Airfoil aerodynamics — camber, chord, thickness: Visual walkthrough

We are reconstructing the airfoil surface, exactly the way the NACA Airfoil Series does it.

Step 1 — Lay down the ruler: the chord

WHAT. Draw a horizontal straight line from the front point (the leading edge, LE) to the back point (the trailing edge, TE). This is the chord line. Its length is the chord $c$ .

WHY. We need a coordinate system before we can place anything. The chord is our $x$ -axis: a variable $x$ that runs from $0$ at the LE to $c$ at the TE. Everything else will be measured as a height above or below this line.

PICTURE. Look at the figure: the white horizontal line is the chord. The little arrow marks $x$ growing to the right. There is no airfoil yet — just the ruler.

Figure — Airfoil aerodynamics — camber, chord, thickness

Step 2 — Draw the skeleton: the mean camber line $y_{c} (x)$

WHAT. At each $x$ , we mark a height $y_{c} (x)$ — the mean camber line, the curved "spine" of the airfoil.

WHY. Real airfoils are not flat. The spine carries all the bend — all the up-or-down curvature of the shape. Later we will wrap flesh symmetrically around this spine, so the spine alone decides how asymmetric (how lifting) the airfoil is. Separating "bend" (spine) from "fatness" (flesh) is the whole trick.

PICTURE. The yellow curve is $y_{c} (x)$ . It starts and ends on the chord (height $0$ at LE and TE) and bows gently upward in between. The tallest gap between yellow and the white chord is the camber.

Step 3 — The slope of the spine, and why we need $θ$

WHAT. At each point of the yellow spine, measure how tilted it is. The tilt is the slope $\frac{d y _{c}}{d x}$ — "how much height you gain per step forward." We package that tilt as an angle $θ$ .

WHY an angle and not just the slope? Because in Step 4 we must add thickness perpendicular to the spine. "Perpendicular" is a direction, and directions are described by angles. The slope is a ratio; to turn a ratio into a direction we ask: which angle has this slope? That question is answered by the arctangent.

PICTURE. The red triangle sits on the spine: horizontal leg $d x$ , vertical leg $d y_{c}$ , hypotenuse lying along the spine. The angle $θ$ between the hypotenuse and the horizontal is what $arctan$ hands back.

Step 4 — Find the perpendicular direction (the unit normal)

WHAT. We need the direction that points straight out of the spine, at $9 0^{\circ}$ to it. Call it the unit normal. We claim it is $(- sin θ, cos θ)$ .

WHY this exact pair? The spine points along itself in direction $(cos θ, sin θ)$ — that is just "go forward by $cos θ$ , up by $sin θ$ ," the two legs of a unit-length step tilted by $θ$ . To turn any direction by $9 0^{\circ}$ counter-clockwise you swap the two numbers and flip the sign of the new first one: $(cos θ, sin θ) \to (- sin θ, cos θ)$ . That rotated arrow is perpendicular — it is the normal.

PICTURE. Yellow spine, blue "along" arrow tangent to it, green "normal" arrow at a clean right angle. The little right-angle box confirms the $9 0^{\circ}$ . Both arrows are the same length ( $= 1$ ).

Step 5 — Wrap the flesh: half-thickness $y_{t} (x)$ along the normal

WHAT. Define $y_{t} (x) \geq 0$ , half the local thickness. From each spine point, step a distance $y_{t}$ along the green normal to get the upper surface point, and $y_{t}$ against it to get the lower point.

WHY half, and why along the normal? The flesh is symmetric about the spine — the same amount above and below — so we add $+ y_{t}$ one way and $- y_{t}$ the other; the total top-to-bottom thickness is $2 y_{t}$ . We move along the normal (not straight up) because that is the honest "thickness perpendicular to the skeleton," which is how NACA distributes it.

PICTURE. From a spine point $P = (x, y_{c})$ : the green normal goes up to the red upper point $U$ , the reversed normal goes down to the red lower point $L$ . The segment $U - L$ has length $2 y_{t}$ and is perpendicular to the yellow spine.

Now write the arithmetic. Start at the spine point $(x, y_{c})$ and add $y_{t}$ times the normal $(- sin θ, cos θ)$ :

Step 6 — The flat-spine limit (why thin airfoils simplify)

WHAT. Suppose the spine is nearly flat: $θ \to 0$ . Then $sin θ \to 0$ and $cos θ \to 1$ , so the formulas collapse to $x_{U} \approx x, y_{U} \approx y_{c} + y_{t}, x_{L} \approx x, y_{L} \approx y_{c} - y_{t} .$

WHY this is the degenerate case that matters. For a symmetric airfoil ( $y_{c} = 0$ , so the spine is the chord, $θ = 0$ exactly) or any thin airfoil, "perpendicular to the spine" and "perpendicular to the chord" become the same direction. Thickness just stacks straight up and down. This is why Thin-Airfoil Theory can treat thickness and camber as simply added — the sideways $sin θ$ nudge vanishes.

PICTURE. Left: steep spine, normals fan out sideways, the $y_{t} sin θ$ shift is visible. Right: flat spine, normals point straight up/down, $x_{U} = x_{L} = x$ . Same $y_{t}$ , different geometry.

Step 7 — Close the loop: LE and TE endpoints

WHAT. At the very nose ( $x = 0$ ) and tail ( $x = c$ ), a well-formed thickness law has $y_{t} \to 0$ . So $U$ and $L$ merge back onto the spine: the top and bottom skins meet.

WHY check the endpoints? An airfoil must be a single closed curve, not two loose ribbons. Zero thickness at LE and TE is what stitches the upper and lower surfaces together into one outline. (The trailing edge closing to a point is also exactly what the Kutta–Joukowski Theorem needs — the flow leaves cleanly from a sharp TE.)

PICTURE. Full airfoil outline: upper (red) and lower (blue) surfaces meeting at LE and TE, yellow spine inside, white chord underneath. The complete NACA-style shape, assembled from Steps 1–5.

The one-picture summary

Everything at once: chord (ruler) → spine $y_{c}$ (bend) → tilt angle $θ = arctan (d y_{c} / d x)$ → normal $(- sin θ, cos θ)$ → step $\pm y_{t}$ along it → closed surface.

Recall Feynman retelling (click to open)

First I drew a straight ruler from nose to tail — that's the chord. Then I drew a gentle wavy spine that stays in the middle of the wing and does all the bending — that's the camber line. To know which way is "sideways-out" of that spine, I looked at a tiny triangle on it: the ratio of its rise to its run is the slope, and asking "which angle has that slope?" — that's arctangent — gives me the spine's tilt $θ$ . Turn the spine's own direction by a quarter turn and I get the arrow pointing straight out of it. Finally I walked a little distance $y_{t}$ out along that arrow to draw the top skin, and the same distance backwards for the bottom skin — so the skin is symmetric about the spine, top-to-bottom thickness $2 y_{t}$ . Because the spine is tilted, stepping out also nudges me sideways, which is why an extra $sin θ$ term sneaks into the $x$ -coordinates. If the spine is flat (a symmetric airfoil) that nudge disappears and thickness just stacks straight up and down. At the nose and tail the thickness shrinks to zero, so the two skins meet and the outline closes into one wing shape.

Recall Quick self-test

Why is there a $y_{t} sin θ$ term in $x_{U}$ ? ::: Because we step perpendicular to a tilted spine, so the perpendicular step has a horizontal component $y_{t} sin θ$ . What is $θ$ when the airfoil is symmetric? ::: $θ = 0$ everywhere (spine = chord), so $sin θ = 0, cos θ = 1$ . What does $arctan$ do here? ::: Converts the spine's slope $d y_{c} / d x$ into its tilt angle. Total thickness in terms of $y_{t}$ ? ::: $2 y_{t}$ .

Connections

NACA Airfoil Series — this decomposition is the NACA 4-digit generator.
Thin-Airfoil Theory — uses the flat-spine limit of Step 6.
Kutta–Joukowski Theorem — needs the closed sharp trailing edge of Step 7.
Boundary Layer & Flow Separation — how the finished thickness shape behaves in real flow.