One time of flight and five ways round

Lambert’s theorem is one of the tidiest results in celestial mechanics. The time to travel between two points on a Keplerian orbit depends only on the semi-major axis, the straight-line distance between the points, and the sum of their distances from the focus — not on the eccentricity, not on the orientation of the ellipse, not on where the perifocus lies. Three lengths and a time, and the shape of the conic drops out.

Read as an inverse problem it becomes the workhorse of interplanetary flight: given two positions and an interval, find the transfer. Stated that way it sounds like it has one answer.

It has one answer only if the vehicle is forbidden to go round.

What the theorem actually fixes #

Write $r_1$ and $r_2$ for the two radii, $c$ for the chord between the two points, and $s = (r_1+r_2+c)/2$ for the semiperimeter. Lambert’s result is that the flight time on an ellipse of semi-major axis $a$ is

$$\sqrt{\frac{\mu}{a^3}}\,t = (\alpha - \sin\alpha) - (\beta - \sin\beta),$$

with

$$\sin\frac{\alpha}{2} = \sqrt{\frac{s}{2a}},\qquad \sin\frac{\beta}{2} = \sqrt{\frac{s-c}{2a}}.$$

Two things about this expression carry the whole of the multi-revolution story.

The first is that $\alpha$ is defined by its sine, and an angle is not determined by its sine. Both $\alpha_0$ and $2\pi - \alpha_0$ satisfy the equation, and both correspond to real arcs through the same two points: one that stays on the near side of the ellipse’s empty focus and one that goes round the far side. These are the short branch and the long branch, and they exist for every $a$ above the minimum.

The second is that there is a smallest possible $a$. The definition of $\alpha$ requires $\sqrt{s/2a}\le 1$, so $a \ge s/2$. That value is the minimum-energy transfer — the ellipse of least semi-major axis, and therefore of least orbital energy, that passes through both points at all. At $a = s/2$ exactly, $\alpha_0 = \pi$ and the two branches coincide, which is why the fold is a fold rather than a crossing.

Adding revolutions #

Nothing in the derivation forbids the vehicle from completing whole circuits before arriving. Each complete revolution adds a full $2\pi$ to the eccentric-anomaly sweep, so the flight time becomes

$$\sqrt{\frac{\mu}{a^3}}\,t_N = 2\pi N + (\alpha - \sin\alpha) - (\beta - \sin\beta).$$

For $N = 0$ the bracket vanishes as $a\to\infty$ on the short branch — the arc straightens towards the parabolic limit — so the time falls to a finite floor. That floor is the fastest any conic can make the journey, and it is what a hyperbolic transfer beats by not being a conic through those two points at all but a different one.

For $N \ge 1$ the bracket does not vanish: it tends to $2\pi N$, and $\sqrt{a^3/\mu}$ grows without limit, so the time rises. Near $a = s/2$ the time is finite and the factor $\sqrt{a^3}$ is small. Between the two, it has a minimum.

That minimum is the whole difference. It sits at $a$ only 2.2 per cent above the minimum-energy value for one revolution and 0.7 per cent for two, because $2\pi N\sqrt{a^3}$ climbs so steeply that the stationary point is pushed hard against the smallest permitted ellipse. On a linear axis in $a$ the fold is invisible; on the logarithmic excess axis of the figure above it is legible, and that is the only reason the figure is drawn that way.

The counting follows. For a flight time $T$, every revolution count whose minimum lies below $T$ contributes two solutions, and $N=0$ always contributes exactly one. So the number of transfers is

$$2N_{\max} + 1,$$

with $N_{\max}$ the largest revolution count whose minimum time is under $T$. For the Earth-to-Mars geometry above and 1,400 days, that is five; at 900 days it is three; below 711 days it is one.

Why this is a practical problem and not a curiosity #

Nothing about the multi-revolution solutions is exotic. Every electric-propulsion mission and every low-thrust trajectory in the inner solar system spends more than a year in transit; every asteroid rendezvous and every sample return does; and the flight times where the counts become plural are exactly the flight times those missions fly.

The trouble is in the solving. Lambert’s problem is inverted numerically, and the standard formulations — Battin’s, Gooding’s, Izzo’s — all reduce it to a one-dimensional root find in some transformed variable. The transformation is chosen so that the time-of-flight function is monotone, because a monotone function has one root and bisection cannot fail on it. The monotonicity is exactly what the multi-revolution case destroys.

So a solver written for the zero-revolution case, handed a long flight time, does one of three things. It converges to the zero-revolution solution and reports it, which is correct and incomplete. It fails to bracket a root and reports failure, on a problem with five answers. Or it converges to whichever multi-revolution branch its initial guess fell nearest, which is correct, incomplete, and indistinguishable from the first case in the output.

What the extra solutions are worth #

The obvious question is whether the multi-revolution transfers are ever cheaper, and the answer is that they often are, for a reason that has nothing to do with the transfer arc and everything to do with the departure.

A transfer’s cost is not its flight time. It is the sum of two velocity differences: the vehicle’s departure speed minus the origin planet’s, and the destination planet’s speed minus the arrival speed. A multi-revolution arc has a semi-major axis close to the minimum-energy value, which means low orbital energy, which means a small departure impulse. The zero-revolution arc that makes the same journey in the same total time has a much larger semi-major axis and needs a much faster departure.

Put concretely: a transfer that leaves slowly, loops the Sun twice and arrives has a lower departure energy than one that leaves fast and arrives directly on the same date. The vehicle pays in patience rather than in propellant, and the rocket equation makes that a very good trade. The trade is not always favourable. A longer flight means more time for the spacecraft to be exposed, more consumables on a crewed vehicle, more operations cost, and — for a sample return — a longer wait for the science. Missions choose differently, and the choice is legitimate. What is not legitimate is choosing without knowing the alternative existed, which is what happens when a solver quietly returns one of five.

The fold, and what it means for a search #

The minimum in each revolution count’s time-of-flight curve is a fold in the mathematical sense: a place where two solution branches meet and annihilate. That gives the search problem its characteristic difficulty.

Approach the minimum from above and the two solutions converge on each other; the derivative of flight time with respect to semi-major axis passes through zero, and any Newton iteration on that variable becomes ill-conditioned exactly there. Approach from below and there is nothing to find.

The practical remedy is to solve for the minimum explicitly rather than to stumble into it. For each revolution count, find $a_{\min,N}$ by solving $\mathrm dt_N/\mathrm da = 0$ — a well-conditioned problem — and then bracket each of the two roots between $s/2$ and $a_{\min,N}$, and between $a_{\min,N}$ and the upper limit. Both brackets are guaranteed to contain exactly one root, and both are monotone inside their bracket. The whole difficulty disappears once the folds are located first.

The geometry hidden in the count #

There is a second way to see why the solution count is odd, and it makes the structure feel less like an accident of the algebra.

Fix the two points and let the transfer angle be $\Delta\theta$. A zero-revolution transfer sweeps $\Delta\theta$; a one-revolution transfer sweeps $\Delta\theta + 2\pi$; and so on. The sweep is the eccentric-anomaly range, and for a given time, a larger sweep must be flown on a smaller, faster orbit. So the revolution counts are ordered by orbit size, tightly packed just above the minimum-energy ellipse, and the branches within each count are the two ways of arranging the same sweep about the empty focus.

The odd one out is the zero-revolution long branch, which has no partner because there is no $N=-1$. That is the whole of the $+1$ in $2N_{\max}+1$.

The packing is worth a number. The one-revolution fold sits at $a = 1.022\,s/2$ and the two-revolution fold at $1.007\,s/2$, and the trend continues: each successive count is squeezed closer to the minimum-energy ellipse than the last. In the limit of many revolutions every solution converges on $a = s/2$ itself, which is the statement that a vehicle with unlimited patience flies the minimum-energy ellipse and simply waits for the destination to come round. That is not a hypothetical regime — it is what a resonant-return trajectory does, and it is why the cheapest transfers in the whole catalogue sit at flight times measured in years.

The floor, and what lies under it #

The zero-revolution short branch has a shortest time, reached as the semi-major axis runs to infinity, and it is worth naming because it bounds everything an ellipse can do.

That limit is the parabolic transfer: the arc through the two points on an orbit of zero total energy. Its flight time is finite and is given in closed form by Euler’s relation, which involves only the two radii and the chord — no semi-major axis at all, because there isn’t one.

Nothing on an ellipse is faster. A vehicle that must make the journey in less time is not on a conic through those two points with negative energy; it is on a hyperbola, and a hyperbola through a given pair of points in a given time exists for any time shorter than the parabolic one.

The cost of that is immediate. A hyperbolic transfer’s departure speed exceeds the local escape speed, so the departure impulse rises steeply as the flight time is cut, and the characteristic energy the launch vehicle has to supply rises with it. That relation — flight time against departure energy — is the vertical structure of every launch-window chart, and it is why those charts have a hard edge at short flight times rather than a gradual one.

So the full family of transfers through two points has three regimes rather than two. Below the parabolic time, hyperbolic arcs, expensive and getting rapidly more so. At the parabolic time, one arc. Above it, ellipses — one for a while, then three, then five, each new pair arriving as the flight time passes another fold, and all of them clustered near the minimum-energy orbit.

The cheapest transfer and the fastest transfer are at opposite ends of the same family, and every mission design is a choice of where on it to sit.

The angle at which the problem stops having an answer #

There is a second way the solution count misbehaves, and it has nothing to do with revolutions. It happens at one particular geometry, and it happens to be the most useful one.

Two points and a central body define a plane — unless the two points and the centre are collinear. If the transfer angle is exactly a hundred and eighty degrees, the two position vectors and the focus lie on one line, and there are infinitely many planes containing that line. Every one of them holds a valid transfer orbit, and they all take the same time.

So at exactly $\Delta\theta = \pi$ the problem is degenerate: the flight time is determined, the semi-major axis is determined, and the orbital plane is not determined at all.

That would be a curiosity except for where it falls. A minimum-energy transfer between two circular coplanar orbits sweeps exactly a hundred and eighty degrees — it is the Hohmann geometry, and it is the transfer every mission is compared against. So the degeneracy sits precisely at the configuration the whole subject is organised around.

In practice the two planets are never exactly coplanar, so the transfer angle is never exactly $\pi$ and the problem is never exactly degenerate. What happens instead is worse for a numerical solver: near the degeneracy the plane is nearly undetermined, so the solution is extremely sensitive to the exact positions, and the required plane change swings wildly as the transfer angle passes through a hundred and eighty degrees.

The consequence appears on every launch-window chart as a narrow ridge of high cost running through the middle of what should be the cheapest region. It is not a physical barrier; it is the transfer plane tipping over, and a trajectory designed just to one side of it is fine while one designed exactly on it is undefined.

The standard remedy is to avoid the ridge rather than to solve on it — to shift the departure date by a few days, which costs almost nothing and moves the transfer angle safely away. It is a rare case in this subject where the right response to a singularity is to step around it.

Where the model stops #

Three things are outside this picture and each matters for a real mission.

The problem is two-body. The two positions are planetary positions, and a planet has a sphere of influence, so the real trajectory is a sequence of conics stitched at those boundaries rather than a single arc from surface to surface. That does not change the count — the heliocentric leg is still a Lambert problem — but it does change what the endpoints are.

The transfer is impulsive, and a low-thrust vehicle does not fly conics at all. The multi-revolution structure survives in a modified form there: a low-thrust spiral naturally makes many revolutions, and the cost is a schedule rather than a manoeuvre, which is a different accounting from the one above.

And nothing here says the transfer is optimal. Lambert’s problem answers “what conic joins these points in this time”; it does not answer “what is the cheapest way to get from planet A to planet B”. The second question is answered by searching over departure and arrival dates — over the two-dimensional surface a porkchop plot draws — and the multi-revolution solutions are extra sheets of that surface rather than extra points on it. Both of the difficulties above have the same practical resolution and it is worth stating, because it is a change in habit rather than in mathematics. A modern solver is expected to enumerate: to return every solution for the requested interval, labelled by revolution count and branch, rather than to return one and leave the caller to wonder. Enumeration is cheap — the folds are located analytically and each bracket contains exactly one root — and it converts a search over a cost surface with hidden sheets into a search over a set of surfaces that are each single-valued. What made that the norm was not a new algorithm but a new class of mission: once flight times of several years became ordinary, a solver that silently returned one of five stopped being adequate.

Where this ladder goes next #

This rung establishes that Lambert’s problem is plural once revolutions are allowed and that the plurality is structured — folds, branches, and a count that is a property of the interval.

Above it lies the search. A trajectory design problem is a global optimisation over departure date, arrival date and revolution count, with a cost surface that has one basin per branch; the algorithms used on it are the ones used on any problem with many local minima, and their success depends on enumerating the branches rather than on hoping to land in the right one.

Beside it lies the same structure in a different guise. The Lambert fold is a saddle-node bifurcation, and the same object appears wherever a solution count changes as a parameter is turned: in the roots of the eighth-degree equation that determines an orbit from three directions, and in the transition from three Lagrange points to five as a mass ratio is varied. A problem whose answer count changes is a problem with a fold in it, and the fold is usually where the interesting design lives.