Question bank — Thrust vector control — single-gimbal, dual-gimbal; TVC angles

False. The engine makes the same magnitude $T$; TVC only rotates that vector. Tilting splits $T$ into a forward part $T\cos \delta$ and a sideways part $T\sin \delta$ — nothing is added.

True. $M=\ell T\sin 0=0$. A thrust pointing straight through the CoM line has no lever arm sideways, so it cannot twist the vehicle.

True. $T\cos 0=T$, the largest $\cos$ can be. Any tilt only lowers the forward push, so straight-back is the most fuel-efficient direction.

True. $M=\ell T\sin \delta$ is linear in $\ell$. A longer lever from CoM to pivot means the same sideways force twists harder — this is Torque and Moment Arm at work.

False. It only holds while $\sin \delta \approx \delta$ (good to ~1% below ~$8^{\circ }$). At $45^{\circ }$, $\sin \delta =0.707$ but $\delta =0.785$ rad — an 11% error. Use the exact $\sin$.

False. It handles pitch and yaw, but the thrust line passes near the roll axis, so the moment arm about roll is ~zero → no roll torque. Roll needs canted engines, differential gimbaling, or Reaction Control System (RCS).

False. ${\delta }_p$ and ${\delta }_y$ are perpendicular components of one physical tilt, so the geometric tilt is ${\delta }_{\text{tot}}=\sqrt{{\delta }_p^2+{\delta }_y^2}$. That magnitude — not the sum — hits the mechanical stop.

False. Loss $=1-\cos \delta \approx {\delta }^2/2$ — it is quadratic, second order in $\delta$. That is exactly why small-angle steering is nearly free while the torque grows linearly.

True. At small $\delta$, you buy a first-order twist ($\propto \delta$) while paying only a second-order fuel cost ($\propto {\delta }^2$). The ratio (benefit/cost) blows up as $\delta \to 0$.

False. Then $\ell =0$, so $M=\ell T\sin \delta =0$ regardless of $\delta$. With no lever arm, the sideways force pushes but cannot twist — see Center of Mass Migration for why $\ell$ changes in flight.

Both $T$ and $\delta$ raise torque ($M=\ell T\sin \delta$), but $T$ is set by the Rocket Thrust Equation and mission profile, not freely dial-able for steering. The controllable knob is $\delta$; that is what the Attitude Control Autopilot moves.

The minus is only a sign convention from the cross product $\vec{r}\times \vec{F}$; it encodes the rotation direction, not a weakening. The magnitude $\ell T\sin \delta$ is the real strength.

The combined tilt is ${\delta }_{\text{tot}}=\sqrt{5^2+5^2}=7.07^{\circ }>6^{\circ }$, so it violates the physical stop and must be clipped (scale both by $6/7.07$). Individual angles being legal is not enough.

Torque gives angular acceleration via $M=I\dot{\omega }$ (rotational Newton's law), not angle directly. You need Rigid Body Rotational Dynamics: integrate $\dot{\omega }$ twice, and $I$ (moment of inertia) matters.

$\sin \delta$ is the sideways (steering) fraction. The loss to forward thrust is $1-\cos \delta$, which is far smaller at small angles (${\delta }^2/2$ vs $\delta$).

RCS burns extra propellant purely for control and adds mass; TVC reuses the main engine's thrust, losing only ~$1\%$ at max angle. TVC is usually far cheaper — RCS shines when the main engine is off or for fine/roll authority.

Because steering needs a sideways force to make torque, and rotating a fixed-length vector gives you exactly that transverse component $T\sin \delta$ while keeping the magnitude fixed.

The transverse component is opposite the tilt angle in the right triangle of thrust components, and opposite/hypotenuse is precisely $\sin \delta$. $\cos \delta$ gives the adjacent (forward) part.

For tiny $\delta$ the sine curve is nearly its own tangent line through the origin, so $\sin \delta \approx \delta$ (the Small-Angle Approximation). A linear law $M\approx \ell T\delta$ is far easier for the controller to invert and stabilize.

Forward thrust $T\cos \delta$ doesn't contain $\ell$ at all. Torque $=\text{force}\times \text{lever arm}$, so only the twisting effect scales with $\ell$; the push does not.

The physical gimbal can only swing so far off-center in any direction; that maximum reach is a radius, and the actual tilt is the vector sum $\sqrt{{\delta }_p^2+{\delta }_y^2}$, which is what reaches that radius.

One dual-gimbal engine gives torque in both planes with a single actuator set — lighter and simpler than mounting and coordinating two separate single-axis engines.

$M=\ell T\sin 90^{\circ }=\ell T$ — the maximum possible torque, but forward thrust $T\cos 90^{\circ }=0$, so you'd get all twist and zero push. Real gimbals never reach this; it's a limiting sanity check.

It diverges: torque $\sim \ell T\delta$ over loss $\sim T{\delta }^2/2$ scales like $2\ell /\delta \to \infty$. Small angles are maximally efficient per unit torque — the mathematical heart of "steering is cheap."

It vanishes: $M=\ell T\sin \delta =0$, so the engine cannot twist the vehicle at all. Center of Mass Migration as propellant drains is a real reason $\ell$ (and thus authority) changes over a burn.

No. $M=\ell \cdot 0\cdot \sin \delta =0$ — with no thrust there is no force to redirect, so no torque. Coasting phases rely on Reaction Control System (RCS) instead.

The first is exactly at the edge (${\delta }_{\text{tot}}=6^{\circ }$, feasible). The second gives ${\delta }_{\text{tot}}=\sqrt{6^2+0.1^2}\approx 6.0008^{\circ }>6^{\circ }$, so it must be clipped — even a whisker of the second axis pushes past the stop.

Not with $\delta =0$ — that gives zero torque, so existing spin continues (Newton: no torque, no change in $\omega$). To halt rotation the autopilot must command a counter-deflection.