Generator

The lambert-arc generator

One pair of points, 260 days, and the conics that fit

One pair of points, 260 days, and the conics that fit. Two positions 1 and 1.524 AU from the Sun, separated by 135°, and the transfer orbits that get from one to the other in 260 days. The short way has semi-major axis 1.2189 AU and eccentricity 0.3244; the long way, sweeping 225° between the same two points in the same 260 days, needs 1.2184 AU and e = 0.2680. Lighter is the minimum-energy transfer, a = s/2 = 1.2161 AU, which takes 244.2 days and is the slowest ellipse rather than the fastest: below it there is no ellipse through these points at all. Its vacant focus is constructed as the intersection of the circles of radius 2a − r about each endpoint, and it lands on the chord, which is what minimum energy means as geometry. Every arc here was solved from Lambert's formula and then checked by integrating Kepler's equation between its own two true anomalies — 260.0000 days against the 260 asked for. The chord is 2.3405 AU and the semiperimeter 2.4322; those two numbers and the semi-major axis fix the time, and nothing else in this drawing does.

Two positions 1 and 1.524 AU from the Sun, separated by 135°, and the transfer orbits that get from one to the other in 260 days. The short way has semi-major axis 1.2189 AU and eccentricity 0.3244; the long way, sweeping 225° between the same two points in the same 260 days, needs 1.2184 AU and e = 0.2680. Lighter is the minimum-energy transfer, a = s/2 = 1.2161 AU, which takes 244.2 days and is the slowest ellipse rather than the fastest: below it there is no ellipse through these points at all. Its vacant focus is constructed as the intersection of the circles of radius 2a − r about each endpoint, and it lands *on the chord*, which is what minimum energy means as geometry. Every arc here was solved from Lambert's formula and then checked by integrating Kepler's equation between its own two true anomalies — 260.0000 days against the 260 asked for. The chord is 2.3405 AU and the semiperimeter 2.4322; those two numbers and the semi-major axis fix the time, and nothing else in this drawing does.

6 essays call lambert-arc. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Every figure listed here is the same construction drawn at different numbers, so a correction to one is a correction to all of them.

Two places and a clock decide the path

The time to fly between two points depends on the semi-major axis, the chord between them, and the sum of their distances — and on nothing else about the orbit. Not the eccentricity, not where periapsis is, not the orientation. Lambert's theorem is why an interplanetary launch date is the root of one equation.

5 transfers through the same two points in the same 1400 days. Time of flight against semi-major axis for every transfer through two points 135° apart at 1 and 1.524 AU, with the revolution count running from 0 to 2. Each count contributes two branches, and for N ≥ 1 the pair folds: the time has a minimum at a = 1.2426 AU for one revolution — only 2.2 per cent above the minimum-energy value of 1.2161, which is why the horizontal axis is the excess over that value and logarithmic — so a flight time above it is met twice and below it not at all. Reading the crossings of the 1400-day line off the drawn curves gives 5 of them — 0 revs high, 1 rev low, 1 rev high, 2 revs low, 2 revs high — which is 2N + 1 with N = 2, and the count is a property of the time rather than of the geometry. That is the practical content: a root-finder started from a single guess returns one of these 5 and gives no sign that the other 4 exist, and the cheapest of them is often not the one nearest the guess.

One time of flight and five ways round

Lambert's theorem says two positions and an interval fix the transfer. Allow the transfer to complete whole revolutions and that stops being true: the flight time folds, one number admits five arcs, and the cheapest of them is usually not the one a solver started nearest to.

A Hohmann transfer, 2.6 to 1 in radius. Two circular orbits and the ellipse that touches both. The first burn raises the far point to the outer orbit; the second, half an orbit later, circularises. The costs are computed from the vis-viva relation.

The cheapest way between two orbits, and why it is so slow

Two burns and a long coast is the least fuel that will move a spacecraft between two circular orbits. It is also, for anything beyond the Moon, an unreasonably long wait.

Lambert's time of flight in the semi-major axis and in Lancaster and Blanchard's variable. The non-dimensional time of flight T = t√(2μ/s³) of the transfer between radii 1 and 1.524 through 135 degrees, against two variables that label the same family of orbits. Left: against the semi-major axis in units of the minimum-energy ellipse's. The family is two curves — the fast branch, which passes the empty focus on one side, and the slow branch, on the other — joined at a = aₘ with T = 1.566, where both have a vertical tangent: the time changes infinitely fast with the semi-major axis there, and every a above aₘ corresponds to two times. Right: against x, the variable of Lancaster and Blanchard, in which a = aₘ/(1 − x²). The same family is one smooth, decreasing curve through T = 1.566 at x = 0 to the parabolic time 0.662 at x = 1, with no tangent vertical and every time given by exactly one x. Nothing about the orbits differs between the panels; only the label on them does.

The variable that makes Lambert's problem easy

Written in the semi-major axis, the time-of-flight equation has two branches joined by a vertical tangent, and a solver near the join struggles. Written in a variable Lancaster and Blanchard chose in 1969, it is one smooth curve — and every trajectory library now solves it that way.

A two-body arc flown in the real field. How far from its target a spacecraft arrives when it flies the velocity Lambert's two-body problem gives, through the Earth's real field with its equatorial bulge, against the angle of the transfer, from a 400 km orbit to a point on a 420 km orbit, on a logarithmic scale. In a pure two-body field every solution arrives within a metre of the target; exactly 180° is left out, where the two points and the Earth's centre are in line and the transfer's plane is not determined. In the real field the miss grows with the transfer: at 51.6° inclination, 4.2 km after 60° and 43.0 km after 300°. An equatorial orbit misses most, because the bulge's extra pull is strongest over the equator and acts on it all the way round; an inclined orbit feels it only part of the time but also has its plane turned.

An answer from two bodies, flown through a flattened Earth

Lambert's problem gives the exact velocity that carries a spacecraft between two points in a given time — exactly, for a spherical Earth. The real Earth bulges at the equator, by a thousandth of its central pull, and a spacecraft flying the exact two-body answer between two points in low orbit arrives tens of kilometres from where it was sent. The repair is to keep the two-body solver and lie to it about the target, and the lie converges by a factor of a hundred each time it is told.

An iteration that slows toward the straight-through transfer. The contraction ratio of the corrected-target iteration — how much each pass shrinks the miss — against the transfer angle, for a transfer between orbits of 51.6° inclination at 400 and 420 km, with the Earth's bulge at its real strength and at 10 and 30 times it, on a logarithmic scale. The iteration converges where the ratio is below 1 (dashed). At the real strength it shrinks the miss 1857-fold per pass at 150°, 83-fold at 178° and 8-fold at 181°: always convergent, ever slower. At 10 times the strength it fails between 181° and 184°; at 30 times it fails between 182.5° and 192°. The slowing is geometric: near 180° the transfer's plane is set by a sideways shift of the aim point, so correcting the bulge's sideways miss tilts the whole arc.

A correction that tilts the arc it meant to aim

A transfer designed with Lambert's two-body solution and flown through the Earth's flattened field misses its target, and the standard repair is to aim at the target minus the miss and solve again. That converges fast — until the transfer angle approaches 180°. There the two end points and the Earth's centre are nearly in line, the arc's plane is decided by a sideways shift of the aim, and correcting the bulge's sideways miss rotates the whole arc. The iteration slows by a factor of a hundred, and with a larger perturbation it runs away. Correcting only within the plane fixes it, at the cost of a plane change paid at arrival.

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