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

This is a worked-examples child of the Lambert topic note. The parent built the machinery (chord $c$, semiperimeter $s$, auxiliary angles $\alpha ,\beta$, Lambert's time-of-flight equation). Here we drive it through every case it can face: short way, long way, minimum-energy, the two-branch ambiguity, a degenerate collinear case, the parabolic escape boundary, a real mission word-problem, and an exam twist.

Before we start, let me re-anchor every symbol so nothing is used unexplained.

Figure: same two arrows ${\vec{r}}_1$ (start) and ${\vec{r}}_2$ (end). The red short arc sweeps $\Delta \theta =90^{\circ }$ (near way); the black long arc sweeps $\Delta \theta =270^{\circ }$ (far way). The angle depends on which way you travel, not just on the arrows.

Figure: the space triangle. Black legs are the radii $r_1$ and $r_2$ from focus $F$; the red side is the chord $c$. Half of the total perimeter $r_1+r_2+c$ is the semiperimeter $s$.

Figure: the two auxiliary sweeps $\alpha$ (wide, red) and $\beta$ (narrow) on the transfer ellipse. Their difference $(\alpha -\sin \alpha )-(\beta -\sin \beta )$ is the swept-area/time between ${\vec{r}}_1$ and ${\vec{r}}_2$.

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The scenario matrix

Every Lambert case falls into one of these cells. The examples below hit all of them.

#	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	Two branches for one $a$	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 (find $\Delta t$ given $a$)	forward direction, no root-find	Ex 9

All examples use canonical units with $\mu =1$ unless stated. Angles in radians inside the equation.

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Ex 1 — Cell A: Short-way ordinary ellipse

Figure: the space triangle for this example. The black arrows are $r_1$ and $r_2$; the red line is the chord $c$ whose length feeds directly into Steps 1–2 below. The dotted arc marks the $90^{\circ }$ transfer angle $\Delta \theta$ we sweep.

1. Chord. $c=\sqrt{1+2.25-2(1)(1.5)\cos 90^{\circ }}=\sqrt{3.25}=1.8028$. Why this step? Everything downstream needs $c$; $\cos 90^{\circ }=0$ kills the cross term.
2. Semiperimeter. $s=\frac{1}{2}(1+1.5+1.8028)=2.1514$. Why? $s$ packages $r_1+r_2$ and $c$ — the only geometry Lambert's theorem lets time depend on.
3. Auxiliary angles. $\sin \frac{\alpha }{2}=\sqrt{s/2a}=\sqrt{2.1514/2.6}=0.9098\Rightarrow \frac{\alpha }{2}=1.1441\Rightarrow \alpha =2.2882$. $\sin \frac{\beta }{2}=\sqrt{(s-c)/2a}=\sqrt{0.3486/2.6}=0.3662\Rightarrow \frac{\beta }{2}=0.3749\Rightarrow \beta =0.7498$. Why? These are the two focus-sweep angles pictured above; short way $\Rightarrow \beta >0$, and $\alpha =2.2882<\pi$ so we are on the fast branch.
4. Time. $\Delta t=a^{3/2}[(\alpha -\sin \alpha )-(\beta -\sin \beta )]$. Numerically $\alpha -\sin \alpha =2.2882-0.7573=1.5309$; $\beta -\sin \beta =0.7498-0.6816=0.0682$. So $\Delta t=1.30^{1.5}(1.5309-0.0682)=1.4822\times 1.4627=2.168$. Why? This is the boxed equation evaluated forward — no root-find needed since $a$ is given.

Recall Verify

Ex 2 will find the minimum-energy time $\Delta t_{\min \text{-en}}\approx 3.402$ at the thinnest ellipse $a=s/2=1.0757$. Our $a=1.30$ is a fatter ellipse, and it sits on the fast branch ($\alpha <\pi$, Ex 3), so it is genuinely faster than the min-energy transfer: $\Delta t=2.168<3.402$ ✓. (Careful: "faster than min-energy" is fine — min-energy is slow, not min-time.) Units: canonical time, dimensionless with $\mu =1$. ✓

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Ex 2 — Cell B: The minimum-energy landmark

1. $a_{\min }$. From $\sin \frac{\alpha }{2}=\sqrt{s/2a}\le 1$ we need $2a\ge s$, so $a_{\min }=s/2=1.0757$. Why? A sine can't exceed 1; equality gives the thinnest allowed ellipse — the extreme case.
2. At $a_{\min }$, $\alpha =\pi$. Then $\sin \frac{\alpha }{2}=1$, so $\alpha /2=\pi /2$, $\alpha =\pi$. Why? This is the forecast answer: the "wide" auxiliary sweep saturates at a half-turn — this is exactly where the fast branch ($\alpha <\pi$) and slow branch ($\alpha >\pi$) meet.
3. $\beta$ at $a_{\min }$. $\sin \frac{\beta }{2}=\sqrt{(s-c)/2a_{\min }}=\sqrt{0.3486/2.1514}=0.4025\Rightarrow \beta =0.8286$. Why? $\beta$ still comes from the same rule; only $a$ changed.
4. Minimum-energy time. $\Delta t_{\min }=a_{\min }^{1.5}[(\pi -\sin \pi )-(\beta -\sin \beta )]=1.0757^{1.5}[(3.1416-0)-(0.8286-0.7358)]$. Here $\beta -\sin \beta =0.8286-0.7358=0.0928$. So $\Delta t_{\min }=1.1158\times (3.1416-0.0928)=1.1158\times 3.0488=3.402$. Why? Plug the landmark values in. Note: this is the minimum-energy (smallest $a$) time, not the minimum-time transfer — see mistake box below.

Recall Verify

$\sin \pi =0$ so the $\alpha$ term is exactly $\pi$. $a_{\min }=s/2$ exactly. Numeric $\Delta t_{\min \text{-en}}=3.402$. ✓ (Checked in VERIFY.) This is the value Ex 1's recall compared against.

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Ex 3 — Cell C: Two branches for one $a$

Figure: same two endpoints and same $a$, two different arcs. The red (slow) ellipse bulges further, sweeping more area, so it takes longer than the black (fast) ellipse. This is the two-branch ambiguity in one picture.

1. The sine ambiguity. $\sin \frac{\alpha }{2}=0.9098$ has two solutions in $(0,\pi )$: $\frac{\alpha }{2}=1.1441$ or $\frac{\alpha }{2}=\pi -1.1441=1.9975$. Why this step? $\sin$ takes the same value at an angle and its supplement — the root of the multiple-branch behaviour.
2. Branch 1 (fast, $\alpha <\pi$). $\alpha =2.2882$ → this is Ex 1 → $\Delta t=2.168$. Why this step? The smaller $\alpha /2=1.1441$ doubles to $\alpha =2.2882<\pi$, which by our definition is the fast branch: a small focus-sweep means a short, tightly curved arc, so the spacecraft covers it in less time. Small $\alpha$ ⇔ short arc ⇔ fast.
3. Branch 2 (slow, $\alpha >\pi$). $\alpha =2(1.9975)=3.9950$. Then $\alpha -\sin \alpha =3.9950-(-0.7573)=4.7523$. With the same $\beta$ term $0.0682$: $\Delta t=1.4822(4.7523-0.0682)=1.4822\times 4.6841=6.943$. Why? Same geometry and same $a$, but the empty focus sits on the far side of the chord → larger sweep $\alpha >\pi$ → longer bulging arc → more time.
4. Which is which? A short requested $\Delta t$ selects the fast branch; a large requested $\Delta t$ selects the slow branch. Why? Newton's method must be seeded near the correct branch or it jumps — this is the parent's warning made concrete.

Recall Verify

Both times share the same $a=1.30$ and same $\beta$; only $\alpha$ flips to its supplement-doubled value. Fast $2.168<$ slow $6.943$. ✓ Both are single-$a$ solutions on opposite sides of the min-time point (checked in VERIFY).

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Ex 4 — Cell D: The long way ($\beta$ flips sign)

Figure: identical endpoints, two travel directions. The red long-way arc ($\Delta \theta =270^{\circ }$) loops the far side; the black short-way arc ($90^{\circ }$) takes the near side. Same chord $c$, but the long arc sweeps more, so it is slower — that difference lives entirely in the sign of $\beta$.

1. Chord (unchanged!). $c=\sqrt{1+2.25-2(1)(1.5)\cos 270^{\circ }}=\sqrt{3.25}=1.8028$ since $\cos 270^{\circ }=0$. Why? $\cos 270^{\circ }=\cos 90^{\circ }=0$ — the chord is blind to which way you loop. So $s=2.1514$ too.
2. $\alpha$ unchanged. $\sin \frac{\alpha }{2}=\sqrt{s/2a}=0.9098\Rightarrow \alpha =2.2882$ (fast branch, $\alpha <\pi$). Why? $\alpha$ depends only on $s,a$, not on direction.
3. $\beta$ FLIPS. Long way ⇒ $\beta \to -\beta$: numerically $\beta =-0.7498$. Why this step? The sign of $\beta$ is the only place "which way around" enters. Going the long way sweeps more angle, so the $\beta$ contribution reverses.
4. Time. $\Delta t=a^{1.5}[(\alpha -\sin \alpha )-(\beta -\sin \beta )]$. Now $\beta -\sin \beta =-0.7498-(-0.6816)=-0.0682$. So $\Delta t=1.4822[1.5309-(-0.0682)]=1.4822\times 1.5991=2.370$. Why? Flipping $\beta$'s sign adds to the time — the long arc genuinely takes longer, as physics demands.

Recall Verify

Forecast answer: no, same chord ≠ same time. Short way $2.168$ < long way $2.370$ for identical $a$. The long way is slower. ✓ (Checked in VERIFY.)

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

1. Chord. $c=\sqrt{1+2.25-2(1)(1.5)\cos 0^{\circ }}=\sqrt{3.25-3}=\sqrt{0.25}=0.5=r_2-r_1$. Why? $\cos 0^{\circ }=1$; the triangle flattens to a straight segment and the chord is just the radial gap.
2. Semiperimeter. $s=\frac{1}{2}(1+1.5+0.5)=1.5$. Why? Still well-defined even though the triangle degenerated.
3. Interpret. $s-c=1.5-0.5=1.0$. The formula still runs; this is a radial (rectilinear-limit) transfer — physically a purely radial burn-and-coast. Auxiliary angles remain real as long as $2a\ge s=1.5$, i.e. $a\ge 0.75$. Why this step? Even degenerate geometry keeps $c$ and $s$ finite, so the machinery does not break — it just describes a thin transfer.

Recall Verify

$c=r_2-r_1=0.5$ exactly when $\Delta \theta =0$. $s=1.5$. $s-c=1.0$. ✓ Chord equals the radial difference — the collapse is consistent.

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

1. Chord (maximal). $c=\sqrt{1+2.25-2(1)(1.5)\cos 180^{\circ }}=\sqrt{3.25+3}=\sqrt{6.25}=2.5=r_1+r_2$. Why? $\cos 180^{\circ }=-1$ adds the cross term, so the straight-line distance is the full sum — points diametrically opposite.
2. Semiperimeter. $s=\frac{1}{2}(1+1.5+2.5)=2.5$. And $s-c=2.5-2.5=0$. Why? $s-c=0$ makes $\sin \frac{\beta }{2}=0\Rightarrow \beta =0$: the $\beta$ term vanishes entirely.
3. Link to Hohmann. With $\Delta \theta =\pi$ and $\beta =0$, this is exactly the Hohmann Transfer geometry — a half-ellipse between opposite points. The chord equals the transfer-ellipse major axis $r_1+r_2=2a_H$, so $a_H=1.25$. Why this step? It shows Lambert contains Hohmann as the special $\Delta \theta =180^{\circ }$ case.

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: The parabolic boundary (limiting $a\to \infty$)

1. Parabolic time formula (the $a\to \infty$ limit). $\Delta t_{\text{par}}=\frac{1}{3}\sqrt{\frac{2}{\mu }}\left[s^{3/2}-(s-c{)}^{3/2}\right]$ Why this step? Take the elliptic time law and let $a\to \infty$; the eccentric-anomaly terms Taylor-expand and the $a$'s cancel, leaving pure geometry. This is the smallest-time single-arc transfer.
2. Plug in. $s=2.1514$, $s-c=0.3486$, $\mu =1$. $s^{1.5}=3.1554$, $(s-c{)}^{1.5}=0.2059$. So $\Delta t_{\text{par}}=\frac{1}{3}\sqrt{2}(3.1554-0.2059)=\frac{1}{3}(1.4142)(2.9495)=1.3903$. Why? Direct evaluation; note $\sqrt{2/\mu }=\sqrt{2}$.
3. Meaning. Any elliptic transfer of this geometry has $\Delta t>\Delta t_{\text{par}}=1.390$. Faster than this needs a hyperbola (excess energy / flyby). Why this step? It caps the fast branch — a hard physical floor. Forecast answer: opening toward parabola makes the fast-branch transfer faster, bottoming out at $\Delta t_{\text{par}}$.

Recall Verify

$\Delta t_{\text{par}}=1.390$. Ex 1 fast branch $2.168>1.390$ ✓ — above the parabolic floor, as every ellipse must be. ✓

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

1. Transfer angle. $\cos \Delta \theta =\frac{{\vec{r}}_1\cdot {\vec{r}}_2}{r_1{r}_2}=\frac{(1)(0)+(0)(1.2)}{(1)(1.2)}=0\Rightarrow \Delta \theta =90^{\circ }$. Why this step? The dot product recovers the swept angle from the two vectors — this is how a real solver reads geometry off state vectors.
2. Chord & $s$. $c=\sqrt{1+1.44-0}=\sqrt{2.44}=1.5620$. $s=\frac{1}{2}(1+1.2+1.5620)=1.8810$. Why? Same law of cosines; $r_2=1.2$ now.
3. Parabolic floor. $s^{1.5}=2.5799$, $(s-c{)}^{1.5}=(0.3190{)}^{1.5}=0.1802$. $\Delta t_{\text{par}}=\frac{1}{3}\sqrt{2}(2.5799-0.1802)=\frac{1}{3}(1.4142)(2.3997)=1.1311$. Why? We need the floor to test feasibility.
4. Feasibility verdict. Requested $\Delta t=1.0<\Delta t_{\text{par}}=1.131$. An ellipse CANNOT do it — this transfer requires a hyperbolic (high-energy) trajectory. The planner must either accept more $\Delta v$ (hyperbola) or relax $\Delta t$ upward past $1.131$. Why this step? This is the real deliverable — feasibility drives launch-window design (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. ✓ (Checked in VERIFY.)

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

1. Chord. $c=\sqrt{4+4-2(2)(2)\cos 60^{\circ }}=\sqrt{8-8(0.5)}=\sqrt{4}=2$. Why? $\cos 60^{\circ }=0.5$; equal radii give the neat $c=2$.
2. Semiperimeter. $s=\frac{1}{2}(2+2+2)=3$. $s-c=1$. Why? Clean numbers by design (that's the exam-twist flavour).
3. Angles. $\sin \frac{\alpha }{2}=\sqrt{3/(2\cdot 2)}=\sqrt{0.75}=0.8660\Rightarrow \frac{\alpha }{2}=\frac{\pi }{3}\Rightarrow \alpha =\frac{2\pi }{3}=2.0944$. $\sin \frac{\beta }{2}=\sqrt{1/4}=0.5\Rightarrow \frac{\beta }{2}=\frac{\pi }{6}\Rightarrow \beta =\frac{\pi }{3}=1.0472$. Why? Every value is an exact special angle — recognisable in an exam. $\alpha =2.0944<\pi$ ⇒ fast branch.
4. Time. $\alpha -\sin \alpha =2.0944-0.8660=1.2284$; $\beta -\sin \beta =1.0472-0.8660=0.1812$. $\Delta t=2^{1.5}(1.2284-0.1812)=2.8284\times 1.0472=2.962$. Why? Forward evaluation — $a$ is given, so we never invert the equation; that's the one-sentence answer.

Recall Verify

$c=2$, $s=3$, $\alpha =2\pi /3$, $\beta =\pi /3$, $\Delta t=2.962$. No Newton iteration — $a$ was supplied, so the RHS is just plugged in. ✓

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Recall — did every cell get covered?

Recall Which example hit each matrix cell?

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: the forward time law here is Kepler's Equation and Time of Flight; velocity recovery uses Lagrange Coefficients (f and g functions); the robust root-solver of choice is the Universal Variable Formulation; feasibility scans feed Porkchop Plots and Launch Windows and Orbital Rendezvous and Targeting; the $\Delta \theta =\pi$ case is the Hohmann Transfer.