Control moment gyroscopes (CMG) — high torque, singularity: Question bank

3.5.49 · D5 · Physics › Guidance, Navigation & Control (GNC) › Control moment gyroscopes (CMG) — high torque, singularity

Shuru karne se pehle, ek common vocabulary banate hain taaki koi bhi symbol unexplained na rahe. Neeche ki picture teen key arrows anchor karti hai — har question ke liye ise dhyan mein rakho.

Figure — Control moment gyroscopes (CMG) — high torque, singularity

$I$ = wheel ka moment of inertia — ek number jo batata hai ki wheel ko spin up karna kitna mushkil hai (bada, bhari rim = bada $I$ ).
$ω$ (ya $ω_{w}$ ) = flywheel ki spin rate rad/s mein — wheel khud kitni tez ghoom raha hai.
$h = I ω \overset{s}{^}$ = ek single wheel ka stored angular momentum — ek arrow jis ki fixed length $h$ hai (spin speed constant hai) aur jis ki direction gimbal ghoomata hai.
$\overset{g}{^}$ = gimbal axis, woh hinge line jis ke around poora spinning wheel tilt hota hai.
$Ω = \dot{δ} \overset{g}{^}$ = gimbal rate — hinge kitni tez ghoomta hai ( $δ$ = gimbal angle).
$τ = Ω \times h$ = spacecraft par output torque; " $\times$ " cross product hai, jo dono arrows ke perpendicular hota hai.
$n$ = cluster mein CMGs ki sankhya (typically 4). Har CMG ek gimbal angle $δ_{i}$ contribute karta hai, toh $n$ hain, $i = 1, \dots, n$ .
$A$ = Jacobian, ek table jis mein $3$ rows hain (ek har spatial axis $x, y, z$ ke liye) aur $n$ columns hain (ek har CMG ke liye). Column $i$ woh arrow hai $\partial h_{i} / \partial δ_{i} = \overset{g}{^}_{i} \times h_{i}$ — "CMG $i$ abhi kis direction mein push kar sakta hai". Toh $A$ ek $3 \times n$ matrix hai.

Definition Pseudo-inverse, null space, aur singularity measure

Torque command karne ke liye hum $τ = - A \dot{δ}$ ko gimbal rates $\dot{δ}$ ke liye solve karte hain. Kyunki $A$ ek $3 \times n$ matrix hai jis mein $n > 3$ hai (equations se zyada unknowns), hum pseudo-inverse use karte hain — dekho Jacobian & pseudo-inverse (Moore–Penrose): $A^{+} = A^{T} (A A^{T})^{- 1} .$ Yahan $A^{T}$ , $A$ hai jis ki rows aur columns swap hain ( $n \times 3$ ), aur product $A A^{T}$ ek choti $3 \times 3$ matrix hai — isliye hum uska determinant le sakte hain. $A^{+}$ $A$ ka "best-fit undo" hai: yeh sabse chhote gimbal rates return karta hai jo wanted torque produce karte hain.

$A$ ka null space — dekho Null space & rank of a matrix — gimbal-rate combinations $\dot{δ}$ ka set hai jahan $A \dot{δ} = 0$ ho: aise motions jo zero torque produce karte hain. $n = 4 > 3$ ke saath yeh non-empty hota hai, jo gimbals ko silently reshuffle karne ke liye null motion room deta hai.

Singular direction $\overset{u}{^}$ woh unit vector hai jahan $\overset{u}{^}^{T} A = 0$ ho: woh output axis jis tak koi gimbal motion pahunch nahi sakta. Yeh $3 \times 3$ matrix $A A^{T}$ ke us eigenvector ke roop mein milta hai jiska eigenvalue zero (ya sabse chota) ho.

Singularity measure $m = det (A A^{T})$ geometrically $A$ ke columns se bane parallelepiped (box) ka volume hai. Jab columns 3-D mein spread out hote hain toh box mota hota hai aur $m$ bada hota hai; jab woh ek plane mein flatten ho jaate hain toh box collapse ho jaata hai aur $m \to 0$ — exactly rank ka loss. Toh $det (A A^{T})$ literally measure karta hai ki reachable directions abhi kitna 3-D volume fill karti hain.

True or false — justify karo

True or false: Ek CMG apna flywheel speed up aur down karke torque create karta hai.

False — yeh ek reaction wheel hai, dekho Reaction wheels — momentum storage & saturation. Ek CMG spin speed

ω

constant rakhta hai aur spin axis tilt karta hai; torque

Ω \times h

hai,

I \overset{ω}{˙}

nahi.

True or false: Kyunki tum pehle se bade momentum ko sirf tilt kar rahe ho, ek chota gimbal motor ek bada torque deliver karta hai.

True — yeh torque amplification hai:

∣ τ ∣ = h \dot{δ}

, aur

h

bada hai, toh gentle

\dot{δ}

bhi bada

τ

deta hai. Momentum store karna sasta hai; use redirect karna leverage deta hai.

True or false: Ek single CMG ka output torque gimbal axis

\overset{g}{^}

ke saath point karta hai.

False — torque

Ω \times h

hai, jo

\overset{g}{^}

ke perpendicular hota hai (aur

h

ke bhi). Motor

\overset{g}{^}

ke around twist karta hai; precession torque sideways deliver karta hai.

True or false: Ek single CMG teen spatial directions mein se kisi bhi direction mein torque produce kar sakta hai.

False — ek CMG sirf us plane mein push kar sakta hai jo uski gimbal axis ke perpendicular hai. Teeno axes cover karne ke liye ek cluster chahiye (usually 4).

True or false: Singularity par CMGs completely band ho jaate hain aur bilkul bhi torque produce nahi kar sakte.

False — woh sirf ek singular direction

\overset{u}{^}

mein torque kho dete hain; baaki har direction abhi bhi reachable hai. Yeh ek missing direction hai, dead cluster nahi.

True or false: Singularity measure

m = det (A A^{T})

exactly zero hota hai jab

A

rank lose karta hai.

True —

det (A A^{T})

A

ke columns se bane box ka volume hai (squared singular values ka product); agar koi bhi singular value zero hit kare, toh box flatten ho jaata hai, rank 3 se neeche aa jaata hai, aur

m \to 0

True or false: CMG flywheel ko energy store karne ke liye spin karta hai, aur woh stored energy hi ship ko torque deta hai.

False — yeh angular momentum store karta hai, torque ke liye energy nahi. Torque us momentum ki direction change karne se aata hai (

\dot{h}

), spin ki kinetic energy se nahi.

True or false: Null motion cluster ke total stored momentum

H

ko change karta hai.

False — null motion

A

ke null space mein rehta hai (zero-torque motions), toh by definition yeh zero net torque produce karta hai aur

H

unchanged rakhta hai; yeh sirf internal gimbal angles reshuffle karta hai.

True or false: CMGs ka ek flat (planar) cluster, sabke vertical gimbal axes ke saath, vertical axis ke around torque kar sakta hai.

False — agar har

\overset{g}{^}_{i}

vertical hai, toh har torque

\overset{g}{^}_{i} \times h_{i}

horizontal hoga. Vertical axis ek structural singularity hai; isliye real clusters 3-D pyramids hote hain.

True or false: Singularity ke paas required gimbal rates

\dot{δ}

chhote ho jaate hain.

False — woh blow up karte hain. Pseudo-inverse

3 \times 3

matrix

A A^{T}

ke nearly-zero eigenvalue se divide karta hai, toh ek tiny commanded torque enormous gimbal rates maangta hai, motors saturate ho jaate hain.

Spot the error

"Kyunki

∣ h ∣

constant hai,

\dot{h} = 0

, toh ek CMG koi torque produce nahi karta." — galti kahan hai?

∣ h ∣

(length) constant hai, lekin

h

ek vector hai jis ki direction rotate hoti hai. Fixed-length rotating vector ka phir bhi

\dot{h} = Ω \times h \neq = 0

hota hai.

"Singularity se bachne ke liye, bas pseudo-inverse

A^{+} = A^{T} (A A^{T})^{- 1}

as usual lo." — kya toot ta hai?

Singularity par

3 \times 3

matrix

A A^{T}

invertible nahi hota (determinant zero, uska box collapse ho gaya hai), toh

A^{+}

exist nahi karta. Tumhe singularity-robust form

A^{T} (A A^{T} + λ I)^{- 1}

chahiye — dekho Singularity-robust inverse & damped least squares.

"Ek elliptic (internal) singularity se nikaalna aasaan hai — bas steer around karo." — ise correct karo.

Yeh ek hyperbolic singularity describe karta hai, jahan nearby gimbal motions singular direction

\overset{u}{^}

se door momentum reduce karte hain, toh tum steer around kar sakte ho. Ek elliptic wala trapped hota hai: har escape route ke liye momentarily

\overset{u}{^}

ki taraf momentum build karna padta hai, toh ordinary steering stuck ho jaati hai; tumhe null motion ya SR-inverse chahiye.

"CMG

i

ke liye Jacobian column sirf uska momentum

h_{i}

hai." — fix karo.

Column

\partial h_{i} / \partial δ_{i} = \overset{g}{^}_{i} \times h_{i}

hai, derivative — woh direction jis mein momentum tip move karta hai jab gimbal ghoomta hai — momentum khud nahi.

"SR-inverse ka damping

λ I

singularity ke paas torque perfectly accurate banata hai." — kya galat hai?

Yeh ulta trade hai:

λ

deliberately ek choti torque error aane deta hai taaki gimbal rates finite rahein. Tum thodi accuracy sacrifice karke safety khareedteho.

"Kyunki torque

Ω \times h

hai, gimbal rate double karne par torque hamesha double ho jaata hai." — yeh kab fail hota hai?

Sirf tab jab

\overset{g}{^} ⊥ h

ho. Generally

∣ τ ∣ = h \dot{δ} sin θ

; agar

\overset{g}{^}

h

ke parallel hone ki taraf swing kare, toh

sin θ

shrink karta hai aur same

\dot{δ}

se kam torque milta hai.

"Singularity matlab sabhi momentum vectors

h_{i}

ek hi direction mein point karte hain." — ise sharpen karo.

Momenta nahi — torque directions

\overset{g}{^}_{i} \times h_{i}

coplanar ho jaate hain, ek common perpendicular

\overset{u}{^}

share karte hain. Equivalently, sabhi

h_{i}

ka $\overset{u}{^}$ par projection same hota hai, jo "sab parallel hain" se ek subtler condition hai.

Why questions

Torque describe karne ke liye simple scalar multiply ki jagah cross product

Ω \times h

kyun choose karte hain?

Kyunki fixed length ka rotating vector hamesha

Ω \times (vector)

ke saath change hota hai — cross product "bina stretching ke rotate karne" ka exact mathematical fingerprint hai, jo ek gimbal

h

ke saath karta hai.

CMG torque amplification kyon deta hai jabki reaction wheel nahi?

Reaction wheel ka torque is par cap hota hai ki motor wheel ko kitni tez spin up kar sakta hai (

I \overset{ω}{˙}

, jahan

I

inertia hai aur

ω

wheel speed). CMG ka torque poore stored

h

par act karta hai use sirf tilt karke, toh ek chota gimbal motor ek bade momentum ka leverage command karta hai.

Hum Jacobian

A

ko partial derivatives

\partial h_{i} / \partial δ_{i}

se kyun banate hain?

Yeh chain rule hai: total torque

- \dot{H} = - \sum (\partial h_{i} / \partial δ_{i}) \dot{δ}_{i}

hai. Har partial ek CMG ki instantaneous push-direction hai, toh

A

collect karta hai "kaun abhi kahan push kar sakta hai".

A A^{T}

ka tiny eigenvalue huge gimbal rates kyun cause karta hai?

Pseudo-inverse

3 \times 3

matrix

A A^{T}

ko invert karta hai, aur near-zero eigenvalue ko invert karne se near-infinite eigenvalue milta hai. Us weak direction ke saath command

1/ ϵ

se scale hoti hai, toh

\dot{δ} \to \infty

jab

ϵ \to 0

Null motion useful kyun hai even though yeh zero torque produce karta hai?

Precisely kyunki yeh torque-free hai (

A

ke null space mein rehta hai), yeh tumhe cluster ko approaching singularity se pre-steer karne deta hai attitude disturb kiye bina — gimbals ko reshape karo pehle, us dead direction ki zaroorat se pehle. Dekho Spacecraft attitude control & slew maneuvers.

CMG ka torque hamesha dono

\overset{g}{^}

aur

h

ke perpendicular kyun hota hai?

Kyunki yeh ek cross product

Ω \times h

hai jisme

Ω ∥ \overset{g}{^}

hai, aur cross product by definition apne dono factors ke perpendicular hota hai.

Clusters sirf 3 axes control karne ke liye 4 CMGs kyun use karte hain?

n = 4 > 3

ke saath extra CMG ek non-trivial null space deta hai, null-motion freedom provide karta hai singularities se steer karne ke liye jo minimal 3-CMG set escape nahi kar sakta.

Damping

λ

actually kaise choose ki jaati hai, aur yeh kya trade off karta hai?

Tum

λ

ko singularity ki closeness ke saath scale karte ho — e.g.

λ = λ_{0} (1 - m / m_{0})

taaki yeh well-conditioned hone par

\approx 0

ho aur sirf wall ke paas grow kare. Badi

λ

gimbal rates ko

\sim 1/ λ

par cap karti hai lekin ek bada torque error

\sim λ

chodti hai; choti

λ

accurate hai lekin huge rates ka risk hai. Yeh classic accuracy-vs-safety knob hai.

Edge cases

Output torque ka kya hota hai jab

\overset{g}{^}

exactly

h

ke parallel ho (

θ = 0

∣ τ ∣ = h \dot{δ} sin 0 = 0

— spin axis ko apne aap ke around tilt karna sirf wheel ko re-spin karta hai aur koi output torque produce nahi karta.

Us instant par jab

\overset{g}{^} ⊥ h

(

θ = 9 0^{\circ}

) ho, kya special hai?

Yeh maximal-torque geometry hai,

∣ τ ∣ = h \dot{δ}

— standard CMG design point, jo har unit gimbal rate par sabse bada torque deta hai.

Agar wheel spin rate

ω

zero tak drift kar jaaye, toh CMG kya ban jaata hai?

h = I ω = 0

ke saath koi stored momentum nahi hai redirect karne ke liye, toh

τ = Ω \times 0 = 0

— "gyroscope" sirf ek dead hinge hai bina kisi gyroscopic torque ke.

Singularity measure

m = det (A A^{T})

ke liye, perfectly conditioned cluster vs. singular cluster mein kya value signal karti hai?

Bada

m

(mota parallelepiped, sabhi singular values healthy) matlab har direction mein torque aasaan hai;

m \to 0

(flattened box) matlab ek direction collapse ho gayi hai aur tum wall par ho.

Do coplanar CMGs ek

\overset{z}{^}

gimbal axis share karte hain

δ_{1} = δ_{2}

ke saath — reachable rank?

Unke do push-columns parallel hain, toh rank

= 1

: woh plane mein sirf ek line ke saath torque kar sakte hain, perpendicular in-plane direction mein nahi (aur

z

ke around kabhi nahi).

SR-inverse

A^{T} (A A^{T} + λ I)^{- 1}

ka limiting behaviour kya hai jab

λ \to 0

Yeh smoothly ordinary pseudo-inverse

A^{+}

tak reduce ho jaata hai jahan

A

full-rank ho — toh SR-inverse meaningfully sirf singularities ke paas differ karta hai, exactly jahan tum chahte ho.

Agar ek commanded torque

τ_{c m d}

entirely singular direction

\overset{u}{^}

ke along ho?

Koi bhi gimbal combination use produce nahi kar sakta:

\overset{u}{^}^{T} A = 0

matlab cluster us direction ke liye blind hai, aur SR-inverse ek chota, damped, inevitably inaccurate response return karega.

Recall Ek-line self-test

Kaun sa single fact ek CMG ki dono amplification aur singularity ko seed karta hai? ::: Torque $Ω \times h$ hai — ek cross product: uski badi length (bade $h$ par leverage) amplify karti hai, aur uski forced perpendicularity $\overset{g}{^}$ se exactly wahi hai jo output directions ko ek plane mein collapse hone deti hai aur ek dead axis create karti hai.