Combined maneuvers — optimal split between plane change and velocity change: Worked examples

This page drills the parent topic until no case can surprise you. Before we start, one promise: every symbol below is already earned in the parent note, but we re-anchor each as it appears so you can read from line one.

The one master formula the parent note built:

Everything on this page is just this formula (and its optimization) meeting every kind of input you can feed it.

The scenario matrix

Think of a "scenario" as a choice of inputs. The inputs are: the two speeds ( $v_{1}$ vs $v_{2}$ ), and the angle $θ$ . Here is every distinct class of case, each with the example that nails it.

Cell	What is special about the inputs	Why it could trip you	Example
A. $θ = 0$ (degenerate angle)	no turn at all	formula must collapse to a plain speed change	Ex 1
B. $v_{1} = v_{2}$ (equal speeds, pure tilt)	only direction changes	must reproduce $2 v sin (θ /2)$	Ex 2
C. $θ = 18 0^{\circ}$ (reverse direction)	fully opposed arrows	$cos$ hits $- 1$ ; speeds add	Ex 3
D. General $v_{1} \neq = v_{2}$ , $0 < θ < 9 0^{\circ}$	the everyday combined burn	mixing magnitude + direction	Ex 4
E. Optimal split, two burns (real mission)	LEO→GEO with inclination	equal-marginal-cost, not equal-angle	Ex 5
F. Limiting case $s \to i$ vs $s \to 0$	put ALL tilt at one end	which endpoint wins, and why	Ex 6
G. Combined vs separate (word problem)	is one throw really cheaper?	triangle inequality made numeric	Ex 7
H. Exam twist: obtuse $θ$ , unequal $v$	$cos θ < 0$	sign of the cosine flips the middle term	Ex 8

Between them, cells A–H cover: zero input, degenerate equal input, the extreme $18 0^{\circ}$ , generic acute, generic obtuse, the full optimization, its two limiting endpoints, and a real design decision. Nothing is left.

Cell A — the zero-angle degenerate case

Forecast: With no turn, the two arrows point the same way. Guess: the answer should just be the difference in their lengths. Write your guess before reading on.

Substitute $θ = 0$ , so $cos θ = 1$ . Why this step? $cos 0 = 1$ is the largest the cosine ever gets, which makes the subtracted middle term as big as possible — that is what "no wasted sideways effort" looks like. $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2}$
Recognise the perfect square. Why this step? $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ . The whole thing under the root is $(v_{2} - v_{1})^{2}$ . $Δ v = (v_{2} - v_{1})^{2} = ∣ v_{2} - v_{1} ∣$
Plug numbers. $Δ v = ∣3.07 - 1.61∣ = 1.46 km/s$

Verify: Units km/s ✅. The answer is exactly the gap between the two speeds — your forecast. Any turn ( $θ > 0$ ) would make $cos θ < 1$ , shrinking the subtracted term and so raising $Δ v$ ; here that penalty is zero.

Cell B — equal speeds, pure tilt

Forecast: Same length arrows, splayed by $3 0^{\circ}$ . The gap between their tips is short-ish but not zero. Will it be more or less than the full $v$ ? Guess.

Figure — Combined maneuvers — optimal split between plane change and velocity change

Figure s01 — Equal-speed pure tilt. The violet arrow is $v_{1}$ and the orange arrow is $v_{2}$ ; both have the same length $v$ , splayed apart by the plane-change angle $θ = 3 0^{\circ}$ . The magenta arrow joining their tips is $Δ v = v_{2} - v_{1}$ . Because the two long sides are equal, the triangle is isosceles, and the magenta side's length works out to exactly $2 v sin (θ /2)$ — the magenta dot marks its midpoint, the "half-base" that the half-angle $θ /2$ controls.

Set $v_{1} = v_{2} = v$ in the master formula. $Δ v = v^{2} + v^{2} - 2 v^{2} cos θ = v 2 - 2 cos θ$ Why this step? Factoring out $v^{2}$ isolates the pure angle penalty $2 - 2 cos θ$ , independent of how fast you go.
Use the half-angle identity $2 - 2 cos θ = 4 sin^{2} (θ /2)$ . Why this step? The identity $1 - cos θ = 2 sin^{2} (θ /2)$ turns the ugly root into a clean sine — and geometrically $sin (θ /2)$ is half the base of the isosceles triangle the two equal arrows make (look at the magenta segment in the figure). $Δ v = v 4 sin^{2} (θ /2) = 2 v sin (θ /2)$
Plug numbers. $θ /2 = 1 5^{\circ}$ , $sin 1 5^{\circ} = 0.2588$ . $Δ v = 2 (7.35) (0.2588) = 3.80 km/s$

Verify: $3.80$ km/s is over half your orbital speed just to tilt $3 0^{\circ}$ — exactly why the parent note screams "turn where slow." Units km/s ✅. Cross-check with master formula: $7.3 5^{2} + 7.3 5^{2} - 2 (7.35)^{2} cos 3 0^{\circ} = 3.80$ ✅.

Cell C — the $18 0^{\circ}$ extreme

Forecast: The arrows point dead against each other. To reverse and speed up, you must cancel your $2.0$ AND build $3.0$ the other way. Guess: the two speeds add.

Substitute $cos 18 0^{\circ} = - 1$ . Why this step? $- 1$ is the smallest cosine can be, so the " $- 2 v_{1} v_{2} cos θ$ " term becomes $+ 2 v_{1} v_{2}$ — the middle term flips from subtracting to adding. This is the worst-case, maximum-cost geometry. $Δ v = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2}$
Recognise the other perfect square. $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ . $Δ v = (v_{1} + v_{2})^{2} = v_{1} + v_{2} = 2.0 + 3.0 = 5.0 km/s$

Verify: Speeds add, matching the forecast. Note Ex 1 ( $θ = 0$ ) gave $∣ v_{2} - v_{1} ∣ = 1.0$ and Ex 3 ( $θ = 18 0^{\circ}$ ) gives $v_{1} + v_{2} = 5.0$ — the two extremes of the master formula, the smallest and largest $Δ v$ possible for these speeds. Every other angle lands between them.

Cell D — the everyday combined burn (acute angle)

Forecast: Somewhere between the pure-speed change ( $2.0$ ) and something bigger from the tilt. Guess a number in km/s.

Combined burn — plug straight into the master formula. $Δ v = 6. 0^{2} + 8. 0^{2} - 2 (6.0) (8.0) cos 4 0^{\circ}$ Why this step? This is the direct third-side length; $cos 4 0^{\circ} = 0.7660$ . $= 36 + 64 - 96 (0.7660) = 100 - 73.54 = 26.46 = 5.14 km/s$
Separate burns — speed change then plane change. Why this step? To measure the penalty of splitting. First a $∣8 - 6∣ = 2.0$ speed change, then a pure tilt at $v_{2} = 8.0$ . The tilt is a pure plane change at speed $8.0$ , so it uses the equal-speed formula $2 v sin (θ /2)$ from Ex 2 — and since the full tilt $θ = 4 0^{\circ}$ is done in this one leg, the half-angle is $θ /2 = 2 0^{\circ}$ . Thus $2 (8.0) sin 2 0^{\circ} = 16 (0.3420) = 5.47$ . $Δ v_{sep} = 2.0 + 5.47 = 7.47 km/s$
Compare. $5.14 < 7.47$ : the single throw saves $2.33$ km/s. Why this step? Because propellant grows exponentially with the Delta-v Budget (see the definition callout above), saving $2.33$ km/s is not a small correction — it can mean a large fraction more payload.

Verify: $5.14 \leq 7.47$ obeys the triangle inequality (the direct side is never longer than the two-leg path). Units km/s ✅. This is the numeric heart of "combine, don't separate."

Cell E — the full optimal split (real mission)

Intuition WHY "equal marginal cost" is the optimum — built from scratch

Imagine you have already chosen some split: $s$ degrees of tilt at the fast burn, $(i - s)$ at the slow burn. Now ask: what happens to the total if I move one tiny extra degree from burn 2 to burn 1? Burn 1's cost goes up by some amount — call it its price per degree, $p_{1}$ . Burn 2's cost goes down by its own price per degree, $p_{2}$ . So the net change in the total is $p_{1} - p_{2}$ per degree moved.

If $p_{1} < p_{2}$ : moving a degree to burn 1 lowers the total — so do it.
If $p_{1} > p_{2}$ : moving a degree back to burn 2 lowers the total — so do that.
You can only stop improving when $p_{1} = p_{2}$ . That equality IS the minimum.

"Price per degree" is exactly what a derivative measures: how fast a quantity changes as you nudge its input. That is why the tool here is $\frac{d Δ v}{d s}$ and not something else — it answers the precise question "how much does cost change per degree of tilt moved?" The parent note computed these derivatives: $\frac{d Δ v _{1}}{d s} = = p_{1} \frac{v _{1, i} v _{1, f} sin s}{Δ v _{1}}, - \frac{d Δ v _{2}}{d s} = = p_{2} \frac{v _{2, i} v _{2, f} sin ( i - s )}{Δ v _{2}}$ Setting the total's derivative to zero, $\frac{d Δ v _{1}}{d s} + \frac{d Δ v _{2}}{d s} = 0$ , is identical to writing $p_{1} = p_{2}$ — the equal-marginal-cost condition. The minus sign on $Δ v_{2}$ is just bookkeeping: raising $s$ lowers burn 2's angle, so its cost falls.

Forecast: Will the tilt split 50/50? Mostly-perigee? Mostly-apogee? Guess before computing.

Figure s02 — Total cost as the split $s$ varies. The horizontal axis is $s$ , the number of degrees of the $28. 5^{\circ}$ tilt assigned to the fast perigee burn (so $28. 5^{\circ} - s$ goes to the slow apogee burn). The violet dashed curve is burn 1's cost $Δ v_{1} (s)$ — it climbs steeply as you load tilt onto the fast burn. The orange dashed curve is burn 2's cost $Δ v_{2} (s)$ — it falls gently as tilt leaves the slow burn. The magenta curve is their sum $Δ v_{t o t} (s)$ ; the navy dot marks its minimum, sitting hard against the left edge at $s \approx 2. 2^{\circ}$ . The picture shows visually why almost all tilt belongs at apogee.

Write each burn as a combined maneuver. Why this step? Each burn is itself a "third side" problem, its own copy of the master formula: $Δ v_{1} (s) = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos s$ $Δ v_{2} (s) = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos (i - s)$
Test "all at apogee" ( $s = 0$ ). Why this step? Apogee has the smallest speeds, so the parent note predicts it should be cheap. $Δ v_{1} (0) = ∣10.15 - 7.73∣ = 2.42, Δ v_{2} (0) = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 28. 5^{\circ} = 1.83$ $Δ v_{tot} (0) = 2.42 + 1.83 = 4.25 km/s$
Test "all at perigee" ( $s = 28. 5^{\circ}$ ). Why this step? To expose the disaster of tilting at high speed. $Δ v_{1} = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos 28. 5^{\circ} = 5.00, Δ v_{2} = ∣3.07 - 1.61∣ = 1.46$ $Δ v_{tot} = 5.00 + 1.46 = 6.46 km/s$
Apply the equal-marginal-cost condition (built in the intuition callout above) to find the true optimum: $\frac{v _{1, i} v _{1, f} s i n s}{Δ v _{1} ( s )} = \frac{v _{2, i} v _{2, f} s i n ( i - s )}{Δ v _{2} ( s )}$ Why this step? At the minimum the two prices per degree are equal. Numerically this gives $s \approx 2. 2^{\circ}$ at perigee, $26. 3^{\circ}$ at apogee, with $Δ v_{tot} \approx 4.24$ km/s.

Verify: All-apogee ( $4.25$ ) already crushes all-perigee ( $6.46$ ) by $2.21$ km/s. The optimal split ( $4.24$ ) shaves only a further $\sim 0.01$ — a whisker. Conclusion: nearly all tilt belongs at apogee; the perigee share is tiny. Units km/s ✅.

Cell F — the two limiting endpoints

Forecast: We already saw all-apogee ( $s = 0$ ) is far cheaper. Guess which endpoint's marginal price is smaller — degrees always migrate toward the cheaper price.

Write the marginal price of each burn. Why this step? From the intuition callout, the price per radian of loading tilt onto a burn is the size of that burn's derivative: $p_{1} (s) = \frac{v _{1, i} v _{1, f} s i n s}{Δ v _{1} ( s )}, p_{2} (s) = \frac{v _{2, i} v _{2, f} s i n ( i - s )}{Δ v _{2} ( s )} (arguments in radians)$
Note the speed products that scale each price. Why this step? The factor out front sets the whole scale. At perigee $v_{1, i} v_{1, f} = 7.73 \times 10.15 = 78.46$ ; at apogee $v_{2, i} v_{2, f} = 1.61 \times 3.07 = 4.943$ . The perigee factor is about $16 \times$ larger — that alone is the reason tilt is expensive at the fast burn.
Evaluate the apogee price at $s = 0$ (so $i - s = 28. 5^{\circ} = 0.4974$ rad). Why this step? This is the price of the last radian of tilt still sitting at apogee, the tilt you'd consider moving to perigee. $p_{2} (0) = \frac{( 1.61 ) ( 3.07 ) s i n ( 0.4974 )}{Δ v _{2} ( 0 )} = \frac{4.943 \times 0.4772}{1.83} = \frac{2.359}{1.83} = 1.289 km/s per radian$
Find where the perigee price catches up — showing the root-finding. At $s = 0$ , $sin s = 0$ , so $p_{1} (0) = 0$ : the very first sliver of tilt at perigee is nearly free. But $p_{1}$ climbs fast because of the $78.46$ factor. We want the $s$ (in radians) where $p_{1} (s) = p_{2}$ has risen to match $1.289$ . Because moving tilt to perigee also slightly changes $p_{2}$ , the exact balance is where $p_{1} (s) = p_{2} (s)$ ; solving numerically by trying values:

$s$ (deg) $s$ (rad) $Δ v_{1} (s)$ $p_{1} (s) = \frac{78.46 sin s}{Δ v _{1}}$

$1. 0^{\circ}$ $0.0175$ $2.421$ $0.566$

$2. 0^{\circ}$ $0.0349$ $2.427$ $1.128$

$2. 2^{\circ}$ $0.0384$ $2.429$ $1.240$

$2. 4^{\circ}$ $0.0419$ $2.431$ $1.352$

Why this step? The table brackets the answer: $p_{1}$ crosses the apogee price ( $\approx 1.28$ – $1.29$ , which itself dips slightly as its own tilt shrinks) between $2. 2^{\circ}$ and $2. 4^{\circ}$ . The full equal-price solution (accounting for $p_{2}$ also moving) lands at $s \approx 2. 2^{\circ}$ — matching the numeric minimization of Ex 5.

$s$ (deg)	$s$ (rad)	$Δ v_{1} (s)$	$p_{1} (s) = \frac{78.46 sin s}{Δ v _{1}}$
$1. 0^{\circ}$	$0.0175$	$2.421$	$0.566$
$2. 0^{\circ}$	$0.0349$	$2.427$	$1.128$
$2. 2^{\circ}$	$0.0384$	$2.429$	$1.240$
$2. 4^{\circ}$	$0.0419$	$2.431$	$1.352$

Verify: The huge speed product at perigee ( $78.46$ vs $4.94$ , a $16 \times$ ratio) means even a couple of degrees there costs as much as many degrees at apogee — so the optimum sits right against the $s = 0$ end, at $s \approx 2. 2^{\circ}$ . The table's crossing between $2. 2^{\circ}$ and $2. 4^{\circ}$ confirms Ex 5's split ✅.

Cell G — combined vs separate, as a word problem

Forecast: The parent note's triangle-inequality argument says combined always wins. Guess the two numbers.

Separate plan. Speed change $∣3.07 - 1.61∣ = 1.46$ , then a pure tilt at $v_{2} = 3.07$ using the equal-speed formula with the full $θ = 28. 5^{\circ}$ (half-angle $14.2 5^{\circ}$ ): $2 (3.07) sin 14.2 5^{\circ} = 6.14 (0.2462) = 1.512$ . $Δ v_{sep} = 1.46 + 1.512 = 2.972 km/s$ Why this step? This is the intern's total — two clean legs.
Combined plan. One angled throw: $Δ v = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 28. 5^{\circ} = 1.83 km/s$ Why this step? The direct third side of the velocity triangle.
Difference. $2.972 - 1.83 = 1.14$ km/s saved by combining. Why this step? By the exponential rocket-equation logic in the Delta-v Budget callout, over a third of the intern's budget vanishes — a large propellant saving.

Verify: $1.83 \leq 2.972$ — triangle inequality holds ✅. Combining saves $1.14$ km/s. Units km/s ✅. (This is exactly the all-apogee burn from Ex 5's step 2, seen from the "should I combine?" angle.)

Cell H — exam twist: obtuse angle, unequal speeds

Forecast: Since $12 0^{\circ} > 9 0^{\circ}$ , the arrows lean away from each other. Guess: bigger than the acute cases — the middle term now adds.

Evaluate $cos 12 0^{\circ} = - \frac{1}{2}$ . Why this step? For obtuse angles the cosine is negative, so $- 2 v_{1} v_{2} cos θ$ becomes $+ 2 v_{1} v_{2} \cdot \frac{1}{2} = + v_{1} v_{2}$ . The classic exam trap is to forget this sign flip and subtract. $Δ v = 4. 0^{2} + 5. 0^{2} - 2 (4.0) (5.0) (- 0.5)$
Simplify. $= 16 + 25 + 20 = 61 = 7.81 km/s$

Verify: $Δ v = 7.81$ lies between the $θ = 0$ value $∣5 - 4∣ = 1$ and the $θ = 18 0^{\circ}$ value $5 + 4 = 9$ — as every combined-maneuver answer must. If a student wrongly used $+ 2 v_{1} v_{2} cos θ$ they'd get $21 = 4.58$ , an impossibly small answer for an obtuse turn — the sanity band exposes the error. Units km/s ✅.

Recall Quick self-test on the matrix

Which cell has the largest possible $Δ v$ for fixed speeds? ::: Cell C, $θ = 18 0^{\circ}$ , giving $v_{1} + v_{2}$ . Which cell gives the smallest? ::: Cell A, $θ = 0$ , giving $∣ v_{2} - v_{1} ∣$ . In Cell E, where does almost all the plane change go? ::: To apogee (the slow burn); only ~2° stays at perigee. Why is the optimum at "equal marginal cost"? ::: If one burn's price per degree were cheaper, you'd move a degree there and lower the total; you can only stop when the two prices match. In Cell H, what is the trap? ::: Forgetting $cos θ < 0$ for obtuse $θ$ , which makes the middle term add.