Exercises — Propulsive forces — thrust misalignment, gimbal angle

No torque. Torque about the CoM is $\vec{\tau }=\vec{r}\times \vec{F}$, and it is nonzero only when the force has a component acting on a lever arm (a perpendicular distance from the CoM to the line of the force). Here the thrust line passes through the CoM, so that perpendicular distance is zero, and $\tau =0$. The engine only translates the rocket forward — no twist. (See Torque and Moment of Inertia.)

The gimbal angle $\delta$ is commanded — the computer deliberately tilts the engine to generate a steering torque. The misalignment ${\delta }_{err}$ is an unintended offset the control system must continuously fight. Both produce the same kind of torque $LT\sin (\cdot )$; the difference is intent, not physics.

Step 1 — convert (WHY: trig needs radians). $\delta =4\times \frac{\pi }{180}=0.06981\text{rad}.$ Step 2 — apply the torque formula. Torque comes from the lateral thrust component $T\sin \delta$ on lever arm $L$: $\tau =LT\sin \delta =15\times 600000\times \sin (0.06981).$ Step 3 — evaluate. $\sin (0.06981)=0.06975$, so $\tau =15\times 600000\times 0.06975\approx 6.28\times 10^5\text{N⋅m}.$ That is the twisting authority available to steer this stage.

The axial (forward) part is $T\cos \delta$; the lost part is $T(1-\cos \delta )$. $\cos (0.06981)=0.99756\Rightarrow \text{loss fraction}=1-0.99756=0.002436=0.244\%.$ $\text{loss}=600000\times 0.002436\approx 1.46\times 10^3\text{N}=1.46\text{kN}.$ A quarter of a percent — negligible — while the torque above is fully usable. This lopsidedness (loss $\propto {\delta }^2$, torque $\propto \delta$) is the whole reason gimballing works.

Step 1 — rotational Newton's law. $\ddot{\theta }=\frac{\tau }{I}=\frac{6.28\times 10^5}{2.5\times 10^6}=0.2513{\text{rad/s}}^2.$ Step 2 — integrate constant acceleration from rest (WHY: constant $\delta$ ⇒ constant $\tau$ ⇒ constant $\ddot{\theta }$): $\dot{\theta }=\ddot{\theta }t=0.2513\times 0.8=0.2010\text{rad/s}.$ Step 3 — convert to degrees/s for intuition: $0.2010\times \frac{180}{\pi }=11.5\text{°/s}.$ Over just 0.8 s the rocket is already turning at ~11.5°/s — this is why real gimbal commands are small and brief, then reversed to stop the rotation. (See Attitude Control & Stability.)

Tilt torque (A): ${\tau }_A=LT\sin \delta$. Offset torque (B): here the thrust is axial ($+x$) but its line misses the CoM by $d$; the perpendicular lever arm is $d$, so ${\tau }_B=dT$ (a pure moment — the axial force itself is the lever-arm force). Set equal: $dT=LT\sin \delta \Rightarrow \boxed{d=L\sin \delta }.$ $T$ cancels — the geometry alone sets the equivalence. Numerically: $d=15\times \sin (2^{\circ })=15\times 0.034899=0.5235\text{m}.$ So a mere ~52 cm sideways mounting error mimics a 2° tilt's twist. (Note: unlike a tilt, an offset costs no axial thrust — the force is still fully forward.)

(a) Disturbance torque. ${\delta }_{err}=0.6\times \frac{\pi }{180}=0.010472\text{rad},\sin {\delta }_{err}=\mathrm{0.010472.}$ ${\tau }_{err}=LT\sin {\delta }_{err}=15\times 600000\times 0.010472\approx 9.42\times 10^4\text{N⋅m}.$ (b) Cancelling command. To null the disturbance, the commanded gimbal must produce an equal-and-opposite torque. Since both use the same $L$ and $T$: $LT\sin {\delta }_c=LT\sin {\delta }_{err}\Rightarrow \sin {\delta }_c=\sin {\delta }_{err}\Rightarrow {\delta }_c={\delta }_{err}=0.6^{\circ }.$ Authority consumed: $\frac{0.6^{\circ }}{5^{\circ }}=0.12=12\%.$ So 12% of the steering budget is permanently spent just holding still — a real cost of misalignment that eats into control margin. (See Attitude Control & Stability.)

Torque scales linearly with $L$ (the $T$ and $\delta$ are unchanged): $\frac{{\tau }_2}{{\tau }_1}=\frac{L_2}{L_1}=\frac{22}{15}=\mathrm{1.467.}$ The same gimbal command produces 47% more torque late in the burn. Meaning: the controller cannot use a fixed relationship between "commanded angle" and "resulting turn" — as $L$ grows, each degree of gimbal bites harder, so the guidance law must adapt its gains through flight. (This is exactly the Center of Mass shift during burn problem TVC systems are tuned around.)

(a) Required angular acceleration. Constant $\ddot{\theta }$ from rest over $t$ reaches rate $\dot{\theta }=\ddot{\theta }t$. Convert target rate to rad/s first: ${\dot{\theta }}_{\text{target}}=10\times \frac{\pi }{180}=0.17453\text{rad/s}.$ $\ddot{\theta }=\frac{{\dot{\theta }}_{\text{target}}}{t}=\frac{0.17453}{1.2}=0.14544{\text{rad/s}}^2.$ (b) Torque needed (rotational Newton's law): $\tau =I\ddot{\theta }=3.2\times 10^6\times 0.14544=4.654\times 10^5\text{N⋅m}.$ (c) Commanded gimbal. Invert $\tau =LT\sin \delta$: $\sin \delta =\frac{\tau }{LT}=\frac{4.654\times 10^5}{18\times 750000}=\frac{4.654\times 10^5}{1.35\times 10^7}=\mathrm{0.034473.}$ $\delta =\arcsin (0.034473)=0.034480\text{rad}=1.975^{\circ }.$ Well within the $\pm 7^{\circ }$ limit — plenty of margin. (Here $\arcsin$ answers the question "which angle has this sine?" — it undoes the $\sin$ that put the lateral thrust component into the torque.) (d) Axial loss at $\delta =0.034480\text{rad}$: $1-\cos \delta =1-0.999406=0.000594=0.0594\%.$ Under a tenth of a percent lost to steer at $\sim 2°$ — the payoff of gimballing in one number.

Reversing the gimbal to $-\delta$ reverses the torque to $-\tau$, giving $\ddot{\theta }=-0.14544{\text{rad/s}}^2$ (same magnitude, opposite sign). Starting from $\dot{\theta }=0.17453\text{rad/s}$ and decelerating to $0$: $t_{\text{stop}}=\frac{\dot{\theta }}{|\ddot{\theta }|}=\frac{0.17453}{0.14544}=1.2\text{s}.$ Symmetric because the deceleration magnitude equals the acceleration magnitude (same $L,T,\delta ,I$), so building up and killing the rate take equal time — the classic bang–bang steering pulse: tilt one way to start the turn, tilt the mirror-image way to stop it. (See Attitude Control & Stability.)