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

3.5.12 · D1 · Physics › Guidance, Navigation & Control (GNC) › Attitude estimation — triad method (two vector measurements)

Is page pe assume kiya gaya hai ki aapne parent topic ki notation pehle kabhi nahi dekhi. Hum har symbol ground up se build karte hain, us order mein jis order mein topic ko unki zaroorat hai.

1. Space mein arrows — ek "vector" kya hota hai

Ek vector bas ek arrow hai: iske paas ek direction hoti hai (kidhar point karta hai) aur ek length hoti hai (kitna lamba hai). Hum ise bold mein likhte hain, jaise $v$ , yeh kehne ke liye ki "yeh ek arrow hai, koi plain number nahi."

Ek arrow ko numbers se likhne ke liye, hum ise teen perpendicular measuring sticks ke set mein daalte hain — x, y, z axes — aur padhte hain ki wo har ek ke saath kitni door tak jaata hai. Wo teen numbers components hain:

$v = (v_{x}, v_{y}, v_{z})$

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

Figure 1 — Ek single arrow $v$ aur uski teen coloured shadow-lengths $v_{x}, v_{y}, v_{z}$ jo x, y, z axes par drop ki gayi hain. Takeaway: ek vector kuch nahi bas teen numbers hain, ek per axis.

Topic ko isko kyun chahiye. TRIAD mein sab kuch — Sun ki direction, magnetic field, spacecraft ke axes — sab arrows hain. Kuch nahi. Agar aap arrows-as-three-numbers se comfortable hain, aap ready hain.

2. Chhoti hat — ek "unit vector" (sirf direction)

Kabhi kabhi hum sirf yeh care karte hain ki ek arrow kis taraf point karta hai, kitna lamba hai nahi. Toh hum ise length $1$ tak shrink kar dete hain. Length-1 arrow ko unit vector kehte hain, aur hum ise ek chhoti hat se mark karte hain: $\hat{v}$ (padho "v-hat").

Kisi bhi arrow ko length 1 tak shrink karne ke liye, use uski apni length se divide karo:

$\hat{v} = \frac{v}{∥ v ∥}$

Yahan $∥ v ∥$ (padho "the norm of v" — double bars ka matlab length hai) 3D Pythagoras rule se nikala jaata hai:

$∥ v ∥ = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$

Topic ko isko kyun chahiye. Parent document mein $\hat{r}_{1}, \hat{b}_{1}$ har jagah likha hai — sab hats. Hat ek promise hai: "is arrow ki length 1 hai, yeh ek direction hai."

3. Do frames — reference vs. body (dono right-handed)

Ek frame ek choice hai ki "kaunsi taraf x hai, kaunsi taraf y hai, kaunsi taraf z hai." Topic do frames use karta hai:

Reference (inertial) frame — sky ka fixed "map." Hum jaante hain ki Sun aur magnetic field yahan almanac/model se kahan point karte hain. Yahan arrows ko letter $\hat{r}$ milta hai (kisi bhi frame quantity par subscript $r$ ).
Body frame — spacecraft se chipka hua, toh yeh rotate hota hai jab spacecraft rotate hota hai. Hamare sensors is frame mein directions report karte hain. Yahan arrows ko letter $\hat{b}$ milta hai (subscript $b$ ).

Figure 2 — Same physical Sun-arrow do baar draw kiya gaya: ek baar fixed reference sky-map par (numbers $\hat{r}_{1}$ ), ek baar tilted spacecraft body frame par (numbers $\hat{b}_{1}$ ). Red arrow woh rotation $A$ hai jo unhe link karta hai. Takeaway: sirf spacecraft ki orientation hi do number-sets ko alag karti hai.

Topic ko isko kyun chahiye. Attitude estimation precisely woh kaam hai jisme hum woh twist dhundhte hain jo in dono frames ko connect karta hai. Do frames nahi, toh problem solve karne ki zaroorat hi nahi.

4. Subscripts $1$ aur $2$ — kaunsa landmark

Subscript bas yeh label karta hai ki kaunsa landmark hai:

$\hat{r}_{1}, \hat{b}_{1}$ = landmark 1 (maano Sun), reference / body coordinates mein.
$\hat{r}_{2}, \hat{b}_{2}$ = landmark 2 (maano magnetic field).

Toh $\hat{b}_{2}$ ka matlab hai: "landmark 2 ki taraf unit vector, jaise body frame mein measure kiya gaya." Bas itna hi.

5. Rotation matrix $A$ — "twist machine"

Ek matrix ek $3 \times 3$ numbers ka grid hai jo ek arrow par act karta hai: ek vector feed karo, yeh ek aur vector wapas deta hai. Woh special matrices jo arrows ko sirf rotate karti hain (kabhi stretch nahi, kabhi squash nahi, kabhi mirror nahi) unhe rotation matrices kehte hain. Topic attitude wali ko $A$ kehta hai.

Yeh ek vector ke reference-frame numbers ( $v_{r}$ ) ko uske body-frame numbers ( $v_{b}$ ) mein map karta hai — exactly wohi $r$ / $b$ subscripts use karke jo humne Section 3 mein set up kiye:

$v_{b} = A v_{r}$

Ise padho: "arrow ke reference-frame numbers lo, unhe twist machine $A$ se guzaaro, aur niklengi body-frame numbers."

Do facts jinka hum baar baar use karenge:

Orthonormal = iske columns unit vectors hain jo mutually perpendicular hain. Aisi matrix ke liye inverse transpose ke barabar hoti hai: $A^{- 1} = A^{T}$ (transpose ke baare mein neeche aur jaankari).
Determinant $= + 1$ = yeh ek proper rotation hai (ek asli turn), mirror-flip nahi.

Topic ko isko kyun chahiye. "Attitude" hai hi yeh matrix $A$ . Ise find karna hi poora goal hai.

6. Transpose $A^{T}$ , identity $I$ , aur inverse $A^{- 1}$

Transpose $A^{T}$ ek matrix ko uske diagonal ke across flip karta hai: row 1 column 1 ban jaata hai, aur aage bhi aise hi.

Identity matrix $I$ "kuch-nahi-karo" matrix hai — diagonal par $1$ s, baaki jagah $0$ s. Kisi bhi vector ko $I$ se multiply karna use seedha wapas de deta hai bina change ke, jaise kisi number ko $1$ se multiply karna:

$I = 100010001, I v = v$

Inverse $A^{- 1}$ "undo" machine hai — $A$ lagao phir $A^{- 1}$ lagao aur aap wapas wahan hain jahan the: $A^{- 1} A = I$ (rotation karo, phir undo karo, aur aisa lagega jaise kuch hua hi nahi).

Final formula mein use kiya gaya jadu: ek rotation matrix ke liye, transpose karna IS inverting karna hai. Grid ko diagonal ke across flip karna rotation ko undo kar deta hai. Isliye parent document hard-to-compute $M_{r}^{- 1}$ ko trivial $M_{r}^{T}$ se replace kar deta hai.

Recall Hum

$M_{r}^{- 1}$ ko $M_{r}^{T}$ se kyun swap kar sakte hain? Kyunki $M_{r}$ orthonormal hai (unit, perpendicular columns), aur orthonormal matrices ke liye $M^{- 1} = M^{T}$ — rotation ko undo karna diagonal ke across flip karne ke barabar hai. ::: Orthonormal matrices ke liye $M^{T} M = I$ , toh $M^{T}$ already hai inverse.

7. Cross product $\times$ — perpendicular-maker

Yeh woh tool hai jo asli kaam karta hai, toh hum ise dhyan se define karte hain aur kehte hain kyun yeh tool aur koi nahi.

Do arrows ka cross product, $a \times b$ likha jaata hai, ek naaya arrow produce karta hai jo dono ke perpendicular hota hai. Components mein:

$a \times b = (a_{y} b_{z} - a_{z} b_{y}, a_{z} b_{x} - a_{x} b_{z}, a_{x} b_{y} - a_{y} b_{x})$

Do properties bahut zyada matter karti hain:

Direction: $a$ aur $b$ wale plane ke perpendicular, us taraf pointing jais taraf aapka right thumb point karta hai jab aapki ungliyan $a$ se $b$ ki taraf curl karti hain (right-hand rule — isliye hi humne insist kiya tha ki dono frames right-handed hain; dekho Cross Product and Right-Handed Frames).
Length: $∥ a \times b ∥ = ∥ a ∥ ∥ b ∥ sin θ$ , jahan $θ$ unke beech ka angle hai.

Figure 3 — Do arrows $a$ (yellow) aur $b$ (blue) ek shaded plane mein lie karte hue, aur unka cross product $a \times b$ (red) dono ke perpendicular seedha bahar nikal raha hai. Takeaway: cross product ek brand-new axis manufacture karta hai jo aapne jo do feed kiye unke right angles par hota hai.

Topic ko isko kyun chahiye. Dono triads ke har second aur third axis ek normalized cross product hai. Yeh woh engine hai jo do arrows ko ek full 3-axis skeleton mein turn karta hai.

8. Triad $M_{r}, M_{b}$ — ek 3-axis skeleton, step by step build kiya gaya

Ab hum actual TRIAD skeleton assemble karte hain. Reference arrows $\hat{r}_{1}, \hat{r}_{2}$ lo aur teen perpendicular unit axes $\hat{t}_{1}^{(r)}, \hat{t}_{2}^{(r)}, \hat{t}_{3}^{(r)}$ banao (hum har axis par superscript $(r)$ use karte hain yeh kehne ke liye ki "reference triad se belong karta hai"):

$\hat{t}_{1}^{(r)} = \hat{r}_{1}$ $\hat{t}_{2}^{(r)} = \frac{r ^ _{1} \times r ^ _{2}}{∥ r ^ _{1} \times r ^ _{2} ∥}$ $\hat{t}_{3}^{(r)} = \hat{t}_{1}^{(r)} \times \hat{t}_{2}^{(r)}$

Har line padho: axis 1 bas trusted pehla arrow hai; axis 2 (normalized) cross product hai, guaranteed axis 1 ke perpendicular; axis 3 pehle do ka cross hai, last perpendicular direction fill in karta hai aur set ko right-handed banata hai.

Identical recipe body arrows $\hat{b}_{1}, \hat{b}_{2}$ ke saath karo (ab superscript $(b)$ ):

$\hat{t}_{1}^{(b)} = \hat{b}_{1}$ $\hat{t}_{2}^{(b)} = \frac{b ^ _{1} \times b ^ _{2}}{∥ b ^ _{1} \times b ^ _{2} ∥}$ $\hat{t}_{3}^{(b)} = \hat{t}_{1}^{(b)} \times \hat{t}_{2}^{(b)}$

Aakhir mein, teen axes ko columns ki tarah stack karo ek matrix banane ke liye. Hum matrix ko ek subscript se likhte hain ( $M_{r}$ reference triad ke liye, $M_{b}$ body triad ke liye) aur uske columns ko matching superscript se:

M_b = \big[\,\hat{\mathbf t}_1^{(b)} \;\; \hat{\mathbf t}_2^{(b)} \;\; \hat{\mathbf t}_3^{(b)}\,\big]$$ Kyunki teen columns right-handed order mein perpendicular unit vectors hain, **$M_r$ aur $M_b$ automatically orthonormal hain aur $\det = +1$** — valid rotations. Yahi baad mein $M^{-1} = M^{\mathsf T}$ trick ko legal banata hai. > [!definition] Triad matrix $M$ > Ek $3\times 3$ grid jiske teen columns ek right-handed set of perpendicular unit vectors $\hat{\mathbf t}_1, \hat{\mathbf t}_2, \hat{\mathbf t}_3$ hain. Picture: arrows ka ek chhota rigid tripod, matrix ki tarah likha hua. $M_r$ sky-map par draw kiya gaya tripod hai; $M_b$ wahi **same physical tripod** spacecraft par draw kiya gaya. **Topic ko isko kyun chahiye.** $M_r$ aur $M_b$ dono **ek hi tripod** describe karte hain, bas do alag frames mein. Jo rotation ek ko doosre mein le jaata hai woh *attitude* hai: $A M_r = M_b$, isliye $A = M_b M_r^{\mathsf T}$. --- ## 9. Symbols ko ek saath rakhna Neeche diya gaya map **bottom-up ek shopping list ki tarah** padho: sabse neeche boxed result (TRIAD attitude) woh hai jo aap cook kar rahe ho; uske upar har box ek ingredient hai jo aapko pehle ready rakhna hai. Har arrow ka matlab hai *"aage wale box se pehle aapko arrow ke peeche wala box chahiye."* Upar se neeche koi bhi path trace karo aur aapko ek valid learning order milega. ```mermaid graph TD A["Vectors as three numbers"] --> B["Unit vectors hat notation"] B --> C["Two right-handed frames reference and body"] C --> D["Rotation matrix A twist machine"] B --> E["Cross product perpendicular maker"] E --> F["Triad matrices Mr and Mb"] D --> F G["Identity and transpose equals inverse for rotations"] --> D F --> H["TRIAD attitude A equals Mb times Mr transpose"] D --> H ``` Toh: pehle arrows aur hats master karo, phir frames, phir do tools (rotation matrices aur cross product), phir unhe triads mein combine karo, aur tabhi attitude read karo. Agar koi bhi box shaky lagta hai, continue karne se pehle uske section par wapas jao. --- ## Equipment checklist Khud test karo — reveal karne se pehle answer zor se bolo. $\hat{\mathbf v}$ mein hat arrow ke baare mein kya promise karta hai? ::: Uski length exactly $1$ hai — yeh ek pure direction hai. Kisi bhi arrow ko unit vector mein kaise shrink karte hain? ::: Use apni length $\lVert\mathbf v\rVert = \sqrt{v_x^2+v_y^2+v_z^2}$ se divide karo. $\hat{\mathbf r}_1$ aur $\hat{\mathbf b}_1$ mein kya farq hai? ::: Same physical direction (landmark 1), lekin fixed reference frame mein vs rotating body frame mein likhi gayi. Frame ke right-handed hone ka kya matlab hai? ::: Right hand ki ungliyan x se y ki taraf curl karti hain, thumb z ke saath point karta hai; equivalently $\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z}$. Matrix $A$ ek vector ke saath *kya* karta hai? ::: Uske reference-frame coordinates ko body-frame coordinates mein rotate karta hai bina length change kiye ($\mathbf v_b = A\mathbf v_r$). Identity matrix $I$ kya hai aur yeh kya karta hai? ::: Diagonal par $1$s aur baaki jagah $0$s wali matrix; yeh kisi bhi vector ko unchanged chhod deti hai ($I\mathbf v=\mathbf v$). Hum $M_r^{-1}$ ki jagah $M_r^{\mathsf T}$ kyun likh sakte hain? ::: Kyunki $M_r$ orthonormal hai, aur orthonormal matrices ke liye transpose inverse ke barabar hoti hai. Teen reference triad axes memory se likho. ::: $\hat{\mathbf t}_1^{(r)}=\hat{\mathbf r}_1$; $\hat{\mathbf t}_2^{(r)}=(\hat{\mathbf r}_1\times\hat{\mathbf r}_2)/\lVert\hat{\mathbf r}_1\times\hat{\mathbf r}_2\rVert$; $\hat{\mathbf t}_3^{(r)}=\hat{\mathbf t}_1^{(r)}\times\hat{\mathbf t}_2^{(r)}$. Cross product $\mathbf a\times\mathbf b$ kis tarah ka result deta hai? ::: Ek naaya arrow dono ke perpendicular, length $\lVert\mathbf a\rVert\lVert\mathbf b\rVert\sin\theta$ ke saath. Cross product ki length kab zero hoti hai, aur yeh TRIAD ko kyun break karta hai? ::: Jab arrows parallel hon ($\sin\theta=0$); us near-zero length se divide karna noise ko catastrophically amplify kar deta hai. Hume ek ki jagah *do* landmarks kyun chahiye? ::: Ek landmark us direction ke around ek spin undetermined chhod deta hai; doosra us last degree of freedom ko lock karta hai. Jab har reveal aasaani se aaye, [[3.5.12 Attitude estimation — triad method (two vector measurements) (Hinglish)|parent topic]] mein derivation par aage badho, aur baad mein noise-optimal upgrades ke liye [[Wahba's Problem]], [[QUEST algorithm]], aur [[Davenport q-method]] dekho.