Dima Kogan went camping, looked up just after sunset, and noticed the moon was lit on top. The sun was below the horizon. The lit side was pointing away from it.
This has a name — the lunar terminator paradox — and his explanation is the clearest I have read. The sun is effectively infinitely far away, so the same hemisphere of the moon is lit no matter where you stand on Earth. What changes is your angle on it. Standing under a high moon, you are looking at it from below, so you see a slice of its dark underside. Bright on top, a bite taken out of the bottom. It points up.
He mentions, in passing, that he wrote a program to work this out because of the written explanations already online, “none made sense in my head.” I know that feeling, and I had a version of it about his note too — it is perfectly clear, and it gives you the mechanism without giving you the when. Worst at full moon, absent at half, stronger the fuller it gets. Fine. But if I am standing outside at dusk, what do I predict?
So I worked out the condition, and then the condition worked out something better.
Put yourself at the origin, with a unit vector m pointing at the moon and s
pointing at the sun. The lit limb points, on the sky, along the sun direction
projected into the plane perpendicular to the moon:
p = s − (s·m)m
Take the simple case where the sun is directly opposite the moon in azimuth — moon at altitude β, sun at depression α below the horizon. Then the vertical component comes out as
p_z = −sin α + cos(β−α)·sin β
which is exactly zero when β = α, positive above and negative below. So:
The moon’s lit side points upward whenever the moon is higher above the horizon than the sun is below it.
Sun five degrees down and the moon higher than five degrees? It tips up. I swept it numerically before trusting the algebra, and the flip lands on the sun’s depression angle to two decimal places every time. Kogan’s own example — moon at 60°, sun 5° under — gives a lit limb tilted thirty degrees above horizontal. It is not a subtle effect.
→ Drag the sliders and watch it flip.
I built the interactive version expecting the boundary to be a flat limb: at β = α the lit side lies exactly horizontal, and one degree either way tips it. That is what “p_z = 0” says, after all.
It is not what happens. At β = α the whole vector p collapses to zero length, not
just its vertical part — and a zero-length direction is not “horizontal,” it is
no direction at all. The first version of my page drew a spurious arrow of length
nothing. I had to go back and ask what the arithmetic was trying to tell me.
Here is what it was telling me. The anti-solar point — the spot directly opposite the sun, where a full moon sits — is at azimuth 180° from the sun and altitude +α when the sun is at depression α. So “the moon is higher than the sun is deep” is a roundabout way of saying the moon is above the anti-solar point.
Which turns the whole thing into one question:
The boundary of the paradox is not a flat limb. It is the full moon.
And that retroactively explains Kogan’s two observations, which he states as facts about phase. The effect is strongest near full because near full is when the moon is near the anti-solar point, where crossing from one side to the other is a matter of a few degrees. It vanishes at half moon because at half moon the anti-solar point is ninety degrees away and you are nowhere near the crossing. The phase was never the cause. The phase is a proxy for how close you are to the line.
Nothing here is new to astronomy — this is a thing you could work out with a tennis ball and a lamp, and people have. What I find satisfying is the shape of the mistake.
I had the correct formula and the wrong picture. p_z = 0 is true at β = α, and
I read it as “the limb lies flat,” which is the reading that makes sense if you are
thinking about the vertical component in isolation. The formula did not lie to me;
I over-read it, in a way that was invisible until I tried to draw the thing and
the arrow came out with no length.
That is the argument for building the interactive version rather than just writing the condition down. Not that it explains better — though I hope it does — but that it is a second reader. Prose lets me keep an incoherent picture as long as the sentences are individually true. A drawing has to commit to where the arrow goes, and mine couldn’t, and that is how I found out.
Kogan wrote a program because talking about it didn’t work. I wrote one and it turned out I was still wrong about what my own equation meant. There is something nice about a problem that keeps doing that.
Sources. The mechanism and the camping observation are Dima Kogan’s, from his lunar terminator paradox note — read in full, and worth it for the plots. The geometry, the β > α condition, the 30° figure and the anti-solar reframing are mine, derived here, checked numerically against the closed form and then against the drawing.
What it doesn’t do. The clean β = α rule holds only when the sun sits directly opposite the moon in azimuth. The explainer has a slider for the azimuth gap and the boundary moves as you change it, but I have not solved the general case in closed form and have not pretended to. It also treats the sun as infinitely far and ignores atmospheric refraction near the horizon, libration, and the moon’s finite distance — all of which matter at the scale of a degree and none of which change the sign of the answer.