Worked examples — Lambert's problem — connecting two positions in given time

3.2.28 · D3 · Physics › Orbital Mechanics & Astrodynamics › Lambert's problem — connecting two positions in given time

Yeh Lambert topic note ka ek worked-examples child hai. Parent note ne machinery banayi thi (chord $c$, semiperimeter $s$, auxiliary angles $\alpha ,\beta$, Lambert's time-of-flight equation). Yahan hum use har us case se guzarte hain jo aa sakta hai: short way, long way, minimum-energy, do-branch ambiguity, ek degenerate collinear case, parabolic escape boundary, ek real mission word-problem, aur ek exam twist.

Shuru karne se pehle, har symbol ko dobara anchor karte hain taaki kuch bhi unexplained na rahe.

Figure: same do arrows ${\vec{r}}_1$ (start) aur ${\vec{r}}_2$ (end). Red short arc sweeps $\Delta \theta =90^{\circ }$ (near way); black long arc sweeps $\Delta \theta =270^{\circ }$ (far way). Angle is par depend karta hai ki tum kis taraf jaate ho, sirf arrows par nahi.

Figure: space triangle. Black legs radii $r_1$ aur $r_2$ hain focus $F$ se; red side chord $c$ hai. Total perimeter $r_1+r_2+c$ ka half hai semiperimeter $s$.

Figure: do auxiliary sweeps $\alpha$ (wide, red) aur $\beta$ (narrow) transfer ellipse par. Unka difference $(\alpha -\sin \alpha )-(\beta -\sin \beta )$ hai ${\vec{r}}_1$ aur ${\vec{r}}_2$ ke beech swept-area/time.

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Scenario matrix

Har Lambert case inhi cells mein se kisi ek mein aata hai. Neeche ke examples sab ko cover karte hain.

#	Case class	Distinguishing feature	Example
A	Short way, ordinary ellipse	$0<\Delta \theta <\pi$, $\beta >0$	Ex 1
B	Minimum-energy landmark	$a=s/2$, $\alpha =\pi$	Ex 2
C	Ek $a$ ke liye do branches	fast vs slow ellipse ($\alpha$ flip)	Ex 3
D	Long way	$\Delta \theta >\pi$, $\beta <0$	Ex 4
E	Degenerate collinear, same side	$\Delta \theta =0$, $c=r_2-r_1$	Ex 5
F	Degenerate collinear, opposite side	$\Delta \theta =\pi$, $c=r_1+r_2$	Ex 6
G	Parabolic boundary (limiting $a\to \infty$)	escape / minimum time	Ex 7
H	Real-world word problem	Earth→Mars-style, prograde	Ex 8
I	Exam twist (given $a$, find $\Delta t$)	forward direction, no root-find	Ex 9

Saare examples canonical units use karte hain jisme $\mu =1$ hai jab tak alag na bataya jaaye. Equation ke andar angles radians mein hain.

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Ex 1 — Cell A: Short-way ordinary ellipse

Figure: is example ke liye space triangle. Black arrows $r_1$ aur $r_2$ hain; red line chord $c$ hai jiska length seedha Steps 1–2 mein jaata hai. Dotted arc $90^{\circ }$ transfer angle $\Delta \theta$ mark karta hai jise hum sweep karte hain.

1. Chord. $c=\sqrt{1+2.25-2(1)(1.5)\cos 90^{\circ }}=\sqrt{3.25}=1.8028$. Yeh step kyun? Aage ka sab kuch $c$ chahta hai; $\cos 90^{\circ }=0$ cross term ko khatam kar deta hai.
2. Semiperimeter. $s=\frac{1}{2}(1+1.5+1.8028)=2.1514$. Kyun? $s$ mein $r_1+r_2$ aur $c$ pack hote hain — sirf yahi geometry hai jis par Lambert's theorem time ko depend hone deta hai.
3. Auxiliary angles. $\sin \frac{\alpha }{2}=\sqrt{s/2a}=\sqrt{2.1514/2.6}=0.9098\Rightarrow \frac{\alpha }{2}=1.1441\Rightarrow \alpha =2.2882$. $\sin \frac{\beta }{2}=\sqrt{(s-c)/2a}=\sqrt{0.3486/2.6}=0.3662\Rightarrow \frac{\beta }{2}=0.3749\Rightarrow \beta =0.7498$. Kyun? Yeh do focus-sweep angles hain jo upar picture mein dikhaye gaye hain; short way $\Rightarrow \beta >0$, aur $\alpha =2.2882<\pi$ toh hum fast branch par hain.
4. Time. $\Delta t=a^{3/2}[(\alpha -\sin \alpha )-(\beta -\sin \beta )]$. Numerically $\alpha -\sin \alpha =2.2882-0.7573=1.5309$; $\beta -\sin \beta =0.7498-0.6816=0.0682$. Toh $\Delta t=1.30^{1.5}(1.5309-0.0682)=1.4822\times 1.4627=2.168$. Kyun? Yeh boxed equation ka forward evaluation hai — koi root-find nahi chahiye kyunki $a$ diya hua hai.

Recall Verify

Ex 2 mein minimum-energy time $\Delta t_{\min \text{-en}}\approx 3.402$ milega thinnest ellipse $a=s/2=1.0757$ par. Hamara $a=1.30$ ek fatter ellipse hai, aur yeh fast branch par hai ($\alpha <\pi$, Ex 3), isliye yeh genuinely min-energy transfer se faster hai: $\Delta t=2.168<3.402$ ✓. (Dhyan raho: "min-energy se faster" theek hai — min-energy slow hoti hai, min-time nahi.) Units: canonical time, dimensionless with $\mu =1$. ✓

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Ex 2 — Cell B: Minimum-energy landmark

1. $a_{\min }$. $\sin \frac{\alpha }{2}=\sqrt{s/2a}\le 1$ se hume $2a\ge s$ chahiye, toh $a_{\min }=s/2=1.0757$. Kyun? Sine 1 se zyada nahi ho sakta; equality se thinnest allowed ellipse milti hai — extreme case.
2. $a_{\min }$ par, $\alpha =\pi$. Tab $\sin \frac{\alpha }{2}=1$, toh $\alpha /2=\pi /2$, $\alpha =\pi$. Kyun? Yeh forecast answer hai: "wide" auxiliary sweep half-turn par saturate hota hai — exactly yahan fast branch ($\alpha <\pi$) aur slow branch ($\alpha >\pi$) milte hain.
3. $a_{\min }$ par $\beta$. $\sin \frac{\beta }{2}=\sqrt{(s-c)/2a_{\min }}=\sqrt{0.3486/2.1514}=0.4025\Rightarrow \beta =0.8286$. Kyun? $\beta$ abhi bhi same rule se aata hai; sirf $a$ badla hai.
4. Minimum-energy time. $\Delta t_{\min }=a_{\min }^{1.5}[(\pi -\sin \pi )-(\beta -\sin \beta )]=1.0757^{1.5}[(3.1416-0)-(0.8286-0.7358)]$. Yahan $\beta -\sin \beta =0.8286-0.7358=0.0928$. Toh $\Delta t_{\min }=1.1158\times (3.1416-0.0928)=1.1158\times 3.0488=3.402$. Kyun? Landmark values plug in karo. Note: yeh minimum-energy (smallest $a$) time hai, minimum-time transfer nahi — neeche mistake box dekho.

Recall Verify

$\sin \pi =0$ toh $\alpha$ term exactly $\pi$ hai. $a_{\min }=s/2$ exactly. Numeric $\Delta t_{\min \text{-en}}=3.402$. ✓ (VERIFY mein check kiya.) Yahi woh value hai jisse Ex 1 ke recall ne compare kiya.

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Ex 3 — Cell C: Ek $a$ ke liye do branches

Figure: same do endpoints aur same $a$, do alag arcs. Red (slow) ellipse zyada bulge karti hai, zyada area sweep karti hai, isliye black (fast) ellipse se zyada time leti hai. Ek hi picture mein two-branch ambiguity.

1. Sine ambiguity. $\sin \frac{\alpha }{2}=0.9098$ ke $(0,\pi )$ mein do solutions hain: $\frac{\alpha }{2}=1.1441$ ya $\frac{\alpha }{2}=\pi -1.1441=1.9975$. Yeh step kyun? $\sin$ ek angle aur uske supplement par same value leta hai — multiple-branch behaviour ki root wahi hai.
2. Branch 1 (fast, $\alpha <\pi$). $\alpha =2.2882$ → yeh Ex 1 hai → $\Delta t=2.168$. Yeh step kyun? Chhota $\alpha /2=1.1441$ double hokar $\alpha =2.2882<\pi$ deta hai, jo hamare definition se fast branch hai: chhota focus-sweep matlab ek chhota, tightly curved arc, toh spacecraft use kam time mein cover karta hai. Chhota $\alpha$ ⇔ short arc ⇔ fast.
3. Branch 2 (slow, $\alpha >\pi$). $\alpha =2(1.9975)=3.9950$. Tab $\alpha -\sin \alpha =3.9950-(-0.7573)=4.7523$. Same $\beta$ term $0.0682$ ke saath: $\Delta t=1.4822(4.7523-0.0682)=1.4822\times 4.6841=6.943$. Kyun? Same geometry aur same $a$, lekin empty focus chord ke doosri taraf baitha hai → bada sweep $\alpha >\pi$ → lamba bulging arc → zyada time.
4. Kaun sa kaun sa hai? Chhota requested $\Delta t$ fast branch select karta hai; bada requested $\Delta t$ slow branch select karta hai. Kyun? Newton's method ko correct branch ke paas seed karna padta hai nahi toh woh jump karta hai — yeh parent ki warning concrete ho gayi.

Recall Verify

Dono times same $a=1.30$ aur same $\beta$ share karte hain; sirf $\alpha$ apne supplement-doubled value par flip karta hai. Fast $2.168<$ slow $6.943$. ✓ Dono min-time point ke opposite sides par single-$a$ solutions hain (VERIFY mein check kiya).

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Ex 4 — Cell D: Long way ($\beta$ sign flip karta hai)

Figure: same endpoints, do travel directions. Red long-way arc ($\Delta \theta =270^{\circ }$) far side loop karta hai; black short-way arc ($90^{\circ }$) near side leta hai. Same chord $c$, lekin long arc zyada sweep karta hai isliye slower hai — woh difference sirf $\beta$ ke sign mein hai.

1. Chord (unchanged!). $c=\sqrt{1+2.25-2(1)(1.5)\cos 270^{\circ }}=\sqrt{3.25}=1.8028$ kyunki $\cos 270^{\circ }=0$. Kyun? $\cos 270^{\circ }=\cos 90^{\circ }=0$ — chord blind hai tum kis taraf loop karte ho. Toh $s=2.1514$ bhi same.
2. $\alpha$ unchanged. $\sin \frac{\alpha }{2}=\sqrt{s/2a}=0.9098\Rightarrow \alpha =2.2882$ (fast branch, $\alpha <\pi$). Kyun? $\alpha$ sirf $s,a$ par depend karta hai, direction par nahi.
3. $\beta$ FLIP HOTA HAI. Long way ⇒ $\beta \to -\beta$: numerically $\beta =-0.7498$. Yeh step kyun? $\beta$ ka sign woh akela jagah hai jahan "kis taraf" aata hai. Long way jaane par zyada angle sweep hota hai, isliye $\beta$ ka contribution reverse hota hai.
4. Time. $\Delta t=a^{1.5}[(\alpha -\sin \alpha )-(\beta -\sin \beta )]$. Ab $\beta -\sin \beta =-0.7498-(-0.6816)=-0.0682$. Toh $\Delta t=1.4822[1.5309-(-0.0682)]=1.4822\times 1.5991=2.370$. Kyun? $\beta$ ka sign flip karna time mein add karta hai — long arc genuinely zyada time leta hai, jaise physics demand karti hai.

Recall Verify

Forecast answer: nahi, same chord ≠ same time. Short way $2.168$ < long way $2.370$ identical $a$ ke liye. Long way slower hai. ✓ (VERIFY mein check kiya.)

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Ex 5 — Cell E: Degenerate collinear, same side ($\Delta \theta =0$)

1. Chord. $c=\sqrt{1+2.25-2(1)(1.5)\cos 0^{\circ }}=\sqrt{3.25-3}=\sqrt{0.25}=0.5=r_2-r_1$. Kyun? $\cos 0^{\circ }=1$; triangle ek straight segment mein flat ho jaata hai aur chord sirf radial gap hai.
2. Semiperimeter. $s=\frac{1}{2}(1+1.5+0.5)=1.5$. Kyun? Degenerate hone ke baad bhi well-defined hai.
3. Interpret. $s-c=1.5-0.5=1.0$. Formula abhi bhi chalta hai; yeh ek radial (rectilinear-limit) transfer hai — physically ek purely radial burn-and-coast. Auxiliary angles real rehte hain jab tak $2a\ge s=1.5$, yaani $a\ge 0.75$. Yeh step kyun? Degenerate geometry bhi $c$ aur $s$ finite rakhti hai, toh machinery nahi tootti — bas ek thin transfer describe karti hai.

Recall Verify

$c=r_2-r_1=0.5$ exactly jab $\Delta \theta =0$ ho. $s=1.5$. $s-c=1.0$. ✓ Chord radial difference ke barabar hai — collapse consistent hai.

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Ex 6 — Cell F: Degenerate collinear, opposite side ($\Delta \theta =\pi$)

1. Chord (maximal). $c=\sqrt{1+2.25-2(1)(1.5)\cos 180^{\circ }}=\sqrt{3.25+3}=\sqrt{6.25}=2.5=r_1+r_2$. Kyun? $\cos 180^{\circ }=-1$ cross term add karta hai, toh straight-line distance poora sum hai — points diametrically opposite hain.
2. Semiperimeter. $s=\frac{1}{2}(1+1.5+2.5)=2.5$. Aur $s-c=2.5-2.5=0$. Kyun? $s-c=0$ se $\sin \frac{\beta }{2}=0\Rightarrow \beta =0$: $\beta$ term poora vanish ho jaata hai.
3. Hohmann se link. $\Delta \theta =\pi$ aur $\beta =0$ ke saath, yeh exactly Hohmann Transfer geometry hai — opposite points ke beech ek half-ellipse. Chord transfer-ellipse major axis $r_1+r_2=2a_H$ ke barabar hai, toh $a_H=1.25$. Yeh step kyun? Yeh dikhata hai ki Lambert Hohmann ko special $\Delta \theta =180^{\circ }$ case ke roop mein contain karta hai.

Recall Verify

$c=r_1+r_2=2.5$ (max chord), $s-c=0\Rightarrow \beta =0$. Hohmann $a_H=(r_1+r_2)/2=1.25$. ✓

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Ex 7 — Cell G: Parabolic boundary (limiting $a\to \infty$)

1. Parabolic time formula ($a\to \infty$ limit). $\Delta t_{\text{par}}=\frac{1}{3}\sqrt{\frac{2}{\mu }}\left[s^{3/2}-(s-c{)}^{3/2}\right]$ Yeh step kyun? Elliptic time law lo aur $a\to \infty$ karo; eccentric-anomaly terms Taylor-expand hote hain aur $a$s cancel ho jaate hain, sirf pure geometry bachti hai. Yeh sabse chhota-time single-arc transfer hai.
2. Plug in karo. $s=2.1514$, $s-c=0.3486$, $\mu =1$. $s^{1.5}=3.1554$, $(s-c{)}^{1.5}=0.2059$. Toh $\Delta t_{\text{par}}=\frac{1}{3}\sqrt{2}(3.1554-0.2059)=\frac{1}{3}(1.4142)(2.9495)=1.3903$. Kyun? Direct evaluation; note karo $\sqrt{2/\mu }=\sqrt{2}$.
3. Meaning. Is geometry ka koi bhi elliptic transfer $\Delta t>\Delta t_{\text{par}}=1.390$ hoga. Is se faster ke liye hyperbola chahiye (excess energy / flyby). Yeh step kyun? Yeh fast branch ko cap karta hai — ek hard physical floor. Forecast answer: parabola ki taraf open hona fast-branch transfer ko faster banata hai, $\Delta t_{\text{par}}$ par bottom out karta hai.

Recall Verify

$\Delta t_{\text{par}}=1.390$. Ex 1 fast branch $2.168>1.390$ ✓ — parabolic floor se upar, jaise har ellipse ko hona chahiye. ✓

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Ex 8 — Cell H: Real-world word problem (prograde transfer)

1. Transfer angle. $\cos \Delta \theta =\frac{{\vec{r}}_1\cdot {\vec{r}}_2}{r_1{r}_2}=\frac{(1)(0)+(0)(1.2)}{(1)(1.2)}=0\Rightarrow \Delta \theta =90^{\circ }$. Yeh step kyun? Dot product do vectors se swept angle recover karta hai — yahi woh tarika hai jisse real solver state vectors se geometry padhta hai.
2. Chord & $s$. $c=\sqrt{1+1.44-0}=\sqrt{2.44}=1.5620$. $s=\frac{1}{2}(1+1.2+1.5620)=1.8810$. Kyun? Same law of cosines; ab $r_2=1.2$ hai.
3. Parabolic floor. $s^{1.5}=2.5799$, $(s-c{)}^{1.5}=(0.3190{)}^{1.5}=0.1802$. $\Delta t_{\text{par}}=\frac{1}{3}\sqrt{2}(2.5799-0.1802)=\frac{1}{3}(1.4142)(2.3997)=1.1311$. Kyun? Feasibility test karne ke liye floor chahiye.
4. Feasibility verdict. Requested $\Delta t=1.0<\Delta t_{\text{par}}=1.131$. Ek ellipse yeh nahi kar sakti — is transfer ke liye hyperbolic (high-energy) trajectory chahiye. Planner ko ya toh zyada $\Delta v$ accept karna hoga (hyperbola) ya $\Delta t$ ko $1.131$ se upar relax karna hoga. Yeh step kyun? Yeh real deliverable hai — feasibility launch-window design drive karti hai (Porkchop Plots and Launch Windows).

Recall Verify

$\Delta \theta =90^{\circ }$, $c=1.562$, $\Delta t_{\text{par}}=1.131$. Requested $1.0<1.131$ ⇒ hyperbolic required. ✓ (VERIFY mein check kiya.)

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Ex 9 — Cell I: Exam twist (forward: given $a$, find $\Delta t$ cleanly)

1. Chord. $c=\sqrt{4+4-2(2)(2)\cos 60^{\circ }}=\sqrt{8-8(0.5)}=\sqrt{4}=2$. Kyun? $\cos 60^{\circ }=0.5$; equal radii se neat $c=2$ milta hai.
2. Semiperimeter. $s=\frac{1}{2}(2+2+2)=3$. $s-c=1$. Kyun? Design se clean numbers hain (yahi exam-twist flavour hai).
3. Angles. $\sin \frac{\alpha }{2}=\sqrt{3/(2\cdot 2)}=\sqrt{0.75}=0.8660\Rightarrow \frac{\alpha }{2}=\frac{\pi }{3}\Rightarrow \alpha =\frac{2\pi }{3}=2.0944$. $\sin \frac{\beta }{2}=\sqrt{1/4}=0.5\Rightarrow \frac{\beta }{2}=\frac{\pi }{6}\Rightarrow \beta =\frac{\pi }{3}=1.0472$. Kyun? Har value ek exact special angle hai — exam mein recognizable. $\alpha =2.0944<\pi$ ⇒ fast branch.
4. Time. $\alpha -\sin \alpha =2.0944-0.8660=1.2284$; $\beta -\sin \beta =1.0472-0.8660=0.1812$. $\Delta t=2^{1.5}(1.2284-0.1812)=2.8284\times 1.0472=2.962$. Kyun? Forward evaluation — $a$ diya hua hai, toh equation kabhi invert nahi karni padti; yahi ek-sentence answer hai.

Recall Verify

$c=2$, $s=3$, $\alpha =2\pi /3$, $\beta =\pi /3$, $\Delta t=2.962$. Koi Newton iteration nahi — $a$ supply kiya gaya tha, toh RHS sirf plug in hua. ✓

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Recall — kya har cell cover hua?

Recall Kaun sa example kis matrix cell se gaya?

A short way ::: Ex 1 B minimum-energy ::: Ex 2 C two branches ::: Ex 3 D long way ::: Ex 4 E collinear same side ($\Delta \theta =0$) ::: Ex 5 F collinear opposite ($\Delta \theta =\pi$, Hohmann) ::: Ex 6 G parabolic boundary ::: Ex 7 H real-world feasibility ::: Ex 8 I exam forward-eval ::: Ex 9

Connections: yahan forward time law hai Kepler's Equation and Time of Flight; velocity recovery use karta hai Lagrange Coefficients (f and g functions); root-solver of choice hai Universal Variable Formulation; feasibility scans feed karte hain Porkchop Plots and Launch Windows aur Orbital Rendezvous and Targeting; $\Delta \theta =\pi$ case hai Hohmann Transfer.