Visual walkthrough — Pitch program — open-loop pitch-over

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Step 1 — What is a "flight-path angle"? Draw the velocity arrow

WHAT. A rocket in flight is at some point in the air, moving in some direction. We draw its velocity as an arrow: length = how fast (call it $v$, the speed), direction = where the motion points right now.

WHY. Before we can talk about "turning," we need a number that says how tilted the motion is. That number is the flight-path angle, written $\gamma$ (Greek letter "gamma"). It is the angle between the velocity arrow and the flat horizon.

PICTURE. Look at figure s01. The horizontal dashed line is the ground/horizon. The blue arrow is the velocity $v$. The pale-yellow wedge between them is $\gamma$.

• $\gamma =90^{\circ }$ → arrow points straight up (just off the pad).
• $\gamma =0^{\circ }$ → arrow points sideways (orbital, the goal).
• $\gamma$ shrinking over time = "the rocket is pitching over."

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Step 2 — What is $\dot{\gamma }$? The dot means "rate of change"

WHAT. We put a dot over a letter to mean "how fast that letter changes each second." So $\dot{\gamma }$ (say "gamma-dot") = how many degrees (or radians) the flight-path angle changes per second.

WHY. The whole ascent is a story of $\gamma$ sliding from $90^{\circ }$ down to $0^{\circ }$. The speed of that slide — $\dot{\gamma }$ — is exactly what a pitch program must control. If $\dot{\gamma }$ is too big we turn over and dive; too small and we never go horizontal.

PICTURE. Figure s02 shows the same blue arrow at two instants one second apart. The yellow wedge got smaller: that shrink, per second, is $\dot{\gamma }$. Because $\gamma$ is decreasing, $\dot{\gamma }$ is a negative number — this sign will reappear in the final formula, so remember it now.

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Step 3 — The forces on the rocket, and which one can bend the path

WHAT. Three forces act on our point-mass rocket: thrust $T$ (push from the engine, along the body), drag $D$ (air resistance, backward), and weight $mg$ (gravity, straight down). Here $m$ = mass, $g$ = gravity's pull per kilogram ($\approx 9.8{\text{m/s}}^2$).

WHY. In an ideal gravity turn the body points exactly along the velocity — the angle of attack $\alpha \approx 0$. That means thrust and drag both lie along the arrow. A force along the arrow can only speed it up or slow it down — it cannot bend it. Only a force sideways to the arrow can turn it. Gravity points straight down, so part of it is sideways to the tilted arrow. Gravity is the only turner.

PICTURE. Figure s03 shows the velocity arrow tilted at angle $\gamma$, with $T$ and $D$ drawn along it (chalk-blue and pink), and weight $mg$ pointing straight down (yellow). Notice weight is the only arrow not lined up with the motion.

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Step 4 — Split gravity into "along" and "across" the velocity

WHAT. We break the downward weight $mg$ into two pieces measured relative to the tilted velocity arrow:

• the part along the arrow (opposing the climb): $mg\sin \gamma$,
• the part across the arrow (bending it): $mg\cos \gamma$.

WHY. We used $\sin$ and $\cos$ — why these? Because $\sin$ and $\cos$ are exactly the tools that answer "how much of a straight-down arrow points in this other direction?" They are the shadow-lengths of the weight onto the two axes we care about. We pick them, not tan or anything else, because we need projections, and sin/cos are the projection ratios of a right triangle.

PICTURE. Figure s04 draws the right triangle formed by weight $mg$ (the down-arrow) and the two axes rotated to match the velocity. The along-piece hugs the arrow; the across-piece sticks out perpendicular. The angle inside the triangle is $\gamma$ (it copies over because both the horizon and the vertical rotate by the same $\gamma$).

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Step 5 — Newton across the path: turning = centripetal-like bending

WHAT. A velocity arrow of length $v$ that swings at rate $\dot{\gamma }$ has its tip moving sideways at speed $v\dot{\gamma }$. Newton says: (mass)×(that sideways acceleration) = (the across force). The only across force is $mg\cos \gamma$, pointing down, which reduces $\gamma$. So: $mv\dot{\gamma }=-mg\cos \gamma$

WHY the minus? The across-piece of gravity points downward and downward makes $\gamma$ shrink (Step 2 told us shrinking = negative $\dot{\gamma }$). So the right side must be negative — we write the minus explicitly.

Term by term:

• $m$ — the rocket's mass (how hard it is to redirect).
• $v\dot{\gamma }$ — how fast the tip of the velocity arrow swings sideways.
• $mg\cos \gamma$ — the across-component of gravity from Step 4.
• $-$ — because that gravity component bends $\gamma$ downward.

PICTURE. Figure s05 shows the velocity arrow sweeping through a small angle in one second; the little arc the tip traces has length $v\dot{\gamma }$, and the yellow across-force $mg\cos \gamma$ points along that arc, pulling the tip down.

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Step 6 — Cancel the mass: the gravity-turn law appears

WHAT. The mass $m$ sits on both sides of $mv\dot{\gamma }=-mg\cos \gamma .$ Divide both sides by $m$ (allowed — $m$ is never zero), then divide by $v$: $\boxed{\dot{\gamma }=-\frac{g\cos \gamma }{v}}$

WHY it's beautiful. The mass vanished. A heavy rocket and a light rocket at the same speed and angle turn at the same rate. The turn is set by only three things: gravity $g$, the current angle $\gamma$, and the current speed $v$.

Term by term:

• $g$ — bigger gravity → faster turn.
• $\cos \gamma$ — near vertical ($\gamma \to 90^{\circ }$) this is ~0, so the turn is barely-there; near horizontal it's ~1, turning is strongest.
• $v$ in the denominator — fast rockets turn slowly; slow, low rockets turn quickly. (This is the why speed matters flip: more inertia, less bendable.)
• $-$ — the arrow is always tipping down.

PICTURE. Figure s06 plots $\dot{\gamma }$ against $\gamma$: a curve that is flat (near zero) at $90^{\circ }$ and steepest near $0^{\circ }$, with three sample rockets at different $v$ showing slower turns for faster $v$.

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Step 7 — The edge cases: vertical start, horizontal end, zero speed

WHAT. We must check every corner so the reader never meets a surprise.

1. Straight up, $\gamma =90^{\circ }$: $\cos 90^{\circ }=0\Rightarrow \dot{\gamma }=0$. The rocket will never turn on its own. This is why a deliberate kick off vertical is needed — you must manually create a small $\cos \gamma \ne 0$ to get the process going.
2. Just after the kick, $\gamma =89^{\circ }$: $\cos 89^{\circ }\approx 0.017$, tiny but non-zero → a slow turn begins, which grows $\cos \gamma$, which grows the turn: self-amplifying.
3. Horizontal, $\gamma =0^{\circ }$: $\cos 0^{\circ }=1$ → maximum turn rate. If the engine is still burning here, gravity keeps pulling the nose below horizon — you must cut off or you dive.
4. Zero speed, $v\to 0$: the formula blows up ($÷0$). Physically this is the instant of liftoff before real speed exists — the gravity-turn model simply doesn't apply until $v$ is safely positive. That's why the kick happens after clearing the tower, not on the pad.

PICTURE. Figure s07 is a four-panel strip: the frozen $90^{\circ }$ rocket, the tiny-kick $89^{\circ }$ rocket, the max-turn $0^{\circ }$ rocket, and the $v\to 0$ "undefined" warning.

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Step 8 — Numbers on the picture (worked checks)

Recall Check the "1/v" claim yourself

Turn rate at $v=60$ divided by turn rate at $v=120$ ::: equals $2$ — halving $v$ doubles $\dot{\gamma }$.

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The one-picture summary

Figure s08 stacks the whole story: the tilted velocity arrow with angle $\gamma$, weight split into along ($mg\sin \gamma$) and across ($mg\cos \gamma$), the mass cancelling, and the boxed result $\dot{\gamma }=-g\cos \gamma /v$ — with the three special angles ($90^{\circ }$ stuck, $89^{\circ }$ nudge, $0^{\circ }$ max) marked along the bottom.

Recall Feynman retelling — say it like a story

A rocket flies with a velocity arrow tilted at angle $\gamma$ above the ground. To turn that arrow you need a force pointing sideways to it. The engine and the air-drag both point along the arrow, so they can only make it longer or shorter — they can't bend it. Gravity, though, always points straight down, and when the arrow is tilted, part of gravity leans across the arrow. That across-part, $mg\cos \gamma$, is the only thing bending the path. Newton says mass times the arrow's sideways swing ($mv\dot{\gamma }$) equals that across-force, and since it bends the arrow down, we put a minus sign. The mass cancels off both sides — light and heavy rockets turn the same — leaving $\dot{\gamma }=-g\cos \gamma /v$. When the rocket is dead vertical, $\cos 90^{\circ }=0$, so nothing turns: that's why we kick it slightly off vertical to wake the process up. Once tilted, the turning grows the tilt, which grows the turning — a gentle avalanche that carries the rocket from straight-up to sideways, all done by gravity, no steering required.

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Related: Gravity turn trajectory · Gravity loss and steering loss · Thrust-to-weight ratio · Attitude control and thrust vectoring (gimbal) · Closed-loop ascent guidance (PEG / IGM)