Gauss's method for Lambert's problem: Visual walkthrough

We are solving Lambert's problem: given two positions and a travel time, find the orbit. Watch the geometry grow one line at a time.

Step 1 — Two dots and a chord (pure geometry, no orbit yet)

WHAT. Put the central body (Sun, Earth, whatever we orbit) at a point called the focus $F$ . Draw an arrow from $F$ to the first position, call it $r_{1}$ ; draw another to the second position, $r_{2}$ . The straight line joining the tips of those two arrows is the chord $c$ .

WHY. Before we know anything about the orbit, we already know these three lengths from the given data. Everything Gauss does is anchored to this triangle, so we draw it first.

PICTURE. Look at the figure. Two arrows leave $F$ ; the dashed line between their tips is $c$ . The angle between the arrows — swept from $r_{1}$ to $r_{2}$ — is the transfer angle $θ$ .

The chord comes from the law of cosines — the rule that finds the third side of a triangle when you know two sides and the angle between them:

$c^{2} = side 1^{2} r_{1}^{2} + side 2^{2} r_{2}^{2} - correction for the angle 2 r_{1} r_{2} cos θ$

And $cos θ$ itself is free from the given vectors via the dot product: $cos θ = \frac{r _{1} \cdot r _{2}}{r _{1} r _{2}} .$

Step 2 — The flat triangle's area

WHAT. Fill in the triangle $F$ –tip $_{1}$ –tip $_{2}$ and measure its area.

WHY. The moving planet does not travel along the chord — it curves. But the flat triangle is the simplest thing we can measure, so we use it as a yardstick and later ask "how much bigger is the true curved region?"

PICTURE. The shaded triangle. Its area is base $\times$ height $/2$ ; taking $r_{1}$ as base, the height is $r_{2} sin θ$ .

Step 3 — The real path is a pie slice, not a triangle

WHAT. Now let the planet actually move on its ellipse from tip $_{1}$ to tip $_{2}$ . The region the arrow $F \to planet$ sweeps is a curved elliptic sector — a pie slice with a bulging crust. Give this region a name: $A_{sector}$ .

WHY. This is where the orbit finally enters. The sector is bigger than the flat triangle because the true path bulges outward (or inward). The amount it differs is the fingerprint of the specific orbit.

PICTURE. The mint pie-slice sits on top of the flat triangle. The extra sliver between the curved arc and the straight chord is the "bulge".

Kepler's second law, written as an equation, says the sweep rate is constant and equals $\frac{1}{2} h$ , where $h = μ p$ is the (constant) angular momentum. Rearranging "area = rate × time": $Δ t = \frac{2 A _{sector}}{h}, h = μ p .$

Step 4 — Define the ratio $y$ : how much fatter is the pie than the triangle?

WHAT. Divide the curved sector area by the flat triangle area. Call it $y$ .

WHY. A raw area depends on units and orbit size — messy. A ratio of two areas is a clean, unit-free number that captures only the shape difference. That is the one quantity Gauss chases.

PICTURE. The two regions side by side; $y$ is literally "green area ÷ blue area".

Recall Why is

$y$ near 1 a good first guess? Guess-when-in-doubt ::: For most transfers the arc is only slightly bulged, so sector ≈ triangle, so $y \approx 1$ starts the iteration close to the answer.

Step 5 — The eccentric anomaly, and where the half-angle comes from

Before the two equations, we need the variable that measures how much of the ellipse the ship covered — and to see why a half-angle $θ /2$ keeps appearing.

WHAT. Draw the ellipse with a helper circle of the same width around it. For any point on the orbit, drop it straight up to the circle; the angle from the ellipse's centre (not the focus!) out to that circle point is the eccentric anomaly $E$ . Call the two tip values $E_{1}$ and $E_{2}$ .

WHY. Time-of-flight through an ellipse is clean only in terms of $E$ (that is Kepler's equation). And the chord joining the two tips, worked out in $E$ , always comes out proportional to $sin (\frac{E _{2} - E _{1}}{2})$ — a half-difference — because a chord of a circle of radius $a$ subtending angle $Δ E$ has length $2 a sin (Δ E /2)$ . That single "chord = $2 a sin (half-angle)$ " identity is the source of every $θ /2$ and every half- $E$ you will see.

PICTURE. The ellipse, its helper circle, the centre $O$ , the focus $F$ , the two eccentric-anomaly angles, and the chord drawn as $2 a sin (Δ E /2)$ .

Step 6 — The two equations, and where they come from

We now have two independent facts about the same orbit, each carrying the variable $x$ . One is pure geometry; one carries the time. The true orbit obeys both at once.

WHY two equations? One unknown pair $(x, y)$ , two independent physical facts (geometry, time) → a determined system. Their simultaneous solution is the orbit.

Step 7 — The constants $ℓ$ and $m$ , motivated

WHAT. Two numbers, computed once from the givens.

WHY. They are just the shape-only and time-only leftovers from the derivation in Step 6, given names so the iteration stays clean.

PICTURE. A label diagram: $ℓ$ tags the triangle, $m$ tags the stopwatch.

Formula The two constants

m=\frac{\mu\,\Delta t^2}{\big(2\sqrt{r_1 r_2}\,\cos(\theta/2)\big)^3}$$ - $\dfrac{r_1+r_2}{\sqrt{r_1 r_2}\cos(\theta/2)}$ in $\ell$ — the **shape** combination: it is the sum-of-radii measured against the chord-scale $\sqrt{r_1r_2}\cos(\theta/2)$ from Step 5. The $-\tfrac12$ centres it so $\ell=0$ marks the reference triangle. - $\cos(\theta/2)$ — the half-angle from Step 5's chord-matching identity, not a coincidence. - $\Delta t^2$ in $m$ — time enters **squared**. This is the fingerprint of Kepler's period law: swept area $\propto \Delta t$, but the length-scale that normalises it $\propto \Delta t^{2/3}$, so once you cube the denominator the packed constant carries $\Delta t^2$. - The denominator $\big(2\sqrt{r_1 r_2}\cos(\theta/2)\big)^3$ — the chord-scale **cubed**, exactly the $\sqrt{p}$-through-Kepler bookkeeping of Step 6.

Step 8 — The iteration as a picture: guess, slide, correct

WHAT. Start at $y = 1$ . Use Equation A to get the $x$ that guess implies. Use Equation B to compute a better $y$ . Repeat.

WHY. We can't solve the two curves' crossing in closed form, so we bounce between them; because $y$ starts near the true value the bounces shrink fast (a contraction map).

PICTURE. A staircase converging onto the crossing point of the two curves.

The step-4 update is Equation B with $y^{2}$ replaced by $m / (ℓ + x)$ : $y - 1 = \frac{m}{y ^{2}} (x - \frac{1}{2} + X) = (ℓ + x) (x - \frac{1}{2} + X) .$

Step 9 — The degenerate and non-elliptic cases you MUST see

WHAT / WHY. Every solver must survive the corners. Here are the ones that bite.

PICTURE. Panels: near- $18 0^{\circ}$ blow-up, the short/long-way branches, the straight-line $θ = 0$ limit, and the parabolic/hyperbolic ( $e \geq 1$ ) regime.

Common mistake Parabolic (

$e = 1$ ) and hyperbolic ( $e > 1$ ) transfers Feels right: we derived everything for an ellipse. Trap: if the required $Δ t$ is very short (a fast, energetic transfer), the connecting orbit is not bound — it is a parabola ( $e = 1$ ) or hyperbola ( $e > 1$ ), for which the eccentric anomaly $E$ and the ellipse-only series $X (x)$ no longer apply (the "helper circle" of Step 5 becomes a helper hyperbola). Lambert's problem still has a solution, but Gauss's specific elliptic series does not converge to it. Fix: the Universal variable formulation handles ellipse, parabola, and hyperbola with one set of equations (Stumpff functions), which is why Battin's method and the Izzo Lambert solver are the modern default. Physically: given $r_{1}, r_{2}$ , short flight times ⇒ hyperbola, one special time ⇒ parabola, longer times ⇒ ellipse.

The one-picture summary

Everything on one canvas: the flat triangle (yardstick), the fatter sector $A_{sector}$ (real path), their ratio $y$ , the two curves whose crossing $y$ satisfies, and the staircase that lands on it. From $y$ we read off $p$ , then the Lagrange f and g functions $f, g, \overset{g}{˙}$ , then the velocities that turn two dots and a stopwatch into a full orbit — the raw material of a Porkchop plot.

Recall Feynman retelling — the whole walkthrough in plain words

I drew two arrows from the Sun to where a spaceship was and where it went, and joined their tips with a straight line — that flat triangle is my ruler. But the ship doesn't fly straight; it curves, sweeping out a fatter pie slice, and I call that curved region the sector. Kepler told me that area is really a clock: how much the ship sweeps tells me how long it took — and the stopwatch time $Δ t$ they gave me tells me how big that pie slice must be. So I compare the pie slice to the flat triangle and call that ratio $y$ . To pin $y$ down I need one more number: how far around the ellipse the ship went, measured by the eccentric-anomaly angle $E$ — and because a chord of a circle is always "twice-radius-times-sine-of-half-the-angle," half-angles $θ /2$ show up everywhere. That gives me two statements about the same orbit: one pure geometry, one carrying the clock. I don't know $y$ yet, so I guess it's about 1 (barely curved), see what arc that implies, check the time, and nudge $y$ until geometry and stopwatch agree. That agreement is the orbit. Watch the edges: opposite sides of the Sun make my numbers explode, forgetting which way around gives the wrong path, perfectly-lined-up points leave no triangle to measure, and a super-fast trip isn't even an ellipse — it's a parabola or hyperbola that needs the universal-variable method instead.