Plane change maneuvers — Δv = 2v·sin(Δi - 2): Visual walkthrough

We are answering one question: if a spacecraft keeps the same speed but points in a slightly different direction, how big a "kick" does that turn cost?

Step 1 — Draw the velocity as an arrow

WHAT: We draw the spacecraft's motion as one arrow $v_{1}$ starting at a point.

WHY: Everything about a plane change is about direction, and the only honest way to talk about direction is to draw the arrow. A single number (speed) throws the direction away — and direction is the whole story.

PICTURE: Look at the cyan arrow. Its length is the speed $v$ (grab this from the Orbital velocity — vis-viva equation). Its tilt on the page is the way the craft is flying right now.

Figure — Plane change maneuvers — Δv = 2v·sin(Δi - 2)

Step 2 — The turn we want: rotate the arrow by $Δ i$

WHAT: We draw a second arrow $v_{2}$ — same starting point, same length, but swung round by the angle $Δ i$ .

WHY the same length? A pure plane change does not speed up or slow down the craft; it only re-aims it. So the orbit's size and shape are untouched, which means

$∣ v_{1} ∣ = ∣ v_{2} ∣ = v$

Here $∣ v_{1} ∣$ means "the length of arrow $v_{1}$ ." Both arrows are exactly $v$ long — only their directions differ by $== Δ i ==$ .

PICTURE: Two arrows fan out from one point like the two hands of a clock, separated by the amber angle $Δ i$ . Same length, different aim.

Step 3 — The kick is the arrow that closes the gap

WHAT: We draw the amber arrow $Δ v$ from the tip of $v_{1}$ to the tip of $v_{2}$ .

WHY: The number that costs fuel is the length of this gap arrow, $Δ v = ∣Δ v ∣$ . That length is what we are hunting for. (See Vector addition and law of cosines for the tip-to-tip subtraction rule.)

PICTURE: Three arrows now form a closed triangle: two equal cyan sides ( $v_{1}$ , $v_{2}$ ) and the amber gap ( $Δ v$ ) joining their tips.

Step 4 — Name the triangle: two equal sides, one known angle

WHAT: We label the triangle: sides $v$ , $v$ , included angle $Δ i$ , and unknown third side $Δ v$ .

WHY this tool and not another? To find a triangle's third side from two sides and the angle between them, the right instrument is the law of cosines. Plain trigonometry ( $sin$ / $cos$ of one angle) only works on right-angled triangles; our apex angle $Δ i$ can be anything from $0°$ to $180°$ , so we need the general law.

$Δ v^{2} = v^{2} + v^{2} - 2 v v cos (Δ i)$

Reading it term by term:

$v^{2} + v^{2}$ — the two equal cyan sides, each squared;
$- 2 v v cos (Δ i)$ — the "correction" that bends a straight sum into a real triangle; it shrinks the result when the angle is small and grows it as the angle opens up;
$Δ v^{2}$ — the squared length of the amber gap, our target.

Tidy the first two terms ( $v^{2} + v^{2} = 2 v^{2}$ ) and factor:

$Δ v^{2} = 2 v^{2} (1 - cos Δ i)$

PICTURE: The same triangle, now fully labelled, with the amber side flagged as the unknown.

Step 5 — Turn the messy $1 - cos$ into a clean square

WHAT: We replace $1 - cos Δ i$ with $2 sin^{2} (Δ i /2)$ :

$Δ v^{2} = 2 v^{2} \cdot 2 sin^{2} (\frac{Δ i}{2}) = 4 v^{2} sin^{2} (\frac{Δ i}{2})$

WHY: We want to take a square root at the end. $Δ v^{2} = 2 v^{2} (1 - cos Δ i)$ has no obvious root. But $4 v^{2} sin^{2} (Δ i /2)$ is a textbook perfect square: $(2 v sin (Δ i /2))^{2}$ . The identity is the key that unlocks the root.

Term by term:

$4 v^{2}$ — comes from $2 \times 2 v^{2}$ ;
$sin^{2} (Δ i /2)$ — the half-angle $Δ i /2$ , squared. The half sneaks in from the identity, not from any guess.

PICTURE: The figure proves the identity geometrically: drop a perpendicular in the isosceles triangle, splitting $Δ i$ into two halves, and the two half-triangles reveal the $sin (Δ i /2)$ pieces directly.

Step 6 — Take the square root

WHAT: Both sides are lengths (never negative), so we take the positive square root:

$Δ v = 2 v sin (\frac{Δ i}{2})$

WHY the half-angle survives: the perpendicular we dropped in Step 5 bisected the apex angle and the base. Each half of the base is $v sin (Δ i /2)$ , and the full amber gap is two of those halves stacked — literally $2 \times v sin (Δ i /2)$ . The geometry hands you the "2" and the "half" for free.

Reading the final formula:

$2$ — two half-bases of the isosceles triangle;
$v$ — the (unchanged) orbital speed you plug in;
$sin (Δ i /2)$ — how much of that speed the sideways turn eats, set by half the turn angle.

Step 7 — Every case: small, medium, and the $180°$ flip

$Δ i \to 0°$ (degenerate, no turn). The two cyan arrows lie on top of each other; the amber gap shrinks to a point. Formula: $Δ v = 2 v sin (0) = 0$ . Free, as it should be — you didn't turn.
$Δ i = 60°$ . $Δ v = 2 v sin 30° = 2 v (0.5) = v$ . The turn costs a whole extra orbital speed. Notice: the amber gap becomes exactly as long as one cyan side (an equilateral triangle!).
$Δ i = 90°$ . $Δ v = 2 v sin 45° = 2 v (0.707) \approx 1.414 v = v 2$ . Makes sense: a right-angle turn between two equal-length arrows gives a diagonal of length $v 2$ .
$Δ i = 180°$ (full reversal, the extreme case). The arrows point exactly opposite. $Δ v = 2 v sin 90° = 2 v$ . You must first cancel all your speed $v$ , then build speed $v$ the other way — total $2 v$ . The formula nails it.

WHY show all of these: so you never meet a $Δ i$ the page didn't cover. The cost rises smoothly from $0$ at $0°$ up to $2 v$ at $180°$ , and it's already as much as your whole speed by just $60°$ .

PICTURE: Four snapshots of the same triangle for $Δ i = 0°, 60°, 90°, 180°$ , with the amber gap growing across the row.

Step 8 — When the two speeds aren't equal (combined burn)

WHAT: Same law of cosines, but with unequal sides:

$Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos Δ i$

$v_{1}$ — speed before; $v_{2}$ — speed after; they differ now;
$cos Δ i$ — still set by the turn angle between them;
when $v_{1} = v_{2} = v$ this collapses straight back to $2 v sin (Δ i /2)$ (try it — Step 5's identity reappears).

WHY it saves fuel: by the triangle inequality, one direct arrow is shorter than two arrows walked back-to-back. Doing the turn and the speed change together is one side of a triangle; doing them separately is two sides — always longer. And since $Δ v \propto v$ , you want to do it where $v$ is smallest: at apoapsis. Every extra bit of $Δ v$ costs exponentially more fuel through the Tsiolkovsky rocket equation.

PICTURE: Two triangles side by side — a stubby "combined" triangle vs a longer two-leg "separate" path — showing the combined route is shorter.

The one-picture summary

Everything above, on one sheet: the isosceles triangle, the bisector that manufactures the half-angle, the labelled sides, and the boxed result.

Recall Feynman: retell the whole walkthrough in plain words

Draw your speed as an arrow. To tilt your orbit, you leave the arrow the same length but swing it sideways by an angle $Δ i$ — imagine the two hands of a clock, both the same length, spread apart by that angle. The push your engine must give is the little arrow that reaches from one clock-hand's tip to the other's. Its length is the fuel cost. Because both hands are equal length, they make a two-equal-sides triangle, so I use the law of cosines to get that gap: $Δ v^{2} = 2 v^{2} (1 - cos Δ i)$ . Then a trig trick, $1 - cos = 2 sin^{2} (half)$ , turns the mess into a perfect square, and the square root pops out cleanly: $Δ v = 2 v sin (Δ i /2)$ . The "half" is real — it's the line I drop straight down the middle of the triangle, splitting the angle in two. Check the ends: no turn costs nothing, a $60°$ turn costs your whole speed, and a full flip costs twice your speed. That's why engineers turn their orbit only where they're crawling slowest.

Active recall

Recall Rebuild each figure from memory

Draw the two equal arrows and the gap ::: two cyan arrows length $v$ from one point, amber gap $Δ v$ tip-to-tip, apex angle $Δ i$ Which law finds the gap and why ::: law of cosines — we know two sides and the angle between them Where does the half-angle come from ::: the perpendicular bisector of the isosceles triangle splits $Δ i$ into two halves $Δ v$ at $Δ i = 60°$ ::: exactly $v$ (equilateral triangle) $Δ v$ at $Δ i = 180°$ ::: $2 v$ (kill $v$ , rebuild $v$ reversed) Formula when the two speeds differ ::: $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos Δ i$

Connections

3.2.22 Plane change maneuvers — Δv = 2v·sin(Δi - 2) (Hinglish) — parent topic
Orbital velocity — vis-viva equation — supplies the $v$ you plug in
Inclination and orbital elements — defines $i$ and $Δ i$
Vector addition and law of cosines — the math engine of the triangle
Hohmann transfer orbit — where the combined burn (Step 8) lives
Apoapsis and periapsis — the slow-point strategy
Tsiolkovsky rocket equation — turns $Δ v$ into propellant mass

Visual walkthrough — Plane change maneuvers — Δv = 2v·sin(Δi - 2)