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

What we decide: at $25^{\circ }$ the small-angle shortcut is no longer trustworthy. Why: the linear law $M\approx \ell T\delta$ is only the first term of $\sin \delta =\delta -{\delta }^3/6+\dots$. Converting first, $\delta =25^{\circ }\times \frac{\pi }{180}=0.4363$ rad, so $\sin \delta =0.4226$, and the linear form overshoots by $\frac{0.4363-0.4226}{0.4226}=3.2\%.$ Answer: use the exact law $\boxed{M=\ell T\sin \delta }$. The small-angle law is only valid to about $\pm 8^{\circ }$.

Resolve the tilted thrust vector along and across the vehicle axis (this is just the trigonometry of a rotated arrow of length $T$, exactly the two dotted arrows in figure s01):

• (a) Axial (useful) thrust: $T_{\text{axial}}=T\cos \delta .$
• (b) Transverse (steering) force: $T_{\perp }=T\sin \delta .$

The steering force $T\sin \delta$ times the arm $\ell$ is exactly the torque $M=\ell T\sin \delta$.

Step 1 — convert the angle. $\delta =4^{\circ }\times \frac{\pi }{180}=0.06981$ rad. Why: the linear law needs radians (our units convention). Step 2 — apply $M\approx \ell T\delta$. $M=5\times (500\times 10^3)\times 0.06981=1.745\times 10^5\text{N⋅m}.$ Why this step: torque is the steering force $T\sin \delta$ acting through the lever $\ell$, i.e. $M=\ell T\sin \delta$; since $4^{\circ }$ is well inside the small-angle window we replace $\sin \delta$ by $\delta$ (error $<0.1\%$), turning the law into the linear product $\ell T\delta$ that we can evaluate directly. We multiply the three given quantities because torque grows with each: more lever, more thrust, or more tilt all twist the vehicle harder. Answer: $M\approx 1.75\times 10^5$ N·m. This torque feeds directly into Rigid Body Rotational Dynamics via $M=I\dot{\omega }$.

Step 1 — needed torque (Newton's rotational law, Torque and Moment Arm): $M=I\dot{\omega }=8.0\times 10^4\times 0.6=4.8\times 10^4\text{N⋅m}.$ Why: to spin the rocket up at $\dot{\omega }$ you must supply exactly the torque $M=I\dot{\omega }$ — this is the rotational twin of $F=ma$. Step 2 — invert the linear law $M=\ell T\delta$: $\delta =\frac{M}{\ell T}=\frac{4.8\times 10^4}{5\times 5\times 10^5}=0.0192\text{rad}.$ Why: we know the torque we need and the fixed hardware ($\ell$, $T$); the only free knob is $\delta$, so we solve the linear law for it by dividing. Step 3 — to degrees: $\delta =0.0192\times \frac{180}{\pi }=1.10^{\circ }.$ Answer: about $1.1^{\circ }$ — small enough that the linear law was legitimate (self-consistent).

Convert first. $\delta =6^{\circ }\times \frac{\pi }{180}=0.10472$ rad; every $\sin$/$\cos$ below uses this radian value (the degree label is only for reading). (a) Axial loss. Fractional loss $=1-\cos \delta$. With $\delta =0.10472$ rad: $\cos \delta =0.99452$, so loss $=0.00548=0.548\%.$ (b) Side force. $T\sin \delta =5\times 10^5\times \sin (0.10472)=5\times 10^5\times 0.10453=5.226\times 10^4$ N $=52.3$ kN. Compare. We spend $0.55\%$ of thrust (about $2.7$ kN of forward push, since $0.00548\times 500\text{ kN}$) and gain $52.3$ kN of steering force — roughly a $19:1$ return. Why it works: the loss $1-\cos \delta \approx {\delta }^2/2$ is second order in $\delta$, while the side force $T\sin \delta \approx T\delta$ is first order. At small $\delta$ a first-order quantity dwarfs a second-order one — that is the whole reason gimbaling is efficient.

Step 1 — combine into one physical tilt (state the assumption). The two deflections are perpendicular components of a single tilt. Assumption used: for small angles, a tilt of ${\delta }_p$ about one axis and ${\delta }_y$ about a perpendicular axis compose to first order into one tilt of magnitude $\sqrt{{\delta }_p^2+{\delta }_y^2}$. Warning on second-order coupling: if ${\delta }_p$ or ${\delta }_y$ grew large (tens of degrees), the two rotations would not commute and the true combined tilt would pick up second-order cross-terms of order ${\delta }_p{\delta }_y$; the clean Pythagorean sum would then be only approximate. Here both angles are $\le 5^{\circ }$, so those cross-terms are $\lesssim 0.4\%$ and we safely use the first-order rule. Because this Pythagorean sum is a linear (degree-homogeneous) operation — no $\sin$/$\cos$ enters — doing it in degrees or radians differs only by the common factor $\frac{\pi }{180}$, which cancels in the comparison ${\delta }_{\text{tot}}$ vs ${\delta }_{\max }$. So we may stay in degrees: ${\delta }_{\text{tot}}=\sqrt{{\delta }_p^2+{\delta }_y^2}=\sqrt{5^2+5^2}=7.071^{\circ }.$ Step 2 — compare to the stop. $7.071^{\circ }>7^{\circ }$ → not reachable, must clip. Step 3 — clip along the same direction. Scale both components by the same factor $k=\frac{{\delta }_{\max }}{{\delta }_{\text{tot}}}=\frac{7}{7.071}=0.9899$ so the tilt direction is preserved while its length lands exactly on the stop: ${\delta }_p^{\prime }={\delta }_y^{\prime }=5\times 0.9899=4.950^{\circ },\sqrt{4.950^2+4.950^2}=7.000^{\circ }.✓$ Answer: clip to $(4.95^{\circ },4.95^{\circ })$.

(a) Early burn. Required torque $M=I\dot{\omega }=1.5\times 10^5\times 1.0=1.5\times 10^5$ N·m. Angle $\delta =\frac{M}{\ell T}=\frac{1.5\times 10^5}{7\times 9\times 10^5}=0.02381$ rad $=1.364^{\circ }$. Comfortable. (b) Late burn. $M=6.0\times 10^4\times 1.6=9.6\times 10^4$ N·m. $\delta =\frac{9.6\times 10^4}{4\times 7\times 10^5}=0.03429$ rad $=1.964^{\circ }$. (c) Margin. Both $1.364^{\circ }$ and $1.964^{\circ }$ sit well under the $8^{\circ }$ stop, so yes — margin remains. But notice the trend: a shorter arm and lower thrust make each degree of gimbal less effective, so the required angle grows even as the vehicle gets lighter. The tightest gimbal demand often occurs late in the burn, exactly when $\ell$ and $T$ have shrunk. This is the synthesis lesson: control authority $=\ell T\delta$ depends on all three factors, and two of them decay during flight.

Set up the relative error. The linear model gives $\ell T\delta$; the truth is $\ell T\sin \delta$. Overprediction fraction: $\epsilon (\delta )=\frac{\ell T\delta -\ell T\sin \delta }{\ell T\sin \delta }=\frac{\delta -\sin \delta }{\sin \delta }=\frac{\delta }{\sin \delta }-1.$ First, a rough locator (with an honest error warning). The series $\frac{\delta }{\sin \delta }=1+\frac{{\delta }^2}{6}+\frac{7{\delta }^4}{360}+\dots$ suggests $\frac{{\delta }^2}{6}\approx 0.05\Rightarrow \delta \approx \sqrt{0.3}=0.5477$ rad. But at this $\delta$ the next term $\frac{7{\delta }^4}{360}=\frac{7(0.5477{)}^4}{360}=0.00175$ is already $\sim 3.5\%$ of the $0.05$ we are matching — so the two-term series is only trustworthy to a couple of percent here and we must not report it as the final answer. It is a starting guess only. Solve exactly (numerically). We solve $\frac{\delta }{\sin \delta }=1.05$. Starting Newton iteration from the guess $0.5477$ rad converges to $\boxed{\delta =0.5401\text{rad}=30.9^{\circ }}$ — about $1.4\%$ below the crude series guess, consistent with the truncation warning above. (Check: $\sin (0.5401)=0.5144$, and $0.5401/0.5144=1.050$ ✓.) Answer: the linear law reaches a $5\%$ overprediction at about $\boxed{31^{\circ }}$. Insight: it stays within $5\%$ all the way to $\sim 31^{\circ }$ — far beyond any real gimbal stop ($\lesssim 8$–$10^{\circ }$). That is why flight software can safely use the linear form: real deflections never come close to where it breaks.

(a) Simplify. Use the half-angle identities $\sin \delta =2\sin \frac{\delta }{2}\cos \frac{\delta }{2}$ and $1-\cos \delta =2{\sin }^2\frac{\delta }{2}$: $\eta =\frac{2\sin \frac{\delta }{2}\cos \frac{\delta }{2}}{2{\sin }^2\frac{\delta }{2}}=\frac{\cos \frac{\delta }{2}}{\sin \frac{\delta }{2}}=\cot \frac{\delta }{2}.$ Since $\cot$ decreases on $(0,\pi )$, $\eta$ falls monotonically as $\delta$ rises — smaller deflections are always more efficient per unit of thrust sacrificed. (b) Values (convert first: $2^{\circ }=0.03491$ rad, $8^{\circ }=0.13963$ rad, so the half-angles are $1^{\circ }$ and $4^{\circ }$). $\eta (2^{\circ })=\cot (1^{\circ })=\cot (0.017453\text{ rad})=\mathrm{57.29.}$ Each unit of thrust lost buys $\approx 57$ units of steering force. $\eta (8^{\circ })=\cot (4^{\circ })=\cot (0.069813\text{ rad})=\mathrm{14.30.}$ Only $\approx 14$ now. Interpret: steering gets less thrifty as you deflect harder — the marginal return drops by a factor of four between $2^{\circ }$ and $8^{\circ }$. Autopilots therefore prefer many small corrections over a few large ones, and hand large-authority needs to the Reaction Control System (RCS).

Exact vs linear torque law ::: $M=\ell T\sin \delta$ exact; $M\approx \ell T\delta$ for $\delta \lesssim 8^{\circ }$. Why steering is cheap ::: side force is $O(\delta )$, thrust loss is $O({\delta }^2)$. Sign of the torque ::: odd in $\delta$ — flip $\delta$ and $M$ reverses sense, same magnitude; $\delta =0$ gives zero torque. Dual-gimbal limit constraint ::: $\sqrt{{\delta }_p^2+{\delta }_y^2}\le {\delta }_{\max }$ (vector magnitude, not sum). How to clip an over-command ::: scale both components by ${\delta }_{\max }/{\delta }_{\text{tot}}$ to keep direction. Where the linear law finally lies by 5% ::: near $\delta \approx 31^{\circ }$, far past any real gimbal stop. Steering efficiency form ::: $\eta =\cot (\delta /2)$, decreasing — small deflections are thriftiest.