Worked examples — Pitch program — open-loop pitch-over

This page is a case-by-case drill for the gravity-turn pitch program from the parent note. We take the one master law and push it into every corner: every angle regime, the two degenerate endpoints ($\gamma =90^{\circ }$ straight up, $\gamma =0^{\circ }$ horizontal), the sign of the turn rate, a real launch word-problem, and an exam-style twist.

The single tool we lean on the whole way:

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The scenario matrix

Every situation this topic can throw at you falls into one of these cells. The worked examples below are each tagged with the cell they cover.

Cell	What makes it special	Example
A. Near-vertical ($\gamma \to 90^{\circ }$)	$\cos \gamma \to 0$, so $\dot{\gamma }\to 0$ — barely turns	Ex 1
B. Mid-turn ($\gamma \approx 45^{\circ }$)	$\cos \gamma$ large — fastest turning	Ex 2
C. Near-horizontal ($\gamma \to 0^{\circ }$)	$\cos \gamma \to 1$, but $v$ huge — turn slows again	Ex 3
D. Speed dependence (fix $\gamma$, vary $v$)	$\dot{\gamma }\propto 1/v$ — the inverse law	Ex 4
E. Degenerate: exactly vertical ($\gamma =90^{\circ }$)	$\dot{\gamma }=0$ — the "stuck" case, why a kick is needed	Ex 5
F. Degenerate: exactly horizontal ($\gamma =0^{\circ }$)	maximum $\cos \gamma =1$; check burnout instant	Ex 6
G. Real-world word problem	convert units, find turn over a time window	Ex 7
H. Exam twist: sign / integration	eliminate time, $dv/d\gamma$, over-vs-under-kick	Ex 8

The figure above plots $\cos \gamma$ across the whole turn. Notice the two ends (vertical and horizontal) are where the raw turning force is smallest and largest respectively — but the actual turn rate also divides by $v$, which is why the real turning bunches up early (cell A→B) when $v$ is still small.

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Worked examples

Ex 1 — Cell A: near-vertical, high speed

Ex 2 — Cell B: mid-turn, fastest bending

Ex 3 — Cell C: near-horizontal, very fast

Ex 4 — Cell D: the inverse-speed law

Ex 5 — Cell E: the degenerate vertical case

Ex 6 — Cell F: the degenerate horizontal case

Ex 7 — Cell G: real-world word problem

Ex 8 — Cell H: exam twist, eliminate time & judge the kick

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Recall Which cell is the "stuck" one, and why?

Cell E, $\gamma =90^{\circ }$ exactly ::: $\cos 90^{\circ }=0$ makes $\dot{\gamma }=0$, so a perfectly vertical rocket never turns — the pitch kick must break this symmetry.

Recall Why does the fastest

turn rate NOT happen at horizontal (where $\cos \gamma$ is biggest)? Because $\dot{\gamma }\propto 1/v$ ::: by the time $\gamma$ is small, $v$ is huge, and the large speed in the denominator overwhelms the maximal $\cos \gamma$.

See also: Gravity turn trajectory · Angle of attack and dynamic pressure (max-Q) · Thrust-to-weight ratio · Gravity loss and steering loss · Tsiolkovsky rocket equation · Closed-loop ascent guidance (PEG / IGM) · Attitude control and thrust vectoring (gimbal).