Given a latitude and a desired inclination, there is exactly one launch azimuth — or none. And before Friendship 7 flew, a trajectory a computer had produced waited on a hand recomputation.
She worked in West Area Computing at Langley, a segregated pool of Black women employed as human computers. She asked to attend the editorial meetings where the engineers argued about the work she had done, was told women did not go, and asked whether there was a law against it. There was not, and she went.
In 1960 her name went on NASA Technical Note D-233, with T. H. Skopinski: Determination of Azimuth Angle at Burnout for Placing a Satellite Over a Selected Earth Position. It was the first report out of her division to carry a woman's name as author. CITED That report is the brick, and this chapter is about what is in it rather than about the film.
The question it answers is concrete. You are at latitude $\varphi$. You want an orbit of inclination $i$. Which way do you point?
$$\sin\beta = \frac{\cos i}{\cos\varphi}$$
$\beta$ measured clockwise from north. Spherical trigonometry: the launch site, the ascending node, and the point of highest latitude form a right spherical triangle, and this is its rule.
Cape Canaveral sits at $28.47°$N. For Friendship 7's orbit, $i = 32.5°$, the formula gives an inertial azimuth of $73.621°$. Then subtract what the Earth is already giving you — the pad is moving east at $465.1\cos\varphi = 409$ m/s — and the azimuth to fly is
burnout speed 7400 m/s → 72.678°
burnout speed 7600 m/s → 72.704°
burnout speed 7800 m/s → 72.729°
Mercury-Atlas 6, as flown → 72.6° COMPUTED
Within a tenth of a degree, and almost independent of the burnout speed. The correction is small and it is not optional: leave it out and you are a degree off, which at orbital distances is a different ocean.
The formula has a boundary, and it is absolute.
$\sin\beta \leq 1$ forces $\cos i \leq \cos\varphi$, so $|i| \geq |\varphi|$. From the Cape you can reach inclination $28.47°$ — due east, $\beta = 90°$ — and nothing lower. Not inefficiently, not expensively. There is no azimuth.
i = 28.47° → sin β = 1.000000 the only solution
i = 28.00° → sin β = 1.004415 no solution
i = 25.00° → sin β = 1.030988 no solution
i = 20.00° → sin β = 1.068966 no solution
COMPUTED
This is the corpus's own shape in the plainest form it takes anywhere in this gallery. The launch azimuth is not chosen. The latitude and the desired inclination leave exactly one option, or none, and which of those it is changes at a point rather than gradually. A plane change in orbit costs propellant proportional to $2v\sin(\Delta i/2)$, which is why the constraint is worth this much arithmetic: for a $5°$ change at orbital speed, roughly $680$ m/s — more than most upper stages have.
The part of the story that belongs in a corpus about verification is the part usually told as a compliment.
Before Friendship 7, the trajectory was computed by an IBM 7090 running the new orbital-mechanics code. Glenn would not fly on it. The request, as it reached her, was to get the girl to check the numbers — and what that meant in practice was that she recomputed the trajectory by hand, on a desk calculator, over days, to see whether the machine agreed with a method anyone could follow.
A new tool produced a result nobody could yet audit. The result was probably right. It was not checked, and the distinction mattered enough that a flight waited on it.
This corpus writes a verify script beside every chapter for the same reason, and it is worth saying plainly that the practice is not new and was not invented here. SHOWN
She also worked the rendezvous problem — two vehicles in different orbits made to arrive at the same place at the same moment — and the Apollo lunar-module return, where the same question is asked with no margin at all. The Presidential Medal of Freedom came in 2015. She died in 2020, a hundred and one years old.
| Operator | In this chapter | In dm³ |
|---|---|---|
| C | the launch site — a latitude, fixed and not negotiable | compression: the constraint |
| K | the desired inclination brought down toward $\varphi$ | approach to the boundary |
| F | $i = \varphi$: $\sin\beta = 1$, and below it there is no azimuth at all | the fold — threshold, not scale SHOWN |
| U | $\beta = 72.6°$ — the one heading the constraint leaves | the branch, verified against the flight COMPUTED |
Every number on this page is produced by book7/ch-katherine-johnson-verify.py. It records in its own closing block what it establishes and what it does not.
T. H. Skopinski and K. G. Johnson, Determination of Azimuth Angle at Burnout for Placing a Satellite Over a Selected Earth Position, NASA TN D-233, Langley, 1960.
K. G. Johnson and J. C. Young, Two Approaches for Obtaining the Braking Ellipse for Return from a Lunar Mission, NASA TN D-3970, 1967.
NASA, Results of the First United States Manned Orbital Space Flight, February 20, 1962, Manned Spacecraft Center — for the flown trajectory parameters.
M. L. Shetterly, Hidden Figures, William Morrow, 2016.
R. R. Bate, D. D. Mueller and J. E. White, Fundamentals of Astrodynamics, Dover, 1971 — chapter 2 for the spherical-triangle derivation.