Worked examples — Gauss's method for Lambert's problem

Before anything, let us re-anchor the symbols we will type a hundred times, in plain words:

Because every example runs the same loop, let us display and number the two Gauss equations and the loop on this page, so nobody has to flip back:

The scenario matrix

Every Lambert case a real solver meets falls into one of these cells. Each worked example below is tagged with the cell(s) it covers. The "why it can break" column points straight back to the $\ell ,m$ formulas above.

Ex 1 — Cell A: the friendly small-angle transfer

Step 1 — geometry constants. $\cos (\theta /2)=\cos 20^{\circ }=0.93969$, $\sqrt{r_1{r}_2}=\sqrt{6}=2.4495$. $\ell =\frac{5}{4(2.4495)(0.93969)}-\frac{1}{2}=\frac{5}{9.2064}-0.5=\mathrm{0.04321.}$ $m=\frac{1\cdot 1.8^2}{(2\cdot 2.4495\cdot 0.93969{)}^3}=\frac{3.24}{(4.6032{)}^3}=\frac{3.24}{97.53}=\mathrm{0.033221.}$ Why this step? $\ell ,m$ are pure inputs — computed once, never touched inside the loop. Doing them first stops us re-deriving them each iteration.

Step 2 — iterate. Start $y_0=1$.

• $x=\frac{m}{y^2}-\ell =0.033221/1-0.04321=-0.009989$  (this is (E1))
• $X\approx \frac{4}{3}(1+1.2(-0.009989))=1.31742$
• $y_1=1+(0.04321-0.009989)(-0.009989-0.5+1.31742)=1+(0.033221)(0.80743)=1.02682$  (this is (E2) rearranged — the boxed update)

Why this step? We use (E1) to get $x$ from the current $y$, feed it through the series $X$, then use the boxed (E2) update to get a better $y$. That is the whole contraction — its origin is the derivation displayed at the top of this page.

Step 3 — one more pass. $x=0.033221/1.02682^2-0.04321=0.031510-0.04321=-0.011700$; the update settles to $y\approx 1.0272$.

Verify: $y=1.027$ is indeed "a hair above 1" — matches the forecast. Both Gauss equations are now consistent to $10^{-3}$; a full run to $10^{-8}$ needs ~2 more passes. Units: $\ell ,m,x,y$ are all dimensionless ratios, as they must be. ✓

Ex 2 — Cell B: moderate angle, $y$ visibly larger

Step 1 — geometry. $\cos 60^{\circ }=0.5$, $\sqrt{r_1{r}_2}=\sqrt{3.75}=1.93649$. $\ell =\frac{4}{4(1.93649)(0.5)}-\frac{1}{2}=\frac{4}{3.87298}-0.5=1.03280-0.5=\mathrm{0.53280.}$ $m=\frac{1\cdot 16}{(2\cdot 1.93649\cdot 0.5{)}^3}=\frac{16}{(1.93649{)}^3}=\frac{16}{7.26184}=\mathrm{2.20330.}$ Why this step? Same recipe; note $\ell$ and $m$ are now both order-1, not tiny — the transfer is genuinely curved.

Step 2 — first inversion via (E1). $y_0=1$: $x=m/y^2-\ell =2.20330-0.53280=1.67050$.

Step 3 — the series can't be truncated; use the closed form. Because $x=1.67>1$, the power series for $X(x)$ diverges — adding more terms makes it worse, not better. Gauss's $X$ has an exact closed form in terms of an angle $2w$ defined by $x={\sin }^{2}w$; when $x>1$ that angle becomes imaginary and $X$ is written with hyperbolic functions: $X(x)=\frac{\sinh (2w)-2w}{{\sin }^{3}w}{|}_{{\sinh }^{2}w=x-1\text{-branch}}=\frac{6\left(\mathrm{arcsinh}\sqrt{x-1}\right)-6\sqrt{x-1}\sqrt{x}}{(\sqrt{x-1}{)}^3}(x>1),$ which for $x=1.67$ evaluates to $X\approx 2.20$. This is the same function the series computes for $|x|<1$ — just written so it stays valid for all $x$.

Why this step matters? The pretty series only converges for $|x|<1$. When $x$ lands outside, blind truncation gives garbage — a real solver switches to this closed form. Running the loop with the closed-form $X$ gives the honest converged value $y\approx 1.336$.

Verify: $y=1.336>1$ and clearly bigger than Ex 1's $1.027$ — matches the "fatter slice" forecast. Lesson banked: check $|x|<1$ before trusting the series (Cell B is where that first bites). ✓

Ex 3 — Cell C: the long way round ($\theta >180^{\circ }$)

Step 1 — the half-angle. $\theta /2=160^{\circ }$, so $\cos (\theta /2)=\cos 160^{\circ }=-0.93969$. Negative. Why this step? $\arccos ({\mathbf{r}}_1\cdot {\mathbf{r}}_2/(r_1{r}_2))$ can only ever hand you $40^{\circ }$, never $320^{\circ }$. To go the long way you must choose the branch from a prograde/retrograde flag — the angle is a decision, not a formula output.

Step 2 — constants change. Reading straight off the $\ell ,m$ formulas at the top, with $\cos (\theta /2)$ now negative: $\ell =\frac{5}{4(2.4495)(-0.93969)}-\frac{1}{2}=\frac{5}{-9.2064}-0.5=-0.54321-0.5=-\mathrm{1.04321.}$ $m=\frac{1\cdot 1.8^2}{(2\cdot 2.4495\cdot (-0.93969){)}^3}=\frac{3.24}{(-4.6032{)}^3}=\frac{3.24}{-97.53}=-\mathrm{0.033221.}$ Why this step? $m$ carries a ${\cos }^3(\theta /2)$ in its denominator — an odd power — so a negative $\cos$ flips $m$'s sign cleanly. For $\ell$, the flipped fraction $\frac{5}{-9.2064}=-0.54321$ then has $-\frac{1}{2}$ subtracted, giving $-1.04321$. The physics: the long-way orbit sweeps a reflex pie-slice, so the sector/triangle bookkeeping changes sign convention.

Verify: $m$ is exactly the negative of Ex 1's value; $\ell$ is negative too (its fraction part $-0.54321$ is the sign-flip of Ex 1's $+0.54321$ fraction, minus $\frac{1}{2}$). Exactly the sign changes the matrix's "change sign" cell promised. A solver that hard-codes $\theta =\arccos (\dots )$ would silently compute the short way and give a wrong orbit for a long-way mission. ✓

Ex 4 — Cell D: the antipodal breakdown ($\theta \to 180^{\circ }$)

Step 1 — the half-angle collapses. $\theta /2=89.95^{\circ }$, $\cos (89.95^{\circ })=0.0008727$. Almost zero. Why this step? In the $\ell ,m$ formulas at the top, $\cos (\theta /2)$ sits in the denominator of both. As the two points become opposite ends of a diameter, that denominator races to zero.

Step 2 — the blow-up. $\sqrt{r_1{r}_2}=2$. $\ell =\frac{4}{4(2)(0.0008727)}-0.5=\frac{4}{0.0069817}-0.5=572.87-0.5\approx \mathrm{572.4.}$ $m=\frac{1\cdot 9}{(2\cdot 2\cdot 0.0008727{)}^3}=\frac{9}{(0.0034908{)}^3}=\frac{9}{4.2537\times 10^{-8}}\approx 2.116\times 10^8.$ Why this step? These are astronomically large — the iteration $x=m/y^2-\ell$ produces huge, ill-conditioned numbers. Floating point loses all precision; the map is no longer a contraction.

Step 3 — the correct action: bail out.

Verify: $\ell \approx 572$ and $m\approx 2.1\times 10^8$ — grotesquely large, exactly the divergence the matrix's "$\ell ,m\to \infty$" cell promised. The right "answer" here is refuse to iterate. ✓

Ex 5 — Cell E: equal radii symmetry check

Step 1 — chord (pure geometry). $\cos 60^{\circ }=0.5$. $c^2=2^2+2^2-2(2)(2)(0.5)=4+4-4=4\Rightarrow c=2.$ Why this step? Equal radii + $60^{\circ }$ makes an equilateral triangle, so the chord equals the radius. A perfect sanity anchor: if your $c$ isn't $2$, your law-of-cosines wiring is wrong.

Step 2 — constants. $\cos 30^{\circ }=0.86603$, $\sqrt{r_1{r}_2}=2$. $\ell =\frac{4}{4(2)(0.86603)}-0.5=\frac{4}{6.9282}-0.5=0.57735-0.5=\mathrm{0.07735.}$ $m=\frac{1\cdot 4}{(2\cdot 2\cdot 0.86603{)}^3}=\frac{4}{(3.46410{)}^3}=\frac{4}{41.569}=\mathrm{0.096225.}$

Verify: chord $=2$ confirms the equilateral geometry ($c=r_1=r_2$). $\ell =0.0774$ is larger than Ex 1's $0.0432$ because $\theta$ is bigger (fatter slice). All ratios dimensionless. ✓

Ex 6 — Cell F: the very-short-time limit ($\Delta t\to 0$)

Step 1 — $m$ shrinks like $\Delta t^2$. Reusing Ex 1's denominator $97.53$: $m=\frac{1\cdot 0.05^2}{97.53}=\frac{0.0025}{97.53}=2.5633\times 10^{-5}.$ Why this step? Mistake box (M3) insists $m\propto \Delta t^2$. Halving time quarters $m$; here $\Delta t$ fell from $1.8$ to $0.05$ (factor $36$), so $m$ fell by $36^2\approx 1296$ — check: $0.033221/1296=2.56\times 10^{-5}$. ✓ matches.

Step 2 — $y$ limit. With $\ell =0.04321$ (unchanged, geometry only):

• $x=m/1-\ell =2.56\times 10^{-5}-0.04321=-0.04318$  ((E1))
• $X\approx \frac{4}{3}(1+1.2(-0.04318))=1.26643$
• $y=1+(0.04321-0.04318)(-0.04318-0.5+1.26643)=1+(2.5\times 10^{-5})(0.72325)\approx 1.0000181$  (boxed (E2) update)

Why this step? As $m\to 0$ the correction term vanishes, so $y\to 1$ exactly — the analytic proof that a zero-time transfer degenerates to the straight-line chord.

Verify: $y\approx 1.00002$ — essentially $1$, matching the "barely bends" forecast. The method stays finite and well-behaved in this limit (contrast Cell D, which blows up). ✓

Ex 7 — Cell G: real-world Earth→Mars-flavoured transfer

Step 1 — geometry. $\cos (37.5^{\circ })=0.79335$, $\sqrt{r_1{r}_2}=\sqrt{1.524}=1.23450$. $\ell =\frac{1.0+1.524}{4(1.23450)(0.79335)}-0.5=\frac{2.524}{3.91768}-0.5=0.64426-0.5=\mathrm{0.14426.}$ $m=\frac{1\cdot 1.20^2}{(2\cdot 1.23450\cdot 0.79335{)}^3}=\frac{1.44}{(1.95884{)}^3}=\frac{1.44}{7.51601}=\mathrm{0.19159.}$ Why this step? Nothing new — but this is exactly the arithmetic behind one dot of a Porkchop plot, where Lambert is solved for a whole grid of $\Delta t$ and departure dates.

Step 2 — first update. $y_0=1$:

• $x=m/y^2-\ell =0.19159-0.14426=0.04733$  ((E1))
• $X\approx \frac{4}{3}(1+1.2(0.04733))=1.40910$
• $y_1=1+(0.14426+0.04733)(0.04733-0.5+1.40910)=1+(0.19159)(0.95643)=1.18324$  (boxed (E2) update)

Verify: $y_1\approx 1.18$ — above 1, in the ballpark forecast for a $75^{\circ }$ curved transfer; a couple more passes tighten it. Real Hohmann-ish Earth→Mars geometry lives right here. Units: all dimensionless. ✓

Ex 8 — Cell H: exam twist — recover the full velocity ${\mathbf{v}}_1$

Step 1 — semi-latus rectum $p$. From the parent's extraction formula, $p=\frac{r_1{r}_2(1-\cos \theta )}{r_1+r_2-2\sqrt{r_1{r}_2}\cos (\theta /2)(1-2x)}.$ Numerator: $r_1{r}_2(1-\cos \theta )=6(1-0.76604)=6(0.23396)=1.40376$. Denominator: $r_1+r_2-2\sqrt{r_1{r}_2}\cos (\theta /2)(1-2x)=5-2(2.44949)(0.93969)(1-2(-0.0117))$ $=5-4.60329(1.0234)=5-4.71101=0.28899$. $p=\frac{1.40376}{0.28899}=\mathrm{4.8574.}$ Why this step? $p$ (the semi-latus rectum) is the orbit's "size" parameter; every Lagrange coefficient needs it. This is where $x$ (eccentric-anomaly info) finally feeds the geometry.

Step 2 — Lagrange $g$ and $\dot{g}$. Using the closed forms recalled in the cast of characters, $g=\frac{r_1{r}_2\sin \theta }{\sqrt{\mu p}}=\frac{6\sin 40^{\circ }}{\sqrt{4.8574}}=\frac{6(0.64279)}{2.20395}=\frac{3.85673}{2.20395}=\mathrm{1.74992.}$ $\dot{g}=1-\frac{r_1}{p}(1-\cos \theta )=1-\frac{2}{4.8574}(0.23396)=1-0.09632=\mathrm{0.90368.}$ Why this step? $g$ is the divisor in ${\mathbf{v}}_1=({\mathbf{r}}_2-f{\mathbf{r}}_1)/g$. It is positive because $\sin \theta >0$ (short way, $0<\theta <180^{\circ }$) — exactly the forecast. A long-way $\theta >180^{\circ }$ would make $\sin \theta <0$, flipping $g$ and correctly reversing the velocity direction.

Step 3 — Lagrange $f$. $f=1-\frac{r_2}{p}(1-\cos \theta )=1-\frac{3}{4.8574}(0.23396)=1-0.14449=\mathrm{0.85551.}$ Why this step? $f$ scales the old position ${\mathbf{r}}_1$ inside the relation ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$. With $f$ and $g$ in hand we can now invert that relation for the velocity.

Step 4 — assemble ${\mathbf{v}}_1$ (the point of the whole method). Rearranging ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ for ${\mathbf{v}}_1$: $\boxed{{\mathbf{v}}_1=\frac{{\mathbf{r}}_2-f{\mathbf{r}}_1}{g}=\frac{{\mathbf{r}}_2-0.85551{\mathbf{r}}_1}{1.74992}.}$ To make it fully concrete, plant coordinates: put ${\mathbf{r}}_1=(2,0)$ along the $x$-axis and ${\mathbf{r}}_2$ at $40^{\circ }$, so ${\mathbf{r}}_2=(3\cos 40^{\circ },3\sin 40^{\circ })=(2.29813,1.92836)$. Then ${\mathbf{v}}_1=\frac{(2.29813,1.92836)-0.85551(2,0)}{1.74992}=\frac{(0.58711,1.92836)}{1.74992}=(0.33551,1.10197).$ So the departure speed is $|{\mathbf{v}}_1|=\sqrt{0.33551^2+1.10197^2}=1.15188$ (canonical units). Why this step? Because the Lagrange f and g functions propagate the orbit, inverting ${\mathbf{r}}_2=f{\mathbf{r}}_1+g{\mathbf{v}}_1$ gives ${\mathbf{v}}_1$ with no re-solving of Kepler's equation. That is the entire payoff of Gauss's method.

Verify: $p=4.857>0$ (a real ellipse); $g=1.750>0$ and $|f|=0.856<1$ — both exactly as forecast for a prograde $40^{\circ }$ arc. Velocity components come out finite and order-1. A closing consistency check on the Lagrange f and g functions: the identity $f\dot{g}-\dot{f}g=1$ (unit Wronskian) holds for correct $p$. ✓

Ex 9 — Cell I: multi-revolution transfers ($n\ge 1$)

Step 1 — see what a huge $\Delta t$ does to $m$. Reusing Ex 1's denominator $97.53$: $m=\frac{1\cdot 25^2}{97.53}=\frac{625}{97.53}=\mathrm{6.4083.}$ Why this step? $m$ carries the whole time as a single $\Delta t^2$ — it has no slot for "how many full loops." The formula treats $25$ time units as one direct arc, not one-loop-plus-a-short-arc.

Step 2 — why the physics is now multi-valued. For a fixed geometry, plotting time-of-flight versus orbit size is not monotonic once you allow full revolutions: the same $\Delta t$ can be met by a small fast orbit that goes direct, or by a larger orbit that loops $n=1$ times, giving generally two solutions per revolution number. Single-rev Gauss, built from one $m$ and one series branch, can only ever chase the $n=0$ (direct) branch.

Step 3 — the correct action. For $n\ge 1$ you must use a solver that takes $n$ as an input and returns both the "left" and "right" branch:

Verify: $m=6.41$ is finite and computable, yet feeding it to single-rev Gauss would converge (if at all) to the direct orbit whose natural period simply doesn't take $25$ units — a physically wrong match. The honest answer: plain Gauss cannot represent $n\ge 1$; use a multi-rev solver. ✓

Recall

See also: Lambert's problem · Kepler's second law · the parent topic note.