Propulsive forces — thrust misalignment, gimbal angle: Worked examples

This page is the drill hall for the parent topic. We will attack the gimbal torque law from every direction: positive tilt, negative tilt, zero tilt, straight-back offset, the degenerate case where the engine sits exactly at the center of mass, the limiting case of a huge tilt, and two "real-life" curveballs. Nothing here uses a symbol we did not earn in the parent note — but let us re-anchor the three that matter most, and pin down our coordinate frame, before we start.

The one formula we exercise, in its full and its small-angle form:

The scenario matrix

Every gimbal problem is one of the cells below. The examples that follow are labelled with the cell they cover, so by the end you will have seen the full board. (The offset symbol $d$ used in Cell D is defined just above.)

Cell	Situation	Sign / regime	What it tests
A	Positive tilt $δ > 0$	$τ < 0$ (nose one way)	Basic torque + direction
B	Negative tilt $δ < 0$	$τ > 0$ (nose other way)	Sign symmetry
C	Zero tilt $δ = 0$	$τ = 0$	Degenerate: no steering
D	Lateral offset $d$ , no tilt	$τ = - d T$ (constant)	Different geometry
E	Engine at CoM, $L = 0$	$τ = 0$ regardless of $δ$	Degenerate lever
F	Large tilt (limit)	$sin δ$ vs $δ$ diverge	Where small-angle breaks
G	Real-world word problem	mixed	Time-to-rate, propellant cost
H	Exam twist: CoM shifts mid-burn	$L$ grows	Changing authority

Let us clear every cell.

Cell A — Positive tilt, basic torque and direction

Worked example A · A commanded right-tilt

A stage has $T = 500 kN$ , lever arm $L = 15 m$ , and the engine is gimballed by $δ = + 2^{\circ}$ . Find the torque magnitude and say which way the nose swings.

Forecast: guess before computing — will the torque be closer to $1 0^{5}$ or $1 0^{6}$ N·m? Which way does the nose go if the exhaust is pushed toward $+ y$ (up in our picture)?

Step 1 — Convert the angle. $δ = 2 \times \frac{π}{180} = 0.03491 rad .$ Why this step? $sin$ only obeys the small-angle rule and gives correct numbers when its input is in radians. Degrees would be silently wrong.

Step 2 — Plug into the magnitude form (minus sign dropped on purpose). $∣ τ ∣ = L T sin δ = 15 \times 500000 \times sin (0.03491) .$ Why this step? We first want the raw turning strength, so we use $∣ τ ∣ = L T sin δ$ — the absolute-value bars remind us we have set the sign aside for now and will restore it in Step 4. We use the exact $sin$ , not the approximation, so we have a reference value.

Step 3 — Evaluate. $sin (0.03491) = 0.03490 \Rightarrow ∣ τ ∣ = 15 \times 500000 \times 0.03490 \approx 2.62 \times 1 0^{5} N\cdotm .$

Step 4 — Restore the sign and read the direction. The signed law gives $τ = - L T sin δ = - 2.62 \times 1 0^{5}$ N·m, i.e. negative → clockwise → nose down. Look at the figure: the thrust pushes the back of the rocket sideways toward $+ y$ ; a force on the tail toward $+ y$ swings the tail up and therefore the nose down, exactly the clockwise sense the minus sign predicts.

Figure A: side view of the rocket with the CoM (yellow dot) at the origin and the pivot (red square) at $x = - L$ . The yellow double arrow marks the lever $L$ . The green arrow is the tilted thrust $T$ ; its small red arrow is the sideways component $T sin δ$ that pushes the tail toward $+ y$ . The yellow arc at the pivot is the gimbal angle $δ$ measured from the dashed axial reference line.

Verify: small-angle cross-check $L T δ = 15 \times 500000 \times 0.03491 = 2.618 \times 1 0^{5}$ N·m — agrees with Step 3 to within 0.01%, confirming the angle is small enough. Units: $m \times N = N\cdotm$ . ✓

Cell B — Negative tilt, sign symmetry

Worked example B · The mirror-image tilt

Same stage ( $T = 500$ kN, $L = 15$ m) but now the engine is tilted the other way, $δ = - 2^{\circ}$ . Compare the signed torque to Cell A.

Forecast: should the magnitude change at all? Should the nose go the same way or the opposite way?

Step 1 — Use the odd symmetry of sine. $sin (- δ) = - sin (δ) .$ Why this step? Instead of recomputing from scratch, we exploit that $sin$ is an odd function — flipping the tilt flips the torque's sign but not its size. This is why one actuator can steer both left and right.

Step 2 — Signed torque. $τ = - L T sin (- 0.03491) = + L T sin (0.03491) = + 15 \times 500000 \times 0.03490 \approx + 2.62 \times 1 0^{5} N\cdotm .$ Why this step? We keep the minus sign of the master law and insert the negative input, so the two negatives multiply to a positive signed torque — this is the whole point of the cell: a genuine sign flip versus Cell A, computed rather than asserted.

Step 3 — Direction. By our convention a positive $τ$ is counter-clockwise = nose up, exactly the mirror of Cell A's nose-down. Why this step? The number alone is meaningless until mapped back to physical rotation through the sign convention fixed at the top — that mapping is what turns " $+ 2.62 \times 1 0^{5}$ " into a steering command.

Verify: magnitude identical to Cell A ( $2.62 \times 1 0^{5}$ N·m), sign reversed. ✓ Symmetry means the controller only needs the magnitude table plus a sign bit.

Cell C — Zero tilt, the degenerate "no steering" case

Cell D — Lateral offset, no tilt (different geometry)

Worked example D · The engine that is straight but shifted

The engine points perfectly straight ( $δ = 0$ ) but its thrust line is offset sideways from the CoM by $d = 0.10 m$ in the $+ y$ direction (a mounting error). $T = 500$ kN. Find the signed torque and which way the nose turns.

Forecast: with $δ = 0$ we just said $τ = 0$ . Is that still true here? What changed?

Step 1 — Recognise it is a different geometry. Here the whole (axial) thrust vector points along $+ x$ but its line of action is shifted to $y = + d$ , so it misses the CoM. The lever arm is now the perpendicular distance $d$ , and the full force $T$ is perpendicular to it. Why this step? Torque is (perpendicular distance) × (force). In Cell A the perpendicular distance was $L$ and only the lateral component $T sin δ$ mattered. Here the perpendicular distance is $d$ and the whole $T$ acts.

Step 2 — Recall what the 2D cross product is, then name the components. Torque is the turning effect of a force, written as the cross product $τ = r \times F$ , where $r$ is the position vector from the CoM to where the force is applied and $F$ is the force. In our flat, two-dimensional picture (everything lives in the $x$ – $y$ plane) this cross product collapses to a single signed number: $τ = r_{x} F_{y} - r_{y} F_{x},$ where $r_{x}, r_{y}$ are the $x$ - and $y$ -parts of $r$ and $F_{x}, F_{y}$ are the $x$ - and $y$ -parts of $F$ . Reading them off this scenario: the force is applied at $r = (0, + d)$ (straight above the CoM by the offset $d$ , with no $x$ -shift because the thrust line crosses the CoM plane there), so $r_{x} = 0, r_{y} = + d$ . The force is purely axial, $F = (T, 0)$ , so $F_{x} = T, F_{y} = 0$ . Why this step? We must earn the formula before using it — $τ = r_{x} F_{y} - r_{y} F_{x}$ is just the 2D shadow of the full $r \times F$ , and pinning down each of the four components is what lets us plug numbers in without guessing signs.

Step 3 — Signed torque. Substituting the four components: $τ = r_{x} F_{y} - r_{y} F_{x} = (0) (0) - (d) (T) = - d T = - 0.10 \times 500000 = - 5.0 \times 1 0^{4} N\cdotm .$ Why this step? We keep the sign so we can state a rotation direction, not just a size.

Step 4 — Direction and contrast. Negative $τ$ → clockwise → nose down: a forward push applied above the CoM tips the nose down, just as pushing the top of a standing pencil topples it away from you. There is no axial loss ( $cos 0 = 1$ ), yet a real, constant disturbance torque exists. This is why offset ( $d$ ) and tilt ( $δ$ ) are two separate failure modes. Why this step? A nonzero torque is only useful (or dangerous) once we know which way it rotates the vehicle — otherwise the controller cannot choose a correcting sign.

Verify: units $m \times N = N\cdotm$ ; magnitude $5.0 \times 1 0^{4}$ N·m. ✓ Note it does not use $sin$ — a classic trap.

Cell E — Engine exactly at the CoM (degenerate lever)

Cell F — Large tilt: where the small-angle rule breaks (and the full domain of $δ$ )

Worked example F · Pushing past the linear regime

Compare the exact torque magnitude with the small-angle estimate at three tilts: $δ = 5^{\circ}, 2 0^{\circ}, 4 5^{\circ}$ . Use $L = 15$ m, $T = 500$ kN. Also report the axial thrust loss. Finally, note what the signed law does across the whole domain $0 \leq δ \leq 2 π$ .

Forecast: at what angle does the cheap approximation $sin δ \approx δ$ start to lie by more than 1%?

Step 1 — Tabulate exact vs approximate. For each $δ$ (in radians) compute $sin δ$ and $δ$ , and the fractional error $(δ - sin δ) / sin δ$ .

$δ$	$δ$ (rad)	$sin δ$	error
$5^{\circ}$	$0.08727$	$0.08716$	$0.13%$
$2 0^{\circ}$	$0.34907$	$0.34202$	$2.06%$
$4 5^{\circ}$	$0.78540$	$0.70711$	$11.1%$

Why this step? It shows why real gimbals stay under ~10°: the linear control math (torque $\propto δ$ ) is trustworthy there and dangerously optimistic beyond.

Step 2 — Axial loss at $4 5^{\circ}$ : $loss = T (1 - cos 4 5^{\circ}) = 500000 \times (1 - 0.70711) = 1.46 \times 1 0^{5}$ N ≈ 29% of thrust gone. At $5^{\circ}$ it is only $T (1 - cos 5^{\circ}) = 1.90 \times 1 0^{3}$ N ≈ 0.38%. Why this step? Steering torque grows like $sin δ$ (first order) but the forward-thrust you sacrifice grows like $1 - cos δ$ (second order); comparing the two numbers at the same angle is the only way to see that a big tilt buys extra turning at a suddenly-steep fuel-efficiency price — the trade-off that sets the hardware limit.

Step 3 — The full domain of $δ$ . The signed law $τ = - L T sin δ$ holds for any angle, not just small ones. Because $sin$ is periodic: torque is zero again at $δ = π$ (engine pointing straight forward — physically impossible for a real nozzle but a valid limit), it reverses sign for $π < δ < 2 π$ (the lateral component now points toward $- y$ ), and repeats every $2 π$ . The peak torque magnitude occurs at $δ = π /2$ ( $9 0^{\circ}$ ), where thrust is entirely lateral and axial thrust is zero. Real hardware lives only in the tiny sliver near $δ = 0$ , but the formula itself is global. Why this step? A "drill hall" must leave no scenario unshown; a reader who meets $δ > 9 0^{\circ}$ in a problem must know the same formula still applies, only with the sign flips just described.

Step 4 — Read the graph. The figure shows $sin δ$ (curve) hugging the straight line $δ$ near the origin then peeling away — the visual reason the approximation is "cheap at small tilt, wrong at large tilt". Why this step? A table of three numbers can be memorised without being understood; seeing the two curves touch at the origin and then visibly separate is what turns "the error grows with angle" into an intuition you can reconstruct, so you never trust the linear rule at a glance beyond where the lines are still on top of each other.

Figure F: two curves against gimbal angle in degrees — the blue $sin δ$ (exact) and the dashed yellow straight line $δ$ (small-angle estimate). Coloured stems at $5^{\circ}$ , $2 0^{\circ}$ , $4 5^{\circ}$ mark the vertical gap between them; the gap is invisible at $5^{\circ}$ (green), small at $2 0^{\circ}$ , and a large 11% at $4 5^{\circ}$ (red), showing exactly where the linear rule stops being trustworthy.

Verify: $sin 2 0^{\circ} = 0.34202$ , error $2.06%$ ; $cos 4 5^{\circ} = 0.70711$ , loss fraction $0.29289$ ; $sin π = 0$ confirms the second zero-torque point. ✓

Cell G — Real-world word problem

Worked example G · How long may we hold the gimbal?

A stage holds $δ = 3^{\circ}$ to correct its attitude. Data: $T = 800$ kN, $L = 20$ m, pitch inertia $I = 3 \times 1 0^{6} kg\cdotm^{2}$ . The guidance limit is a pitch rate of $1 0^{\circ} / s$ . Starting from rest, how long may the gimbal stay held before that rate is reached?

Forecast: seconds, or tens of seconds?

Step 1 — Torque magnitude. $δ = 3^{\circ} = 0.05236$ rad. $∣ τ ∣ = L T sin δ = 20 \times 800000 \times sin (0.05236) = 20 \times 800000 \times 0.05234 = 8.37 \times 1 0^{5} N\cdotm .$ Why this step? We only need the size of the turning effect to find how fast the rate builds, so we use the magnitude form $∣ τ ∣ = L T sin δ$ (minus sign dropped on purpose); direction does not affect the time to reach a given rate.

Step 2 — Angular acceleration. $\ddot{θ} = \frac{∣ τ ∣}{I} = \frac{8.37 \times 1 0 ^{5}}{3 \times 1 0 ^{6}} = 0.2791 rad/s^{2} .$ Why this step? Attitude evolves via $τ = I \ddot{θ}$ ; dividing torque by inertia is the bridge from force-world to rotation-world, and it is the only way to turn newton-metres into a rate of turning.

Step 3 — Convert the rate limit to radians. $1 0^{\circ} / s = 0.1745 rad/s$ . Why this step? $\ddot{θ}$ came out in rad/s²; the target rate must be in the same radian units or the division in Step 4 mixes incompatible quantities.

Step 4 — Time from rest ( $\dot{θ} = \ddot{θ} t$ ): $t = \frac{θ ˙ _{m a x}}{θ ¨} = \frac{0.1745}{0.2791} = 0.625 s .$ Why this step? From rest with constant acceleration, rate grows linearly in time, so the time to hit a rate limit is simply the limit divided by the acceleration — the fastest possible check that a held gimbal is safe.

Verify: in $0.625$ s the rate reaches $0.2791 \times 0.625 = 0.1745$ rad/s $= 1 0^{\circ} / s$ . ✓ This is why real gimbal pulses last a fraction of a second — the authority is enormous.

Cell H — Exam twist: the CoM shifts during the burn

Worked example H · Same command, growing lever

Early in the burn the CoM-to-pivot lever is $L_{1} = 18 m$ ; late in the burn, after propellant has drained forward, it is $L_{2} = 24 m$ . The gimbal command and thrust are unchanged: $δ = 2^{\circ}$ , $T = 600$ kN. By what factor does the control torque grow, and what is each value?

Forecast: guess the ratio $τ_{2} / τ_{1}$ from the numbers before computing.

Step 1 — Note what changes. Only $L$ changes; $T$ and $δ$ are fixed, so torque scales linearly with $L$ : $∣ τ ∣ \propto L$ . Why this step? In $∣ τ ∣ = L T sin δ$ , with two of the three factors frozen the ratio collapses to $L_{2} / L_{1}$ — recognising this saves us from re-doing the trig and exposes the underlying proportionality that the exam is really testing.

Step 2 — Early torque magnitude. $δ = 2^{\circ} = 0.03491$ rad, $sin δ = 0.03490$ . $∣ τ_{1} ∣ = 18 \times 600000 \times 0.03490 = 3.769 \times 1 0^{5} N\cdotm .$ Why this step? We compute one concrete anchor value first so the ratio in Step 4 has physical numbers behind it, not just an abstract fraction — and so we can sanity-check the units (N·m).

Step 3 — Late torque magnitude. $∣ τ_{2} ∣ = 24 \times 600000 \times 0.03490 = 5.026 \times 1 0^{5} N\cdotm .$ Why this step? Evaluating the late-burn case explicitly, with the same $T$ and $δ$ but larger $L$ , isolates the CoM shift as the sole cause of the change — the pedagogical heart of the twist.

Step 4 — Ratio. $\frac{∣ τ _{2} ∣}{∣ τ _{1} ∣} = \frac{24}{18} = 1.333.$ Why it matters: the same stick-deflection produces 33% more turning late in the burn. The controller must reduce gimbal command over time or it will over-rotate — this is the CoM-shift problem in one line.

Verify: $∣ τ_{2} ∣/∣ τ_{1} ∣ = 24/18 = 4/3 = 1.3333$ ; $∣ τ_{1} ∣ = 3.769 \times 1 0^{5}$ , $∣ τ_{2} ∣ = 5.026 \times 1 0^{5}$ N·m. ✓

Recall check

Recall Which cell am I in?

Engine points straight but is bolted 8 cm to the side of the CoM — which formula? ::: Cell D, lateral offset: $τ = - d T$ (no $sin$ , no $L$ ); sign gives the rotation direction. Same gimbal command gives more torque late in the flight — why? ::: Cell H: CoM moves forward, lever $L$ grows, and $∣ τ ∣ \propto L$ . Tilt is $- 4^{\circ}$ instead of $+ 4^{\circ}$ — what changes? ::: Cell B: same magnitude, torque sign flips (sine is odd) → nose turns the opposite way. Why can't an engine mounted exactly at the CoM steer? ::: Cell E: $L = 0$ , so $τ = L T sin δ = 0$ for any tilt — pure translation. At $4 5^{\circ}$ tilt, roughly how much thrust is lost? ::: Cell F: $T (1 - cos 4 5^{\circ}) \approx 29%$ — that is why big tilts are avoided. Where else is the torque zero besides $δ = 0$ ? ::: Cell F: at $δ = π$ (engine straight forward); the sign then reverses for $π < δ < 2 π$ .