Bi-elliptic transfer — when it wins over Hohmann: Worked examples

This page is the drill-ground for the parent topic. We take the machinery it built — the Vis-viva equation and the three-burn recipe — and grind it against every kind of case the topic can throw at you: small ratios where the detour is folly, big ratios where it wins, the $r_{b} \to \infty$ limit, the exact crossover zone, inward transfers, and the hidden cost of time.

Everything here rests on two formulas the parent already earned. Let me restate them so no symbol appears un-anchored.

The scenario matrix

Every question about "Hohmann vs bi-elliptic" lands in one of these cells. The examples below are labelled with the cell they hit, so together they tile the whole space.

Cell	What makes it special	Expected winner	Example
A — small ratio	$R < 11.94$ : detour is waste	Hohmann	Ex 1
B — guaranteed-win ratio	$R > 15.58$ : detour pays off	Bi-elliptic	Ex 2
C — the grey zone	$11.94 < R < 15.58$ : depends on $r_{b}$	either — must compute	Ex 3
D — the $r_{b} \to \infty$ limit	bi-parabolic best case	limit value	Ex 4
E — degenerate input	$r_{2} = r_{1}$ ( $R = 1$ ): no transfer needed	zero $Δ v$	Ex 5
F — sign trap	is a burn a push or a brake?	check signs	Ex 6
G — real-world word problem	LEO → far station, real units	numeric $Δ v$	Ex 7
H — exam twist / time cost	fuel wins but time loses	tradeoff verdict	Ex 8
I — inward transfer	$R < 1$ : target inside start, reversed	mirror of outward	Ex 9

Case A — small ratio, Hohmann wins

Worked example Example 1 —

$R = 6$ (Cell A) Start circular at $r_{1} = 1$ , target circular at $r_{2} = 6$ . Try an intermediate apoapsis $r_{b} = 100$ . Which transfer is cheaper?

Forecast: $6 < 11.94$ , so guess Hohmann wins and the detour is pure waste. Let's prove it.

Step 1 — circular speeds. $v_{c 1} = 1/1 = 1$ , $v_{c 2} = 1/6 = 0.40825$ . Why this step? Every transfer starts and ends on a circle; these are the speeds we must match.

Step 2 — Hohmann ellipse. Its semi-major axis is the average of the two circle radii: $a_{H} = (1 + 6) /2 = 3.5$ . Why this step? The single Hohmann ellipse touches $r_{1}$ at periapsis and $r_{2}$ at apoapsis, so its size is fixed by both.

Step 3 — Hohmann burns. Burn A at $r_{1}$ (speed up onto the ellipse): $2/1 - 1/3.5 - 1 = 1.71429 - 1 = 0.30931$ . Burn B at $r_{2}$ (speed up from ellipse-apoapsis to circular): $1/6 - 2/6 - 1/3.5 = 0.40825 - 0.04762 = 0.40825 - 0.21822 = 0.19003$ . Total Hohmann $= 0.49934$ . Why this step? Each $Δ v$ is (speed I want) − (speed I have) at that point.

Step 4 — bi-elliptic ellipses. $a_{1} = (1 + 100) /2 = 50.5$ , $a_{2} = (6 + 100) /2 = 53$ . Why this step? Ellipse 1 stretches from $r_{1}$ (periapsis) out to $r_{b}$ (apoapsis); ellipse 2 stretches from $r_{2}$ (periapsis) out to the same $r_{b}$ . Each semi-major axis is the average of its own two apsidal radii — we need both to feed vis-viva.

Step 5 — bi-elliptic burns. Burn 1 at $r_{1}$ : $2/1 - 1/50.5 - 1 = 1.98020 - 1 = 0.40719$ . Burn 2 at $r_{b} = 100$ : $2/100 - 1/53 - 2/100 - 1/50.5 = 0.00113 - 0.00020 = 0.03364 - 0.01407 = 0.01958$ . Burn 3 at $r_{2}$ (brake, magnitude): $2/6 - 1/53 - 1/6 = 0.31447 - 0.40825 = 0.56078 - 0.40825 = 0.15253$ . Total bi-elliptic $= 0.57930$ . Why this step? Each of the three burns is again (speed I want) − (speed I have) at that radius: Burn 1 climbs onto ellipse 1 at $r_{1}$ , Burn 2 switches from ellipse 1 to ellipse 2 out at $r_{b}$ , Burn 3 drops from ellipse 2 onto the target circle at $r_{2}$ .

Verify: $0.49934 < 0.57930$ ⇒ Hohmann wins by about $0.08$ . Sanity: burn 1 (climb to $r_{b} = 100$ ) alone cost $0.407$ , already almost as much as the entire Hohmann. The tiny middle burn ( $0.02$ ) could never repay that. Matches the parent's rule $R < 11.94$ . ✅

Case B — guaranteed-win ratio, bi-elliptic wins

Worked example Example 2 —

$R = 20$ (Cell B) $r_{1} = 1$ , $r_{2} = 20$ , $r_{b} = 100$ .

Forecast: $20 > 15.58$ , so guess bi-elliptic wins for a suitable $r_{b}$ .

Step 1 — circular speeds. $v_{c 1} = 1/1 = 1$ , $v_{c 2} = 1/20 = 0.22361$ . Why this step? These are the two speeds we must start from and land on; every burn is measured against them.

Step 2 — Hohmann. $a_{H} = (1 + 20) /2 = 10.5$ . Burn A: $2 - 1/10.5 - 1 = 1.90476 - 1 = 0.38014$ . Burn B: $1/20 - 2/20 - 1/10.5 = 0.22361 - 0.00476 = 0.22361 - 0.06901 = 0.15460$ . Total Hohmann $= 0.53474$ . Why this step? This single-ellipse transfer (semi-major axis = average of the two circle radii) is the baseline we're trying to beat.

Step 3 — bi-elliptic. $a_{1} = 50.5$ , $a_{2} = (20 + 100) /2 = 60$ . Burn 1: $2 - 1/50.5 - 1 = 0.40719$ . Burn 2: $2/100 - 1/60 - 2/100 - 1/50.5 = 0.00333 - 0.00020 = 0.05774 - 0.01407 = 0.04367$ . Burn 3 (brake): $2/20 - 1/60 - 1/20 = 0.08333 - 0.22361 = 0.28868 - 0.22361 = 0.06507$ . Total bi-elliptic $= 0.51593$ . Why this step? Same recipe as Ex 1, but now $r_{2}$ is much larger, so the Hohmann's own second burn is expensive — that's the room bi-elliptic exploits.

Verify: $0.51593 < 0.53474$ ⇒ bi-elliptic wins by $\approx 0.019$ . Notice Hohmann's Burn B ( $0.155$ ) is much pricier than bi-elliptic's Burn 3 ( $0.065$ ): the detour trades one expensive circularization for a cheap far-out reshape. Matches $R > 15.58$ . ✅

The figure below plots exactly this competition across all ratios: the cyan curve is Hohmann's total $Δ v$ , the amber curve is bi-elliptic's (taken at very large $r_{b}$ ), both as functions of $R$ .

Figure — Bi-elliptic transfer — when it wins over Hohmann

Read it left to right. On the left ( $R = 6$ , our Ex 1) the amber curve sits above cyan — bi-elliptic loses. The two dashed white lines mark the parent's thresholds $R = 11.94$ and $R = 15.58$ : this is the "grey zone" where the curves nearly overlap. To the right of $15.58$ (our Ex 2 dot at $R = 20$ ) the amber curve has dipped below cyan — and that gap is the fuel saving. The single picture encodes the entire crossover rule.

Case C — the grey zone

Worked example Example 3 —

$R = 13$ , two choices of $r_{b}$ (Cell C) $r_{1} = 1$ , $r_{2} = 13$ . Between $11.94$ and $15.58$ the winner depends on $r_{b}$ . Test $r_{b} = 30$ then $r_{b} = 500$ .

Forecast: guess the small $r_{b} = 30$ might still lose to Hohmann (not far enough out), while the large $r_{b} = 500$ wins.

Step 1 — Hohmann baseline. $a_{H} = (1 + 13) /2 = 7$ . Burn A: $2 - 1/7 - 1 = 1.85714 - 1 = 0.36277$ . Burn B: $1/13 - 2/13 - 1/7 = 0.27735 - 0.01099 = 0.27735 - 0.10483 = 0.17252$ . Total Hohmann $= 0.53529$ . Why this step? We need the two-burn benchmark first, so we know exactly what number the detour must undercut.

Step 2 — bi-elliptic with $r_{b} = 30$ . $a_{1} = (1 + 30) /2 = 15.5$ , $a_{2} = (13 + 30) /2 = 21.5$ . Burn 1: $2 - 1/15.5 - 1 = 1.93548 - 1 = 0.39121$ . Burn 2: $2/30 - 1/21.5 - 2/30 - 1/15.5 = 0.02016 - 0.00214 = 0.14200 - 0.04624 = 0.09576$ . Burn 3 (brake): $2/13 - 1/21.5 - 1/13 = 0.10736 - 0.27735 = 0.32766 - 0.27735 = 0.05031$ . Total $= 0.53728$ . Why this step? At only $r_{b} = 30$ the middle burn ( $0.096$ ) isn't cheap enough — we didn't go far enough into the "calm zone."

Step 3 — bi-elliptic with $r_{b} = 500$ . $a_{1} = (1 + 500) /2 = 250.5$ , $a_{2} = (13 + 500) /2 = 256.5$ . Burn 1: $2 - 1/250.5 - 1 = 1.99601 - 1 = 0.41280$ . Burn 2: $2/500 - 1/256.5 - 2/500 - 1/250.5 = 0.0001010 - 0.0000009 = 0.01005 - 0.00097 = 0.00908$ . Burn 3 (brake): $2/13 - 1/256.5 - 1/13 = 0.15000 - 0.27735 = 0.38730 - 0.27735 = 0.10995$ . Total $= 0.53183$ . Why this step? We push $r_{b}$ from $30$ all the way to $500$ so the middle burn happens far deeper in the "calm zone" — watch it drop from $0.096$ to $0.009$ . That is the whole bi-elliptic idea working, and it flips the verdict for the same $R$ .

Verify: With $r_{b} = 30$ : $0.53728 > 0.53529$ → Hohmann barely wins. With $r_{b} = 500$ : $0.53183 < 0.53529$ → bi-elliptic wins. Same $R$ , opposite verdicts — exactly what "grey zone" means. ✅

Case D — the $r_{b} \to \infty$ (bi-parabolic) limit

Worked example Example 4 — pushing

$r_{b}$ to infinity, $R = 15$ (Cell D) $r_{1} = 1$ , $r_{2} = 15$ . Take $r_{b}$ enormous ( $1 0^{6}$ ) to approximate the Bi-parabolic transfer limit.

Forecast: the middle burn should shrink to essentially zero; the total should approach a fixed limiting number.

Step 1 — the limit formula. As $r_{b} \to \infty$ , $a_{1}, a_{2} \to \infty$ , so $1/ a \to 0$ . Vis-viva at $r_{1}$ gives $2/ r_{1}$ (escape-ish speed), and Burn 2 → 0. The known closed form is: $Δ v_{bipar} = \frac{2 μ}{r _{1}} - \frac{μ}{r _{1}} + \frac{2 μ}{r _{2}} - \frac{μ}{r _{2}}$ Why this step? Each ellipse becomes a parabola (energy $\to 0$ ); its speed at any radius $r$ is exactly $2 μ / r$ . So Burn 1 = parabolic − circular at $r_{1}$ , Burn 3 = parabolic − circular at $r_{2}$ , Burn 2 vanishes.

Step 2 — plug in. $Δ v_{bipar} = (2 - 1) + (2/15 - 1/15) = 0.41421 + (0.36515 - 0.25820) = 0.41421 + 0.10696 = 0.52117$ . Why this step? We evaluate the closed form so we have a target number the finite- $r_{b}$ calculation must converge to.

Step 3 — numerically check with $r_{b} = 1 0^{6}$ . $a_{1} \approx 500000.5$ , $a_{2} \approx 500007.5$ . Burn 1: $2 - 1/500000.5 - 1 \approx 0.41421$ . Burn 2: $\approx 2 \cdot 1 0^{- 6}$ -scale difference $\approx 0.0000068$ (essentially 0). Burn 3 (brake): $2/15 - 1/500007.5 - 1/15 \approx 0.36515 - 0.25820 = 0.10695$ . Total $\approx 0.52117$ . Why this step? Plugging a huge but finite $r_{b}$ back in confirms the limit formula wasn't a leap of faith — the numbers actually land on it.

Step 4 — Hohmann comparison for $R = 15$ . $a_{H} = (1 + 15) /2 = 8$ . Burn A: $2 - 1/8 - 1 = 1.875 - 1 = 0.36931$ . Burn B: $1/15 - 2/15 - 1/8 = 0.25820 - 0.00833 = 0.25820 - 0.09129 = 0.16691$ . Total Hohmann $= 0.53622$ . Why this step? To judge whether the bi-parabolic limit is worth it we must line it up against the same- $R$ Hohmann baseline, computed by the identical two-burn recipe.

Verify: finite- $r_{b}$ total ( $0.52117$ ) matches the closed-form limit ( $0.52117$ ) to 4 decimals, and Burn 2 → 0. Hohmann for $R = 15$ is $0.53622 > 0.52117$ , so even the limit bi-elliptic beats it — consistent with $R = 15$ sitting just below $15.58$ yet still winnable at large $r_{b}$ . ✅

Case E — degenerate input ( $R = 1$ )

Worked example Example 5 — same orbit,

$R = 1$ (Cell E) $r_{1} = r_{2} = 1$ . What does every formula say?

Forecast: you're already there — every $Δ v$ should collapse to zero, or reveal a pointless "climb and return."

Step 1 — Hohmann. $a_{H} = (1 + 1) /2 = 1$ . Burn A: $2 - 1 - 1 = 0$ . Burn B: $1 - 2 - 1 = 0$ . Total $= 0$ . Why this step? When start and target coincide the Hohmann ellipse degenerates into the circle itself — no burn needed.

Step 2 — bi-elliptic, $r_{b} = 100$ . $a_{1} = a_{2} = 50.5$ . Burn 1: $2 - 1/50.5 - 1 = 0.40719$ . Burn 2: $2/100 - 1/50.5 - 2/100 - 1/50.5 = 0$ (identical ellipses, apoapsis speeds equal). Burn 3 (brake): $2/1 - 1/50.5 - 1/1 = 0.40719$ . Total $= 0.81438$ .

Verify: Hohmann costs 0 (correct — no transfer needed). The bi-elliptic still "flings out and comes back," burning $0.407$ up and $0.407$ down for nothing. This is the degenerate limit of "detour is waste": $R = 1 < 11.94$ , and the waste is total. ✅

Case F — the sign trap

Worked example Example 6 — is each burn a push or a brake?

$R = 20, r_{b} = 100$ (Cell F) Reuse Ex 2's orbit. For each burn, state whether you accelerate (prograde, +) or decelerate (retro, brake), and confirm the arithmetic sign inside our magnitude.

Forecast: guess burns 1 and 2 are pushes (raising the orbit), burn 3 is a brake — the classic mistake is calling burn 3 a push.

Step 1 — Burn 1 at $r_{1}$ . Ellipse-1 speed $= 2 - 1/50.5 = 1.40719$ vs circular $1$ . Speed-I-want > speed-I-have ⇒ prograde push. $Δ v_{1} = + 0.40719$ . Why this step? To climb outward you add energy → speed up at periapsis.

Step 2 — Burn 2 at $r_{b}$ . Ellipse-2 apoapsis speed $= 2/100 - 1/60 = 0.05774$ vs ellipse-1 apoapsis speed $= 2/100 - 1/50.5 = 0.01407$ . Want > have ⇒ prograde push. $Δ v_{2} = + 0.04367$ . Why this step? Raising periapsis from $r_{1}$ to $r_{2}$ means speeding up at the far apoapsis.

Step 3 — Burn 3 at $r_{2}$ . You arrive at ellipse-2's periapsis, moving at $2/20 - 1/60 = 0.28868$ , but the target circle needs only $1/20 = 0.22361$ . Have > want ⇒ retro-burn / brake. $Δ v_{3} = 0.22361 - 0.28868 = - 0.06507$ ; we report $∣Δ v_{3} ∣ = 0.06507$ . Why this step? This kills the "burn 3 is acceleration" error — see the [!mistake] in the parent.

Verify: signs are $+, +, -$ . Total magnitude $0.40719 + 0.04367 + 0.06507 = 0.51593$ , matching Ex 2 exactly. The brake at $r_{2}$ is real: you are always faster than circular at the periapsis of an inbound ellipse. ✅

Case G — a real-world word problem

Worked example Example 7 — LEO to a distant relay, real units (Cell G)

A probe is in a circular low-Earth orbit at $r_{1} = 7000$ km. It must reach a circular orbit at $r_{2} = 140000$ km ( $R = 20$ ). Earth's $μ = 398600 km^{3} / s^{2}$ . Use $r_{b} = 700000$ km. Compare bi-elliptic vs Hohmann in real km/s.

Forecast: $R = 20 > 15.58$ ⇒ bi-elliptic should win. We just scale our normalized numbers by the real velocity unit.

Step 1 — the speed scale. $v_{c 1} = μ / r_{1} = 398600/7000 = 56.943 = 7.546 km/s$ . Why this step? This is the LEO circular speed and the multiplier that turns our dimensionless $Δ v$ into km/s (since we normalized $r_{1} = 1, μ = 1$ ).

Step 2 — normalized ratios. $R = 140000/7000 = 20$ , $r_{b} / r_{1} = 700000/7000 = 100$ . Identical to Ex 2! So the dimensionless totals carry over: bi-elliptic $= 0.51593$ , Hohmann $= 0.53474$ . Why this step? Physics only cares about ratios; same $R$ and same $r_{b} / r_{1}$ ⇒ same normalized answer.

Step 3 — rescale to km/s. Bi-elliptic $= 0.51593 \times 7.546 = 3.893 km/s$ . Hohmann $= 0.53474 \times 7.546 = 4.035 km/s$ .

Verify: units check — dimensionless × km/s = km/s. ✅ Bi-elliptic saves $4.035 - 3.893 = 0.142 km/s \approx 142 m/s$ , a genuine fuel win at $R = 20$ . Sanity: LEO circular speed $7.546$ km/s is the well-known real value (~7.5 km/s at 630 km altitude), so the scale is right. ✅

Case H — the exam twist: fuel wins, time loses

Worked example Example 8 — the tradeoff verdict,

$R = 18$ (Cell H) $r_{1} = 1$ , $r_{2} = 18$ , $r_{b} = 400$ . (a) Show bi-elliptic wins on $Δ v$ . (b) Estimate the time penalty using transfer-time reasoning and give the mission verdict.

Forecast: guess bi-elliptic wins fuel (since $18 > 15.58$ ) but loses badly on time.

Step 1 — Hohmann $Δ v$ . $a_{H} = (1 + 18) /2 = 9.5$ . Burn A: $2 - 1/9.5 - 1 = 1.89474 - 1 = 0.37650$ . Burn B: $1/18 - 2/18 - 1/9.5 = 0.23570 - 0.00585 = 0.23570 - 0.07646 = 0.15924$ . Total Hohmann $= 0.53574$ . Why this step? This is the baseline we must beat — the straight two-burn transfer.

Step 2 — bi-elliptic $Δ v$ . $a_{1} = (1 + 400) /2 = 200.5$ , $a_{2} = (18 + 400) /2 = 209$ . Burn 1: $2 - 1/200.5 - 1 = 1.99501 - 1 = 0.41244$ . Burn 2: $2/400 - 1/209 - 2/400 - 1/200.5 = 0.0002144 - 0.0000156 = 0.01465 - 0.00395 = 0.01070$ . Burn 3 (brake): $2/18 - 1/209 - 1/18 = 0.10633 - 0.23570 = 0.32608 - 0.23570 = 0.09038$ . Total bi-elliptic $= 0.51352$ . Why this step? $0.51352 < 0.53574$ ⇒ bi-elliptic saves $\approx 0.022$ in $Δ v$ ; the far-out reshape ( $r_{b} = 400$ ) made the middle burn small.

Step 3 — transfer times. Time on half an ellipse $= π a^{3} / μ$ . Hohmann (one half-ellipse, $a_{H} = 9.5$ ): $t_{H} = π 9. 5^{3} = π 857.375 = π \cdot 29.281 = 91.99$ (time units). Bi-elliptic (two half-ellipses, $a_{1} = 200.5$ then $a_{2} = 209$ ): $t_{bi} = π 200. 5^{3} + π 20 9^{3} = π \cdot 2839.6 + π \cdot 3021.5 = 8920 + 9492 = 18412$ . Why this step? Each burn-to-burn coast is a half orbit; bi-elliptic strings together two enormous ones, so its clock runs far longer.

Verify: fuel ratio bi/Hohmann $= 0.51352/0.53574 = 0.959$ (about 4% cheaper). Time ratio $= 18412/91.99 \approx 200 \times$ longer. Verdict: for a probe with fuel to spare and a deadline, Hohmann wins the mission despite losing on paper — this is the hidden-cost lesson from the parent. ✅

Case I — inward transfer ( $R < 1$ )

Worked example Example 9 — dropping

inward, $r_{2} < r_{1}$ (Cell I) Now the target is inside the start: $r_{1} = 1$ , $r_{2} = 1/20 = 0.05$ (so $R = r_{2} / r_{1} = 0.05$ ). Does the same machinery work in reverse?

Forecast: guess the whole story mirrors an outward transfer — every formula is symmetric in $r_{1} \leftrightarrow r_{2}$ , so an inward Hohmann costs the same $Δ v$ as the outward one between the same two radii, just with the burns' push/brake roles swapped.

Step 1 — the symmetry. A transfer between radii $p$ and $q$ uses the same ellipse ( $a = (p + q) /2$ ) whichever direction you travel — you just run it forwards or backwards. So $Δ v$ between $r_{1} = 1$ and $r_{2} = 0.05$ equals $Δ v$ between $1$ and $20$ scaled: work it directly to be safe. Why this step? The vis-viva formula only sees the two radii and $a$ , not the direction of travel, so a mirror argument is legitimate — but we still verify numerically.

Step 2 — Hohmann, inward. $a_{H} = (1 + 0.05) /2 = 0.525$ . Burn A at $r_{1} = 1$ (now a brake — we're dropping in): $2/1 - 1/0.525 - 1/1 = 0.09524 - 1 = 0.30861 - 1 = - 0.69139$ , magnitude $0.69139$ . Burn B at $r_{2} = 0.05$ (also a brake into the tighter, faster circle): $1/0.05 - 2/0.05 - 1/0.525 = 4.47214 - 38.09524 = 4.47214 - 6.17213 = - 1.69999$ , magnitude $1.69999$ . Total Hohmann $= 2.39138$ . Why this step? Dropping deep into the well makes the inner circle fast, so the numbers are large — but the recipe (want − have at each apsis) is unchanged; only the signs flip to brakes.

Step 3 — should you ever go bi-elliptic inward? The crossover rule uses $R = r_{2} / r_{1}$ . For an inward transfer $R < 1$ , which is far below $11.94$ , so bi-elliptic never helps going in — flinging further out first would be absurd when your goal is deeper down. Why this step? It answers the "does the rule apply in reverse?" question: the threshold is stated for the larger/smaller ratio, and inward transfers sit permanently in Hohmann's kingdom.

Verify: the inward Hohmann total $2.39138$ is huge, dominated by the final brake $1.70$ into the fast inner orbit — physically sensible (low orbits are fast, so matching them from above needs a big slow-down). And $R = 0.05 < 11.94$ confirms bi-elliptic is never the answer inward. ✅

Recall Quick self-test on the matrix

Which cell is " $R = 1$ "? ::: Cell E — degenerate, Hohmann costs zero, bi-elliptic wastes two climbs. At the same $R$ in the grey zone, can two different $r_{b}$ give opposite winners? ::: Yes (Cell C, Ex 3): $r_{b} = 30$ favored Hohmann, $r_{b} = 500$ favored bi-elliptic. In the $r_{b} \to \infty$ limit, which burn vanishes? ::: The middle burn (Burn 2) → 0; each ellipse becomes a parabola. If two problems share the same $R$ and same $r_{b} / r_{1}$ , are their normalized $Δ v$ equal? ::: Yes — physics depends only on ratios (Ex 7). For an inward transfer ( $R < 1$ ), is bi-elliptic ever worth it? ::: No — $R < 1 < 11.94$ , so Hohmann always wins going in (Ex 9).