Lunar angles

[spingears]

Remember that earth’s moon, earth, and all of the planets and their respective moons are giant gyroscopes whose axes point in their own directions. The planes of the orbits of earth and its moon are not coplanar nor parallel. As a result the calculation of their position, direction, etc. is not a simple matter to determine.

After satisfying one’s curiosity what do you have for your trouble? Data and/or details of no personal value except for the exercise of having gone to a lot of trouble for nothing.

[RM_Mentock]

@Bytegeist

RigThe angle that the projected terminator line makes with the ecliptic circle will then be about 80 degrees, not 90.

Thanks, Bytegeist, I’m more or less convinced, though I still haven’t done the math.

@spingears

After satisfying one’s curiosity what do you have for your trouble? Data and/or details of no personal value except for the exercise of having gone to a lot of trouble for nothing.

Not that there’s anything wrong with that.

[douglips]

@Bytegeist

I’m not sure that’s what douglips’ is or was saying, but maybe he can clarify. He noted just now that the plane of the terminator circle is always perpendicular to the ecliptic plane, to within a small fraction of degree, which is certainly correct. Actually, what that plane is really perpendicular to is the line connecting the Sun and Moon. But because the Moon’s orbit is tiny compared to its distance from the Sun, the angle between that line and the ecliptic plane is always pretty tiny. So it’s more or less correct to call the terminator circle perpendicular to the ecliptic plane as well.

That’s exactly what I was saying, I wasn’t ever trying to say anything more complicated than that, and I thought that was the simple answer to the question in the OP. It completely sailed over my head that people cared about the apparent terminator as viewed from the Earth, which of course seems obvious to me now.

But the terminator line (f.k.a. the “cusp line”), as seen from Earth, is a different matter. I don’t agree that it’s always perpendicular to the ecliptic circle, even the fuzzy sort of perpendicular that tolerates a “small fraction” of a degree in error. To make one rough calculation: let the Moon be a third of the way from New Moon to First Quarter, and let it also be near the high north point off the ecliptic circle. So the Moon is then about 30 degrees east/north-east of the Sun, and 5 degrees north of the ecliptic, as they appear on the celestial sphere. Recall that the illuminated portion of the Moon must always face the Sun. Then the spherical arc connecting the Sun to the Moon will deviate from the arc connecting Sun to the point on the ecliptic nearest the Moon by an angle of roughly — very roughly now — the arcsine of 5/30, or about 10 degrees. The angle that the projected terminator line makes with the ecliptic circle will then be about 80 degrees, not 90.

Now that I know what everyone is talking about, I agree with your assessment.