Worked examples — Thrust vector control — single-gimbal, dual-gimbal; TVC angles

This page is the drill floor for the parent topic. We will hit every kind of case the gimbal maths can throw at you: small angles, big angles, zero deflection, the degenerate "engine dead" case, both single- and dual-gimbal geometry, sign flips (steer left vs right), a real-world word problem, and an exam twist that catches most students. Every number you see below is machine-checked — see the closing "Machine-check summary" box.

Before we start, here are the two engine formulas the whole page runs on, plus three auxiliary results derived from them:

Recall The two core formulas (+ three helpers)

The two engines (everything below is one of these, or a rearrangement):

• Engine 1 — exact torque magnitude (turning strength from tilting the engine): $|M|=\ell T\sin \delta$. The signed torque (which way it spins) is $M=-\ell T\sin \delta$ in the frame of the next box — see the note on signs just below.
• Engine 2 — small-angle torque magnitude (the linearised form the autopilot uses): $|M|\approx \ell T\delta$, with $\delta$ in radians.

Signed vs magnitude — read this once: most examples quote the magnitude $|M|=\ell T\sin \delta$ (how hard it turns). Only when the direction matters (Example 4) do we carry the minus sign from the cross product, giving the signed $M=-\ell T\sin \delta$. Same physics; the sign just names which way the nose swings.

Three helper results (each follows from the engines / basic geometry, not new physics):

• Axial (useful, forward) thrust: $T_{\text{axial}}=T\cos \delta$ — the leftover of the same tilt.
• Fractional thrust lost to steering: $1-\cos \delta \approx {\delta }^2/2$ — Taylor expansion of the line above.
• Dual-gimbal total tilt: ${\delta }_{\text{tot}}=\sqrt{{\delta }_p^2+{\delta }_y^2}$ — Pythagoras on two perpendicular tilts (the two angles ${\delta }_p,{\delta }_y$ are defined in the box below).

Here $T$ is thrust magnitude, $\ell$ is the moment arm (CoM-to-gimbal distance), and $\delta$ is the tilt angle. All are built from scratch in the parent note.

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Sign convention used on this page

Everything below uses one fixed body frame so the signs never surprise you:

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The scenario matrix

Think of this table as a checklist. Each row is a class of problem that behaves differently. The last column names the example that covers it — so when you finish, every cell is ticked.

#	Case class	What makes it special	Covered by
A	Small angle, forward	$\sin \delta \approx \delta$ valid; solve linearly	Example 1
B	Large angle (>10°)	must use exact $\sin \delta$, not $\delta$	Example 2
C	$\delta =0$ (degenerate)	zero torque, max thrust — the baseline	Example 3
D	Sign / direction	steer left vs right, sign of the torque	Example 4
E	Dual-gimbal feasible	vector sum inside the limit	Example 5
F	Dual-gimbal clipped	vector sum exceeds limit → scale down	Example 6
G	Limiting value	thrust loss at the mechanical hard stop	Example 7
H	Real-world word problem	dead engine (T→0), what breaks	Example 8
I	Exam twist	degrees-vs-radians trap; roll authority = 0	Example 9

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Example 1 — Case A: small angle, solve linearly

Forecast: guess — will $\delta$ be a fraction of a degree, a few degrees, or tens of degrees?

1. Torque needed: $M=I\dot{\omega }=1.5\times 10^5\times 0.4=6.0\times 10^4\text{N⋅m}$. Why this step? Newton's rotational law $M=I\dot{\omega }$ links the want (spin-up rate) to the cause (torque). See Rigid Body Rotational Dynamics.
2. Invert the small-angle law: $\delta =\frac{M}{\ell T}=\frac{6.0\times 10^4}{5\times 1.0\times 10^6}=0.012\text{rad}$. Why this step? We assume small angle so torque is linear, giving one clean division. We'll justify the assumption in Verify.
3. Convert to degrees: $\delta =0.012\times \frac{180}{\pi }\approx 0.687^{\circ }$. Why this step? Engineers read gimbal limits in degrees; it lets us sanity-check against a typical $\pm 8^{\circ }$ range.

Verify: $0.012$ rad is tiny, so small-angle holds beautifully ($\sin 0.012=0.0119997$, error $<0.003\%$). Plug back: $\ell T\sin \delta =5\times 10^6\times \sin (0.012)=5.9998\times 10^4\approx M$. ✓ Units: $\text{N⋅m}$ throughout.

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Example 2 — Case B: large angle, exact law required

Forecast: at 20°, is the small-angle formula off by <1%, ~2%, or >5%?

1. Convert to radians: $20^{\circ }=20\times \frac{\pi }{180}=0.3491\text{rad}$. Why? $\sin$ and the linear form both need radians for a fair comparison.
2. Exact torque: $M_{\text{exact}}=\ell T\sin 20^{\circ }=5\times 10^6\times 0.3420=1.710\times 10^6\text{N⋅m}$. Why? Beyond ~10° the approximation $\sin \delta \approx \delta$ breaks; the true transverse thrust is $T\sin \delta$.
3. Linear estimate: $M_{\text{lin}}=\ell T\delta =5\times 10^6\times 0.3491=1.745\times 10^6\text{N⋅m}$. Why? This is what a naive autopilot using the linear law would think it commanded.
4. Error: $\frac{M_{\text{lin}}-M_{\text{exact}}}{M_{\text{exact}}}=\frac{1.745-1.710}{1.710}\approx 2.0\%$.

Verify: the linear form always over-estimates because $\delta >\sin \delta$. A 2% error at 20° is the takeaway — at the real-world $8^{\circ }$ the same error would be under $0.4\%$. Units cancel in the ratio. ✓

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Example 3 — Case C: the degenerate baseline, $\delta =0$

Forecast: which is max and which is zero?

1. Torque: $M=\ell T\sin 0=0\text{N⋅m}$. Why? $\sin 0=0$: no tilt means no transverse force, so no moment. The thrust line runs straight through... actually through the gimbal, then along $-x_b$ — it passes the CoM with zero lever, so no turning.
2. Axial thrust: $T_{\text{axial}}=T\cos 0=1.0\times 10^6\text{N}$ — the full thrust. Why? $\cos 0=1$: nothing is spilled sideways. This is the fuel-efficient cruise state.

Verify: $\delta =0$ is exactly the state the parent's Mistake #1 highlights: forward thrust is maximized here, not by increasing $\delta$. Torque $0$, thrust $100\%$. ✓

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Example 4 — Case D: sign / direction of the torque

Forecast: same size, opposite sign — or something else?

The figure below draws the fixed body frame from the "Sign convention" box: the white line is the vehicle axis $x_b$, the yellow dot is the CoM, the blue dot is the gimbal pivot at $(-\ell ,0)$. The red thrust arrow is the $+\delta$ case (exhaust thrown up-and-back, so $F_y>0$) and its red curved arrow shows the nose pitching down. The green thrust arrow is the $-\delta$ case, mirror-imaged, with a green curved arrow showing the nose pitching up. Read the two curved arrows: same size, opposite spin.

1. Magnitude both cases: $|M|=\ell T\sin 4^{\circ }=6\times 8\times 10^5\times 0.06976=3.348\times 10^5\text{N⋅m}$. Why? $\sin$ of $|\delta |$ sets the size; the two deflections are mirror images.
2. Signs (from the convention box): with $\vec{r}=(-\ell ,0)$, the planar torque is $M=-\ell T\sin \delta$. A $+\delta$ gives $F_y=T\sin (+\delta )>0$, so $M<0$ → nose-down (red). A $-\delta$ flips $F_y$, so $M>0$ → nose-up (green). Why? The sign encodes which way it spins — match it to the red vs green curved arrows in the figure.
3. Conclusion: equal-and-opposite torques, $\mp 3.348\times 10^5\text{N⋅m}$.

Verify: $\sin (-4^{\circ })=-\sin (4^{\circ })$, so the torques are exact negatives. This is why a single gimbal can command pitch in both directions from one actuator. ✓

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Example 5 — Case E: dual-gimbal, feasible command

Forecast: $6+8=14>12$ — is it therefore infeasible? (Careful!)

The figure plots the command as a point in a deflection plane: the blue arrow is the yaw leg ${\delta }_y=8^{\circ }$ (horizontal), the yellow arrow is the pitch leg ${\delta }_p=6^{\circ }$ (vertical, at right angles), and the red arrow is their vector sum — the true nozzle tilt. The green dashed quarter-circle is the mechanical limit at radius $12^{\circ }$; the command point sits inside it, so it is reachable.

1. Combine as a vector, not a sum: ${\delta }_{\text{tot}}=\sqrt{6^2+8^2}=\sqrt{100}=10^{\circ }$. Why? Pitch and yaw tilts are perpendicular components of one physical cone-tilt of the nozzle (the two legs meet at a right angle, the hypotenuse is the true tilt). They add by Pythagoras, not linearly.
2. Compare to limit: $10^{\circ }<12^{\circ }$ → feasible. ✓
3. Note the trap: the naive linear sum $14^{\circ }$ would have wrongly rejected a perfectly legal command.

Verify: $6$-$8$-$10$ is a scaled 3-4-5 triangle, so ${\delta }_{\text{tot}}=10^{\circ }$ exactly, comfortably under $12^{\circ }$. ✓

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Example 6 — Case F: dual-gimbal, clipped command

Forecast: legal or must-be-scaled?

1. Total tilt: ${\delta }_{\text{tot}}=\sqrt{10^2+10^2}=\sqrt{200}=14.142^{\circ }$. Why? Same vector-sum reason as Example 5.
2. Check: $14.142^{\circ }>12^{\circ }$ → illegal, the gimbal can't reach that far. Why? The nozzle physically hits its bearing stop at $12^{\circ }$ of tilt.
3. Clip while keeping direction: scale both by $k=\frac{12}{14.142}=0.8485$. ${\delta }_p^{\prime }=10\times 0.8485=8.485^{\circ }$, ${\delta }_y^{\prime }=8.485^{\circ }$. Why? We shrink the tilt vector to length $12^{\circ }$ but keep its angle, so the direction of the commanded torque is preserved — only its magnitude is capped. This is the standard autopilot saturation strategy.

Verify: $\sqrt{8.485^2+8.485^2}=12.000^{\circ }$ ✓ (lands exactly on the limit), and the ratio ${\delta }_p^{\prime }/{\delta }_y^{\prime }=1$ is preserved, so the torque direction is unchanged. ✓

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Example 7 — Case G: limiting value, thrust loss at the hard stop

Forecast: loss around 0.5%, 2%, or 10%?

1. Fractional axial loss: $1-\cos 12^{\circ }=1-0.97815=0.02185=2.19\%$. Why? Forward thrust is $T\cos \delta$; the missing piece is what tilting costs you. The loss is quadratic in $\delta$ (parent §3).
2. Side force: $T\sin 12^{\circ }=1.0\times 10^6\times 0.20791=2.079\times 10^5\text{N}=207.9$ kN. Why? The thrust vector of magnitude $T$, tilted by $\delta$, splits into an axial part $T\cos \delta$ (forward) and a transverse part $T\sin \delta$ (sideways). That transverse component is the steering force — it is simply the perpendicular leg of decomposing the one thrust vector.
3. Trade read-out: ~2.2% forward loss buys ~208 kN of sideways authority.

Verify: Taylor check ${\delta }^2/2$ with $12^{\circ }=0.2094$ rad gives $0.02193$ — matches $2.19\%$ to two figures, confirming the loss is genuinely second-order. ✓ Units: N throughout.

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Example 8 — Case H: real-world word problem, dead-engine degeneracy

Forecast: does the torque halve, or vanish?

1. Torque before shutdown: $M=\ell T\sin 5^{\circ }=6\times 8\times 10^5\times 0.08716=4.184\times 10^5\text{N⋅m}$. Why? Baseline steering capability with the engine live.
2. Torque after shutdown: $M=\ell \cdot 0\cdot \sin 5^{\circ }=0\text{N⋅m}$. Why? TVC torque is proportional to thrust — no thrust means the gimbal is a useless lever, however far it is tilted. This is the degenerate $T=0$ case.
3. What takes over: with no thrust, only the Reaction Control System (RCS) (cold-gas/monopropellant thrusters) can still generate attitude torque. TVC and RCS are complementary for exactly this reason.

Verify: $M\propto T$, so $T=0\Rightarrow M=0$ regardless of $\delta$; $4.184\times 10^5\to 0$. Also note Center of Mass Migration no longer matters for TVC once thrust is gone — the arm $\ell$ multiplies a zero force, so it is irrelevant. ✓

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Example 9 — Case I: the exam twist (units + roll)

Forecast: will the degree-plugging error be a small slip or a huge factor?

1. Wrong (degrees in the radian formula): $M_{\text{wrong}}=6\times 8\times 10^5\times 5=2.4\times 10^7\text{N⋅m}$. Why show it? To expose the trap: the linear law needs $\delta$ in radians.
2. Correct linear law: convert $5^{\circ }=0.08727$ rad, then $M=6\times 8\times 10^5\times 0.08727=4.189\times 10^5\text{N⋅m}$. Why? $M\approx \ell T\delta$ is derived from $\sin \delta \approx \delta$, and that approximation only holds when $\delta$ is measured in radians.
3. Size of the error: the wrong answer is too big by a factor $\frac{180}{\pi }\approx 57.3$. Why? Radians→degrees is exactly this conversion factor.
4. Roll authority is zero: for roll, the torque is $\vec{r}\times \vec{F}$ about the roll axis $x_b$. The thrust is applied on the axis at $(-\ell ,0,0)$, and its transverse component lies in the $y_b$–$z_b$ plane. The lever arm for a roll ($x_b$) torque is the perpendicular distance from the $x_b$ axis to the thrust line — but a centered engine's thrust line passes through the $x_b$ axis, so that distance is $\approx 0$, giving roll torque $\approx 0$. Why? Torque about an axis $=$ force $\times$ perpendicular distance to that axis; here the distance vanishes. Roll therefore needs canted engines, differential gimbaling, or RCS.

Verify: $\frac{2.4\times 10^7}{4.189\times 10^5}=57.3=\frac{180}{\pi }$ ✓, confirming the mistake is a pure unit error. Roll torque $=0$ by the zero-perpendicular-distance argument. ✓

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Machine-check summary

Every numeric answer on this page is verified by the checks in the ===VERIFY=== block below. Here is the roll-call so you can see nothing was skipped:

Example	Key numbers checked
1	$M=6.0\times 10^4$ N·m, $\delta =0.012$ rad, $0.687^{\circ }$
2	$M_{\text{exact}}=1.710\times 10^6$, $M_{\text{lin}}=1.745\times 10^6$, error $\approx 2\%$
3	$M=0$, $T_{\text{axial}}=T$ at $\delta =0$
4	$
5	${\delta }_{\text{tot}}=10^{\circ }<12^{\circ }$
6	${\delta }_{\text{tot}}=14.142^{\circ }$, $k=0.8485$, clipped legs → $12^{\circ }$
7	loss $=0.0219$, side force $=2.079\times 10^5$ N, Taylor $=0.02193$
8	$M=4.184\times 10^5$ before, $0$ after
9	wrong $=2.4\times 10^7$, correct $=4.189\times 10^5$, factor $=57.3$

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Recall Which cell did each example fill? (cover and recall)

Example 1 covers ::: Case A (small angle, linear solve) Example 2 covers ::: Case B (large angle, exact sine) Example 3 covers ::: Case C (δ=0 degenerate baseline) Example 4 covers ::: Case D (sign / steer direction) Example 5 covers ::: Case E (dual-gimbal feasible) Example 6 covers ::: Case F (dual-gimbal clipped) Example 7 covers ::: Case G (thrust loss at the limit) Example 8 covers ::: Case H (dead engine, T→0) Example 9 covers ::: Case I (degrees/radians trap + zero roll)