Propulsive forces — thrust misalignment, gimbal angle: Visual walkthrough

We assume nothing. If you have never seen an arrow that means "force", a lever arm, a sine, or a cross product — good. We define each one the moment it is needed, and not before.

Step 1 — What "thrust" even is (an arrow)

WHAT we set up: a rocket lying on its side, nose pointing right. WHY: we need a fixed picture to hang every later idea on. Everything — angles, levers, torques — will be measured against this rocket's own body. PICTURE: the thrust arrow starts at the engine (back, left) and points forward (right). Its size $T$ is just the arrow's length.

We draw two reference lines through the rocket:

the roll axis (the long centre line, nose-to-tail) — call it the $x$ -direction,
the lateral direction (straight across) — call it the $y$ -direction.

$T = (forward part T_{x}, sideways part T_{y})$

Here $T_{x}$ is "how much of the push goes forward" and $T_{y}$ is "how much leaks sideways". When the engine points perfectly straight, $T_{y} = 0$ and all of $T$ is forward.

Step 2 — The centre of mass, the one point that matters

WHAT: we place the CoM and mark it as our origin — the point $(0, 0)$ . WHY: a force pushing exactly through the CoM can only slide the rocket, never spin it. A force that misses the CoM can spin it. So the CoM is the pivot every twist is measured around. Putting it at the origin makes the arithmetic clean. PICTURE: the CoM is a dot near the front-middle. The engine pivot sits behind it, a distance $L$ down the axis.

$r = (engine is behind CoM - L, on the axis 0)$

$r$ = the arrow from the CoM to where the force is applied (the pivot).
$- L$ = the pivot's forward-coordinate. It is negative because "behind" is the $- x$ side.
$0$ = the pivot sits on the centre line, no sideways offset.

This length $L$ is the lever arm: the reach from pivot-point to push-point. Remember it — it is the "long handle" that makes small pushes into big turns.

Step 3 — Tilt the engine: splitting the thrust arrow

WHAT: rotate the thrust arrow by $δ$ and split it into its forward and sideways parts. WHY: torque only cares about the sideways part $T_{y}$ — that is the piece that pushes off-centre. To find $T_{y}$ we must resolve the tilted arrow into components. This is exactly what sine and cosine are for.

$T = (forward: survives T cos δ, sideways: does the turning T sin δ)$

$T cos δ$ = forward thrust. At $δ = 0$ , $cos 0 = 1$ , so it equals the full $T$ — nothing lost.
$T sin δ$ = sideways thrust. At $δ = 0$ , $sin 0 = 0$ , so no side-push — the rocket flies straight. Tilt it, and this term grows.

Step 4 — What "torque" is: the twisting number

WHAT: combine the lever arm $r$ (Step 2) with the force $T$ (Step 3) into one twisting number. WHY: we want a single quantity that says "how hard, and which way, does this spin the rocket?" In flat 2D, that number comes from the cross product, which in a plane collapses to:

$τ = r_{x} T_{y} - r_{y} T_{x}$

$r_{x} T_{y}$ = (forward reach) × (sideways push) — this is the real twister.
$r_{y} T_{x}$ = (sideways reach) × (forward push) — zero here, since $r_{y} = 0$ .

PICTURE: the sideways thrust at the end of the long lever swings the tail one way, so the nose swings the other — the rocket pivots about the CoM dot.

Plug in $r = (- L, 0)$ and $T = (T cos δ, T sin δ)$ :

$τ = (- L) (T sin δ) - (0) (T cos δ)$

The second term dies. What remains:

The forward part $T cos δ$ contributed nothing to the twist — it points straight along the lever, so it only shoves the rocket forward. Only the sideways part turns it. That single fact is the whole derivation.

Step 5 — Why the turn is "cheap": the small-angle picture

WHAT: compare the turn we gain against the forward-push we lose, as the tilt grows. WHY: this comparison is the entire justification for TVC. If losses grew as fast as the turn, gimballing would be a bad deal. They don't — losses are a squared term, which stays tiny while the linear turn shoots up. PICTURE: two curves out of the origin. The torque curve $sin δ$ rises like a straight line; the loss curve $1 - cos δ$ hugs the floor before it lifts.

$grows fast (linear) τ \approx L T δ grows slowly (squared) loss = T (1 - cos δ) \approx \frac{1}{2} T δ^{2}$

At $δ = 5^{\circ}$ : you get $sin 5^{\circ} = 0.0872$ of full sideways authority, but lose only $1 - cos 5^{\circ} = 0.0038$ — that's 0.38% of thrust. Big turn, near-zero cost.

Step 6 — Every case: which way does it spin, and the degenerate ones

WHAT: nail down the sign (turn direction) and check the boundary cases so no scenario surprises us. WHY: a control law that turns the wrong way, or blows up at $δ = 0$ , is worse than useless. We must show the formula behaves everywhere.

$δ > 0$ (nozzle swung toward $+ y$ ): $sin δ > 0$ , so $τ = - L T sin δ < 0$ . The tail is shoved to $+ y$ , so the nose swings toward $- y$ — a negative (clockwise) rotation. The minus sign is just this bookkeeping.
$δ < 0$ (nozzle swung the other way): $sin δ < 0$ , so $τ > 0$ . Nose swings the opposite way. The controller steers both directions simply by choosing the sign of $δ$ .
$δ = 0$ (perfectly straight): $sin 0 = 0 \Rightarrow τ = 0$ . Pure forward flight, no spin. The formula degenerates gracefully.
$δ = 9 0^{\circ}$ (a nonsense extreme): $sin 9 0^{\circ} = 1$ , all thrust sideways, $cos 9 0^{\circ} = 0$ , zero forward push. Maximum twist, no acceleration — which is exactly why real gimbals are hard-limited to a few degrees.
$L = 0$ (engine at the CoM): $τ = 0$ no matter how you tilt. No lever, no twist — you'd just slide sideways. This is the Attitude Control & Stability reason engines sit far behind the CoM.
Pure offset, no tilt ( $δ = 0$ but pivot shifted sideways by $d$ ): now $r$ has a $y$ -part, and the forward thrust gets a lever. Torque becomes $τ = d T$ — a different geometry, same principle: a force missing the CoM twists the body.

Step 7 — From torque to motion: the rocket actually pitches

WHAT: convert the twist $τ$ into an actual pitch-rate change $\ddot{θ}$ . WHY: the torque is only useful if it moves the nose. This last step closes the loop from "engine tilt" to "rocket points somewhere new."

$\ddot{θ} = \frac{τ}{I} = - \frac{L T δ}{I}$

$\ddot{θ}$ = how fast the pitch-rate is building (angular acceleration).
$I$ = moment of inertia — a big $I$ (long heavy rocket) turns sluggishly.
A held gimbal ⇒ constant $\ddot{θ}$ ⇒ the nose keeps pitching over faster and faster.

The one-picture summary

Everything on one canvas: the rocket, the CoM dot, the lever arm $L$ , the tilted thrust split into a surviving forward part $T cos δ$ and a turning sideways part $T sin δ$ , and the resulting spin about the CoM.

$τ = - L T sin δ \approx - L T δ ⟹ \ddot{θ} = \frac{τ}{I}$

Recall Feynman: the whole walkthrough in plain words

Picture the rocket on its side, nose to the right. There's one magic balance point, the centre of mass — spin the rocket and it spins around this dot. The engine sits well behind that dot, at the end of a long handle we called $L$ .

Normally the engine pushes straight up the middle, so the push runs right through the balance point and the rocket just flies forward. Now tilt the engine a hair. The push arrow is now slanted, so we chop it into two: a big piece still going forward, and a small piece leaking sideways. The forward piece runs along the handle and only shoves — it can't twist anything. But the sideways piece, out at the end of that long handle, swings the tail across, so the nose swings the other way. How hard? Handle length times push times the sideways fraction: $L \times T \times sin δ$ .

The beautiful trick: tilt a little and the sideways fraction grows almost as fast as the tilt, but the forward push you sacrifice barely dips at all (it dips as the square of a tiny number, which is tinier still). So you buy a strong turn for almost no cost. Tilt the other way and it turns the other way. Point the engine dead straight, or put it right at the balance point, and there's no turn at all. That is thrust vector control, start to finish.

Recall

Which part of a tilted thrust actually turns the rocket?

The sideways part,

T sin δ

; the forward part

T cos δ

runs along the lever and only pushes.

Why does the cross product reduce to

τ = r_{x} T_{y} - r_{y} T_{x}

in 2D?

In a flat plane the spin is a single number, and only the perpendicular pairing of reach and push contributes.

What makes gimballing "cheap"?

Turn grows like

δ

(linear) but thrust loss grows like

δ^{2}

(second order), so tiny tilt = big turn, negligible loss.

What does the minus sign in

τ = - L T sin δ

mean?

A direction label: positive

δ

swings the tail one way, spinning the nose the other (clockwise here).

What happens to

τ

L = 0

Zero — engine at the CoM has no lever arm, so no twist regardless of tilt.

Torque from a pure sideways offset

d

with no tilt?

τ = d T

— the forward thrust now misses the CoM and twists.