Visual walkthrough — Forces on a rocket in flight — thrust, aerodynamic (normal, axial), gravity

We assume nothing beyond: an arrow can represent a force (longer = stronger, pointing the way it pushes), and forces add tip-to-tail like a walk.

Step 0 — The two arrows you must never confuse

WHAT: we mark the nose direction and the travel direction on one picture, with the positive sense of $\alpha$ drawn as a curved arrow. WHY: every force below is measured relative to one of these two lines, so we must fix them — and their sign convention — first. PICTURE: the two arrows below; the curved orange arrow shows the positive (counter-clockwise) sense of $\alpha$, from $\hat{v}$ up to $\hat{b}$.

See Angle of Attack and Stability for why $\alpha$ also decides whether the rocket wants to straighten out or tumble.

Step 1 — The air pushes back with ONE force

WHAT: draw the single aerodynamic resultant $\vec{R}$ acting on the rocket. WHY: if we start from the honest single force, the four component names ($L,D,A,N$) become just two different shadows of the same arrow — and the conversion formulas fall out for free. PICTURE: one plum arrow $\vec{R}$ pushing on the tilted body.

Step 2 — First shadow: lift and drag (the WIND frame)

Line up your ruler with the velocity $\hat{v}$. We need a second direction at right angles to it, and we must fix which right angle — otherwise the sign of lift is ambiguous.

Now drop the single force $\vec{R}$ onto $\hat{v}$ and $\hat{n}$.

WHAT: we split $\vec{R}$ into two arrows measured off the velocity line, with the perpendicular side and signs pinned down. WHY these two and not others? Because $D$ and $L$ depend cleanly on airspeed: each is $\frac{1}{2}\rho v^{2}S$ times a coefficient. That is the natural frame for aerodynamics. PICTURE: teal $\hat{v}$ line, orange $D$ backward along it, teal $L$ square to it on the $+90^{\circ }$ side.

Step 3 — Second shadow: axial and normal (the BODY frame)

Now line your ruler with the body axis $\hat{b}$ instead. Build the body's own perpendicular unit arrow $\hat{p}$ using the same $+90^{\circ }$ counter-clockwise rule: $\hat{p}$ is $\hat{b}$ rotated left by $90^{\circ }$. (Using the identical turning rule for $\hat{n}$ and $\hat{p}$ is what will make the rotation in Step 4 clean.) Drop the same force $\vec{R}$ onto these.

WHAT: the very same arrow $\vec{R}$, re-shadowed onto the nose direction. WHY: the structure — fins, skin, the joints between stages — feels force along and across the tube, not along the wind. $A$ compresses the rocket; $N$ tries to bend and rotate it. This is the frame an engineer sizing the airframe needs. PICTURE: the same plum $\vec{R}$, now with orange $A$ down the body and teal $N$ across it.

Both frames describe one arrow: $-D\hat{v}+L\hat{n}=A\hat{b}+N\hat{p}$. The frames differ by exactly the angle $\alpha$ — because the body is tilted from the velocity by $\alpha$. That single fact is the whole conversion.

Step 4 — Rotating one frame onto the other

The four building blocks (all from "$\hat{b}$ is $\hat{v}$ turned by $+\alpha$"). Two unit arrows separated by an angle have a dot product equal to the cosine of that angle. Working out each pair: $\hat{b}\cdot \hat{v}=\cos \alpha ,\hat{b}\cdot \hat{n}=-\sin \alpha ,\hat{p}\cdot \hat{v}=\sin \alpha ,\hat{p}\cdot \hat{n}=\cos \alpha .$

WHAT: we project the vector sum onto the tilted body arrows by taking dot products. WHY trig, and why exactly $\cos$ and $\sin$? Because "how much of one unit arrow points along another" is the cosine of the angle between them; a perpendicular pair gives sine. That is precisely the question "what fraction of $L$ or $D$ lands on the body axis?"

Axial — project $\vec{R}$ onto $\hat{b}$, one term at a time:

This is the axial component measured with $A$ positive down the nose $+\hat{b}$. Because the air's resultant actually pushes backward (it opposes forward flight), its along-body part comes out negative in this convention. The parent note reports the magnitude of the retarding axial load — flip the sign (i.e. define positive $A$ pointing aft, the direction the force truly points) and you get the familiar

i.e. how hard the air presses backward along the tube. Both terms are positive here because both drag and the tilted lift push the airframe aft-along-the-body.

Normal — project $\vec{R}$ onto $\hat{p}$, one term at a time:

Here the signs come straight out with no flip: lift's cross-body share adds, drag's cross-body share subtracts. That minus is real — the tilt makes drag rob a little from the normal force.

PICTURE: the tilted-rectangle projection, each dashed drop-line labelled with its $\cos$ or $\sin$ leg.

Step 5 — Edge cases: $\alpha =0$, negative $\alpha$, and large $\alpha$

Case A — $\alpha =0$ (nose points where you go). The body ruler and velocity ruler coincide. $\cos 0=1$, $\sin 0=0$: $A=D(1)+L(0)=D,N=L(1)-D(0)=L.$ Axial equals drag and normal equals lift only here.

Case B — $\alpha <0$ (nose below velocity). Since $\sin (-\alpha )=-\sin \alpha$, every $\sin$ term flips sign: $A=D\cos \alpha -L\sin |\alpha |,N=L\cos \alpha +D\sin |\alpha |.$ Lift now pulls the other way across the body, so its axial share subtracts and the drag term adds to $N$. Nothing new to memorise — the sine simply changes sign.

Case C — large $\alpha$ (past $90^{\circ }$). When $\alpha >90^{\circ }$, $\cos \alpha$ turns negative: $A=D\cos \alpha +L\sin \alpha$ can go negative, meaning the air now pushes forward along the body relative to our aft-positive convention (the rocket is flying nearly sideways — a tumbling or re-entry attitude). The formula still holds; it just reports the reversed axial load honestly.

WHAT: the general formulas specialised and sign-checked. WHY: so no reader hits an attitude — nose-down, or tumbling — that the algebra hasn't already covered. PICTURE: the two rulers coincident ($\alpha =0$), with a small inset showing the nose-below-velocity ($\alpha <0$) mirror.

Step 6 — Adding thrust and gravity: the two equations of motion

Now assemble ALL the forces on the flight-path axes (along $\hat{v}$ and across $\hat{n}$). We need one more angle and two more symbols:

Why gravity splits into $-mg\sin \gamma$ and $-mg\cos \gamma$ (a projection, exactly like Step 4). Weight is one arrow of length $mg$ pointing straight down. To use it in flight-path axes we must ask the same two questions as before: how much of "straight down" lies along $\hat{v}$, and how much across it (along $\hat{n}$)?

Set up the geometry: $\hat{v}$ points at angle $\gamma$ above horizontal, so the downward direction sits at angle $\gamma$ below $\hat{v}$ on the retarding side. Projecting the down-arrow:

Multiply by the length $mg$:

• Along the path: $-mg\sin \gamma$. At $\gamma =90^{\circ }$ (straight up) this is $-mg$ — gravity fights the climb hardest; at $\gamma =0$ (level) it is $0$ — gravity does nothing to your speed. For a dive ($\gamma <0$), $\sin \gamma <0$ so the term becomes positive — gravity now speeds you up, exactly right.
• Across the path: $-mg\cos \gamma$. At $\gamma =0$ (level) this is $-mg$ — all of gravity bends the path downward; at $\gamma =90^{\circ }$ it is $0$ — straight up, gravity does not curve you. This cross-path term is what tips a rocket over in a Gravity Turn Trajectory.

PICTURE (gravity projection): the down-arrow $mg$ dropped onto the $\gamma$-tilted $\hat{v},\hat{n}$ cross, with its $\sin \gamma$ and $\cos \gamma$ legs.

Now add the other players, each projected onto $\hat{v}$ and $\hat{n}$:

• Thrust $T$ points along the body $\hat{b}$, tilted from $\hat{v}$ by $\alpha$ → gives $T\cos \alpha$ along path, $T\sin \alpha$ across path (same $\cos /\sin$ split as Step 4).
• Aerodynamics in wind axes is already aligned: $-D$ along path, $+L$ across path.
• Gravity: $-mg\sin \gamma$ along path, $-mg\cos \gamma$ across path (just derived).

WHAT: two projections of $\vec{F}=m\vec{a}$. WHY two? A velocity can change in size (first line — speeding up: here $v$ is the speed, the length of $\vec{v}$, and $dv/dt$ is how fast that length grows) or in direction (second line — turning: $vd\gamma /dt$ is the rate at which the path bends). One equation per way-to-change.

(Remember the mass $m$ in every term is the instantaneous mass $m(t)=m_0-\dot{m}t$ — see Variable Mass Systems and the Tsiolkovsky Rocket Equation.)

The one-picture summary

One tilted arrow $\vec{R}=-D\hat{v}+L\hat{n}=A\hat{b}+N\hat{p}$, two rulers ($\hat{v}$ and $\hat{b}$) $\alpha$ apart, thrust down the body, gravity straight down — every formula on this page is a shadow of this one drawing.

Quick self-test