# How does the length of the minor semi-axis of the moon's disc's inner terminator vary through the cycle?

The curve that separates the bright and dark parts of the moon's disc as viewed from the earth is called the inner lunar terminator, which is semi-elliptical in shape. We can model the disc as a unit circle centred on the origin, in which the inner terminator runs from (0,1) to (0,-1) and its major semi-axis always has length 1.

How does its minor semi-axis vary during the lunar cycle? Please assume that orbits are circular.

In other words, if the point halfway along the inner terminator is X, what is the equation for the motion of X back and forth along the line-segment from (-1,0) to (1,0)?

Note that the shape of the visible disc is not a lune, except in the two trivial cases.

• If the lunar period is p, and X starts at the origin at t=0 and moves right, perhaps the motion is harmonic and the answer is that X = $\sin(\frac{2\pi t}{p})$, which implies that the shape of the moon's disc changes fastest at half moon? But are things that simple?
– user15430
Jan 7, 2017 at 18:40
• This reads like a homework problem, but one of the moon phase videos at svs.gsfc.nasa.gov/4537 may help you figure it out. Jan 7, 2017 at 19:28
• It's not a homework problem. But it's true that a good guess at the equation could probably be made by spending a few hours stepping through one of the NASA videos and repeatedly measuring the semi-axis on the screen.
– user15430
Jan 7, 2017 at 20:12
• A sine function would oscillate both ways. The lunar terminator does not do that. Jan 8, 2017 at 19:57
• @MikeG - Good point! As you say in your answer, we need to flip from sin to -sin at full and new moon.
– user15430
Jan 10, 2017 at 1:05

The moon is rotating wrt to the sun. As the moon rotates, the longitude of the terminator increases linearly. Treating the moon as being at "great distance, so the moon appears as an orthographic projection, the position of the middle of the terminator varies sinusoidally, it moves fastest when the moon is at first or third quarter.

In the orthographic projection, the longitude and latitude $(\lambda,\varphi)$ are mapped to (x,y) by \begin{align} x &= R\,\cos\varphi \sin\left(\lambda - \lambda_0\right) \\ y &= R\big(\cos\varphi_0 \sin\varphi - \sin\varphi_0 \cos\varphi \cos\left(\lambda - \lambda_0\right)\big) \end{align}

For points on the equator, and treating the central longitude $\lambda_0$ as the lunar prime meridian, and a scale of R=1, that simplifies to $x = \sin(\lambda)$. As the terminator moves, its longitude increases linearly, so x varies sinusoidally.

Correcting for elliptical orbits, librations of the moon and a general perspective view would only marginally alter this.

• Many thanks for this, but please can you expand it. I don't understand how you arrive at the conclusion that it varies sinusoidally, or why, having assumed the moon is at great distance, you then say you can introduce a general perspective view, which would put the observer only a small distance away, without changing the sinusoidal nature of the motion.
– user15430
Jan 7, 2017 at 23:15
• Extended. I didn't say that the general perspective view wouldn't change the motion, but it doesn't change it substantially. Our distance from the moon is much greater than it's radius. Jan 7, 2017 at 23:33
• Thanks. I understand the answer now and have accepted it. It's as if a plate is rotating at constant speed on a large table and we are looking at it from the table's edge. Which is close to a definition of sinusoidal.
– user15430
Jan 8, 2017 at 3:27
• I think there are discontinuities. Jan 8, 2017 at 12:27

As this NASA visualization video shows, the terminator moves almost sinusoidally, always from lunar east to west, never from west to east. We see the sunrise half of the terminator in the waxing phases and the sunset half in the waning phases. At the new and full phases, our view of the terminator switches from one half to the other, making the position vs. time function a discontinuous, piecewise-sinusoidal sawtooth wave.