Reaction wheels — momentum management, zero-crossing: Worked examples

3.5.48 · D3 · Physics › Guidance, Navigation & Control (GNC) › Reaction wheels — momentum management, zero-crossing

Yeh page ek drill deck hai. Parent note mein poori kahani hai; lekin yeh page apne aap mein complete hai, toh tum sirf yahan bhi drill kar sakte ho. Hum wheel problem ke har possible case ko cover karte hain — dono signs, zero aur saturated wheels, degenerate geometry, limiting behaviour, ek real-world word problem, aur ek exam twist.

Shuru karne se pehle, paanch symbols jo hum baar baar use karenge — har ek simple words mein define hai, koi prior page assume nahi kiya:

Definition Paanch quantities jo hum juggle karte hain

Spacecraft ko ek bada box (the body) samjho jisme ek chota flywheel (the wheel) spin kar raha hai.

$Ω_{w}$ = wheel spin rate radians per second (rad/s) mein — chota flywheel kitni tezi se ghoomta hai. Ek radian ≈ 57°; "rad/s" = har second mein kitne radians turn hote hain.
$ω_{b}$ = body angular rate (rad/s) — poora box kitna slowly rotate kar raha hai. Yahi cheez hai jo hum ultimately control karna chahte hain (telescope kahan point kar raha hai).
$I_{w}$ = wheel ki spin inertia ( $kg\cdotm^{2}$ ) — flywheel ki spin hone ki resistance.
$I_{b}$ = body ki inertia ( $kg\cdotm^{2}$ ) usi axis ke baare mein — bade box ki turn hone ki resistance. Yeh $I_{w}$ se bahut zyada badi hoti hai (box apne chote wheel se kaafi bhaari hota hai).
$τ$ = torque (units $N\cdotm$ = newton-metre) = ek twist. Torque angular momentum ka rate of change hai: $τ = \dot{H} = \frac{d H}{d t}$ . Chota dot matlab "per second."

Combined: $H_{w} = I_{w} Ω_{w}$ = wheel angular momentum (units $N\cdotm\cdots$ ) — wheel mein kitna spinning-oomph stored hai. Isi tarah $H_{b} = I_{b} ω_{b}$ body ke liye.

Woh ek pair of equations jisse neeche sab kuch nikalta hai:

Scenario matrix

Har reaction-wheel numerical problem in cells mein se ek hai. Neeche ke examples mein cell(s) ka label lagaa hai jo woh hit karta hai.

Cell	Kya special hai	Example
A. Positive slew	Wheel ek taraf spin karo, body doosri taraf ghoomti hai ( $ω_{b}$ ka sign)	Ex 1
B. Negative / reverse slew	Opposite sign — check karo ki minus sach mein flip hota hai	Ex 2
C. Constant torque → saturation	$τ$ = const, linear growth, finite time	Ex 3
D. Sign of external torque	Torque wheel ko help karta hai vs fight karta hai (bias direction)	Ex 4
E. Zero-crossing / degenerate $Ω_{w} = 0$	Wheel zero se guzarta hai; stiction; bias kyun	Ex 5
F. Geometry limit: $m \times B$	Angle $θ$ ; $sin θ$ ; $θ \to 0$ degenerate case	Ex 6
G. Real-world word problem	Do torques combine karo ek orbit mein	Ex 7
H. Exam twist: multi-wheel array	Momentum share karo taaki koi wheel zero cross na kare	Ex 8

Worked examples

Ex 1 — Positive slew (cell A)

Worked example Wheel spin karo, body turn karo

$I_{b} = 500 kg\cdotm^{2}$ , $I_{w} = 0.05 kg\cdotm^{2}$ . Wheel jaata hai $0 \to + 200 rad/s$ . Koi external torque nahi, shuru mein rest par. Body rate $ω_{b}$ nikalo.

Forecast: aage padhne se pehle $ω_{b}$ ka sign aur rough size guess karo. (Hint: wheel ki inertia body se 10 000× chhoti hai.)

Step 1 — Total momentum 0 par fixed hai. Yeh step kyun? Conservation of Angular Momentum kehta hai ki koi external torque nahi toh total $H$ apni starting value se kabhi nahi badhega, jo $0$ hai (sab kuch rest par). $I_{b} ω_{b} + I_{w} Ω_{w} = 0$

Step 2 — Body rate ke liye solve karo. Yeh step kyun? Hume akela $ω_{b}$ chahiye; algebraically isolate karo. $ω_{b} = - \frac{I _{w} Ω _{w}}{I _{b}} = - \frac{0.05 \times 200}{500} = - 0.02 rad/s$

Verify: Wheel ne positive spin kiya, body rate negative hai → dono oppose karte hain, bilkul jaisa master relation ka minus sign demand karta hai. Size check: body spin mein $I_{b} / I_{w} = 1 0^{4}$ times bhaari hai, toh $∣ ω_{b} ∣ = 200/1 0^{4} = 0.02$ . ✓ Units: $\frac{kg\cdotm ^{2} \cdot rad/s}{kg\cdotm ^{2}} = rad/s$ . ✓

Ex 2 — Reverse slew (cell B)

Ex 3 — Constant torque to saturation (cell C)

Worked example Wheel "full" hone mein kitna time lagega?

Ek constant disturbance $τ_{ext} = + 5 \times 1 0^{- 5} N\cdotm$ (maano ek fixed Solar Radiation Pressure offset). Wheel saturate hota hai $H_{m a x} = I_{w} Ω_{m a x} = 0.05 \times 600 = 30 N\cdotm\cdots$ par. Wheel bias $H_{w} (0) = 0$ se start karta hai. Saturation tak kitna time?

Forecast: days ya hours? Compute karne se pehle guess karo.

Step 1 — Constant torque integrate karo. Yeh step kyun? Master relation: $H_{w} (t) = H_{w} (0) + \int_{0}^{t} τ d t^{'}$ . Ek constant integral ke andar $τ \cdot t$ deta hai — ek seedhi line. $H_{w} (t) = τ_{ext} t$

Neeche ki picture yeh seedhi line plot karti hai: horizontal axis time in days hai, vertical axis stored wheel momentum $H_{w}$ in $N\cdotm\cdots$ hai. Blue line steadily climb karti hai; dashed red line redline $H_{m a x} = 30$ hai; yellow dot woh instant mark karta hai jab dono milte hain — saturation.

Step 2 — Redline ke equal set karo aur $t$ ke liye solve karo. Yeh step kyun? Saturation woh instant hai jab $H_{w}$ , $H_{m a x}$ tak pahunch jaata hai. $t_{sat} = \frac{H _{m a x}}{τ _{ext}} = \frac{30}{5 \times 1 0 ^{- 5}} = 6 \times 1 0^{5} s \approx 6.94 days$

Verify: Units: $\frac{N\cdotm\cdots}{N\cdotm} = s$ . ✓ Scaling se cross-check: ek chhota torque $2 \times 1 0^{- 5}$ deta $30/ (2 \times 1 0^{- 5}) = 1.5 \times 1 0^{6} s \approx 17.4$ d; yahan torque $2.5 \times$ bada hai, toh time $17.4/2.5 = 6.94$ d hona chahiye. ✓ Consistent.

Ex 4 — External torque ka sign aur bias direction (cell D)

Worked example Kya disturbance wheel ki help karta hai ya fight karta hai?

Wheel bias $Ω_{bias} = + 400 rad/s$ par run kar raha hai, toh $H_{w} (0) = 0.05 \times 400 = + 20 N\cdotm\cdots$ . Redline $H_{m a x} = + 30$ , aur negative redline $- 30 N\cdotm\cdots$ . Ek Gravity Gradient Torque deta hai $τ_{ext} = - 1 \times 1 0^{- 4} N\cdotm$ (bias se opposite sign). Saturation tak kitna time lagega, aur woh kaun sa redline hit karega?

Forecast: kya ek negative torque ek positive wheel ko $+ 30$ ya $- 30$ ki taraf drive karta hai?

Step 1 — Sahi signs ke saath accumulation likho. Yeh step kyun? Is cell ka poora point signs hain; inhe rakho. $H_{w} (t) = + 20 + (- 1 \times 1 0^{- 4}) t = 20 - 1 0^{- 4} t$

Step 2 — Dhundo kahan yeh redline hit karta hai. Yeh step kyun? $H_{w}$ decrease ho raha hai, toh yeh negative redline $- 30$ ki taraf ja raha hai. $- 30 = 20 - 1 0^{- 4} t \Rightarrow t = \frac{20 - ( - 30 )}{1 0 ^{- 4}} = \frac{50}{1 0 ^{- 4}} = 5 \times 1 0^{5} s \approx 5.79 days$

Verify: Yeh total $50 N\cdotm\cdots$ ka span travel karta hai ( $+ 20$ se neeche $- 30$ tak). Negative torque positive bias se fight karta hai — pehle yeh wheel ko zero ki taraf unwind karta hai, zero cross karta hai (⚠️ Ex 5 dekho), phir negatively pile up karta hai. Isi liye bias direction jaane-maane secular disturbance के against choose ki jaati hai, taaki disturbance wheel ko zero se door le jaaye, iske through nahi. ✓ Units check pehle jaisa. ✓

Ex 5 — Zero-crossing aur stiction (cell E, degenerate $Ω_{w} = 0$ )

Worked example Wheel zero par atka jaata hai

Ex 4 continue karo: wheel $Ω_{w} = 0$ se neeche jaata ja raha hai. Fine-pointing ko sirf $τ_{cmd} = 3 \times 1 0^{- 4} N\cdotm$ ka control torque chahiye. Bearing ki static (stick) friction zero speed ke paas (Bearing Friction and Stiction) tak $τ_{stick} = 8 \times 1 0^{- 4} N\cdotm$ hold kar sakti hai. Kya motor wheel ko zero se smoothly move kar sakta hai?

Forecast: kya wheel zero se glide karega ya stick karega?

Step 1 — Command torque ko stiction torque se compare karo. Yeh step kyun? $Ω_{w} = 0$ par wheel tabhi move hoga jab applied torque static friction se zyada ho; usse neeche yeh lock ho jaata hai. $τ_{cmd} = 3 \times 1 0^{- 4} N\cdotm < τ_{stick} = 8 \times 1 0^{- 4} N\cdotm$

Step 2 — Degenerate case interpret karo. Yeh step kyun? Kyunki command < stiction, wheel $Ω_{w} = 0$ par hang karta hai. Pointing error badhta rehta hai jab tak accumulated demand $8 \times 1 0^{- 4}$ se zyada na ho, phir wheel snap karta hai free — ek impulsive jitter.

Neeche ki picture yeh dikhati hai. Horizontal axis = time (s), vertical axis = wheel speed $Ω_{w}$ (rad/s). Dashed blue line ideal smooth glide hai jo seedha zero se guzarti hai; solid red line reality hai — yeh yellow zero-line par flat ho jaati hai (wheel stuck hai), phir suddenly neeche snap karti hai (woh jerk jo telescope ko hila deta hai).

Step 3 — Fix: zero se door bias karo. Yeh step kyun? Agar hum wheel ko $Ω_{bias}$ par hold karein jahan friction chhoti aur unidirectional ho, toh wheel kabhi sign-flip nahi dekhta. $Ω_{w} > 0$ par sirf steady kinetic friction act karta hai, jise motor aasani se overcome karta hai.

Verify: Ratio $τ_{stick} / τ_{cmd} = 8/3 \approx 2.67 > 1$ stick confirm karta hai. Agar $τ_{cmd}$ $10 \times 1 0^{- 4} N\cdotm$ hota, toh ratio $8/10 = 0.8 < 1$ aur yeh glide through karta — yeh dikhata hai ki threshold exactly $τ_{cmd} = τ_{stick}$ hai. ✓

Ex 6 — Magnetorquer geometry aur $θ \to 0$ limit (cell F)

Worked example Dipole aur field ke beech angle ke against torque

Ek magnetorquer magnetic dipole produce karta hai $m = 40 A\cdotm^{2}$ . Earth ka field Magnetorquers and Earth's Magnetic Field hai $B = 3 \times 1 0^{- 5} T$ . Torque hai $τ = m B sin θ$ , jahan $θ$ angle hai $m$ aur $B$ ke beech. $θ = 9 0^{\circ}$ , $3 0^{\circ}$ , aur $0^{\circ}$ ke liye $τ$ nikalo.

Forecast: kaun sa angle zero torque deta hai, aur kyun?

Step 1 — Yaad karo $sin θ$ kyun aata hai. Yeh step kyun? Torque cross product hai $τ = m \times B$ ; cross product ki magnitude hoti hai $∣ m ∣∣ B ∣ sin θ$ . Sirf woh part jo $m$ mein perpendicular hai $B$ ke — woh twist karta hai; parallel part kuch nahi karta.

Neeche ki picture yeh concretely dikhati hai. Blue arrow $B$ hai (field, horizontal axis "along B" ke along point karta hai); red arrow $m$ hai (dipole) yellow angle $θ$ se tilt hua; green dashed arrow perpendicular part $m sin θ$ hai — yahi piece twisting karta hai. Jab $θ \to 0$ red arrow blue par flat ho jaata hai aur green piece shrink ho ke kuch nahi ho jaata.

Step 2 — Teeno angles plug karo. $θ = 9 0^{\circ} : τ = 40 \times 3 \times 1 0^{- 5} \times sin 9 0^{\circ} = 1.2 \times 1 0^{- 3} N\cdotm$ $θ = 3 0^{\circ} : τ = 1.2 \times 1 0^{- 3} \times sin 3 0^{\circ} = 1.2 \times 1 0^{- 3} \times 0.5 = 6 \times 1 0^{- 4} N\cdotm$ $θ = 0^{\circ} : τ = 1.2 \times 1 0^{- 3} \times sin 0^{\circ} = 0 N\cdotm$

Verify: $θ = 0$ par dipole $B$ ke along hai — kuch perpendicular nahi, zero torque. Yahi woh degenerate case hai jo tumhe orbit ke rotate hone ka wait karwata hai $B$ ke liye pehle us axis ke along dump karne se. Max $9 0^{\circ}$ par, exactly cross-product peak. ✓ Units: $A\cdotm^{2} \cdot T = A\cdotm^{2} \cdot \frac{kg}{A\cdots ^{2}} = N\cdotm$ . ✓

Ex 7 — Real-world word problem (cell G)

Worked example Ek orbit mein ek telescope

LEO mein ek Earth-imaging satellite ko Aerodynamic Drag Torque $τ_{1} = + 8 \times 1 0^{- 6} N\cdotm$ (hamesha same sign, ram-facing) aur ek averaged Solar Radiation Pressure torque $τ_{2} = + 4 \times 1 0^{- 6} N\cdotm$ (same axis, same sign) milta hai. Ek orbit $T = 5400 s$ (90 min) ka hota hai. Ek orbit mein aur ek din mein kitna wheel momentum accumulate hoga?

Forecast: roughly kitne $N\cdotm\cdots$ per orbit — $0.01$ , $0.1$ , ya $1$ ke kareeb?

Step 1 — Do constant torques add karo (same axis, same sign). Yeh step kyun? Same axis par torques scalars ki tarah add hote hain; dono wheel ko same taraf push karte hain, toh yeh reinforce karte hain — worst case. $τ_{tot} = 8 \times 1 0^{- 6} + 4 \times 1 0^{- 6} = 1.2 \times 1 0^{- 5} N\cdotm$

Step 2 — Ek orbit par integrate karo. Yeh step kyun? Time $T$ par constant torque deta hai $Δ H = τ_{tot} T$ . $Δ H_{orbit} = 1.2 \times 1 0^{- 5} \times 5400 = 6.48 \times 1 0^{- 2} N\cdotm\cdots$

Step 3 — Ek din tak scale karo. Yeh step kyun? $86400/5400 = 16$ orbits per day. $Δ H_{day} = 6.48 \times 1 0^{- 2} \times 16 = 1.0368 N\cdotm\cdots/day$

Verify: $H_{m a x} = 30 N\cdotm\cdots$ ke saath, saturation time $\approx 30/1.0368 \approx 28.9$ days — toh daily dump ek huge margin rakhta hai. ✓ Magnitude $\sim 0.065 N\cdotm\cdots/orbit$ forecast bracket se match karta hai (" $0.1$ ke kareeb"). ✓

Ex 8 — Exam twist: 4-wheel array, no zero-crossing (cell H)

Worked example Momentum share karo taaki koi zero cross na kare

Ek maneuver ko ek axis ke along body momentum change chahiye $Δ H_{body} = 12 N\cdotm\cdots$ . Do wheels us axis par aligned hain, har ek $H_{i} (0) = + 5 N\cdotm\cdots$ par biased, redline $\pm 30$ . Demand wheels par negative hai (wheels ko total $12 N\cdotm\cdots$ give up karna hai body turn karne ke liye). Kya hum ise split kar sakte hain taaki koi bhi wheel $Ω_{w} = 0$ cross na kare? Compare karo: (a) ek wheel sab kuch kare, (b) equally split karo.

Forecast: kaun si strategy dono wheels ko safely positive rakhegi?

Step 1 — Option (a): ek wheel poora $- 12$ absorb kare. Yeh step kyun? Naive single-wheel plan ko zero line ke against test karo. $H_{1} = 5 - 12 = - 7 N\cdotm\cdots (< 0!)$ Yeh wheel zero cross karta hai — stiction jitter (Ex 5). ✗

Step 2 — Option (b): $- 12$ ko $- 6$ each split karo. Yeh step kyun? Control Moment Gyroscopes (CMG)-style redundancy: demand distribute karo taaki har wheel zero ke ek side par rahe. $H_{1} = 5 - 6 = - 1, H_{2} = 5 - 6 = - 1 N\cdotm\cdots$ Dono zero cross karte hain! ✗ Equal split kaafi nahi hai kyunki demand ( $12$ ) zero se pehle total positive bias headroom ( $5 + 5 = 10$ ) se zyada hai.

Step 3 — Sahi fix: pehle bias badhao. Yeh step kyun? $12$ total remove karne ke baad dono wheels ko $\geq 0$ rakhne ke liye, har ek ko bias $\geq 6$ chahiye. $H_{i} (0) = + 7$ set karo; tab $H_{i} = 7 - 6 = + 1 > 0$ . ✓

Verify: Required minimum bias per wheel = $Δ H_{body} / (n wheels) = 12/2 = 6$ , toh bias $> 6$ chahiye; $+ 7$ kaam karta hai, aur dono redline $+ 30$ se neeche rehte hain. Yahi exactly design rule hai: biases set karo taaki koi wheel maneuver ke dauran zero cross na kare. ✓

Recall Self-test (koi line click karo apna answer reveal karne ke liye)

Wheel spins $+ 150 rad/s$ , $I_{w} = 0.05$ , $I_{b} = 500$ — body rate? ::: $ω_{b} = - 0.05 \times 150/500 = - 0.015 rad/s$ Constant $τ = 1 0^{- 4} N\cdotm$ , $H_{m a x} = 30$ — saturate hone mein kitna time? ::: $t = 30/1 0^{- 4} = 3 \times 1 0^{5} s \approx 3.47$ days Magnetorquer $m = 40 A\cdotm^{2}$ , $B = 3 \times 1 0^{- 5} T$ , angle $θ = 0^{\circ}$ — torque? ::: $τ = m B sin 0^{\circ} = 0 N\cdotm$ (dipole field ke parallel hai, twist karne ke liye kuch perpendicular nahi) Wheel ko zero se door kyun bias karte hain? ::: bearing stiction ke sign-flip se bachne ke liye jo $Ω_{w} = 0$ par jitter cause karta hai

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