Intuition Why a whole page of examples?
The four 3DOF rules from the parent note look simple, but the sign of every term flips as the rocket climbs, tips over, coasts, or dives. This page walks a matrix of every case — every flight-path-angle quadrant, thrust on/off, mass changing, degenerate zeros, and a couple of exam-style twists — so you never meet a scenario you haven't already solved once.
Before we start, one reminder of the notation we will lean on (every symbol earned in the parent ):
Definition The five state quantities
V = speed = length of the velocity arrow (metres per second, always ≥ 0 ).
γ (Greek "gamma") = flight-path angle = the angle the velocity arrow makes with the horizon. γ > 0 means climbing, γ = 0 means flying flat, γ < 0 means descending.
x = downrange distance (how far along the ground).
h = altitude (how high).
m = mass (shrinks as fuel burns).
A dot on top (V ˙ ) means "rate of change per second" — how fast that quantity is changing right now .
Definition The four forces (and two engine numbers) we will plug in
T = thrust = the forward push the engine makes, measured in newtons (N) or kilonewtons (1 kN = 1000 N ). It acts along the velocity arrow (zero angle of attack).
D = drag = the air's backward friction on the vehicle. It always opposes the velocity arrow, so it enters with a minus sign.
L = lift = a force perpendicular to the velocity (sideways to the arrow). For a plain ballistic rocket L = 0 ; wings or a flared body make it non-zero.
g = local gravitational acceleration = 9.81 m/s 2 near the ground — how hard weight pulls each kilogram straight down.
I s p = specific impulse (seconds) = a fuel-efficiency score for the engine; bigger means more push per kilogram of propellant burned.
g 0 = standard gravity = 9.80665 ≈ 9.81 m/s 2 = a fixed reference constant used only to convert I s p into an exhaust speed. It is NOT the local g that pulls the rocket — it never changes with altitude; it is just the bookkeeping number baked into the definition of I s p .
Every 3DOF problem is really one of these cells. The examples below are labelled [C1], [C2], … so you can see each cell is covered.
Cell
Case class
What is special
Key term that changes sign / vanishes
C1
Steep climb , 0 < γ < 9 0 ∘
thrust on, gravity slows & turns
sin γ > 0 , cos γ > 0
C2
Vertical , γ = 9 0 ∘
local accel, drag opposes climb
cos γ = 0 ⇒ γ ˙ = 0
C3
Level flight , γ = 0 ∘
gravity does not slow speed
sin γ = 0 , cos γ = 1 (fastest turn)
C4
Descending / diving , γ < 0 ∘
gravity now adds speed
sin γ < 0 ⇒ − g sin γ > 0
C5
Coast / ballistic , T = 0
engine off; drag − D / m still slows, or D = 0 ⇒ pure projectile
thrust term vanishes; drag term survives unless D = 0
C6
Mass changing , m ˙ = 0
same thrust, rising acceleration
m shrinks over the burn
C7
Degenerate speed , V → 0
turning rate blows up
γ ˙ = − g cos γ / V → ∞
C8
Word problem (gravity turn)
initial pitch-over
small γ , uses full set
C9
Exam twist (lift added)
L = 0 curves path up
normal equation full form
We now hit each cell.
Worked example Example 1 — Instantaneous acceleration climbing at
γ = 4 0 ∘
T = 180 kN , D = 25 kN , m = 9000 kg , γ = 4 0 ∘ , g = 9.81 . Find V ˙ .
Forecast: guess — will V ˙ be bigger or smaller than the level-flight value? (Gravity is stealing some speed, so it must be smaller than if γ were 0 .)
Write the tangential rule: V ˙ = m T − D − g sin γ .
Why this step? Only the along-velocity forces change the length of the velocity arrow. Thrust pulls it longer, drag and the along-track slice of gravity shrink it.
Net propulsive term: 9000 180000 − 25000 = 9000 155000 = 17.22 m/s 2 .
Why this step? Thrust minus drag is the force that survives before gravity; dividing by m turns force into acceleration (Newton).
Gravity along-track slice: g sin 4 0 ∘ = 9.81 × 0.6428 = 6.306 m/s 2 .
Why this step? At γ = 4 0 ∘ only the fraction sin 4 0 ∘ of weight points backward along the motion — that is the piece that slows the arrow.
Combine: V ˙ = 17.22 − 6.31 = 10.92 m/s 2 .
Why this step? The tangential rule is a sum of accelerations along the arrow: the propulsive push (step 2) minus the gravity slice (step 3). Subtracting them gives the single net rate at which the speed changes this instant.
Verify: units kg N = m/s 2 ✓. It is positive (still accelerating) and smaller than the pure 17.22 , matching the forecast ✓.
Worked example Example 2 — Straight up, and does it turn?
T = 180 kN , D = 25 kN , m = 9000 kg , γ = 9 0 ∘ , V = 120 m/s , L = 0 . Find V ˙ and γ ˙ .
Forecast: at γ = 9 0 ∘ , all of gravity now opposes the climb — so V ˙ should drop by the full g . And a vertical rocket should keep going vertical, so γ ˙ = 0 .
sin 9 0 ∘ = 1 , so V ˙ = 9000 155000 − 9.81 ( 1 ) = 17.22 − 9.81 = 7.41 m/s 2 .
Why this step? Straight up, "backward along velocity" is straight down — the whole weight fights the motion. This is the local acceleration form of the vertical-climb equation; integrating this along a drag-free vertical burn is what leads to the Tsiolkovsky Rocket Equation Δ V formula — but that integral Δ V result and this instantaneous V ˙ are two different quantities, not the same expression.
cos 9 0 ∘ = 0 , so γ ˙ = − V g cos γ = − 120 9.81 × 0 = 0 rad/s .
Why this step? Perpendicular to a vertical velocity, gravity has no sideways slice — nothing turns the arrow.
Verify: at γ = 9 0 ∘ the along-track gravity is maximal and the turning gravity is zero — the two extremes of the "S-along, C-across" rule ✓. A vertical climb stays vertical ✓.
How to read figure s01. The horizontal axis is the flight-path angle γ sweeping from level (0 ∘ ) to vertical (9 0 ∘ ); the vertical axis is a gravity acceleration in m/s 2 . The amber curve is g sin γ — the slice of gravity that acts along the velocity and appears in the speed equation V ˙ . The cyan curve is g cos γ — the slice that acts across the velocity and drives the turning equation γ ˙ . Notice they cross at 4 5 ∘ and trade places: at the left edge (level, C3) turning is maximal, at the right edge (vertical, C2) slowing is maximal. Each curve is literally the number you multiply by in the corresponding worked example.
Worked example Example 3 — Flying flat: no speed loss, hardest bend
V = 250 m/s , γ = 0 ∘ , L = 0 , g = 9.81 . Find V ˙ contribution from gravity, and γ ˙ .
Forecast: flying level, gravity is straight down , exactly perpendicular to a horizontal velocity — so it steals no speed but bends the path the most .
Gravity along-track: − g sin 0 ∘ = − 9.81 × 0 = 0 .
Why this step? Horizontal motion means "down" makes a 9 0 ∘ angle with the velocity — no along-track component.
Turning rate: γ ˙ = − V g cos 0 ∘ = − 250 9.81 × 1 = − 0.03924 rad/s .
Why this step? cos 0 ∘ = 1 : the entire weight now acts sideways to the motion, so the path bends at its maximum rate for this speed.
In degrees: − 0.03924 × π 180 = − 2.24 8 ∘ / s .
Why this step? The formula produces radians per second (the natural unit of γ ˙ ), but a reader thinks in degrees. Multiplying by 180/ π converts consistently so the two forms describe the same physical turn rate — mixing rad and deg is a classic source of wrong answers.
Verify: compare with the parent's level-vs-vertical statement — turning is fastest at γ = 0 , zero at γ = 9 0 ∘ ✓ (Example 2 gave γ ˙ = 0 ).
Worked example Example 4 — Coasting downhill at
γ = − 3 0 ∘
Coast phase: T = 0 , D = 12 kN , m = 9000 kg , γ = − 3 0 ∘ , g = 9.81 . Find V ˙ .
Forecast: pointing down, gravity now pulls forward along the motion — it should speed the rocket up , partly cancelled by drag.
V ˙ = 9000 0 − 12000 − g sin ( − 3 0 ∘ ) .
Why this step? Thrust is off (cell C5 overlaps here); we keep the full formula and let the sign of γ do the work.
sin ( − 3 0 ∘ ) = − 0.5 , so − g sin ( − 3 0 ∘ ) = − 9.81 ( − 0.5 ) = + 4.905 m/s 2 .
Why this step? A negative γ flips the gravity term positive — this is gravity helping , the mathematical reason a diving vehicle accelerates.
Drag term: 9000 − 12000 = − 1.333 m/s 2 .
Why this step? With the engine off, drag is the only along-track force slowing the arrow; it enters as − D / m because drag always opposes the velocity. Computing it separately keeps the helping (gravity) and hurting (drag) pieces visible before we add them.
Combine: V ˙ = − 1.333 + 4.905 = + 3.572 m/s 2 .
Why this step? The tangential rule sums the along-track accelerations: the drag slice (step 3) plus the now-positive gravity slice (step 2). Their sum is the single net rate of speed change.
Verify: engine off, yet V ˙ > 0 — only possible because γ < 0 let gravity add energy; drag alone would give − 1.333 ✓. Units m/s² ✓.
− g sin γ as always negative
Why it feels right: in a climb it was a loss.
The fix: the term is − g sin γ . When γ < 0 , sin γ < 0 , so the whole term becomes positive — gravity is now a friend. Always substitute the signed angle.
Worked example Example 5 — Engine-off, drag-free: is it a parabola?
T = 0 , D = 0 , γ = 4 5 ∘ , V = 200 m/s , g = 9.81 . Find V ˙ , γ ˙ , x ˙ , h ˙ .
Forecast: with no thrust and no drag this is just a thrown ball — the classic projectile. (Compare with Example 4, which was also thrust-off but had drag − D / m still slowing the vehicle; setting D = 0 here removes that term too.)
V ˙ = 0 − 0 − g sin 4 5 ∘ = − 9.81 × 0.7071 = − 6.936 m/s 2 .
Why this step? With both T = 0 and D = 0 , only gravity's along-track slice remains; climbing, it slows the speed. (In Example 4 the extra − D / m term survived because there D = 0 .)
γ ˙ = − V g cos 4 5 ∘ = − 200 9.81 × 0.7071 = − 0.03468 rad/s .
Why this step? Gravity's sideways slice tips the arrow down — the ball arcs over.
x ˙ = V cos 4 5 ∘ = 200 × 0.7071 = 141.4 m/s , h ˙ = V sin 4 5 ∘ = 141.4 m/s .
Why this step? Speed splits into horizontal and vertical by cos /sin of γ — the numbers you integrate to plot the arc.
Verify: horizontal acceleration should be zero for a drag-free projectile. Check x ¨ = d t d ( V cos γ ) = V ˙ cos γ − V sin γ γ ˙ = ( − 6.936 ) ( 0.7071 ) − ( 141.4 ) ( − 0.03468 ) = − 4.905 + 4.905 = 0 ✓. That exact cancellation is the parabola — horizontal speed is constant. Beautiful.
How to read figure s02. The axes are the ground frame: horizontal is downrange x (m), vertical is altitude h (m). The cyan curve is the coast trajectory — a perfect parabola. The amber arrows drawn along it are the velocity arrows at three instants; watch how their horizontal reach stays the same length while their vertical reach shrinks, tips over, then grows downward. That constant horizontal component is exactly the x ¨ = 0 result we just verified — the picture and the algebra say the same thing.
Worked example Example 6 — Same thrust, later in the burn
Engine keeps T = 180 kN , D = 25 kN , γ = 4 0 ∘ (same as Example 1) but the tank has emptied so m = 5000 kg . Also find m ˙ if I s p = 280 s , g 0 = 9.81 . Find V ˙ and m ˙ .
Forecast: less mass, same push → higher acceleration than Example 1's 10.92 .
V ˙ = 5000 180000 − 25000 − 9.81 sin 4 0 ∘ = 31.0 − 6.306 = 24.69 m/s 2 .
Why this step? Newton's a = F / m : shrink m and the same net force yields more a . This is why burnout acceleration is brutal.
Mass rate: m ˙ = − I s p g 0 T = − 280 × 9.81 180000 = − 65.53 kg/s .
Why this step? Here g 0 is the fixed reference constant baked into the definition of I s p (not the local gravity that pulls the rocket). The product I s p g 0 is the exhaust speed in m/s; thrust divided by exhaust speed gives kilograms of propellant leaving per second — negative because mass only ever falls.
Verify: Example 1 had m = 9000 and got 10.92 ; here m = 5000 gives 24.69 > 10.92 , matching the forecast ✓. m ˙ < 0 (mass only ever decreases) ✓.
Worked example Example 7 — Right at liftoff,
V almost zero
Vehicle barely moving, V = 5 m/s , γ = 1 0 ∘ , L = 0 , g = 9.81 . Find γ ˙ . Then discuss V → 0 .
Forecast: dividing by a tiny V makes γ ˙ huge — the path direction whips around when the rocket is nearly stationary.
γ ˙ = − V g cos 1 0 ∘ = − 5 9.81 × 0.9848 = − 5 9.663 = − 1.933 rad/s (≈ − 110. 7 ∘ / s ).
Why this step? The normal equation has V in the denominator; a small speed with a non-zero cos γ numerator makes the ratio explode — the direction of an almost-stationary arrow can swing wildly.
Limit: as V → 0 + , γ ˙ → − ∞ (for cos γ = 0 ).
Why this step? Physically the model breaks — a truly motionless point mass has no well-defined "direction of velocity" to turn. Real launches sit near-vertical (cos γ ≈ 0 ) to keep this term small, which is exactly why a rocket goes up first before pitching.
Verify: the danger only exists off-vertical: at γ = 9 0 ∘ , cos γ = 0 so γ ˙ = 0 even at V = 0 — no blow-up ✓ (consistent with Example 2). This is the mathematical justification for the Gravity Turn Trajectory starting from vertical.
Worked example Example 8 — How long to tip from
8 9 ∘ to 8 8 ∘ ?
Just after liftoff a rocket is at γ = 8 9 ∘ , climbing at a steady V = 100 m/s (treat as roughly constant over this brief instant), L = 0 . Estimate the time to lose 1 ∘ of flight-path angle by gravity alone.
Forecast: near vertical cos γ is tiny, so the turn is slow — this gentle start is the whole point of a gravity turn.
γ ˙ = − V g cos 8 9 ∘ = − 100 9.81 × 0.017452 = − 1.712 × 1 0 − 3 rad/s .
Why this step? Gravity is the only bender; near vertical its sideways slice cos 8 9 ∘ is minute.
Convert 1 ∘ to radians: 1 ∘ = 0.017453 rad .
Why this step? γ ˙ is in rad/s, so the target change must be in radians too — units must match.
Time = ∣ γ ˙ ∣ ∣Δ γ ∣ = 1.712 × 1 0 − 3 0.017453 = 10.19 s .
Why this step? Here γ ˙ is negative (the angle is shrinking) and we are asked "how long to lose one degree" — a duration, which must be positive. Taking the absolute value of both the change and the rate strips the direction sign and leaves the elapsed time , since time = size-of-change ÷ size-of-rate.
Verify: about ten seconds to shed just one degree — the pitch-over really is gentle, exactly why no control effort is wasted ✓. Uses the full flight-path-angle geometry.
Worked example Example 9 — Add lift to the normal equation
m = 7000 kg , V = 220 m/s , γ = 3 0 ∘ , g = 9.81 , and now a lift L = 90 kN acts perpendicular to velocity (upward). Find γ ˙ . Compare with L = 0 .
Forecast: lift pushes the arrow's tip up , fighting gravity's downward bend — γ ˙ should be less negative (maybe even positive) than the lift-free case.
Full normal equation: γ ˙ = mV L − m g cos γ .
Why this step? Perpendicular forces set the turning rate; lift and the sideways gravity slice both live here. This is the general form of the γ ˙ rule before we assumed L = 0 in every earlier example.
Gravity slice: m g cos 3 0 ∘ = 7000 × 9.81 × 0.8660 = 59470 N .
Why this step? cos 3 0 ∘ picks the fraction of weight acting across the velocity — the downward-bending force.
Numerator: L − m g cos γ = 90000 − 59470 = 30530 N (net upward ).
Why this step? Lift (up) minus gravity's cross-slice (down) is the net perpendicular force. It came out positive, so the surviving push is upward.
γ ˙ = 7000 × 220 30530 = 1 , 540 , 000 30530 = + 0.01982 rad/s (≈ + 1.13 6 ∘ / s ).
Why this step? Dividing the net perpendicular force by mV turns it into a turning rate (the normal-equation form). Positive now — lift overpowers gravity, so the path curves upward instead of down.
Lift-free comparison: γ ˙ = − V g cos 3 0 ∘ = − 220 9.81 × 0.8660 = − 0.03862 rad/s .
Why this step? Setting L = 0 recovers the equation used in every earlier example, so we can see exactly what the lift changed.
Verify: the sign flipped from negative (curving down, lift-free) to positive (curving up) once lift beat gravity's slice — matching the forecast ✓. Numerically + 0.01982 vs − 0.03862 rad/s : adding lift did not just soften the turn, it reversed its direction. This is the term dropped in a ballistic rocket but central to winged vehicles and to 6DOF Rigid-Body Dynamics , where the attitude that generates L becomes its own set of equations.
Recall Which term changes sign when the rocket dives?
The gravity term in the speed equation, − g sin γ ::: for γ < 0 , sin γ < 0 so − g sin γ > 0 — gravity adds speed in a dive.
Recall Why is
γ ˙ = 0 for a vertical rocket even at V = 0 ?
Because γ ˙ = − g cos γ / V and cos 9 0 ∘ = 0 ::: the numerator is zero, killing the blow-up; no sideways gravity means nothing turns the arrow.
Recall Same thrust, half the mass — what happens to
V ˙ ?
It (roughly) doubles ::: a = F / m ; shrinking m raises acceleration, which is why burnout is violent.
Recall What is
g 0 and how is it different from g ?
g 0 = 9.80665 m/s 2 is a fixed reference constant used only inside I s p g 0 to get exhaust speed ::: g is the local gravity actually pulling the rocket and can vary with altitude; g 0 never changes.
Recall In a thrust-off coast, when does the
− D / m term disappear?
Only when D = 0 (drag-free) ::: with T = 0 the drag term still slows the vehicle unless drag itself is zero, which is the pure-projectile subcase that recovers the parabola.
Mnemonic Sign compass for gravity
Climb → gravity SLOWS (− g sin γ < 0 ). Dive → gravity SPEEDS (− g sin γ > 0 ). Level → gravity only TURNS (max γ ˙ ). Vertical → gravity only SLOWS (no turn).