Reaction wheels — momentum management, zero-crossing: Question bank

3.5.48 · D5 · Physics › Guidance, Navigation & Control (GNC) › Reaction wheels — momentum management, zero-crossing

Definition Is page par use hone wale symbols (ek baar define, freely use karo)

Neeche sab kuch ek chosen spin axis ke baare mein measure kiya gaya hai, isliye saari quantities plain signed numbers hain (scalars us axis ke along) — jab tak arrow na dikhe.

$I_{b}$ = body inertia: spacecraft body ko us axis ke baare mein spin karna kitna mushkil hai (kg·m², ek fixed number jo satellite ke mass layout se set hota hai).
$I_{w}$ = wheel inertia: flywheel ko usi axis ke baare mein spin karna kitna mushkil hai (kg·m², ek chhota fixed number).
$ω_{b}$ = body angular rate: body us axis ke baare mein kitni tez ghoomti hai (rad/s). Sign convention: right hand se, fingers rotation ke along curl karo; agar thumb $+$ axis ki taraf point kare, toh $ω_{b} > 0$ . Hum $+$ ko "left" aur $-$ ko "right" bol rahe hain — sirf labels hain.
$Ω_{w}$ = wheel spin rate: flywheel usi axis ke baare mein kitni tez ghoomta hai, same sign rule.
$τ_{m}$ = motor torque: woh internal twist jo wheel ka motor wheel par lagata hai (aur, reaction se, body par ulta twist). Yeh $τ_{m} = I_{w} \dot{Ω}_{w}$ follow karta hai.
$τ_{ext}$ = external torque: bahar se aane wala real twist (drag, gravity gradient, solar pressure, magnetorquer, thruster). Sirf yahi total momentum badalta hai.
$H_{tot}$ = total angular momentum us axis ke baare mein (units kg·m²/s = N·m·s): body ka momentum aur wheel ka momentum ka sum. Yeh woh bookkeeping quantity hai jo conservation pin karti hai.
$H_{w}$ = wheel angular momentum us axis ke baare mein ( $H_{w} = I_{w} Ω_{w}$ , units N·m·s): flywheel mein kitna rotational "oomph" stored hai. Yahi cheez redline tak fill hoti hai aur ise dump karte hain. Jab yeh general direction mein ho toh vector $H_{w}$ likhte hain.
$Ω_{m a x}$ = wheel redline speed: wheel physically jitni maximum speed reach kar sakta hai; corresponding stored momentum hai $H_{m a x} = I_{w} Ω_{m a x}$ .
$m$ = magnetic dipole moment ek magnetorquer ka (A·m²): coil mein current chala ke produce hone wala commandable "bar-magnet strength and direction"; zyada current → zyada $m$ .
$H_{target}$ = desired wheel momentum: woh bias value jo controller chahta hai ki har wheel ka momentum $H_{w}$ wahan baitha rahe (usually ek non-zero bias, zero nahin), taaki dump law $H_{w}$ ko us taraf nudge kare.

Formula Woh ek relation jis par sab kuch tika hai

Do momenta ko axis ke along add karo total pane ke liye. Plain numbers ke roop mein add kyun? Angular momentum hai (inertia)×(rate), aur ek shared axis ke along yeh scalars ki tarah combine hote hain: $H_{tot} = I_{b} ω_{b} + I_{w} Ω_{w} = I_{b} ω_{b} + H_{w}$ Koi external torque nahin hone par, $H_{tot}$ change nahin ho sakta, isliye uska time-derivative zero hai: $\frac{d}{d t} (I_{b} ω_{b} + I_{w} Ω_{w}) = 0 \Rightarrow I_{b} \overset{ω}{˙}_{b} = - I_{w} \dot{Ω}_{w}$ Minus sign kyun hai, slogan nahin: sum ek constant par pinned hai, isliye agar ek term badhta hai toh doosra exactly utna hi kam hona chahiye — wheel ka momentum gain body ke momentum ki keemat par hota hai, aur vice versa. Yahi forced trade-off hai reaction. External torque hone par right-hand side $0$ ki jagah $τ_{ext}$ ho jaata hai.

Conservation of Angular Momentum se directly do pillars milte hain:

Internal actuators momentum ko body aur wheel ke beech shuffle karte hain — sum $H_{tot}$ fixed rehta hai.
External torques sum ko change karte hain: $\frac{d H _{tot}}{d t} = τ_{ext}$ .

Neeche ki picture exactly yahi trade-off dikhati hai — total bar ki height kabhi nahin badlti; sirf body aur wheel ke beech ka split badlta hai.

Figure — Reaction wheels — momentum management, zero-crossing

Neeche ka almost har trap inhi do pillars ka ek chupaaya hua violation hai.

True or false — justify (sahi ya galat — justify karo)

A reaction wheel can hold a satellite pointed against a constant solar-pressure torque forever, using no propellant.

False. Yeh pointing kuch time ke liye hold karta hai, lekin ek constant-sign external torque bina bound ke integrate hota rehta hai (

H_{w} = \int τ_{ext} d t

), isliye wheel finite time mein saturate ho jaata hai aur phir control kho deta hai. Wheel ek buffer hai, sink nahin.

Spinning a reaction wheel changes the total angular momentum of the spacecraft.

False. Motor torque

τ_{m}

internal hai, isliye yeh sirf momentum ko wheel aur body ke beech move karta hai;

H_{tot}

fixed rehta hai. Sirf ek external actuator (magnetorquers ya thrusters) total ko change kar sakta hai.

If the wheel accelerates, the body must decelerate about the same axis (or reverse).

True, jab koi external torque nahin lagta:

I_{b} \overset{ω}{˙}_{b} = - I_{w} \dot{Ω}_{w}

, isliye dono rates hamesha opposite signs ke saath change hoti hain — sum pinned hai, isliye wheel ka gain exactly body ka loss hai.

A wheel sitting at exactly zero speed is the safest, most benign state for fine pointing.

False. Zero speed sabse bura jagah hai: bearing friction wahan sign reverse karti hai —

Ω_{w} = 0

ke aas-paas sticking (static) friction direction flip karti hai, isliye wheel briefly lock hoti hai phir jhatke se free hoti hai (figure s02 mein friction curve yeh discontinuity dikhata hai — Bearing Friction and Stiction dekho), aur motor cogging rotational averaging ke bina sabse kharab hoti hai. Isse jo jitter hoti hai woh precise pointing barbad kar deti hai, isliye wheels ko zero se door bias mein rakha jaata hai.

Momentum dumping is needed only if the spacecraft is actively slewing a lot.

False. Ek bilkul still, non-maneuvering satellite bhi external secular torques se wheel momentum accumulate karta rehta hai (gravity gradient, solar pressure, drag). Dumping environment se drive hota hai, tumhare maneuvers se nahin.

A magnetorquer can dump wheel momentum in any direction you like.

False. Torque hai

τ = m \times B

, jo hamesha Earth ke field

B

ke perpendicular hota hai; momentum ka woh component jo

B

ke along hai woh us waqt touch nahin kiya ja sakta. Tum

B

ke orbit ke saath rotate hone par depend karte ho taaki eventually sab directions cover ho sakein.

In a 4-wheel pyramid you can arrange biases so that no single wheel ever crosses zero during a maneuver.

True. Redundancy se freedom milti hai ki total momentum wheels ke beech kaise distribute ho; tum required momentum ko is taraf route kar sakte ho ki har wheel zero ke ek hi side par rahe (figure s03 mein pyramid schematic chaar tilted axes dikhata hai jo yeh freedom dete hain), stiction jitter bilkul avoid karte hue.

Thrusters and magnetorquers both change

H_{tot}

, so they are interchangeable for dumping.

Half true. Dono external torque lagate hain aur isliye dono dump kar sakte hain, lekin thrusters propellant kharch karte hain aur kisi bhi direction mein act kar sakte hain, jabki magnetorquers propellant-free hain lekin

B

ke perpendicular tak constrained hain aur sirf wahan useful hain jahan field exist karta ho (LEO). Trade-off, equivalence nahin, yahi point hai.

With three orthogonal wheels, spinning up one wheel only ever affects the body about that wheel's axis.

Aisa generally False hai. Jab wheel aur body axes perfectly aligned nahin hain (ya body pehle se rotate ho rahi ho), ek spinning wheel gyroscope ki tarah act karta hai: uska momentum

H_{w}

body rate ke saath cross karke teesre axis ke baare mein ek cross-coupling torque

ω_{b} \times H_{w}

deta hai (figure s04 do input arrows aur perpendicular output dikhata hai). Real 3-axis controllers ko yeh coupling compensate karni padti hai.

Spot the error (galti pakdo)

"To turn the satellite left, spin the wheel left."

Wrong sign. Hamare right-hand convention (

+

= "left") se,

\overset{ω}{˙}_{b} = - (I_{w} / I_{b}) \dot{Ω}_{w}

: wheel ko

+

("left") sense mein spin karne se body

-

("right") sense mein ghoomti hai. Body ko left point karne ke liye wheel ko right spin karna padega (target se door).

H_{w} (t) = H_{w} (0) + τ_{ext} t

works for any external torque."

Sirf tab valid hai jab

τ_{ext}

constant ho. Generally

H_{w} (t) = H_{w} (0) + \int_{0}^{t} τ_{ext} d t^{'}

; linear-in-

t

form constant integrand ka special case hai.

"Since friction is tiny, we can command the wheel through zero and ignore stiction."

Wrong: zero speed ke paas static (stick) friction fine pointing ke liye needed tiny torque se zyada ho sakti hai, isliye wheel stick karti hai, pointing error badhta hai, phir motor free hoti hai aur snap karta hai — ek impulsive disturbance. Average par small, lekin jis instant matter karta hai us par catastrophic.

"The magnetorquer dump law

m = - k (H_{w} - H_{target}) \times B

directly sets the momentum to target."

Nahin — yeh dipole

m

set karta hai (coil ka magnetic moment), jo ek torque

m \times B

produce karta hai jo sirf time ke saath

H_{w}

ko bias

H_{target}

ki taraf drive karta hai, aur sirf

B

ke perpendicular directions mein. Yeh ek slow feedback nudge hai, instantaneous reset nahin.

"Because

τ_{m} = I_{w} \dot{Ω}_{w}

, a bigger wheel inertia

I_{w}

always means you can slew faster."

Error: ek given motor torque

τ_{m}

ke liye,

\dot{Ω}_{w} = τ_{m} / I_{w}

, isliye bada

I_{w}

dheere spin up hota hai. Bada

I_{w}

zyada momentum store karta hai lekin apne aap faster body acceleration nahin deta — woh reaction

τ_{m} / I_{b}

se set hoti hai.

"If the body is held perfectly still, no angular momentum is stored anywhere."

Wrong. Body ko external torque ke against still rakhne se wheel ko woh sab absorb karna padta hai:

I_{w} \dot{Ω}_{w} = τ_{ext}

, isliye wheel ka stored momentum exactly isliye badhta hai kyunki body still rakhी gayi hai.

Why questions (kyun ke sawaal)

Why is the effect called a reaction wheel?

Motor torque

τ_{m}

jo wheel ko spin karta hai woh body par wapas react karta hai (Newton's third law / momentum conservation), aur body par yahi reaction desired rotation produce karta hai — body wheel ke reaction mein move hoti hai.

Why must desaturation use an external actuator rather than a cleverer wheel command?

Wheels sirf

H_{tot}

ko internally redistribute karte hain; koi bhi internal cleverness total nahin badal sakti. Accumulated momentum hatane ke liye total ko change karna padta hai, jo sirf external torque

τ_{ext}

kar sakta hai.

Why do operators deliberately keep wheels at a non-zero bias speed?

Zero-crossing region se bachne ke liye jahan bearing friction reverse hoti hai aur cogging sabse kharab hoti hai;

H_{w}

ko bias value

H_{target}

ki taraf regulate karne se friction unidirectional rehti hai, fine pointing ke dauran stick-slip jitter eliminate hoti hai.

Why does a constant, tiny external torque eventually saturate a wheel even though it is far smaller than the motor's capability?

Kyunki yeh constant-sign hai: uska integral hamesha ke liye linearly badhta rehta hai, isliye torque chahe kitna bhi chhota ho,

H_{w}

diye gaye waqt mein inevitably

H_{m a x} = I_{w} Ω_{m a x}

reach kar leta hai. Saturation accumulation ke baare mein hai, instantaneous magnitude ke baare mein nahin.

Why can magnetorquer dumping be slow and require spreading over the orbit?

Available torque

m B sin θ

(jahan

m

dipole magnitude hai,

θ

m

aur

B

ke beech ka angle hai) bahut chhota hota hai kyunki Earth ka field LEO mein weak hai, aur yeh sirf

B

ke perpendicular act karta hai; tumhe

B

ke orbit ke saath rotate hone ka wait karna padta hai taaki pehle na touch hone wala momentum component reach ho sake.

Why does the wheel absorb external angular momentum when the controller holds the body fixed?

\overset{ω}{˙}_{b} \approx 0

hone par, momentum balance

\frac{d}{d t} (I_{b} ω_{b} + I_{w} Ω_{w}) = τ_{ext}

reduce hoke

I_{w} \dot{Ω}_{w} = τ_{ext}

ban jaata hai — injected external momentum ka har bit wheel mein jaata hai.

Why are reaction wheels preferred over thrusters for years-long precise pointing?

Yeh sirf electric power use karke re-point karte hain (solar → motor), routine attitude changes ke liye koi propellant nahin kharchte, jabki thrusters finite fuel jalate hain aur khatam ho jaate. Momentum internally borrow kiya jaata hai, bahar nahin phinka jaata.

Edge cases

What happens the instant a wheel reaches saturation (

Ω_{w} = Ω_{m a x}

, i.e. stored momentum

H_{m a x} = I_{w} Ω_{m a x}

) while an external torque still pushes the same way?

Wheel aur tez nahin ghoom sakta, isliye yeh aur momentum absorb nahin kar sakta; excess ab body ko torque karta hai, aur attitude control tab tak kho jaata hai jab tak momentum dump nahin hota. Saturation = buffering capacity ka end.

If the external torque is zero everywhere, does the wheel ever need dumping?

Nahin —

τ_{ext} = 0

hone par total constant hai aur wheel sirf maneuvers ke dauran body ke saath momentum trade karta hai, baad mein bias par wapas aa jaata hai. Dumping sirf isliye required hai kyunki real environments net momentum inject karte hain.

At the instant

m

is parallel to

B

, how much dumping torque is available?

Zero, kyunki

m \times B = 0

jab yeh parallel hon (

sin θ = 0

). Us geometry mein kisi bhi axis ke baare mein koi torque produce nahin hota — geometry change hone ka wait karna padta hai.

What if the momentum to be dumped lies entirely along

B

at this moment?

Magnetorquer us waqt use touch nahin kar sakta (yeh sirf

B

ke perpendicular act karta hai). Tumhe

B

ke orbit ke saath reorient hone ka wait karna padta hai taaki woh component perpendicular aur dumpable ho sake.

For a wheel regulated toward zero (no bias, so

H_{target} = 0

), what is the failure mode as

Ω_{w} \to 0

Yeh stick-slip regime mein enter karta hai: static friction fine-pointing torque command se zyada ho sakti hai, wheel hang karta hai, error badhta hai, phir impulsively free hota hai — exactly finest pointing ke dauran jitter inject karta hai.

In the limiting case

I_{w} ≪ I_{b}

(a small wheel on a big satellite), what body rate results from a fixed wheel speed?

Ek tiny sa:

ω_{b} = - (I_{w} / I_{b}) Ω_{w} \to 0

jaise ratio chhota hota hai. Chhote wheels fine, slow control authority dete hain — achha resolution, limited slew rate aur limited momentum storage.

When wheel and body axes are misaligned (or the body already rotates), why does spinning one wheel disturb other axes?

Ek spinning wheel momentum

H_{w}

carry karta hai; body rotation ke saath combine hokar yeh ek gyroscopic cross-coupling torque

ω_{b} \times H_{w}

produce karta hai jo ek teesre axis ke along point karta hai (same physics jo Control Moment Gyroscopes (CMG) exploit karte hain). Ek achhe Attitude Determination and Control System (ADCS) / controller ko use model aur cancel karna padta hai.

Recall Pure bank ka ek-line summary

Internal wheels momentum shuffle karte hain (conservation, minus sign, reaction); external actuators hi single tarika hain total badalne ka (dumping); zero speed dangerous hai (stiction jitter), isliye wheels bias par rehte hain; aur misaligned spinning wheels control axes ko gyroscopically cross-couple karte hain.