Lambert's problem — connecting two positions in given time: Visual walkthrough

Step 1 — Draw the two points and the triangle they make

Figure — Lambert's problem — connecting two positions in given time

Look at the figure. Three labelled things:

$r_{1}$ — the arrow from the Sun $F$ to the start point $P_{1}$ . Its length is $r_{1} = ∣ r_{1} ∣$ .
$r_{2}$ — the arrow from $F$ to the arrival point $P_{2}$ . Its length is $r_{2} = ∣ r_{2} ∣$ .
$Δ θ$ — the angle at the Sun between those two arrows. This is the transfer angle: how far around the sky the spacecraft travels.

The straight line from $P_{1}$ to $P_{2}$ is the chord — the shortcut across, not the path flown:

$c = ∣ r_{2} - r_{1} ∣ = law of cosines on the triangle r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos Δ θ$

Step 2 — Bundle the three lengths into the semiperimeter

$s = \frac{1}{2} (r_{1} + r_{2} + c)$

Step 3 — Meet the two angles $α$ and $β$

$sin \frac{α}{2} = \frac{s}{2 a}, sin \frac{β}{2} = \frac{s - c}{2 a}$

Intuition WHERE these two equalities come from (the right-triangle picture)

These are not guessed — they fall out of the ellipse's own geometry. On any ellipse, the distance from a point to the focus is $r = a (1 - e cos E)$ , where $E$ is that point's eccentric anomaly (its "clock angle"). Write this at both ends and also write the chord $c$ using the coordinates $(a cos E, b sin E)$ of the two points. Grinding the algebra (a standard half-angle regrouping) turns those into $r_{1} + r_{2} = 2 a (1 - cos \frac{α}{2} cos \frac{β}{2}), c = 2 a sin \frac{α}{2} cos \frac{β}{2} (short-way sign)$ with $α = E_{2} + E_{1}$ and $β = E_{2} - E_{1}$ . Add and subtract these two: $s = \frac{1}{2} (r_{1} + r_{2} + c) = 2 a sin^{2} \frac{α}{2}, s - c = 2 a sin^{2} \frac{β}{2},$ using $1 - cos x = 2 sin^{2} \frac{x}{2}$ . Take square roots and you land exactly on the two boxed equalities. The figure draws each as a right triangle: vertical leg $s$ (or $s - c$ ), hypotenuse $2 a$ , so the enclosed angle is $α /2$ (or $β /2$ ) and its sine is $s /2 a$ (or $(s - c) /2 a$ ).

Step 4 — Kepler's clock, applied at both ends and subtracted

$= n \frac{μ}{a ^{3}} Δ t = (E_{2} - e sin E_{2}) - (E_{1} - e sin E_{1})$

Step 5 — The miracle: eccentricity cancels

Intuition WHY the eccentricity cancels (the crucial algebra, shown)

Take Step 4's right side, $(E_{2} - E_{1}) - e (sin E_{2} - sin E_{1})$ . The first bracket is just $β$ . For the second, use the sum-to-product identity $sin E_{2} - sin E_{1} = 2 cos \frac{E _{2} + E _{1}}{2} sin \frac{E _{2} - E _{1}}{2} = 2 cos \frac{α}{2} sin \frac{β}{2}$ . Now we need the value of $e cos \frac{α}{2}$ — and Step 3's geometry hands it to us. From $r_{1} + r_{2} = 2 a (1 - cos \frac{α}{2} cos \frac{β}{2})$ combined with $r_{1} r_{2}$ relations, the ellipse identity reduces to $e cos \frac{α}{2} = cos \frac{β}{2}$ — a clean, geometry-only value for $e cos \frac{α}{2}$ ; the eccentricity is pinned, not free. Substitute it: $e (sin E_{2} - sin E_{1}) = 2 (e cos \frac{α}{2}) sin \frac{β}{2} = 2 cos \frac{β}{2} sin \frac{β}{2} = sin β .$ The $e$ has vanished, swallowed into $sin β$ . Doing the same bookkeeping on the $α$ piece gives $sin α$ , and the whole right side collapses to $(α - sin α) - (β - sin β)$ .

$\frac{μ}{a ^{3}} Δ t = β (E_{2} - E_{1}) - = s i n β via e c o s \frac{α}{2} = c o s \frac{β}{2} e (sin E_{2} - sin E_{1}) ⟶ (α - sin α) - (β - sin β)$

Step 6 — The smallest possible orbit (degenerate edge)

A sine can never exceed $1$ . So in $sin (α /2) = s /2 a$ we need $\frac{s}{2 a} \leq 1 ⟹ a \geq \frac{s}{2} .$

The smallest allowed value is the minimum-energy semi-major axis: $a_{m i n} = \frac{s}{2}, at which sin \frac{α}{2} = 1 \Rightarrow α = π .$

Step 7 — Which way around? Branches: $β$ sign, $α$ branch, and multi-rev

Definition The

$α$ -branch rule (two ellipses per $a$ ) For a given $a > a_{m i n}$ , the equation $sin \frac{α}{2} = s /2 a$ has two solutions in $[0, 2 π]$ : the principal one $α_{0} = 2 arcsin s /2 a$ (which lies in $[0, π]$ ), and its supplement $α = 2 π - α_{0}$ .

Use $α = α_{0}$ when the transfer's swept angle keeps the arc's far point (relative to the focus geometry) on the near side — the lower-time solution.
Use $α = 2 π - α_{0}$ for the other ellipse (focus on the opposite side of the chord) — the higher-time solution.

WHY the switch happens: both ellipses have the same $a$ and pass through both points, but their empty focus sits on opposite sides of the chord. $arcsin$ only ever returns the principal value in $[0, π]$ , so it can name only one of the two — the other is recovered as $2 π - α_{0}$ . You choose based on which $Δ t$ the two branches give: they bracket the minimum-time transfer at $a_{m i n}$ .

The figure shows both arcs sharing one green chord, with the long-way arc bulging the opposite side and $β$ marked negative.

Worked check — short-way numbers line up

Canonical units, $μ = 1$ : $r_{1} = 1$ , $r_{2} = 1.5$ , $Δ θ = 9 0^{\circ}$ .

Chord: $c = 1 + 2.25 - 0 = 3.25 \approx 1.8028$ .
Semiperimeter: $s = \frac{1}{2} (1 + 1.5 + 1.8028) = 2.1514$ .
Minimum-energy: $a_{m i n} = s /2 = 1.0757$ .
At $a = 1.30$ (principal $α$ -branch): $sin (α /2) = 2.1514/2.6 = 0.9096 \Rightarrow α = 2.2810$ rad; $sin (β /2) = 0.3486/2.6 = 0.3662 \Rightarrow β = 0.7503$ rad. Then $Δ t = a^{3/2} [(α - sin α) - (β - sin β)] \approx 1.06.$

So the direct short-way, principal-branch transfer at $a = 1.30$ gives $Δ t \approx 1.06$ . To hit the parent note's target $Δ t = 1.2$ you nudge $a$ (the root-find of Step 5); the value lands near $a \approx 1.25$ . The point stands: one scalar $a$ tunes the time. (The recovered $v_{1}, v_{2}$ then follow from the Lagrange Coefficients (f and g functions), feeding Orbital Rendezvous and Targeting and Porkchop Plots and Launch Windows.)

The one-picture summary

Recall Feynman retelling — say it back in plain words

I put the Sun at a point $F$ and draw two arrows to where I start ( $P_{1}$ ) and where I must end ( $P_{2}$ ). Those two arrow-tips plus the Sun make a flat triangle. I measure the long sides ( $r_{1}, r_{2}$ ) and the shortcut across ( $c$ ), and I add them all up and halve it to get one tidy number $s$ . Now I guess a size for the ellipse I'll fly — call its half-width $a$ . From the ellipse's own geometry (distance-to-focus is $a (1 - e cos E)$ ) I build two angles: $α$ , the sum of the two endpoint clock-angles $E_{1} + E_{2}$ , and $β$ , their difference $E_{2} - E_{1}$ , and I find $sin \frac{α}{2} = s /2 a$ and $sin \frac{β}{2} = (s - c) /2 a$ . I then use Kepler's steady clock at both ends of the trip and subtract, so all that's left is the time difference — the actual flight time, equal to $n Δ t$ . When I substitute the sum and difference angles and use $sin E_{2} - sin E_{1} = 2 cos \frac{α}{2} sin \frac{β}{2}$ together with the geometry fact $e cos \frac{α}{2} = cos \frac{β}{2}$ , the eccentricity cancels and I'm left with $μ Δ t = a^{3/2} [(α - sin α) - (β - sin β)]$ . The right side is just a number that depends on my guess $a$ , so I keep adjusting $a$ until the time comes out to the target. If I want the cheapest trip I use the thinnest legal ellipse, $a = s /2$ . If I fly the long way round I flip the sign of $β$ ; if the transfer angle is exactly $18 0^{\circ}$ then $β = 0$ and I must hand-pick the plane; if I want the other ellipse of the same size I swap $α$ for $2 π - α$ ; and if I have loads of time I can add whole loops with $2 π N$ . This is all for a bound elliptic orbit — a parabola or hyperbola would swap those sines for their straight-line or hyperbolic cousins. Out pops the orbit — and from it the velocities and the fuel.

Recall Test yourself

Which three scalars does Lambert's time of flight depend on? ::: $a$ , the chord $c$ , and $r_{1} + r_{2}$ (equivalently $s$ ) — never the eccentricity separately. How do $α, β$ relate to the eccentric anomalies? ::: $α = E_{2} + E_{1}$ (the sum) and $β = E_{2} - E_{1}$ (the difference). What geometry fact makes $e$ cancel? ::: $e cos \frac{α}{2} = cos \frac{β}{2}$ , which turns $e (sin E_{2} - sin E_{1})$ into $sin β$ . Why does $a \geq s /2$ ? ::: Because $sin (α /2) = s /2 a$ and a sine cannot exceed 1, forcing $s /2 a \leq 1$ . What tells short-way from long-way in the formula? ::: The sign of $β$ : positive for $Δ θ < π$ , flipped negative for $Δ θ > π$ . What happens at $Δ θ = 18 0^{\circ}$ ? ::: $s - c = 0$ , so $β = 0$ and the transfer plane becomes undefined — you must supply it. How do multi-revolution solutions enter? ::: Add $2 π N$ inside the bracket for $N$ extra full loops; each $N$ has two $α$ -branches.