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Edit: Geometric answer from Math.stackexchange

I am trying to understand where the circumpolar equation comes from.

$\delta \geq \frac{\pi}{2} - l$

I can understand that the sort of "rotation" of the reference frame that one experiences in horizon coordinates. I'm not sure where to go from the fact that a star's lowest altitude must be greater than 0.

enter image description here

It is pretty simple to show that $a = l$ and $c = \frac{\pi}{2} - l$. $a$ in this case is the altitude of the north celestial pole. What I need to do is show that the angle of the cone made by a circumpolar star with $\delta + l \geq \frac{\pi}{2}$.

You can deduce that $a_{min} \geq 0$ for a star to be considered circumpolar. I just need to do this geometrically in terms of its declination from the celestial equator.

Edit: I think my latest attempt has found a solution.

enter image description here

For a star to be circumpolar, the star must, at the very least, come above the horizon.

$0 \leq a \leq \frac{\pi}{2}$

In the celestial sphere coordinate system, $c$ can be found to be

$c = \frac{\pi}{2} - \delta$

In the horizon coordinate system, $c$ can be found to be

$c = l - a_{min}$

Therefore

$l - a_{min} = \frac{\pi}{2} - \delta$ $\\$ $l + \delta = \frac{\pi}{2} + a_{min}$

$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.

$l + \delta = \frac{\pi}{2}$

$\delta$ must be in the cone made by the star's rotation about its declination angle. Therefore

$\frac{\pi}{2} - l \leq \delta$

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  • $\begingroup$ Can you please in this case and always define the quantities you're talking about? $\endgroup$ Jan 25, 2019 at 10:50
  • $\begingroup$ @AtmosphericPrisonEscape What do you mean by quantities? I have no physical quantities that I desire. I mean, I can make some up. Like my latitude is 40N and need to know if Mizar is circumpolar with a declination of ~55N. I am just looking for a geometric derivation of how this equation works. I'll add some visuals that I made up. $\endgroup$
    – abyssmu
    Jan 25, 2019 at 14:22
  • $\begingroup$ What you've just partially done with your edit. Clarify what is $\delta$, what is $l$. Astronomy is a huge field, people are not bound to know immediately that you're talking about celestial coordinates. This way will also help you to get answers faster, when people know what you're asking about. $\endgroup$ Jan 25, 2019 at 15:23
  • $\begingroup$ Gotcha. Thanks! I'll keep that in mind for the future. $\endgroup$
    – abyssmu
    Jan 25, 2019 at 15:28

1 Answer 1

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There is nothing wrong with your explanation. You could add some explanatory notes like $a_{min} = l-c$ and $a_{max}=l+c$. Also, I suggest that you put the equation $l+\delta = \frac{\pi}2-a_{min}$ on a separate line.

It is very good to figure out these equations on your own.

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Here is an alternative explanation with less algebra:

  • If you walk outside at night, you will notice that the north star has an elevation of $l$ degrees if your latitude is $l$.
  • The north star does not move.
  • The angular distance between stars does not change (in the short term!).
  • In order for the a star to be circumpolar, it needs to be within $l$ degrees of the north star (otherwise it will dip below the horizon when its hour angle is 12 hours.)
  • So, the angle $\theta$ between the north star to a circumpolar star needs to be less than $l$. That angle is $\theta=90^\circ-\delta = \pi/2 -\delta$ where $\delta$ is the declination of the star.
  • Finally, $$ \begin{align} \theta &\leq l\\ \pi/2 -\delta &\leq l \\ \pi/2 &\leq \delta + l \\ \pi/2 - l &\leq \delta. \end{align} $$
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