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    Re: Going Loony with Lunars
    From: Frank Reed
    Date: 2017 Dec 12, 12:32 -0800

    Ed, two weeks ago, you wrote:
    "I had thought that when the moon's height is greater than the other body's height and subsequent corrections for HP and refraction etc. elevate the moon still higher and corrections lower the other body, I should see a cleared lunar to be a bit larger than its pre-cleared value. But this is generally [not] what I find. I have many worked many examples where the change was NEGATIVE. That is, the cleared distance is less than the pre-cleared. In fact, the change only goes positive when the moon's height is significantly higher than the other body.  Why is this? It seems counterintuitive."

    It's easiest to puzzle out things like this by considering simple geometries when the two bodies are on the same vertical circle, which is equivalent to saying that their azimuths are equal or opposite. The process of clearing takes out refraction and parallax and yields the geocentric distance. That implies that the clearing process lowers both bodies by a small amount (their individual refractions) and raises the Moon by a large amount. To put numbers on it, the individual refractions will be on the order of 1-3' except when the bodies are very low in the sky while the parallax in altitude of the Moon is HP·cos(h) where HP is, as usual, the almanac tabulated value of the Moon's horizontal parallax and h is just the observed altitude of the Moon. This parallax in altitude for the Moon will average nearly a degree when the Moon is low in the sky, falling below half a degree when the Moon's altitude is above 60°.

    When the Moon and the Sun (or other body) are on the same side of the sky, one above the other, things are simple. In the clearing process, the Moon goes up by a large amount, and both bodies fall by a small amount. This implies that if the Moon is below the Sun, the cleared distance will be smaller than the original (comparing with pre-cleared distance --center-to-center), and if the Moon is above the Sun, the cleared distance is greater. This fits your intuitive expectations, right? When the bodies are aligned like this, the math is incredibly simple: LDclear = LDpre +/- HP·cos(h) + ref. In words, you add or subtract the Moon's parallax in altitude (subtract if the Moon is below the Sun) and add the absolute difference in refraction between the two bodies.

    Now consider the case where the Sun and Moon are on opposite azimuths. In that case, both bodies are lowered slightly by the clearing process due to refraction while the Moon is elevated by a large amount (because we are "taking out" its parallax in altitude). That means that the Moon will be pushed towards the Sun in the great majority of cases. Thus when the Sun and Moon are on opposite azimuths, the cleared distance is less than the original even when the Moon is very low in the sky. The simple equation in this case is LDclear = LDpre - HP·cos(h) + ref.

    It's not hard to see that there must be some transitional circumstance when the two bodies are not on the same vertical circle. If we limit ourselves to the cases where refraction is minimal (and treating the other body's parallax as negligible), we can figure out what's going on if we use the geometry implied by series solutions to the lunars clearing problem. It's all about those corner cosines that I first detailed for NavList (back when it was a mailing "list") way back in 2004. The cleared distance is given by:
      LDclear = LDpre - HP·cos(h)·A + (small refraction correction)
    where A is a "percentage" telling us how much of the parallax "acts" along the lunar arc. It's a number between -1 and +1 and is equal to the "corner cosine" --the cosine of the angle at the Moon corner in the Moon-Zenith-Sun triangle. Now in this slightly approximate model where we are ignoring refraction, we can see quite plainly when the cleared distance is greater and when it's smaller. It's driven entirely by the sign of A which is determined by whether that corner angle at the Moon is greater or less than 90°. This is something that you can visually predict when you shoot a lunar, just by looking at the sky! Next time you spot the Moon in the sky, find a likely star to use for a lunar observation, and picture the great circle arcs that make up the Moon-Zenith-Sun triangle. If the angle in the corner at the Moon is less than 90°, which means that the lunar arc climbs in altitude away from the Moon, then the clearing process will reduce the distance. Also from this analysis, it's plain that the cleared distance will be very close to the pre-cleared distance in those cases where the lunar "arc" leaves the Moon in a nearly horizontal direction.

    Wait, wait, wait... while editing the last paragraph, I think I have hit on the best way to think about this: if the lunar arc climbs away from the Moon, then the clearing process decreases the distance. Note that this is the initial angle leaving the Moon --completely analogous to the initial great circle course from one location on the Earth to another. That should resolve 99% of the cases you have examined. Of course, you're looking at textbook puzzles, which means you don't have the sky in front of you to see the geometry. If it's not obvious how the bodies are positioned relative to each other, you might want to simulate those textbook cases you've assembled using something like Stellarium to visualize them better.

    Frank Reed

       
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