Is lunar elevation at a given location for a given day unimodal:

Unimodal function

In other words, once the moon's elevation reaches a minimum (which may be above or below the horizon depending on if the moon is circumpolar), does it increase continuously to its maximum, or can it decrease briefly and then increase again (resulting in a local minimum that's not a global minimum) before reaching the maximum.

And also similar to the above as it goes from the maximum elevation (which may also be above or below the horizon depending on whether the moon rises at all on a given day).

Why this is important: I think I found a bug in libnova that assumes the moon is circumpolar if it is above the horizon when it's due north. This is untrue

My fix does not assume this, but does assume that lunar elevation is unimodal.

If that's untrue, I'll need to further fix my fix.

EDIT: Here's how far I've gotten:

  • The moon's declination and hour angle (at a given location) are both functions of time. Thus, we can implicitly consider the moon's declination to be a function of its hour angle. This only works because the moon never experiences retrograde motion. Bodies that experience retrograde motion could have two declinations for a given hour angle, so the declination would no longer be a function of the hour angle.

  • Using the standard formula, we see the moon's elevation is:

$$ \sin ^{-1}(\sin (\text{lat}) \sin (\text{dec}(\text{ha}))+\cos (\text{ha}) \cos (\text{lat}) \cos (\text{dec}(\text{ha}))) $$

where dec(ha) is the declination at hour angle, ha is the hour angle, and lat is the latitude.

  • Since the moon's elevation is a differentiable function (except potentially at the poles and equator), we find the derivative with respect to the hour angle:

$$ \frac{\text{dec}'(\text{ha}) (\sin (\text{lat}) \cos (\text{dec}(\text{ha}))-\cos (\text{ha}) \cos (\text{lat}) \sin (\text{dec}(\text{ha})))-\sin (\text{ha}) \cos (\text{lat}) \cos (\text{dec}(\text{ha}))}{\sqrt{1-(\sin (\text{lat}) \sin (\text{dec}(\text{ha}))+\cos (\text{ha}) \cos (\text{lat}) \cos (\text{dec}(\text{ha})))^2}} $$

  • The moon's elevation will reach a min/max (in the sense of unimodality) when this derivative is 0. That occurs when:

$ \sin (\text{ha})\text{ = } \text{dec}'(\text{ha}) (\tan (\text{lat})-\cos (\text{ha}) \tan (\text{dec}(\text{ha}))) $

So the question becomes: can this equation have more than two solutions for 0 < ha < 2*Pi

Which I hope someone answers at: https://math.stackexchange.com/questions/587136/


1 Answer 1


Yes, it is unimodal.

Earth's movement is a simple rotation, thus if the Moon were fixed on sky it will describe a circumference like the fixed stars.

From that basic movement we need to introduce Moon's own movement on the fixed stars sphere, which makes Moon not to complete the circumference in 1 day. Instead, Moon performs an arc similar to that of the fixed stars but shorter.

This circumference-like movement does not wobble, so it has just one maximun and one minimum in elevation.

  • $\begingroup$ Could you provide a more mathematical explanation? I agree the circumference-like movement (fixed declination) doesn't wobble, but, what if, combined with the moon's own declination change, it does wobble? $\endgroup$
    – user21
    Commented Nov 22, 2013 at 18:49
  • $\begingroup$ I never said fixed declination. In fact, it is not, due to the Moon's proper motion on the orbit. But since that is a movement on a great circle (the orbit's projection on sky sphere) and is composed with a movement on a constant declination circle (the Earth's rotation) you get a wavy movement. If the sky sphere were projected into a plane that movement would be a trocoid. There is no wobbling there, besides the own's of the Moon moving. $\endgroup$
    – Envite
    Commented Nov 23, 2013 at 0:27
  • $\begingroup$ Even after looking at en.wikipedia.org/wiki/Trochoid I'm not sure I understand. Could you provide some mathematical formulas to confirm this? $\endgroup$
    – user21
    Commented Nov 24, 2013 at 4:14
  • $\begingroup$ I'm sorry but I can not. I have my degree notes buried deeply under a ton of other papers $\endgroup$
    – Envite
    Commented Nov 24, 2013 at 13:29
  • $\begingroup$ Bummer. After what happened with the "circumpolar" answer, I'm suspicious of any verbal arguments that aren't backed up with math. $\endgroup$
    – user21
    Commented Nov 24, 2013 at 17:10

You must log in to answer this question.