Plane change maneuvers — Δv = 2v·sin(Δi - 2): Worked examples

Intuition What this page is

The parent note built the formula $Δ v = 2 v sin (Δ i /2)$ from an isosceles velocity triangle. Here we stress-test it against every case the universe can throw at you: tiny angles, giant angles, the exact $180°$ reversal, the degenerate $0°$ "no-op", different-speed combined burns, a slow-point strategy, and a full exam twist. Every number is machine-checked at the bottom.

Before any symbol: $v$ is your speed (how fast, a plain number in km/s). $v$ (arrow on top) is your velocity — speed plus the direction you point. $Δ i$ is the angle you rotate that direction by. $Δ v$ is the length of the shove (also km/s) needed to do the rotation.

How the figures are named. Each figure filename ends in -s01, -s02, and so on — these are just figure numbers in reading order (s01 = "step figure 1", s02 = "step figure 2", …). When the text says "look at figure s03", scroll to the third embedded picture. Nothing cryptic: s0N = the $N$ -th figure on this page.

The scenario matrix

Every plane-change problem lives in one of these boxes (this is a table, not a figure). The examples below are labelled with the box they hit — together they fill the whole grid.

Cell	Case class	What's special	Example
A	Degenerate: $Δ i = 0°$	no turn, shove must be zero	Ex 1
B	Small angle $Δ i \leq 5°$	$sin$ almost linear; nudge	Ex 2
C	Mid angle ( $30°$ – $60°$ )	the "expensive turn" regime	Ex 3
D	Exactly $Δ i = 60°$	the famous $Δ v = v$	Ex 4
E	Boundary $Δ i = 90°$	$Δ v / v = 2$ exactly	Ex 5
F	Large angle ( $90°$ – $180°$ )	costs more than the orbit speed	Ex 6a
G	Limiting: $Δ i = 180°$	full reversal, $Δ v = 2 v$	Ex 6b
H	Slow-point strategy	pick where to burn (min $v$ )	Ex 7
I	Combined burn, $v_{1} \neq = v_{2}$	general law of cosines	Ex 8
J	Real-world word problem	fuel mass via Tsiolkovsky	Ex 9
K	Exam twist: solve for $Δ i$	invert the formula (arcsin)	Ex 10

Prerequisites we lean on: Vector addition and law of cosines (the triangle), Orbital velocity — vis-viva equation (gives $v$ ), Apoapsis and periapsis (the slow point), Tsiolkovsky rocket equation (turns $Δ v$ into fuel). Parent: Plane change maneuvers ($\Delta v = 2v\sin(\Delta i/2)$).

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

Figure s01 (the first embedded picture above) plots $Δ v / v = 2 sin (Δ i /2)$ against $Δ i$ from $0°$ to $180°$ . Read it like a menu: every worked example below is a labelled dot on this curve (the letters B, C, D, E, F, G match the cell labels). Notice it starts at zero (Ex 1), is nearly a straight line for small angles (the dashed blue line, Ex 2), passes through $1$ at $60°$ (Ex 4), crosses $2$ at $90°$ (Ex 5), and climbs to $2$ at $180°$ (Ex 6b). Come back to figure s01 after each example and find its dot.

Cell A — the degenerate no-turn

Worked example Ex 1 ·

$Δ i = 0°$ (dot at the origin of figure s01) A satellite is already in the plane it wants. Someone still runs the formula with $Δ i = 0°$ , $v = 7.7$ km/s. What $Δ v$ ?

Forecast: should it cost anything to not turn? Guess before reading — and find where $Δ i = 0°$ sits on figure s01 (the far-left edge).

Figure s02 shows the velocity triangle for four turn angles. In the leftmost, $v_{1}$ and $v_{2}$ lie on top of each other — there is no triangle, so its third side (the pink $Δ v$ ) has length zero.

Plug in: $Δ v = 2 \cdot 7.7 \cdot sin (0°/2) = 2 \cdot 7.7 \cdot sin (0°)$ . Why this step? Direct substitution to check the formula behaves at its floor.
$sin (0°) = 0$ , so $Δ v = 15.4 \cdot 0 = 0$ km/s. Why this step? No rotation means $v_{2} = v_{1}$ , so the triangle collapses to a single arrow and its "third side" has length zero.

Verify: the velocity vector doesn't move, so $Δ v = v_{2} - v_{1} = 0$ . A zero shove for a zero turn — the formula passes its sanity floor. ✔

Cell B — the small-angle nudge

Worked example Ex 2 ·

$Δ i = 2°$ , GPS orbit (dot B on figure s01) $v = 3.9$ km/s. Change inclination by a mere $2°$ . What is $Δ v$ ? Is $2°$ "negligible"?

Forecast: $2°$ sounds trivial — guess whether $Δ v$ is under $50$ m/s, then check how low dot B sits on figure s01.

Half-angle first: $Δ i /2 = 1°$ . In radians $1° = π /180 \approx 0.01745$ rad. Why this step? $sin$ of a small angle in radians is almost equal to the angle itself ( $sin θ \approx θ$ ), which is why the curve looks straight near the origin — see the dashed blue line on figure s01.
$sin (1°) \approx 0.017452$ . Why this step? We need the actual sine, but the small-angle picture tells us to expect roughly $0.01745$ .
$Δ v = 2 \cdot 3.9 \cdot 0.017452 = 7.8 \cdot 0.017452 \approx 0.1361$ km/s $= 136.1$ m/s. Why this step? Multiply out; even a $2°$ tilt costs $136$ m/s — not negligible at all.

Verify: the small-angle approximation predicts $Δ v \approx 2 v \cdot (Δ i /2 in rad) = v \cdot Δ i_{rad} = 3.9 \cdot 0.034907 = 0.1361$ km/s. Matches to the digit. Units: (km/s) $\times$ (dimensionless) $=$ km/s. ✔

Cell C — the mid-angle expensive turn

Worked example Ex 3 ·

$Δ i = 45°$ , LEO (dot C on figure s01) $v = 7.7$ km/s. Rotate the plane by $45°$ . What fraction of the orbital speed does this cost?

Forecast: guess the ratio $Δ v / v$ — bigger or smaller than $0.5$ ? Then eyeball dot C's height on figure s01.

Figure s03 draws the isosceles velocity triangle: two equal sides of length $v$ with the apex angle $Δ i = 45°$ between them, and the pink base $Δ v$ closing it. The dashed line is the perpendicular bisector that splits the apex into two $22.5°$ halves — this is where the half-angle comes from.

Half-angle: $45°/2 = 22.5°$ . Why this step? The isosceles-triangle geometry always uses the half angle (perpendicular-bisector split of the apex, dashed line in figure s03).
$sin (22.5°) \approx 0.38268$ . Why this step? This is the ratio (half-base)/(side) in the velocity triangle.
$Δ v = 2 \cdot 7.7 \cdot 0.38268 \approx 5.893$ km/s. Ratio $Δ v / v = 2 sin (22.5°) \approx 0.7654$ . Why this step? Multiply; $45°$ already costs 76.5% of the whole orbital speed.

Verify: on figure s01, $Δ i = 45°$ sits well above the halfway height — matches $0.765$ . Also $Δ v < 2 v = 15.4$ (never exceeds the reversal cost). ✔

Cell D — the famous 60°

Worked example Ex 4 ·

$Δ i = 60°$ , LEO (dot D on figure s01) $v = 7.7$ km/s. The textbook "shock" case. Show $Δ v = v$ exactly.

Forecast: the parent claims $Δ v = v$ exactly — that means dot D lands right on the $Δ v / v = 1$ line of figure s01. Believe it before checking.

Half-angle: $60°/2 = 30°$ . Why this step? Half of the rotation, as always.
$sin (30°) = \frac{1}{2}$ exactly (a $30$ – $60$ – $90$ triangle). Why this step? This is the one special value that makes the algebra collapse.
$Δ v = 2 v \cdot \frac{1}{2} = v = 7.7$ km/s. Why this step? The $2$ and the $\frac{1}{2}$ cancel — turning $60°$ costs a whole extra orbit's worth of speed.

Verify: in Ex 3 ( $45°$ ) we got $0.765 v$ ; a bigger turn ( $60°$ ) should cost more, and $1.0 v > 0.765 v$ . Monotone increasing — consistent with the rising s01 curve. On figure s01 dot D sits exactly on the pink $Δ v / v = 1$ line. ✔

Cell E — the boundary at 90°

Worked example Ex 5 ·

$Δ i = 90°$ , LEO (dot E on figure s01) $v = 7.7$ km/s. A right-angle plane change — the exact border between "mid" and "large". Show $Δ v / v = 2$ .

Forecast: turning a full right angle — guess if it costs more or less than $1.4 v$ , then locate dot E on figure s01.

Figure s04 shows the velocity triangle for $Δ i = 90°$ : because the apex is a right angle, $v_{1} ⊥ v_{2}$ , so $Δ v$ is the hypotenuse of a right triangle with two legs of length $v$ . Pythagoras alone gives the answer.

Right-angle apex means we can use Pythagoras: $Δ v = v^{2} + v^{2} = v 2$ . Why this step? When the two velocity arrows are perpendicular the law of cosines' $cos (90°) = 0$ term vanishes, leaving plain Pythagoras — a lovely cross-check on the formula.
Formula route: $Δ v = 2 v sin (45°) = 2 v \cdot \frac{2}{2} = v 2$ . Why this step? $sin (45°) = \frac{2}{2}$ , and the two halves combine — the formula agrees with Pythagoras exactly.
Numbers: $Δ v = 7.7 2 \approx 10.89$ km/s, ratio $2 \approx 1.4142$ . Why this step? A $90°$ turn already costs $1.41 \times$ the orbit speed.

Verify: two independent methods (Pythagoras and the formula) both give $v 2$ . On figure s01 dot E sits between dot D ( $1.0$ ) and the $180°$ ceiling ( $2.0$ ), at height $1.414$ . ✔

Cells F & G — large angle and the 180° limit (two separate cases)

Worked example Ex 6b ·

$Δ i = 180°$ (cell G, dot G on figure s01) $v = 5$ km/s. A full reversal of direction — the limiting case.

Forecast: guess the exact cost of flipping your velocity end-for-end, then check that dot G is the highest point on figure s01.

Half-angle $= 180°/2 = 90°$ , and $sin (90°) = 1$ . Why this step? Reversing direction is the extreme: the triangle flattens into a straight line (rightmost in figure s05), base $= 2 v$ .
$Δ v = 2 \cdot 5 \cdot 1 = 10$ km/s $= 2 v$ . Why this step? You must kill all your speed ( $- 5$ ) and rebuild it the other way ( $+ 5$ ): total $10$ .

Verify: the maximum of $2 sin (Δ i /2)$ over $[0°, 180°]$ is at $180°$ where it equals $2$ . So $Δ v \leq 2 v$ always — Ex 6a ( $8.66 < 10$ ) obeys this ceiling. This also kills the common mistake of using $sin (Δ i)$ : that would give $sin (180°) = 0$ , absurdly claiming a free reversal. ✔

Cell H — the slow-point strategy

Worked example Ex 7 · Same

$30°$ turn, two places in an ellipse An elliptical orbit has periapsis speed $v_{p} = 10$ km/s and apoapsis speed $v_{a} = 1.6$ km/s (see Apoapsis and periapsis). You need a $30°$ plane change. Compare the cost at each end.

Forecast: same angle, same formula — guess how many times cheaper apoapsis is. (Hint: on figure s01 the angle-dot is the same, so only $v$ can differ.)

Figure s06 sketches the ellipse: a big velocity arrow at periapsis (fast, near Earth) and a small one at apoapsis (slow, far out). Turning the big arrow by $30°$ needs a big shove; turning the small arrow by the same $30°$ needs a small shove — because $Δ v \propto v$ .

Common factor: $2 sin (30°/2) = 2 sin (15°) \approx 2 \cdot 0.258819 = 0.517638$ . Why this step? The angle-part is identical at both ends; only $v$ differs, since $Δ v \propto v$ .
At periapsis: $Δ v_{p} = 0.517638 \cdot 10 \approx 5.176$ km/s. Why this step? Fast point → expensive rotation.
At apoapsis: $Δ v_{a} = 0.517638 \cdot 1.6 \approx 0.828$ km/s. Why this step? Slow point → cheap rotation.
Ratio $= v_{p} / v_{a} = 10/1.6 = 6.25 \times$ cheaper at apoapsis. Why this step? Because $Δ v$ is linear in $v$ , the saving is exactly the speed ratio.

Verify: $Δ v_{p} /Δ v_{a} = 5.176/0.828 = 6.25 = v_{p} / v_{a}$ . The mantra "turn slow to spend less" is literally the ratio of speeds. ✔

Cell I — combined burn with different speeds

Worked example Ex 8 · GTO apogee: speed change

and plane change at once At apogee you have $v_{1} = 1.6$ km/s (on the transfer ellipse) and want the circular geostationary speed $v_{2} = 3.07$ km/s, while also turning the plane by $Δ i = 28.5°$ (a Cape Canaveral launch). Find $Δ v$ for the combined burn, and compare to doing them separately.

Forecast: guess whether combining beats separate — and by how much. Then look at figure s07 and predict which pink path is shorter.

Figure s07 contrasts two paths from the same start: the combined triangle (one short pink diagonal, sides $v_{1}$ and $v_{2}$ with the $28.5°$ angle between them) versus the separate L-shaped path (speed change first, then a plane-change leg). The diagonal is visibly shorter — that's the triangle inequality doing your fuel budgeting.

Combined (general law of cosines, unequal sides): $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos Δ i$ Why this step? When the two velocity arrows have different lengths, the isosceles trick fails; we need the full law of cosines.
Numbers: $v_{1}^{2} = 2.56$ , $v_{2}^{2} = 9.4249$ , $2 v_{1} v_{2} = 9.824$ , $cos (28.5°) \approx 0.87882$ . Why this step? Assemble the three pieces of the triangle.
$Δ v = 2.56 + 9.4249 - 9.824 \cdot 0.87882 = 11.9849 - 8.6357 = 3.3492 \approx 1.830$ km/s. Why this step? The single diagonal shove of the triangle.
Separate: first speed change $∣ v_{2} - v_{1} ∣ = 1.47$ km/s, then pure plane change at $v_{2}$ : $2 \cdot 3.07 \cdot sin (14.25°) = 6.14 \cdot 0.246154 \approx 1.5114$ km/s. Sum $= 1.47 + 1.5114 = 2.981$ km/s. Why this step? Two sequential arrows laid end to end.

Verify: combined $1.830 <$ separate $2.981$ . The triangle inequality guarantees the direct side is shorter than the two-leg path — combining saves $\approx 1.15$ km/s here. ✔

Cell J — the real-world fuel word problem

Worked example Ex 9 · How much propellant for a

$10°$ change? A $2000$ kg satellite at $v = 7.5$ km/s must change inclination by $10°$ . Its thruster has exhaust velocity $v_{e} = 3.0$ km/s. How much fuel mass does this burn?

Forecast: $10°$ is small — guess if fuel is under $100$ kg. (On figure s01 the $Δ v / v$ dot is low; but Tsiolkovsky bends it exponentially into fuel.)

$Δ v = 2 \cdot 7.5 \cdot sin (5°) = 15 \cdot 0.0871557 \approx 1.3073$ km/s. Why this step? Half-angle $= 5°$ ; even $10°$ needs $1.3$ km/s.
Mass ratio via Tsiolkovsky rocket equation: $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{1.3073/3.0} = e^{0.43578} \approx 1.5463$ . Why this step? Fuel grows exponentially with $Δ v$ , not linearly — this is why plane changes hurt.
Final (dry-side) mass: $m_{f} = m_{0} /1.5463 = 2000/1.5463 \approx 1293.4$ kg. Why this step? $m_{0} = 2000$ kg is the full wet mass before the burn.
Fuel burned: $m_{0} - m_{f} = 2000 - 1293.4 \approx 706.6$ kg. Why this step? The difference is propellant expelled.

Verify: $706.6$ kg is 35% of the whole satellite for a mere $10°$ nudge. The exponential mapping means small $Δ v$ still eats large fuel fractions. ✔

Cell K — the exam twist: invert for $Δ i$

Worked example Ex 10 · "You have

$Δ v = 2$ km/s to spend at $v = 7.7$ km/s. What's the biggest plane change?" Solve for $Δ i$ , not $Δ v$ .

Forecast: guess whether the answer is bigger or smaller than $15°$ . (On figure s01, find the height $Δ v / v = 2/15.4 \approx 0.13$ and read left to the angle — that's the inverse move.)

Rearrange $Δ v = 2 v sin (Δ i /2)$ for the sine: $sin (\frac{Δ i}{2}) = \frac{Δ v}{2 v} = \frac{2}{2 \cdot 7.7} = \frac{2}{15.4} \approx 0.12987$ Why this step? Isolate the unknown inside the sine — algebra, no geometry lost.
Undo the sine with $arcsin$ (the "which angle has this sine?" question): $\frac{Δ i}{2} = arcsin (0.12987) \approx 7.462°$ Why this step? $arcsin$ is the exact inverse of $sin$ on $[0°, 90°]$ , and half-angles here never leave that range (since $Δ i \leq 180°$ ), so it's unambiguous.
Double it: $Δ i = 2 \cdot 7.462° \approx 14.92°$ . Why this step? Recover the full rotation from its half.

Verify: plug back — $2 \cdot 7.7 \cdot sin (14.92°/2) = 15.4 sin (7.46°) = 15.4 \cdot 0.12984 \approx 2.00$ km/s. Round-trips to the given budget. And $14.92° < 15°$ , matching a $Δ v$ that's only $13%$ of $2 v$ . ✔

Active recall

Recall Did you cover every cell?

Which example handled each: degenerate $0°$ ? the $90°$ boundary? the exact $180°$ ceiling? unequal-speed combined burn? inverting for $Δ i$ ? Degenerate 0° ::: Ex 1 (Δv = 0) 90° boundary (Δv/v = √2) ::: Ex 5 (Pythagoras cross-check) 180° reversal (ceiling Δv = 2v) ::: Ex 6b Combined burn with v₁ ≠ v₂ ::: Ex 8 (general law of cosines) Solving for Δi from a Δv budget ::: Ex 10 (arcsin, then double) Why is apoapsis the cheap place? ::: Δv ∝ v, so minimum speed → minimum cost (Ex 7)

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