Visual walkthrough — Combined maneuvers — optimal split between plane change and velocity change

We will assume you know only three things: what an arrow ("vector") is, that speed is "how fast" and direction is "which way," and the idea of a right-angle triangle. Everything else we build.

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Step 1 — Draw the two velocities as arrows

WHAT. A spacecraft has, at one instant, a velocity: an arrow whose length is the speed and whose direction is the way it is moving. Before the burn the arrow is ${\vec{v}}_1$; after the burn it is ${\vec{v}}_2$. We draw them tail-to-tail (both starting at the same dot).

WHY tail-to-tail. Because we care about the difference in direction between them. The little wedge of angle where the two tails meet — call it $\theta$ — is exactly the plane-change angle, the amount the direction of motion swings. If we drew them tip-to-tail we'd hide that wedge; tail-to-tail puts it in plain view.

PICTURE. Blue arrow ${\vec{v}}_1$, orange arrow ${\vec{v}}_2$, the green wedge $\theta$ between them.

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Step 2 — The burn is the arrow that closes the gap

WHAT. The engine's job is to convert ${\vec{v}}_1$ into ${\vec{v}}_2$. The push it must deliver is whatever arrow, added to ${\vec{v}}_1$, lands you exactly on ${\vec{v}}_2$. That arrow is the third side of the triangle, drawn from the tip of ${\vec{v}}_1$ to the tip of ${\vec{v}}_2$.

WHY subtraction. "What must I add to ${\vec{v}}_1$ to get ${\vec{v}}_2$?" is literally ${\vec{v}}_2-{\vec{v}}_1$. Written out:

$\underset{\text{the burn}}{\underbrace{\overrightarrow{\Delta v}}}=\underset{\text{where we want to be}}{\underbrace{{\vec{v}}_2}}-\underset{\text{where we started}}{\underbrace{{\vec{v}}_1}}$

Every symbol earned: $\overrightarrow{\Delta v}$ is the arrow the rocket must supply; its length $\Delta v$ (no arrow) is the fuel-hungry number we want to minimize.

PICTURE. The red arrow $\overrightarrow{\Delta v}$ bridging the two tips, closing the triangle.

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Step 3 — Length² of an arrow is the arrow dotted with itself

WHAT. We need the length of $\overrightarrow{\Delta v}$, but arrows don't have a length button — we get length through the dot product, a machine that eats two arrows and spits out a number.

WHY the dot product. For any arrow $\vec{a}$, the dot product with itself, $\vec{a}\cdot \vec{a}$, equals $|\vec{a}{|}^2$ — its length squared. This is the only clean way to turn "arrow" into "a plain number we can do algebra on." So we compute $\Delta v^2$ first, then square-root at the end.

$\Delta v^2=\overrightarrow{\Delta v}\cdot \overrightarrow{\Delta v}=({\vec{v}}_2-{\vec{v}}_1)\cdot ({\vec{v}}_2-{\vec{v}}_1)$

WHY we may expand it like ordinary algebra. The dot product obeys two rules that look exactly like the rules of multiplication: it distributes over addition ($\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\cdot \vec{b}+\vec{a}\cdot \vec{c}$) and it is symmetric ($\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$). These two facts — together called linearity — are what let us FOIL the bracket. Distributivity is not magic: geometrically, the dot product $\vec{a}\cdot \vec{b}$ is "length of $\vec{a}$ times the shadow (projection) of $\vec{b}$ onto $\vec{a}$," and shadows of a sum add up exactly like the vectors do (line up two arrows head-to-tail and their combined shadow is just the two shadows placed end to end). So:

$({\vec{v}}_2-{\vec{v}}_1)\cdot ({\vec{v}}_2-{\vec{v}}_1)={\vec{v}}_2\cdot {\vec{v}}_2-{\vec{v}}_2\cdot {\vec{v}}_1-{\vec{v}}_1\cdot {\vec{v}}_2+{\vec{v}}_1\cdot {\vec{v}}_1$

The two middle terms are equal (symmetry), so they merge into $-2{\vec{v}}_1\cdot {\vec{v}}_2$:

$\Delta v^2=\underset{=v_2^2}{\underbrace{{\vec{v}}_2\cdot {\vec{v}}_2}}-\underset{\text{the cross term}}{\underbrace{2{\vec{v}}_1\cdot {\vec{v}}_2}}+\underset{=v_1^2}{\underbrace{{\vec{v}}_1\cdot {\vec{v}}_1}}$

PICTURE. The projection (shadow) picture that justifies distributivity, plus the reminder $\vec{a}\cdot \vec{a}=|\vec{a}{|}^2$.

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Step 4 — The cross term hides the angle

WHAT. The only mysterious piece left is ${\vec{v}}_1\cdot {\vec{v}}_2$. By the definition above, it equals $v_1{v}_2\cos \theta$. Substitute:

$\Delta v^2=v_1^2+v_2^2-2v_1{v}_2\cos \theta$

WHY $\cos \theta$ and not $\sin \theta$. Cosine measures alignment. When $\theta =0$ the two arrows point the same way, $\cos \theta =1$, and the cross term is at its biggest — the arrows help each other. When $\theta =90^{\circ }$, $\cos \theta =0$, they share nothing and the cross term vanishes. Cosine is exactly the "how much do these two agree" dial, which is precisely what a difference of arrows needs to know.

Taking the square root (length is never negative, so we keep the $+$ root):

This is the Law of Cosines wearing a spacesuit — the same identity for the third side of any triangle.

PICTURE. The finished triangle with all three sides and the angle labelled, and the boxed formula mapped onto the sides.

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Step 5 — Edge case: no turn ($\theta =0$)

WHAT. Set $\theta =0$. Then $\cos 0=1$, and

$\Delta v=\sqrt{v_1^2+v_2^2-2v_1{v}_2}=\sqrt{(v_1-v_2{)}^2}=|v_2-v_1|.$

WHY it must be this. With no direction change, both arrows lie on the same line; the triangle collapses to a straight segment. The "difference" is just how much longer one is than the other — a pure speed change. Our formula reproduces this exactly. ✅

PICTURE. The triangle flattened onto one line; $\Delta v$ is the gap between the two tips.

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Step 6 — Edge case: the right-angle turn ($\theta =90^{\circ }$)

WHAT. Set $\theta =90^{\circ }$. Then $\cos 90^{\circ }=0$, the whole cross term vanishes, and

$\Delta v=\sqrt{v_1^2+v_2^2}.$

WHY it must be this. With a perpendicular turn, the two velocity arrows share nothing in common direction — one runs purely "east," the other purely "north." The triangle they close is a plain right triangle, so $\Delta v$ is just its hypotenuse: the familiar $\sqrt{v_1^2+v_2^2}$ (Pythagoras). This is the exact moment where $\cos \theta =0$ removes the correction term and the law of cosines degenerates back into Pythagoras — proof that the cosine term is precisely the "how aligned" adjustment and disappears when there is no alignment to speak of.

PICTURE. The right-angle triangle with legs $v_1$, $v_2$ and hypotenuse $\Delta v$.

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Step 7 — Edge case: pure turn, equal speeds ($v_1=v_2=v$)

WHAT. Set $v_1=v_2=v$ (tilt the orbit but keep the same speed):

$\Delta v=\sqrt{v^2+v^2-2v^2\cos \theta }=v\sqrt{2-2\cos \theta }.$

WHY the half-angle identity — derived, not quoted. We want to simplify $2-2\cos \theta$. Look at the isosceles triangle: two equal sides $v$ with the wedge $\theta$ between them. Drop a line from the apex straight down the middle. It does two things at once: it bisects the wedge into two angles of $\theta /2$, and it cuts the base $\Delta v$ exactly in half (isosceles symmetry). In the little right triangle that results, the side opposite the $\theta /2$ angle is half the base, and the hypotenuse is $v$. By the definition of sine ("opposite over hypotenuse"):

$\sin \left(\frac{\theta }{2}\right)=\frac{\frac{1}{2}\Delta v}{v}⟹\Delta v=2v\sin \left(\frac{\theta }{2}\right).$

That geometric fact is the half-angle identity in disguise: squaring both sides, $\Delta v^2=4v^2{\sin }^2(\theta /2)$, and comparing with $\Delta v^2=v^2(2-2\cos \theta )$ forces $2-2\cos \theta =4{\sin }^2(\theta /2)$, i.e. $1-\cos \theta =2{\sin }^2(\theta /2)$. We derived the identity from the picture rather than memorizing it.

THE STING. At $\theta =60^{\circ }$, $\sin (30^{\circ })=\frac{1}{2}$, so $\Delta v=2v\cdot \frac{1}{2}=v$ — you must spend your entire orbital speed again just to turn $60^{\circ }$. This is why the parent note screams "turn where slow": $\Delta v$ scales straight with $v$.

PICTURE. Isosceles triangle, the dropped bisector, the two right triangles with $\theta /2$.

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Step 8 — Degenerate case: complete reversal ($\theta =180^{\circ }$)

WHAT. Point the after-arrow exactly backwards, $\theta =180^{\circ }$, $\cos 180^{\circ }=-1$:

$\Delta v=\sqrt{v_1^2+v_2^2+2v_1{v}_2}=\sqrt{(v_1+v_2{)}^2}=v_1+v_2.$

WHY it must be this. The two arrows now point in opposite directions. To flip from moving one way at $v_1$ to moving the other way at $v_2$, you must first kill all of $v_1$, then build all of $v_2$ — the costs simply add. The minus sign in the formula became a plus because $\cos$ flipped sign, and the triangle again collapses to a line, but this time the tips are on opposite sides of the shared tail. This is the worst possible turn.

PICTURE. Both arrows on one line pointing opposite ways; $\Delta v$ spans tip to tip = $v_1+v_2$.

Recall Why every sign is now covered

$\cos \theta$ runs smoothly from $+1$ (aligned, cheapest) through $0$ (perpendicular, Pythagoras) to $-1$ (reversed, most expensive). Because we let $\theta$ take any value $0^{\circ }$–$180^{\circ }$, and length is always the positive root, there is no input the boxed formula can't handle. Zero turn, right-angle turn, half turn, full reversal — all fall out of the one triangle.

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Step 9 — From one burn to the optimal split

WHAT — the setup. A real transfer (see Hohmann Transfer Orbit) has two burns, and each burn is its own triangle exactly like Step 4. To avoid clashing with the "initial" subscript, we write the total plane-change angle as $\Phi$ (capital phi). First we name everything:

Applying the boxed formula to each burn:

$\Delta v_1(s)=\sqrt{v_{1,\text{in}}^2+v_{1,\text{out}}^2-2v_{1,\text{in}}{v}_{1,\text{out}}\cos s},\Delta v_2(s)=\sqrt{v_{2,\text{in}}^2+v_{2,\text{out}}^2-2v_{2,\text{in}}{v}_{2,\text{out}}\cos (\Phi -s)}$

The total fuel is the sum of the two red-arrow lengths:

$\Delta v_{\text{tot}}(s)=\Delta v_1(s)+\Delta v_2(s).$

WHY a derivative. To find the cheapest split we ask: at the best $s$, nudging it either way must not help. That is exactly the condition "slope $=0$," i.e. $\frac{d\Delta v_{\text{tot}}}{ds}=0$.

WHY the derivative looks like $\frac{v_{\text{in}}{v}_{\text{out}}\sin \theta }{\Delta v}$ — shown, not asserted. Take $\Delta v_1(s)=\sqrt{g(s)}$ where the inside is $g(s)=v_{1,\text{in}}^2+v_{1,\text{out}}^2-2v_{1,\text{in}}{v}_{1,\text{out}}\cos s$. The chain rule for a square root says $\frac{d}{ds}\sqrt{g}=\frac{g^{\prime }(s)}{2\sqrt{g}}$. Only the $\cos s$ carries $s$, and $\frac{d}{ds}(-2v_{1,\text{in}}{v}_{1,\text{out}}\cos s)=+2v_{1,\text{in}}{v}_{1,\text{out}}\sin s$ (since $\frac{d}{ds}\cos s=-\sin s$, the two minus signs cancel). So:

$\frac{d\Delta v_1}{ds}=\frac{2v_{1,\text{in}}{v}_{1,\text{out}}\sin s}{2\sqrt{g(s)}}=\frac{v_{1,\text{in}}{v}_{1,\text{out}}\sin s}{\Delta v_1}.$

For burn 2 the inside angle is $(\Phi -s)$, and $\frac{d}{ds}\cos (\Phi -s)=+\sin (\Phi -s)$ (chain rule brings a $-1$ from the inner $-s$, flipping the sign again), which lands with an overall minus:

$\frac{d\Delta v_2}{ds}=-\frac{v_{2,\text{in}}{v}_{2,\text{out}}\sin (\Phi -s)}{\Delta v_2}.$

Adding them and setting the total slope to zero gives the optimal-split condition:

$\boxed{\frac{v_{1,\text{in}}{v}_{1,\text{out}}\sin s}{\Delta v_1(s)}=\frac{v_{2,\text{in}}{v}_{2,\text{out}}\sin (\Phi -s)}{\Delta v_2(s)}}$

HOW you actually solve it. Notice $\Delta v_1$ and $\Delta v_2$ each contain $s$ inside a square root, so this equation cannot be untangled into a clean "$s=\dots$" by algebra — it is transcendental. You solve it numerically:

1. Pick the four speeds from the Vis-viva Equation and the total tilt $\Phi$.
2. Sweep $s$ from $0$ to $\Phi$ in small steps, computing $\Delta v_{\text{tot}}(s)$ each time (or, equivalently, checking which $s$ makes the two sides of the boxed condition equal).
3. The $s$ giving the smallest total is the optimal split. A quick shortcut: because burn 2 sits at the big, slow orbit its $v\cdot v$ product is tiny, so its marginal cost is tiny — the balance point lands with almost all the tilt at burn 2 (apoapsis) and only a whisker at burn 1.

PICTURE. The $U$-shaped total-cost curve $\Delta v_{\text{tot}}(s)$ with its minimum marked far toward "all at apoapsis."

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The one-picture summary

Everything above compressed: two velocity arrows tail-to-tail, the red closing arrow, the law-of-cosines label, and the four cases ($\theta =0,90^{\circ },60^{\circ },180^{\circ }$) marching along the bottom to show cost climbing from $|v_2-v_1|$ to $v_1+v_2$.

Recall Feynman retelling — the whole walkthrough in plain words

Draw your speed as an arrow before the burn (blue) and after the burn (orange), both starting from the same spot. The little wedge between them is how much you're turning. The push your rocket must give is the arrow (red) that reaches from the tip of blue to the tip of orange — and only its length costs fuel. To get that length we use the triangle rule (law of cosines): $\Delta v^2=v_1^2+v_2^2-2v_1{v}_2\cos \theta$. The $\cos \theta$ is a "how aligned are the arrows" dial: same way ($\theta =0$) means the cost is just the speed difference; square-on ($\theta =90^{\circ }$) gives plain Pythagoras; opposite ways ($\theta =180^{\circ }$) means the costs just add. Equal-speed turns simplify to $2v\sin (\theta /2)$, which says a $60^{\circ }$ turn costs your whole speed again — that's why turning is a nightmare when you're fast. With two burns you slide the split $s$ until the marginal cost is equal at both (a curve you minimize by sweeping, since algebra can't crack it), and since the far burn is slow, nearly all the turning happens out there. One picture, one triangle, one rule — that's the entire chapter.