Bi-elliptic transfer — when it wins over Hohmann: Visual walkthrough

Step 1 — Two circles, and what "orbit speed" even means

WHAT. We draw the whole problem: a heavy body (a planet or star) at the centre, a small circle of radius $r_{1}$ where we start, and a big circle of radius $r_{2}$ where we want to end up. Both circles lie in the same flat plane — the orbits are coplanar, so no tilting is needed.

WHY. Before any burns, we must agree on a single fact: how fast do you move on a circle of radius $r$ ? Everything downstream is a comparison of speeds, so we need the speed rule first.

PICTURE. Look at Step 1. The inner circle carries a fast blue arrow (long); the outer circle carries a slow blue arrow (short). Farther out = slower. That single truth is the engine of the whole trick.

Step 2 — Why a single burn can't teleport you between circles

WHAT. We ask: can one engine push move you from the inner circle straight onto the outer circle? No. A burn changes your speed at the spot you're standing; it does not change your radius instantly. So a burn bends your circle into a stretched loop — an ellipse — that carries you outward over time.

WHY. This is the reason transfers have shape. You are forced to travel on curved bridges (ellipses), and each ellipse touches the two circles at two special points. Naming those points is Step 3.

PICTURE. Step 2 shows a single prograde nudge (pink arrow, "speed up") at the inner circle. The circle immediately blooms into a yellow ellipse whose near end still kisses $r_{1}$ but whose far end swings out to a new, larger radius.

Step 3 — The one number that names an ellipse: $a$

WHAT. We measure an ellipse by a single length: put periapsis distance $r_{peri}$ and apoapsis distance $r_{apo}$ end to end and take the average.

WHY. Vis-viva needs one number to describe the whole orbit, and that number is the semi-major axis $a$ . It secretly stores the orbit's total energy (see Semi-major axis and orbital energy): bigger $a$ = higher, more energetic orbit. Every burn's cost is really "how much did I move $a$ ?"

PICTURE. Step 3 lays the two apsidal radii on one straight line through the centre. Their full span is $r_{peri} + r_{apo}$ ; halve it and you land on $a$ , marked at the ellipse's waistline.

Now every orbit — circle or ellipse — can be fed into one master speed rule.

Step 4 — The master ruler: vis-viva reads speed off any orbit

WHAT. We state the one equation that gives your speed at any radius $r$ on any orbit whose size is $a$ .

WHY. Every burn is "speed I have on the new orbit" minus "speed I had on the old orbit," both measured at the same spot $r$ . If we have one formula that reads speed from $(r, a)$ , then every $Δ v$ is just two readings subtracted. This is the whole toolkit — we never need anything else.

PICTURE. Step 4 is a dial: fix the radius $r$ (where you stand), and vis-viva returns your speed. The same standing spot on two different ellipses gives two different needle positions; a burn is the jump between them.

Step 5 — The three burns, drawn on one map

WHAT. We now lay all three impulses of the bi-elliptic route onto one figure and name the two bridging ellipses.

Ellipse 1 runs from $r_{1}$ (periapsis) out to $r_{b}$ (apoapsis), an intermediate radius chosen bigger than $r_{2}$ . Its size: $a_{1} = \frac{r _{1} + r _{b}}{2}$ .
Ellipse 2 runs from $r_{2}$ (periapsis) out to that same far point $r_{b}$ (apoapsis). Its size: $a_{2} = \frac{r _{2} + r _{b}}{2}$ .

WHY. We deliberately overshoot the target ( $r_{b} > r_{2}$ ) so that the reshaping burn happens way out where Step 1 promised speeds are tiny. The cost of that middle burn is what we're trying to starve.

PICTURE. Step 5 shows the full trajectory: burn 1 (pink, speed up) at $r_{1}$ launching yellow ellipse 1; burn 2 (yellow, gentle) at the far point $r_{b}$ lifting onto blue ellipse 2; burn 3 (pink, brake) at $r_{2}$ settling onto the outer circle.

Each burn is now just two vis-viva readings at the same $r$ , subtracted.

Step 6 — Watch the middle burn vanish (the far-out limit)

WHAT. We push $r_{b}$ larger and larger and watch the three burns respond.

WHY. To feel the trick you must see the trend: burn 2 collapses toward zero, burn 3 shrinks toward its own limit, while burn 1 keeps climbing. Their tug-of-war decides everything.

PICTURE. Step 6 plots each burn's size against $r_{b}$ . Burn 2 (blue) dives to zero; burn 3 (pink) flattens; burn 1 (yellow) rises. The sum (white) sags to a minimum near the bi-parabolic limit $r_{b} \to \infty$ .

Step 7 — The crossover: when does the detour finally pay? (deriving 11.94 and 15.58)

WHAT. We overlay the total bi-elliptic $Δ v$ against the two-burn Hohmann transfer cost, sweeping the ratio $R = r_{2} / r_{1}$ , and find exactly where the two are equal.

WHY. This is the payoff picture — it converts all the geometry into a single yes/no rule. To get the famous numbers we must solve $Δ v_{H} (R) = Δ v_{bi} (R)$ .

HOW the numbers arise. Work in normalized units $μ = 1, r_{1} = 1$ , so $r_{2} = R$ . Two natural comparisons give the two thresholds:

Threshold $R \approx 15.58$ (the guaranteed-win number). Compare Hohmann against the best-case bi-parabolic limit ( $r_{b} \to \infty$ , from Step 6). Setting the two $Δ v$ expressions equal produces one transcendental equation in $R$ alone; solving it numerically gives $R^{⋆} \approx 15.58$ . Above this, even the best Hohmann loses to the best bi-elliptic, so bi-elliptic wins for some finite $r_{b}$ .
Threshold $R \approx 11.94$ (the never-below number). Now ask the opposite: below what $R$ can no finite $r_{b}$ ever beat Hohmann? This is the largest $R$ at which $\frac{d}{d r _{b}} Δ v_{bi}_{r_{b} = r_{2}} \geq 0$ — i.e. starting to overshoot raises the cost. That boundary condition solves to $R \approx 11.94$ .
Between them ( $11.94 < R < 15.58$ ): overshooting helps, but only enough to win if $r_{b}$ is large enough — the winner depends on your chosen $r_{b}$ (and your time budget).

PICTURE. Step 7 plots both curves against $R$ . Below $R \approx 11.94$ Hohmann's line sits lower; above $R \approx 15.58$ the bi-elliptic line dips under. The grey band between is the "it depends on $r_{b}$ " zone.

Step 8 — Edge & degenerate cases (never get caught out)

WHAT + PICTURE. Step 8 collects the corner cases side by side; the left panel shows the key one algebraically and visually.

$r_{b} = r_{2}$ (no overshoot). Then $a_{2} = \frac{r _{2} + r _{b}}{2} = \frac{r _{2} + r _{2}}{2} = r_{2}$ — ellipse 2's size equals the outer circle's radius, so ellipse 2 is the outer circle. Its periapsis speed already equals circular speed, so

$Δ v_{2} = μ (\frac{2}{r _{b}} - \frac{1}{a _{2}}) - μ (\frac{2}{r _{b}} - \frac{1}{a _{1}}), and burn 3 vanishes: \frac{μ}{r _{2}} - μ (\frac{2}{r _{2}} - \frac{1}{r _{2}}) = 0.$ The transfer collapses exactly into a Hohmann (one bridging ellipse, two burns). So Hohmann is the " $r_{b} = r_{2}$ " member of the same family. The left panel of Step 8 draws this collapse.

$r_{b} \to \infty$ (bi-parabolic): burn 2 $\to 0$ , transfer time $\to \infty$ . Lowest possible $Δ v$ for the given $R$ , but never physically reached.
$r_{1} = r_{2}$ ( $R = 1$ ): start and end circles coincide, no transfer needed, all burns → 0.
Small $R$ (e.g. $R = 6$ ): burn 1's extra climb dwarfs any burn-2 saving — the detour is pure waste (parent Example 2).

The one-picture summary

One frame holds the whole story: two circles, two bridging ellipses meeting at the far point $r_{b}$ , three labelled burns (big prograde, tiny reshape, prograde-then-brake), and the crossover strip telling you when it beats Hohmann.

Recall Feynman retelling — the walkthrough in plain words

We started by noticing one thing: the farther out you orbit, the slower you drift (Step 1) — and we got its speed rule by setting the orbit's "size" equal to its radius. A rocket burn only changes your speed, so it bends your circle into a stretched loop, an ellipse, that carries you outward (Step 2). We measured each ellipse with one number — the average of its near and far distances, called $a$ (Step 3); push the far end to infinity and that loop opens into a parabola. One master rule, vis-viva, reads your speed from where you are and how big your orbit is (Step 4). Then we drew the bi-elliptic route: fling far past the target to a point $r_{b}$ , gently nudge your near-side up to the target, then coast back and brake onto the circle (Step 5) — the sign of each $Δ v$ tells you which way to fire, its size tells you the fuel. Because everything is sluggish out at $r_{b}$ , that middle nudge costs almost nothing, and the farther out you fling, the cheaper it gets — though burn 1 grows to pay for it, so you pick $r_{b}$ by how much time you can stand (Step 6). Racing the total against the simple two-burn Hohmann and solving where they tie gives the famous numbers: below $R \approx 12$ Hohmann always wins, above $R \approx 16$ bi-elliptic always wins (Step 7). And we tidied the corners: shrink the fling back to the target and, since $a_{2} = r_{2}$ , burn 2 and burn 3 collapse — you literally become Hohmann again (Step 8). Same physics, one family, one dial called $r_{b}$ .

Recall Quick self-test

Why is burn 2 nearly free? ::: It happens at huge radius $r_{b}$ , where the shared $2/ r_{b}$ term dominates both vis-viva speeds, so they almost cancel. What does $r_{b} = r_{2}$ reduce the bi-elliptic transfer to? ::: An ordinary Hohmann transfer — because $a_{2} = r_{2}$ , so ellipse 2 is the outer circle and burns 2, 3 collapse. Where does the number 15.58 come from? ::: Setting Hohmann's $Δ v$ equal to the best-case bi-parabolic ( $r_{b} \to \infty$ ) $Δ v$ and solving for $R$ . Above what $R$ is bi-elliptic guaranteed to win? ::: $R > 15.58$ . Is burn 3 prograde or retrograde? ::: Retrograde — you arrive at periapsis too fast and must brake.