full precision passes into the estimate.