De Laval nozzle geometry — conical, bell (Rao contour), 80% bell: Visual walkthrough

This page rebuilds the single most important number in the parent note — the divergence correction factor

$λ = \frac{1 + c o s α}{2}$

— from absolutely nothing. No integrals memorised, no symbol used before it is drawn. If you have never seen a cosine or an integral sign, start at line one and you will still arrive at the finish.

We assume only what Converging-Diverging Nozzle Basics already gave you: hot gas is squeezed through a narrow throat and then allowed to spread out through a widening cone, coming out fast. Our one question for this whole page:

When the gas fans outward instead of shooting straight back, how much push do we lose?

Step 1 — What "thrust" even is, as a picture

WHAT. Thrust is the backward-shove the engine gets from throwing gas out the back. If the gas leaves straight backward, all of that shove pushes the rocket straight forward. That is the ideal.

WHY start here. Before we can measure a loss, we need the full amount to compare against. So we first draw the perfect case: every gas particle moving parallel to the centre line.

PICTURE. Look at figure s01. The dashed horizontal line is the nozzle axis (the centre line the rocket flies along). The pale-yellow arrows are gas particles leaving. In the ideal case they all point straight left — perfectly along the axis.

Figure — De Laval nozzle geometry — conical, bell (Rao contour), 80% bell

Step 2 — A real cone makes the gas fan out

WHAT. In a straight-walled conical nozzle, the walls flare outward at a fixed angle. The gas follows the walls, so it does not leave straight back — it leaves tilted outward.

WHY it matters. A tilted arrow has two parts: one part pointing straight back (useful) and one part pointing sideways (wasted). We must split every arrow into these two parts.

PICTURE. In figure s02 the two chalk-blue walls open like a funnel. A single gas arrow at the edge leaves along the wall — tilted by an angle we will call $α$ away from the axis.

Step 3 — Splitting one tilted arrow: meet the cosine

WHAT. Take one gas arrow tilted by angle $θ$ from the axis (I use $θ$ for a general arrow; the wall's arrow is the special case $θ = α$ ). We keep only its forward-pointing part.

WHY the cosine and not something else. We need a machine that eats an angle and gives back "what fraction of this arrow points forward". Draw the arrow as the slanted side of a right triangle whose bottom side lies along the axis. The forward part is that bottom side. The rule "bottom side ÷ slanted side" for a right triangle is exactly what $cos θ$ means — cosine is defined as adjacent over hypotenuse. That is precisely the question we are asking, so cosine is the right tool, not sine (which would give the sideways part) and not tangent.

PICTURE. Figure s03: the pink arrow of length $1$ (one unit of speed) is the tilted gas. Drop it onto the axis. The yellow segment underneath, of length $cos θ$ , is the forward part we keep.

Check the extremes with the picture in mind:

$θ = 0$ (straight back): $cos 0 = 1$ → keep everything. ✓
$θ = 90°$ (straight sideways): $cos 90° = 0$ → keep nothing. ✓

WHAT. A cone's exit is a full ring. Gas near the centre leaves almost straight ( $θ$ near $0$ ); gas at the very edge leaves at the full wall angle ( $θ = α$ ). Every angle from $0$ up to $α$ is present.

WHY we can't just use $cos α$ . Using $cos α$ alone would pretend every arrow leaves at the edge angle. That over-counts the loss. We must average $cos θ$ over the whole spray of arrows — and the average must be fair: it must weight each angle by how much gas actually leaves at that angle.

PICTURE. Figure s04 shows the exit as a shallow spherical cap (imagine the gas leaving through a curved dome). A thin ring on that dome sits at angle $θ$ . More gas flows through the bigger rings near the rim than through the tiny rings near the centre — this is the weighting we must respect.

Step 5 — Adding up infinitely many rings: the integral

WHAT. "Average of $cos θ$ , weighted by ring size $sin θ$ " = (sum of $cos θ \times ring area$ ) ÷ (sum of ring areas). The integral sign $\int$ is just "add up infinitely many thin rings".

WHY an integral. The angle changes smoothly from $0$ to $α$ — there are not 5 or 50 rings but a continuous fan of them. An ordinary sum can't handle "infinitely many, each infinitely thin"; the integral is the tool built exactly for a smooth continuous sum. Each ring at angle $θ$ has area proportional to $2 π R sin θ d θ$ ( $d θ$ = an infinitely thin slice of angle).

PICTURE. Figure s05 stacks the thin rings from $θ = 0$ (centre) to $θ = α$ (rim), each shaded by its weight $sin θ$ — faint near the centre, bold near the rim.

Cancelling $2 π R$ :

$λ = \frac{\int _{0}^{α} cos θ sin θ d θ}{\int _{0}^{α} sin θ d θ}$

Step 6 — Doing the two sums (both are areas under simple curves)

WHAT. Evaluate the top and bottom sums.

WHY these particular antiderivatives. Integrating is "un-differentiating": we ask what function, when its slope is taken, gives back $sin θ cos θ$ (or $sin θ$ )?

The slope of $\frac{1}{2} sin^{2} θ$ is $sin θ cos θ$ → so the top sum is $[\frac{1}{2} sin^{2} θ]_{0}^{α}$ .
The slope of $- cos θ$ is $sin θ$ → so the bottom sum is $[- cos θ]_{0}^{α}$ .

PICTURE. Figure s06 shows both curves over $0$ to $α$ ; the shaded areas beneath them are exactly these two sums.

Plugging the limits in (at $θ = 0$ : $sin 0 = 0$ , $cos 0 = 1$ ):

$top = \frac{1}{2} sin^{2} α - \frac{1}{2} sin^{2} 0 = \frac{1}{2} sin^{2} α$ $bottom = (- cos α) - (- cos 0) = 1 - cos α$

$λ = \frac{\frac{1}{2} s i n ^{2} α}{1 - c o s α}$

Step 7 — The clean-up that reveals the famous formula

WHAT. Simplify using one identity so no messy squares remain.

WHY this identity. We want $(1 - cos α)$ to appear on top so it cancels the bottom. The identity $sin^{2} α = 1 - cos^{2} α = (1 - cos α) (1 + cos α)$ does exactly that — it factors the top into a piece matching the bottom and a leftover piece.

$λ = \frac{\frac{1}{2} ( 1 - c o s α ) ( 1 + c o s α )}{1 - c o s α} = \frac{1 + c o s α}{2}$

The $(1 - cos α)$ cancels top and bottom, leaving the result the parent note quoted.

Term-by-term sanity in three cases (cover them all):

$α \to 0$ (a perfectly straight pipe): $cos 0 = 1$ , so $λ = \frac{1 + 1}{2} = 1$ . No fanning, no loss. ✓
$α = 15°$ (a typical cone): $cos 15° = 0.9659$ , so $λ = \frac{1.9659}{2} = 0.983$ → a 1.7% loss. ✓
$α \to 90°$ (walls flat open, gas sprays sideways): $cos 90° = 0$ , so $λ = \frac{1 + 0}{2} = 0.5$ → half the thrust gone. ✓

The one-picture summary

Figure s07 puts the whole chain on one board: one tilted arrow → split by $cos θ$ → averaged over rings weighted by $sin θ$ → the two areas → the boxed answer, with the $15°$ number pinned on.

Recall Feynman retelling — say it back in plain words

The engine wants to throw gas straight backward. A cone can't do that: it fans the gas out into a spray. Any sprayed arrow only helps a little — just the part pointing backward, and that part is exactly $cos θ$ of it, because "backward-part over whole-arrow" is what cosine means on a right triangle. Different bits of gas leave at different tilts, from zero at the centre to the full wall angle $α$ at the rim, and there's more gas at the rim (bigger rings, weighted by $sin θ$ ). So I average $cos θ$ fairly across all those rings — that averaging is the integral. The two sums come out to $\frac{1}{2} sin^{2} α$ on top and $1 - cos α$ on the bottom, one identity cancels the mess, and out drops $λ = \frac{1 + c o s α}{2}$ : the fraction of thrust that survives the fanning. Straight pipe → $1$ , a $15°$ cone → $0.983$ , walls thrown wide open → $0.5$ . And the punchline: bend the rim back to straight and you get that lost $1.7%$ back for free — that's the bell.

Recall Quick self-check

Why do we average $cos θ$ instead of just using $cos α$ ? ::: Because gas leaves at every angle from $0$ to $α$ , not only at the rim; using $cos α$ alone would overstate the loss. Why is each ring weighted by $sin θ$ ? ::: A ring at angle $θ$ has radius $\propto sin θ$ , so bigger (outer) rings carry more gas and must count more in the average. What does $λ = 1$ mean physically? ::: All gas leaves parallel to the axis ( $α \to 0$ ) — zero divergence loss.