Exercises — Pitch program — open-loop pitch-over

Open-loop: the commanded pitch angle is a fixed function of time ${\theta }_c(t)$, decided before flight, not recomputed from where the rocket actually is. (An attitude loop still tracks that command — open-loop refers to the command itself.) What turns the rocket: gravity. After one small initial kick, gravity's sideways component $g\cos \gamma$ bends the velocity vector; the engine stays pointed along the body axis with near-zero angle of attack.

$\cos 90^{\circ }=0$, so $\dot{\gamma }=-\frac{g\cdot 0}{v}=0.$ A perfectly vertical rocket has zero turn rate — gravity is exactly opposite the velocity, none of it is sideways. This is why you need a kick: without tipping slightly off vertical, $\cos \gamma$ stays $0$ and the rocket never begins to turn.

$\cos 89^{\circ }=0.017452$. $\dot{\gamma }=-\frac{9.8\times 0.017452}{120}=-0.001425\text{rad/s}.$ Convert: $1\text{rad}=57.2958^{\circ }$, so $-0.001425\times 57.2958=-0.0817^{\circ }/\text{s}.$ The rocket turns only about $0.082^{\circ }$ each second — barely at all. That is why the initial kick must actively start the turn.

$\dot{\gamma }=-\frac{9.8\times 0.017452}{60}=-0.002850\text{rad/s}=-0.1633^{\circ }/\text{s}.$ Exactly twice the L2.1 rate, because $\dot{\gamma }\propto 1/v$ and we halved $v$. A slow rocket gravity-turns faster, so it needs a smaller initial kick.

$\cos 45^{\circ }=0.707107$. $\dot{\gamma }=-\frac{9.8\times 0.707107}{800}=-0.008662\text{rad/s}=-0.4962^{\circ }/\text{s}.$ Even though $v$ is much larger, $\cos 45^{\circ }$ is far bigger than $\cos 89^{\circ }$, so the turn rate is higher than at kick-over. The turn accelerates as $\gamma$ drops — until $v$ grows enough to slow it down again.

Write the rate as a function of $v$: $\dot{\gamma }(v)=-\frac{g\cos \gamma }{v}$. Replace $v$ by $v/2$: $\dot{\gamma }(v/2)=-\frac{g\cos \gamma }{v/2}=2\left(-\frac{g\cos \gamma }{v}\right)=2\dot{\gamma }(v).$ Halving $v$ doubles the magnitude — confirmed. For $4v$: $\dot{\gamma }(4v)=\frac{1}{4}\dot{\gamma }(v)=\frac{1}{4}(-0.001425)=-0.000356\text{rad/s}$.

The sideways component is $g\cos \gamma$. We want $g\cos \gamma =\frac{1}{2}g$, i.e. $\cos \gamma =\frac{1}{2}\Rightarrow \gamma =\arccos (0.5)=60^{\circ }.$ ($\arccos$ asks "which angle has this cosine?" — it undoes $\cos$.) At $\gamma =60^{\circ }$, half of gravity is busy turning the velocity and half is fighting the climb along the path. As $\gamma$ keeps dropping toward $0^{\circ }$, $\cos \gamma \to 1$ and all of gravity turns the arrow — but by then $v$ is huge, so the actual turn rate is still modest.

$\cos 88.5^{\circ }=0.0261769$, $\cos 89.5^{\circ }=0.0087265$. $\dot{\gamma }(88.5^{\circ })=-\frac{9.8\times 0.0261769}{100}=-0.0025653\text{rad/s},$ $\dot{\gamma }(89.5^{\circ })=-\frac{9.8\times 0.0087265}{100}=-0.0008552\text{rad/s}.$ The ratio is $0.0261769/0.0087265=3.0$ — the $88.5^{\circ }$ kick turns three times faster right from the start. Because $\cos \gamma$ is very small and steep near $90^{\circ }$, a tiny angle change swings the initial rate a lot; that seeds a self-amplifying bend (more turn → bigger $\cos \gamma$ → more turn), so the burnout angle ends up wildly different. This sensitivity is exactly why the pitch-over must be tuned carefully.

Setup (WHAT): point mass, speed $v$, flight-path angle $\gamma$, thrust $T$ along $\vec{v}$ (zero angle of attack), gravity $mg$ down. Split forces into "along $\vec{v}$" (tangential) and "across $\vec{v}$" (normal). Normal balance (WHY): the only force that can rotate the velocity vector is the piece of gravity perpendicular to it. From the triangle in figure s01, that piece is $mg\cos \gamma$, pointing to bend $\vec{v}$ downward. Newton's law across the path (centripetal form) is $mv\dot{\gamma }=-mg\cos \gamma .$ The left side $mv\dot{\gamma }$ is mass $\times$ (rate of turning of the arrow $\times$ speed) — the sideways acceleration. The minus sign: gravity turns $\gamma$ down. Cancel $m$ (WHY): mass appears on both sides, so it drops out — the turn rate does not depend on how heavy the rocket is, only on $g$, $v$, $\gamma$: $\dot{\gamma }=-\frac{g\cos \gamma }{v}.■$

Convert the target: $-0.1^{\circ }/\text{s}=-0.1/57.2958=-0.00174533\text{rad/s}$. Solve the formula for $v$: $v=-\frac{g\cos {\gamma }_0}{\dot{\gamma }}=-\frac{9.8\times \cos 89^{\circ }}{-0.00174533}.$ With $\cos 89^{\circ }=0.0174524$: $v=\frac{9.8\times 0.0174524}{0.00174533}=97.98\text{m/s}\approx 98\text{m/s}.$ So the kick should be commanded when the vehicle is near $98\text{m/s}$ to hit that turn rate.

Eliminate time (WHY): we want how speed depends on angle, not on the clock, so divide the two rate equations — the $dt$ cancels: $\frac{dv}{d\gamma }=\frac{\dot{v}}{\dot{\gamma }}=\frac{(T-mg\sin \gamma )/m}{-g\cos \gamma /v}.$ Write $T/m=ng$ (since $n=T/(mg)$):

$dv/d\gamma =0$ requires the numerator $v(n-\sin \gamma )=0$. Since $v\ne 0$: $n-\sin \gamma =0\Rightarrow \sin \gamma =n=\mathrm{1.5.}$ But $\sin \gamma \le 1$ always, and here $n=1.5>1$, so there is no such $\gamma$ — the numerator never vanishes. Physically: because thrust exceeds weight everywhere along the path ($n>1$), the along-path acceleration $g(n-\sin \gamma )$ stays positive at every angle ($\sin \gamma \le 1<n$), so the rocket is always speeding up. Only if $n<1$ (thrust below weight) could $\sin \gamma =n$ have a solution, marking an angle where climbing losses exactly eat the thrust and $v$ momentarily stops rising.

$\dot{\gamma }$ at $\gamma =90^{\circ }$ ::: Zero — no sideways gravity, no turn without a kick. Effect of halving $v$ on turn rate ::: Doubles it ($\dot{\gamma }\propto 1/v$). Does mass appear in $\dot{\gamma }$? ::: No — it cancels in the normal force balance. Sign of $\dot{\gamma }$ during ascent ::: Negative — $\gamma$ falls from $90^{\circ }$ toward $0^{\circ }$.