Worked examples — Attitude estimation — triad method (two vector measurements)

3.5.12 · D3 · Physics › Guidance, Navigation & Control (GNC) › Attitude estimation — triad method (two vector measurements)

Yeh page TRIAD parent note ka drill ground hai. Parent ne recipe banai; yahan hum ise har tarah ke input ke against run karte hain jo method ko mil sakta hai — clean rotations, noisy sensors, degenerate (near-parallel) geometry, sign traps, ek real Sun/mag word problem, aur ek exam twist.

Shuru karne se pehle, poore machine ka ek reminder ek saanss mein: hum do reference vectors se ek right-handed triad banate hain (teen mutually perpendicular unit axes — ek chhota coordinate cross), phir wahi triad do body measurements se banate hain, har triad ko ek $3\times 3$ matrix ($M_r$ aur $M_b$) ke columns mein stack karte hain, aur attitude hai $A=M_b{M}_r^{\mathsf{T}}$. Perpendicular-axis idea ke liye Cross Product and Right-Handed Frames dekho aur $A$ mein $\det =+1$ kyun hona chahiye iske liye Rotation Matrices and SO(3) dekho.

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Scenario matrix

Har TRIAD problem inhi cells mein se kisi ek mein land karta hai. Neeche ke examples ko unke cell label ke saath mark kiya gaya hai.

Cell	Kya badlta hai	Danger / lesson
A. Clean, aligned	true rotation, no noise, axis-aligned vectors	zero distraction ke saath mechanics seekho
B. Clean, tilted axis	true rotation ek tilted axis ke baare mein	non-integer entries; machine ko test karo
C. Noisy slot-2	sensor-2 angle kuch degrees off	pura point — angle discard, error chhoti rehti hai
D. Sign / handedness trap	slot 1 ↔ slot 2 swap karo, ya left-handed pair do	$A$ badlta hai; $\det$ phir bhi $+1$ hona chahiye
E. Degenerate: near-parallel	do vectors ke beech $\theta \to 0^{\circ }$	$\sin \theta \to 0$, noise blow up hoti hai — refuse karo
F. Degenerate: exactly parallel	$\theta =0^{\circ }$ ya $180^{\circ }$	cross product $0$ hai; attitude undetermined
G. Real word problem	Sun sensor + magnetometer, given numbers	slot 1 = accurate sensor choose karo
H. Exam twist	$A$ se rotation angle recover karo	trace se $\theta$ read out karta hai

Hum A–H order mein karte hain.

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Cell A — Clean, aligned (zero noise ke saath mechanics)

Forecast: Compute karne se pehle guess karo — true rotation $z$ ke baare mein quarter-turn hai. Tumhare hisaab se $A$ ke $(1,1)$ aur $(2,1)$ entries kaisi dikhni chahiye? (Hint: standard $R_z(90^{\circ })$.)

1. Reference triad. ${\hat{\mathbf{t}}}_1^{(r)}=(1,0,0)$. ${\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2=(1,0,0)\times (0,1,0)=(0,0,1)$, already unit, to ${\hat{\mathbf{t}}}_2^{(r)}=(0,0,1)$. ${\hat{\mathbf{t}}}_3^{(r)}={\hat{\mathbf{t}}}_1^{(r)}\times {\hat{\mathbf{t}}}_2^{(r)}=(1,0,0)\times (0,0,1)=(0,-1,0)$. Yeh step kyun? Slot 1 trusted vector rakhta hai; slot 2 ka cross product automatically usके perpendicular hota hai; slot 3 right-handed box close karta hai.

   $M_r=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right]$

2. Body triad, same recipe. ${\hat{\mathbf{t}}}_1^{(b)}=(0,1,0)$; $(0,1,0)\times (-1,0,0)=(0,0,1)={\hat{\mathbf{t}}}_2^{(b)}$; $(0,1,0)\times (0,0,1)=(1,0,0)={\hat{\mathbf{t}}}_3^{(b)}$.

   $M_b=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$ Yeh step kyun? Dono triads same physical cross describe karte hain — ek reference words mein, ek body words mein.

3. Assemble $A=M_b{M}_r^{\mathsf{T}}$: $A=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ Yeh step kyun? $A{M}_r=M_b$ column by column; $M_r$ orthonormal $\Rightarrow M_r^{-1}=M_r^{\mathsf{T}}$.

Verify: $A{\hat{\mathbf{r}}}_1=(0,1,0)={\hat{\mathbf{b}}}_1$ ✅ aur $A{\hat{\mathbf{r}}}_2=(-1,0,0)={\hat{\mathbf{b}}}_2$ ✅; $\det A=+1$ ✅. Yeh exactly $R_z(90^{\circ })$ hai.

Neeche figure — kya dekhna hai: teen colored arrows reference triad hain. Cyan arrow slot 1 hai (${\hat{\mathbf{t}}}_1={\hat{\mathbf{r}}}_1$, exactly rakha gaya), amber arrow slot 2 hai (${\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2$, cyan ke perpendicular), aur white arrow slot 3 hai (${\hat{\mathbf{t}}}_1\times {\hat{\mathbf{t}}}_2$, right-handed box close karta hai). Notice karo ki teeno $90^{\circ }$ par milte hain — woh orthogonality hi $M_r^{-1}=M_r^{\mathsf{T}}$ banati hai aur hence $A$ ek clean rotation hoti hai.

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Cell B — Clean, tilted axis (non-integer entries)

Forecast: Rotation poori tarah $xy$-plane mein hai, to $\hat{\mathbf{z}}$ untouched rehna chahiye — expect karo ki $A$ ki third row/column $(0,0,1)$ ho.

1. Reference triad — Cell A jaisa hi: $M_r=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right]$. Yeh step kyun? Reference vectors unchanged hain, to reference frame identical hai; sirf body side move karti hai.

2. Body triad. ${\hat{\mathbf{t}}}_1^{(b)}=(\frac{1}{2},\frac{\sqrt{3}}{2},0)$. ${\hat{\mathbf{b}}}_1\times {\hat{\mathbf{b}}}_2=(0,0,\frac{1}{4}+\frac{3}{4})=(0,0,1)$ (already unit) $={\hat{\mathbf{t}}}_2^{(b)}$. ${\hat{\mathbf{t}}}_3^{(b)}={\hat{\mathbf{t}}}_1^{(b)}\times (0,0,1)=(\frac{\sqrt{3}}{2},-\frac{1}{2},0)$. Yeh step kyun? Do in-plane vectors ka cross $z$ ke along point karta hai — woh axis jiske perpendicular dono sensors lie karte hain.

3. Assemble $A=M_b{M}_r^{\mathsf{T}}$: $A=\left[\begin{array}{ccc}\frac{1}{2}& -\frac{\sqrt{3}}{2}& 0\\ \frac{\sqrt{3}}{2}& \frac{1}{2}& 0\\ 0& 0& 1\end{array}\right]=R_z(60^{\circ }).$

Verify: $A{\hat{\mathbf{r}}}_1=(\frac{1}{2},\frac{\sqrt{3}}{2},0)={\hat{\mathbf{b}}}_1$ ✅ aur $A{\hat{\mathbf{r}}}_2=(-\frac{\sqrt{3}}{2},\frac{1}{2},0)={\hat{\mathbf{b}}}_2$ ✅; $\det A=+1$ ✅. Upar define ki gayi trace-to-angle identity use karke, $\mathrm{tr}(A)=\frac{1}{2}+\frac{1}{2}+1=2$, to $\theta =\arccos \left(\frac{2-1}{2}\right)=\arccos (0.5)=60^{\circ }$ ✅ — truth se match karta hai. (Quadrant caveat yaad rakhna: trace akela $+60^{\circ }$ aur $-60^{\circ }$ mein fark nahi karta; positive off-diagonal pattern counter-clockwise sense confirm karta hai. Hum ise Cell H mein poori tarah push karte hain.)

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Cell C — Noisy slot-2 (pura point)

Forecast: TRIAD ${\hat{\mathbf{b}}}_1$ ko exactly rakhta hai aur sirf cross product ki direction use karta hai. To kya tumhare hisaab se $A$ mein error $2^{\circ }$ ke aaspaas hogi, ya bahut chhoti? Padhne se pehle guess karo.

1. Slot 1 exact hai. ${\hat{\mathbf{t}}}_1^{(b)}={\hat{\mathbf{b}}}_1=(0,1,0)$ — noise se untouched. Yeh step kyun? Jo accurate sensor ne ${\hat{\mathbf{b}}}_1$ produce kiya, use perfectly honor kiya gaya hai.

2. Slot 2 — cross product, phir normalize. ${\hat{\mathbf{b}}}_1\times {\hat{\mathbf{b}}}_2=(0,1,0)\times (-0.99939,-0.03490,0)=(0,0,0.99939)$. Iska norm $0.99939$ hai, to normalize karne par ${\hat{\mathbf{t}}}_2^{(b)}=(0,0,1)$ milta hai exactly (working precision tak). Yeh step kyun? $2^{\circ }$ tilt cross product ki length change karta hai ($\sin 88^{\circ }$ vs $\sin 90^{\circ }$), lekin normalizing length erase kar deta hai. Sirf direction $\hat{\mathbf{z}}$ survive karta hai — aur woh direction barely hili kyunki dono vectors abhi bhi $xy$-plane mein hain.

3. Slot 3 aur assemble. Cell A jaisa identical: ${\hat{\mathbf{t}}}_3^{(b)}=(1,0,0)$, jisse same $A=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ milta hai.

Verify: Recovered $A$ abhi bhi $A{\hat{\mathbf{r}}}_1=(0,1,0)={\hat{\mathbf{b}}}_1$ exactly satisfy karta hai aur $A{\hat{\mathbf{r}}}_2=(-1,0,0)$ ✅; $\det A=+1$ ✅. Slot 2 mein $2^{\circ }$ noise ne zero attitude error produce kiya kyunki yeh in-plane tilt thi — exactly TRIAD ka design intent: sensor-2 angle discard. (Generally slot-2 noise mein residual first-order nahi, second-order hota hai.) Isliye zyada accurate sensors jaise Sun sensor slot 1 mein jaate hain. Ek method ke liye jo dono measurements optimally weight karta hai, Wahba's Problem aur QUEST algorithm dekho.

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Cell D — Sign / handedness trap (order matters)

Forecast: Parent note ne warn kiya tha ki TRIAD asymmetric hai. Guess: same $A$, ya different $A$?

1. Reference triad (swapped). ${\hat{\mathbf{t}}}_1^{(r)}=(0,1,0)$; $(0,1,0)\times (1,0,0)=(0,0,-1)={\hat{\mathbf{t}}}_2^{(r)}$; $(0,1,0)\times (0,0,-1)=(-1,0,0)={\hat{\mathbf{t}}}_3^{(r)}$. Yeh step kyun? Cross product ka sign flip ho gaya Cell A ke versus kyunki order flip hua — ${\hat{\mathbf{r}}}_1^{\prime }\times {\hat{\mathbf{r}}}_2^{\prime }=-({\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2)$.

   $M_r^{\prime }=\left[\begin{array}{ccc}0& 0& -1\\ 1& 0& 0\\ 0& -1& 0\end{array}\right]$

2. Body triad (swapped). ${\hat{\mathbf{t}}}_1^{(b)}=(-1,0,0)$; $(-1,0,0)\times (0,1,0)=(0,0,-1)={\hat{\mathbf{t}}}_2^{(b)}$; $(-1,0,0)\times (0,0,-1)=(0,-1,0)={\hat{\mathbf{t}}}_3^{(b)}$.

   $M_b^{\prime }=\left[\begin{array}{ccc}-1& 0& 0\\ 0& 0& -1\\ 0& -1& 0\end{array}\right]$

3. Assemble $A^{\prime }=M_b^{\prime }{M}_r^{\prime \mathsf{T}}$: $A^{\prime }=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$

Verify: Yahan data noiseless tha, to swapping se Cell A jaisa same $A^{\prime }$ milta hai — achha, dono truth se agree karte hain. Mappings check karo: $A^{\prime }{\hat{\mathbf{r}}}_1^{\prime }=(-1,0,0)={\hat{\mathbf{b}}}_1^{\prime }$ ✅ aur $A^{\prime }{\hat{\mathbf{r}}}_2^{\prime }=(0,1,0)={\hat{\mathbf{b}}}_2^{\prime }$ ✅. Bhi $\det A^{\prime }=+1$ ✅ aur $A^{\prime }{A}^{\prime \mathsf{T}}=I$ ✅, abhi bhi proper rotation. Lesson: $A$ matrix slot ordering ke beech sirf tab differ kar sakta hai jab noise ho (tab do orderings disagree karte hain, aur tumhe trusted sensor slot 1 mein dalna hoga). Left-handed pair dena TRIAD ko break nahi karega — cross product phir bhi automatically right-handed triad build karta hai, guarantee karta hai ki $\det A=+1$ regardless.

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Cell E — Degenerate: near-parallel (noise blows up)

Forecast: Cross-product magnitude $\sin \theta$ hai. $\theta =3^{\circ }$ ke saath yeh chhota hai. Guess: kya ek chhoti input error chhoti rehti hai, ya amplify ho jaati hai?

Pehle, error precisely name karo. $\epsilon$ (radians mein) ko ${\hat{\mathbf{r}}}_2$ mein ek tiny out-of-plane angular error hone do — yani, true ${\hat{\mathbf{r}}}_2$ $xy$-plane mein lie karta hai, lekin noisy measurement angle $\epsilon$ se us plane se bahar tip hui hai (size $\approx \epsilon$ ka ek small $z$-component). Hum track karte hain ki slot-2 unit vector kitna swing karta hai.

1. Cross-product magnitude denominator set karta hai. $\Vert {\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2\Vert =\sin \theta =\sin 3^{\circ }\approx 0.05234$. Yeh step kyun? Slot 2 hai $({\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2)$ is magnitude se divided. Ek tiny denominator numerator mein jo bhi hai use magnify karta hai.

2. In-plane wobble harmless kyun hai lekin out-of-plane slip nahi — geometry. Yaad karo ${\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2$ do vectors ke spanned plane ke perpendicular hai, length $\sin \theta$ ke saath. Agar ${\hat{\mathbf{r}}}_2$ sirf $xy$-plane ke andar wiggle kare, to spanned plane abhi bhi $xy$-plane hai, isliye cross product abhi bhi $\pm \hat{\mathbf{z}}$ ke along point karta hai — iska direction unchanged hai, sirf length breathe karti hai (aur normalize karna length erase kar deta hai, jaise Cell C mein). Lekin agar ${\hat{\mathbf{r}}}_2$ plane se $\epsilon$ se bahar tip ho, to woh numerator ke perpendicular ko sidewise $\approx \epsilon$ amount se drag karta hai (ek absolute nudge), jabki jo vector hum divide karte hain woh sirf $\sin \theta$ lamba hai. To unit normal ki fractional swing hai $\text{tilt of }{\hat{\mathbf{t}}}_2\approx \frac{\epsilon }{\sin \theta }.$ Yeh step kyun? Yahi amplification hai: fixed absolute nudge $\epsilon$ ko chhoti length $\sin \theta$ se divide karna use multiply karta hai.

3. Numbers daalo. $\theta =3^{\circ }$ ke saath, factor hai $1/\sin 3^{\circ }\approx 19.1$. Ek modest $\epsilon =0.5^{\circ }$ out-of-plane slip $\approx 0.5^{\circ }\times 19.1\approx 9.5^{\circ }$ attitude error ban jaata hai.

Verify: Amplification $=1/\sin 3^{\circ }=19.107$ ✅; $0.5^{\circ }\times 19.107\approx 9.55^{\circ }$ ✅. Rule: achha separation chahiye (best near $90^{\circ }$, jahan $1/\sin 90^{\circ }=1$, koi amplification nahi). Agar tumhare do directions close hain, TRIAD galat tool hai — Kalman Filter — Attitude jaisa filtered estimator ya optimal Davenport q-method use karo.

Neeche figure — kya dekhna hai: cyan curve amplification factor $1/\sin \theta$ hai separation angle $\theta$ ke versus. Dekho yeh $\theta \to 0^{\circ }$ par upar rocket karta hai ($3^{\circ }$ par amber marker gain $\approx 19$ par baitha hai), aur white marker $\theta =90^{\circ }$ par flatten ho jaata hai — sweet spot jahan errors unmagnified pass hoti hain.

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Cell F — Degenerate: exactly parallel (undetermined)

Forecast: Dono vectors ek hi line ke along hain, to kya attitude fully known hai, ya ek degree of freedom lost hai?

1. Slot 2 compute karo. ${\hat{\mathbf{r}}}_1\times {\hat{\mathbf{r}}}_2=(1,0,0)\times (1,0,0)=(0,0,0)$. Yeh step kyun? Parallel vectors ka cross product zero hota hai ($\sin 0^{\circ }=0$).

2. Normalize → division by zero. ${\hat{\mathbf{t}}}_2=0/0$ undefined hai. Triad build nahi ho sakta. Yeh step kyun? Geometrically, ek direction sirf 3 mein se 2 degrees of freedom fix karta hai — us line ke baare mein spin invisible hai, isliye sirf in do collinear vectors use karne wala koi bhi method use nahi kar sakta.

Verify: $\Vert (1,0,0)\times (-1,0,0)\Vert =\sin 180^{\circ }=0$ ✅ (anti-parallel utna hi degenerate hai). Rule: TRIAD ko do non-parallel directions chahiye — yahi reason hai ki parent ne "do independent directions" par stress kiya tha.

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Cell G — Real word problem (slot 1 wisely choose karo)

Forecast: Kaunsa measurement slot 1 mein jaata hai? Aage padhne se pehle guess karo.

1. Slots accuracy se assign karo. Slot 1 = Sun sensor ($0.1^{\circ }$, sabse trusted). To ${\hat{\mathbf{r}}}_1={\hat{\mathbf{s}}}_{\text{ref}}=(1,0,0)$, ${\hat{\mathbf{r}}}_2={\hat{\mathbf{m}}}_{\text{ref}}=(0,0,1)$; ${\hat{\mathbf{b}}}_1=(0,0,1)$, ${\hat{\mathbf{b}}}_2=(0,-1,0)$. Yeh step kyun? Slot 1 exactly honor kiya jaata hai, isliye accurate sensor ki info kabhi degrade nahi hoti.

2. Reference triad. ${\hat{\mathbf{t}}}_1^{(r)}=(1,0,0)$; $(1,0,0)\times (0,0,1)=(0,-1,0)={\hat{\mathbf{t}}}_2^{(r)}$; $(1,0,0)\times (0,-1,0)=(0,0,-1)={\hat{\mathbf{t}}}_3^{(r)}$. $M_r=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]$

3. Body triad. ${\hat{\mathbf{t}}}_1^{(b)}=(0,0,1)$; $(0,0,1)\times (0,-1,0)=(1,0,0)={\hat{\mathbf{t}}}_2^{(b)}$; $(0,0,1)\times (1,0,0)=(0,1,0)={\hat{\mathbf{t}}}_3^{(b)}$. $M_b=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$

4. Assemble $A=M_b{M}_r^{\mathsf{T}}$: $A=\left[\begin{array}{ccc}0& -1& 0\\ 0& 0& -1\\ 1& 0& 0\end{array}\right]$

Verify: $A{\hat{\mathbf{r}}}_1=(0,0,1)={\hat{\mathbf{b}}}_1$ ✅ (Sun exactly match kiya); $A{\hat{\mathbf{r}}}_2=(0,-1,0)={\hat{\mathbf{b}}}_2$ ✅; $\det A=+1$ ✅ aur $A{A}^{\mathsf{T}}=I$ ✅, ek proper rotation. $A$ ko onboard propagation ke liye quaternion mein convert karna Quaternion Attitude Kinematics ka kaam hai.

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Cell H — Exam twist ($A$ se angle read karo)

Forecast: $A$ ki trace (diagonal entries ka sum) rotation angle encode karta hai. Kaun sa angle $\mathrm{tr}=2$ deta hai — aur kya tum trace se akele bata sakte ho ki yeh clockwise ghuma ya counter-clockwise?

1. Trace kyun, koi ek entry kyun nahi? Angle $\theta$ se koi bhi proper rotation ke eigenvalues $1,e^{i\theta },e^{-i\theta }$ hote hain; unka sum (trace) $1+2\cos \theta$ hai, axis se independent. Ek single matrix entry axis aur angle mix karti hai, isliye woh $\theta$ isolate nahi kar sakti; trace kar sakta hai. (Yeh identity upar ke definition callout mein stated hai.) Yeh step kyun? Hum $A$ se woh ek scalar choose karte hain jo sirf $\theta$ par depend karta hai.

2. $\arccos$ se invert karo. Cell B se, $\mathrm{tr}(A)=\frac{1}{2}+\frac{1}{2}+1=2$. Tab $\theta =\arccos \left(\frac{\mathrm{tr}(A)-1}{2}\right)=\arccos \left(\frac{2-1}{2}\right)=\arccos (0.5)=60^{\circ }.$ Yeh step kyun? $\arccos$ yeh sawaal hai "kaunsa angle is cosine ka hai?" — yeh cosine ko undo karta hai.

3. Quadrant ambiguity ka saamna karo. Kyunki $\cos \theta =\cos (-\theta )$ aur $\arccos$ sirf $[0^{\circ },180^{\circ }]$ output karta hai, trace $\theta =60^{\circ }$ deta hai lekin nahi bata sakta ki turn $+60^{\circ }$ (counter-clockwise) tha ya $-60^{\circ }$ (clockwise) — dono ka trace $2$ hai. Tie break karne ke liye tum off-diagonal part padhte ho: skew piece $A-A^{\mathsf{T}}$ axis ko uske sign ke saath carry karta hai. Yahan $A_{21}-A_{12}=\frac{\sqrt{3}}{2}-(-\frac{\sqrt{3}}{2})=\sqrt{3}>0$, jo sense ko $+\hat{\mathbf{z}}$ ke baare mein counter-clockwise pin karta hai. Yeh step kyun? $\cos$ even hai, isliye sign lose ho jaata hai; antisymmetric part use restore karta hai.

Verify: $\cos 60^{\circ }=0.5$ aur $1+2(0.5)=2=\mathrm{tr}(A)$ ✅. Recovered magnitude $=60^{\circ }$, jo truth se match karta hai jo humne hide ki thi; positive skew entry $+\hat{\mathbf{z}}$ sense confirm karti hai ✅.

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Active Recall

Recall Kaunsi cell TRIAD ko break karti hai, aur kyun?

Cells E aur F — near-parallel aur exactly-parallel inputs. F mein cross product $0$ hai (division by zero); E mein $\sin \theta$ se divide karna noise ko $1/\sin \theta$ se multiply karta hai. Fix: achha angular separation demand karo, ideally $90^{\circ }$ ke paas.

Recall Cell C mein slot-2 sensor

$2^{\circ }$ off tha. Kitna attitude error aaya, aur kyun? Essentially zero yahan — kyunki cross product ko normalize karna uski length throw away kar deta hai ($\sin$ angle ki), sirf direction rakhta hai, jo in-plane tilt ke liye barely hili. TRIAD slot-2 angle discard karne ke liye designed hai.

Recall

$\mathrm{tr}(A)=1+2\cos \theta$ kya batata hai, aur kya hide karta hai? Yeh rotation angle ka magnitude deta hai $\theta =\arccos ((\mathrm{tr}A-1)/2)$ se ::: lekin iska sign nahi — clockwise vs counter-clockwise ke liye off-diagonal (skew) part chahiye, kyunki $\cos$ even hai.

Recall

$0.1^{\circ }$ Sun sensor aur $2^{\circ }$ magnetometer ke liye slot assignment? Sun sensor slot 1 mein (exactly honored), magnetometer slot 2 mein (sirf uski direction use hoti hai) ::: kyunki slot 1 ka measurement perfectly preserve hota hai jabki slot 2 ka precise angle discard ho jaata hai.