Visual walkthrough — Combined maneuvers — optimal split between plane change and velocity change

3.2.23 · D2 · Physics › Orbital Mechanics & Astrodynamics › Combined maneuvers — optimal split between plane change and

Hum assume karenge ki tumhe sirf teen cheezein pata hain: ek arrow ("vector") kya hota hai, speed matlab "kitni tez" aur direction matlab "kis taraf," aur right-angle triangle ka idea. Baaki sab hum khud banayenge.

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Step 1 — Do velocities ko arrows ki tarah draw karo

KYA. Ek spacecraft ke paas, ek instant mein, ek velocity hoti hai: ek arrow jiska length speed hai aur jiska direction wo taraf hai jis taraf wo move kar raha hai. Burn se pehle arrow ${\vec{v}}_1$ hai; burn ke baad wo ${\vec{v}}_2$ hai. Hum inhe tail-to-tail draw karte hain (dono same dot se shuru hote hain).

Tail-to-tail kyun. Kyunki hum un dono ke beech direction ka difference jaanna chahte hain. Chhota sa wedge wala angle jahan dono tails milte hain — use $\theta$ bolo — exactly plane-change angle hai, matlab direction of motion kitna swing karta hai. Agar hum tail-to-tip draw karte to wo wedge chhup jaata; tail-to-tail mein wo seedha dikhta hai.

PICTURE. Blue arrow ${\vec{v}}_1$, orange arrow ${\vec{v}}_2$, unke beech green wedge $\theta$.

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Step 2 — Burn wo arrow hai jo gap close karta hai

KYA. Engine ka kaam hai ${\vec{v}}_1$ ko ${\vec{v}}_2$ mein convert karna. Jo push use deni hogi wo wahi arrow hai jo ${\vec{v}}_1$ mein add hone ke baad tumhe exactly ${\vec{v}}_2$ pe pahuncha de. Woh arrow triangle ki teesri side hai, ${\vec{v}}_1$ ki tip se ${\vec{v}}_2$ ki tip tak.

Subtraction kyun. "${\vec{v}}_1$ mein kya add karun taaki ${\vec{v}}_2$ mile?" literally ${\vec{v}}_2-{\vec{v}}_1$ hai. Likha hua:

$\underset{\text{the burn}}{\underbrace{\overrightarrow{\Delta v}}}=\underset{\text{where we want to be}}{\underbrace{{\vec{v}}_2}}-\underset{\text{where we started}}{\underbrace{{\vec{v}}_1}}$

Har symbol kamaaya hua hai: $\overrightarrow{\Delta v}$ wo arrow hai jo rocket supply karta hai; uski length $\Delta v$ (bina arrow ke) wo fuel-hungry number hai jise hum minimize karna chahte hain.

PICTURE. Red arrow $\overrightarrow{\Delta v}$ do tips ko bridge karta hua, triangle close karta hua.

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Step 3 — Ek arrow ka Length² matlab arrow ka apne aap se dot product

KYA. Humein $\overrightarrow{\Delta v}$ ki length chahiye, lekin arrows mein koi "length button" nahi hota — hum length dot product se nikalte hain, ek aisi machine jo do arrows khaati hai aur ek number ugalti hai.

Dot product kyun. Kisi bhi arrow $\vec{a}$ ke liye, apne aap se dot product, $\vec{a}\cdot \vec{a}$, equals $|\vec{a}{|}^2$ hota hai — uski length squared. Yahi ek clean tarika hai "arrow" ko "ek plain number jis pe algebra ho sake" mein baadalna. To pehle $\Delta v^2$ compute karte hain, phir end mein square-root lete hain.

$\Delta v^2=\overrightarrow{\Delta v}\cdot \overrightarrow{\Delta v}=({\vec{v}}_2-{\vec{v}}_1)\cdot ({\vec{v}}_2-{\vec{v}}_1)$

Hum ise ordinary algebra ki tarah expand kyun kar sakte hain. Dot product ke do rules hain jo exactly multiplication ke rules jaisi dikhti hain: wo addition pe distribute hota hai ($\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\cdot \vec{b}+\vec{a}\cdot \vec{c}$) aur wo symmetric hota hai ($\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$). Yahi do facts — milke linearity kehlate hain — jo hume bracket ko FOIL karne dete hain. Distributivity magic nahi hai: geometrically, dot product $\vec{a}\cdot \vec{b}$ hai "$\vec{a}$ ki length times $\vec{b}$ ka shadow (projection) $\vec{a}$ pe," aur ek sum ka shadow exactly arrows ki tarah add hota hai (do arrows head-to-tail rakh do aur unka combined shadow do shadows end-to-end hoga). To:

$({\vec{v}}_2-{\vec{v}}_1)\cdot ({\vec{v}}_2-{\vec{v}}_1)={\vec{v}}_2\cdot {\vec{v}}_2-{\vec{v}}_2\cdot {\vec{v}}_1-{\vec{v}}_1\cdot {\vec{v}}_2+{\vec{v}}_1\cdot {\vec{v}}_1$

Do middle terms equal hain (symmetry), to wo merge ho jaate hain $-2{\vec{v}}_1\cdot {\vec{v}}_2$ mein:

$\Delta v^2=\underset{=v_2^2}{\underbrace{{\vec{v}}_2\cdot {\vec{v}}_2}}-\underset{\text{the cross term}}{\underbrace{2{\vec{v}}_1\cdot {\vec{v}}_2}}+\underset{=v_1^2}{\underbrace{{\vec{v}}_1\cdot {\vec{v}}_1}}$

PICTURE. Projection (shadow) picture jo distributivity justify karta hai, aur reminder $\vec{a}\cdot \vec{a}=|\vec{a}{|}^2$.

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Step 4 — Cross term mein angle chhupa hai

KYA. Abhi bacha hua mysterious piece sirf ${\vec{v}}_1\cdot {\vec{v}}_2$ hai. Upar ki definition se, wo equals hai $v_1{v}_2\cos \theta$. Substitute karo:

$\Delta v^2=v_1^2+v_2^2-2v_1{v}_2\cos \theta$

$\cos \theta$ kyun, $\sin \theta$ kyun nahi. Cosine alignment measure karta hai. Jab $\theta =0$ dono arrows same direction mein point karte hain, $\cos \theta =1$, aur cross term sabse bada hota hai — arrows ek doosre ki help karte hain. Jab $\theta =90^{\circ }$, $\cos \theta =0$, wo kuch share nahi karte aur cross term zero ho jaata hai. Cosine exactly woh "dono kitna agree karte hain" wala dial hai, jo precisely ek difference of arrows ko chahiye.

Square root lete hain (length kabhi negative nahi hoti, to $+$ root rakhte hain):

Yeh Law of Cosines hai spacesuit pahne hua — kisi bhi triangle ki teesri side ke liye same identity.

PICTURE. Complete triangle jisme teenon sides aur angle label hain, aur boxed formula sides pe mapped hai.

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Step 5 — Edge case: koi turn nahi ($\theta =0$)

KYA. $\theta =0$ set karo. To $\cos 0=1$, aur

$\Delta v=\sqrt{v_1^2+v_2^2-2v_1{v}_2}=\sqrt{(v_1-v_2{)}^2}=|v_2-v_1|.$

Yahi hona chahiye kyun. Bina direction change ke, dono arrows ek hi line pe hain; triangle ek straight segment mein collapse ho jaata hai. "Difference" bas itna hai ki ek doosre se kitna lamba hai — ek pure speed change. Hamara formula exactly yahi reproduce karta hai. ✅

PICTURE. Triangle ek line pe flatten ho gaya; $\Delta v$ do tips ke beech ka gap hai.

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Step 6 — Edge case: right-angle turn ($\theta =90^{\circ }$)

KYA. $\theta =90^{\circ }$ set karo. To $\cos 90^{\circ }=0$, poora cross term vanish ho jaata hai, aur

$\Delta v=\sqrt{v_1^2+v_2^2}.$

Yahi hona chahiye kyun. Perpendicular turn ke saath, do velocity arrows mein common direction kuch bhi nahi — ek purely "east" jaata hai, doosra purely "north." Jo triangle banta hai wo ek seedha right triangle hai, to $\Delta v$ bas uska hypotenuse hai: jaana-paahchaana $\sqrt{v_1^2+v_2^2}$ (Pythagoras). Yeh exactly woh moment hai jab $\cos \theta =0$ correction term remove kar deta hai aur law of cosines wapas Pythagoras ban jaata hai — proof ki cosine term precisely "kitna aligned" wala adjustment hai aur disappear ho jaata hai jab koi alignment hi nahi hoti.

PICTURE. Right-angle triangle jisme legs $v_1$, $v_2$ hain aur hypotenuse $\Delta v$ hai.

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Step 7 — Edge case: pure turn, equal speeds ($v_1=v_2=v$)

KYA. $v_1=v_2=v$ set karo (orbit tilto but same speed rakho):

$\Delta v=\sqrt{v^2+v^2-2v^2\cos \theta }=v\sqrt{2-2\cos \theta }.$

Half-angle identity kyun — derive kiya hua, quote nahi. Hum $2-2\cos \theta$ simplify karna chahte hain. Isosceles triangle dekho: do equal sides $v$ aur unke beech wedge $\theta$. Apex se seedha ek line beech mein drop karo. Wo do kaam karta hai: wedge ko bisect karta hai do $\theta /2$ angles mein, aur base $\Delta v$ exactly half cut karta hai (isosceles symmetry). Jo chhota right triangle banta hai usme $\theta /2$ angle ke opposite side half base hai, aur hypotenuse $v$ hai. Sine ki definition se ("opposite over hypotenuse"):

$\sin \left(\frac{\theta }{2}\right)=\frac{\frac{1}{2}\Delta v}{v}⟹\Delta v=2v\sin \left(\frac{\theta }{2}\right).$

Yeh geometric fact hi half-angle identity hai disguise mein: dono sides square karo, $\Delta v^2=4v^2{\sin }^2(\theta /2)$, aur $\Delta v^2=v^2(2-2\cos \theta )$ se compare karo to force hota hai $2-2\cos \theta =4{\sin }^2(\theta /2)$, yaani $1-\cos \theta =2{\sin }^2(\theta /2)$. Humne identity picture se derive ki, memorize nahi ki.

STING. $\theta =60^{\circ }$ pe, $\sin (30^{\circ })=\frac{1}{2}$, to $\Delta v=2v\cdot \frac{1}{2}=v$ — tumhe sirf $60^{\circ }$ turn karne ke liye apni poori orbital speed phir se kharchi karni padti hai. Isliye parent note chillata hai "slow hone pe turn karo": $\Delta v$ directly $v$ ke saath scale karta hai.

PICTURE. Isosceles triangle, dropped bisector, do right triangles $\theta /2$ ke saath.

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Step 8 — Degenerate case: complete reversal ($\theta =180^{\circ }$)

KYA. After-arrow ko exactly peeche point karo, $\theta =180^{\circ }$, $\cos 180^{\circ }=-1$:

$\Delta v=\sqrt{v_1^2+v_2^2+2v_1{v}_2}=\sqrt{(v_1+v_2{)}^2}=v_1+v_2.$

Yahi hona chahiye kyun. Ab do arrows opposite directions mein point karte hain. $v_1$ pe ek taraf move karne se $v_2$ pe doosri taraf move karne tak flip karne ke liye, pehle poora $v_1$ kill karna hoga, phir poora $v_2$ build karna hoga — costs simply add ho jaati hain. Formula mein minus sign plus ban gaya kyunki $\cos$ ka sign flip hua, aur triangle phir ek line mein collapse ho gaya, lekin is baar tips shared tail ke opposite sides pe hain. Yeh sabse worst possible turn hai.

PICTURE. Dono arrows ek line pe opposite directions mein; $\Delta v$ tip se tip tak = $v_1+v_2$.

Recall Har sign ab kyun covered hai

$\cos \theta$ smoothly run karta hai $+1$ se (aligned, sabse sasta) through $0$ (perpendicular, Pythagoras) to $-1$ (reversed, sabse mehnga) tak. Kyunki humne $\theta$ ko koi bhi value $0^{\circ }$–$180^{\circ }$ lene diya, aur length hamesha positive root hoti hai, koi bhi input nahi hai jo boxed formula handle na kar sake. Zero turn, right-angle turn, half turn, full reversal — sab ek hi triangle se nikal aate hain.

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Step 9 — Ek burn se optimal split tak

KYA — setup. Ek real transfer (dekho Hohmann Transfer Orbit) mein do burns hote hain, aur har burn apna triangle hai exactly Step 4 ki tarah. "Initial" subscript se clash avoid karne ke liye, hum total plane-change angle ko $\Phi$ (capital phi) likhte hain. Pehle sab kuch name karte hain:

Boxed formula har burn pe apply karo:

$\Delta v_1(s)=\sqrt{v_{1,\text{in}}^2+v_{1,\text{out}}^2-2v_{1,\text{in}}{v}_{1,\text{out}}\cos s},\Delta v_2(s)=\sqrt{v_{2,\text{in}}^2+v_{2,\text{out}}^2-2v_{2,\text{in}}{v}_{2,\text{out}}\cos (\Phi -s)}$

Total fuel do red-arrow lengths ka sum hai:

$\Delta v_{\text{tot}}(s)=\Delta v_1(s)+\Delta v_2(s).$

Derivative kyun. Sabse sasta split dhundne ke liye hum puchte hain: best $s$ pe, ise kisi bhi taraf nudge karna help nahi karna chahiye. Yahi exactly "slope $=0$" condition hai, yaani $\frac{d\Delta v_{\text{tot}}}{ds}=0$.

Derivative $\frac{v_{\text{in}}{v}_{\text{out}}\sin \theta }{\Delta v}$ jaisi kyun dikhti hai — dikhaya gaya, assert nahi kiya. Lo $\Delta v_1(s)=\sqrt{g(s)}$ jahan inside hai $g(s)=v_{1,\text{in}}^2+v_{1,\text{out}}^2-2v_{1,\text{in}}{v}_{1,\text{out}}\cos s$. Square root ke liye chain rule kehta hai $\frac{d}{ds}\sqrt{g}=\frac{g^{\prime }(s)}{2\sqrt{g}}$. Sirf $\cos s$ mein $s$ hai, aur $\frac{d}{ds}(-2v_{1,\text{in}}{v}_{1,\text{out}}\cos s)=+2v_{1,\text{in}}{v}_{1,\text{out}}\sin s$ (kyunki $\frac{d}{ds}\cos s=-\sin s$, do minus signs cancel ho jaate hain). To:

$\frac{d\Delta v_1}{ds}=\frac{2v_{1,\text{in}}{v}_{1,\text{out}}\sin s}{2\sqrt{g(s)}}=\frac{v_{1,\text{in}}{v}_{1,\text{out}}\sin s}{\Delta v_1}.$

Burn 2 ke liye inside angle $(\Phi -s)$ hai, aur $\frac{d}{ds}\cos (\Phi -s)=+\sin (\Phi -s)$ (chain rule inner $-s$ se $-1$ laata hai, sign phir flip hoti hai), jo overall minus ke saath aata hai:

$\frac{d\Delta v_2}{ds}=-\frac{v_{2,\text{in}}{v}_{2,\text{out}}\sin (\Phi -s)}{\Delta v_2}.$

Dono add karo aur total slope zero set karo to milti hai optimal-split condition:

$\boxed{\frac{v_{1,\text{in}}{v}_{1,\text{out}}\sin s}{\Delta v_1(s)}=\frac{v_{2,\text{in}}{v}_{2,\text{out}}\sin (\Phi -s)}{\Delta v_2(s)}}$

Actually solve kaise karte hain. Notice karo ki $\Delta v_1$ aur $\Delta v_2$ mein $s$ ek square root ke andar hai, to yeh equation algebra se clean "$s=\dots$" mein nahi tutega — yeh transcendental hai. Ise numerically solve karte hain:

1. Char speeds Vis-viva Equation se aur total tilt $\Phi$ lo.
2. $s$ ko $0$ se $\Phi$ tak chhote steps mein sweep karo, har baar $\Delta v_{\text{tot}}(s)$ compute karo (ya equivalently, dekhte jao kaunsa $s$ boxed condition ke do sides equal banata hai).
3. Jo $s$ smallest total de woh optimal split hai. Ek quick shortcut: kyunki burn 2 bade, slow orbit pe hota hai uska $v\cdot v$ product chhota hota hai, to uski marginal cost chhoti hoti hai — balance point land karta hai almost sab tilt burn 2 (apoapsis) pe aur burn 1 pe sirf ek whisker.

PICTURE. $U$-shaped total-cost curve $\Delta v_{\text{tot}}(s)$ jiska minimum "all at apoapsis" ki taraf far mark kiya hua hai.

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Ek-picture summary

Upar sab kuch compressed: do velocity arrows tail-to-tail, red closing arrow, law-of-cosines label, aur char cases ($\theta =0,90^{\circ },60^{\circ },180^{\circ }$) neeche march karte hue cost $|v_2-v_1|$ se $v_1+v_2$ tak badhti dikhaate hue.

Recall Feynman retelling — poora walkthrough plain words mein

Apni speed ko ek arrow ki tarah draw karo burn se pehle (blue) aur burn ke baad (orange), dono same spot se shuru. Unke beech ka chhota wedge batata hai ki tum kitna turn kar rahe ho. Jo push tumhara rocket dega wo woh arrow (red) hai jo blue ki tip se orange ki tip tak pahunchta hai — aur sirf uski length pe fuel lagta hai. Woh length paane ke liye hum triangle rule (law of cosines) use karte hain: $\Delta v^2=v_1^2+v_2^2-2v_1{v}_2\cos \theta$. $\cos \theta$ ek "arrows kitne aligned hain" wala dial hai: same way ($\theta =0$) matlab cost bas speed difference hai; square-on ($\theta =90^{\circ }$) plain Pythagoras deta hai; opposite ways ($\theta =180^{\circ }$) matlab costs bas add ho jaati hain. Equal-speed turns simplify ho jaate hain $2v\sin (\theta /2)$ mein, jo kehta hai $60^{\circ }$ turn tumhari poori speed phir se kharcha karta hai — isliye turning ek nightmare hai jab tum fast ho. Do burns ke saath split $s$ slide karo jab tak marginal cost dono pe equal na ho (ek curve jo sweep se minimize hoti hai, kyunki algebra crack nahi kar sakta), aur kyunki far burn slow hai, almost sab turning wahan bahar hoti hai. Ek picture, ek triangle, ek rule — yahi poora chapter hai.