Circumpolar Equation Derivation

Question

Edit: Another algebraic solution from Astronomy.stackexchange

I am trying to understand how the circumpolar star equation works. Although it is astronomy, this part is pretty much just geometry. The equation is this

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

The setup is for an observer at latitude $l$, stars whose declination $\delta$ fit this equation are circumpolar. What I cannot get past is how to get from the star's lowest altitude $a$ to there. *Declination is measured from the equator and altitude is measured from the horizon for those who don't know." I know that the angle of the cone made by the rotating star is within the $\frac{\pi}{2}$ limitation. I just can't express it in an algebraic way.

This shows what is meant by circumpolar stars.

This shows the sort of rotation undergone by the celestial sphere due to the observer's latitude. NCP and SCP are North and South Celestial Poles.

In the second picture, it is easy to prove that the altitude $a$ is equal to longitude $l$ and that $c = \frac{\pi}{2} - l$. What I need to know is how to go about proving that a star of declination $\delta$ will always have an altitude $a \geq 0$.

Edit: Included pictures in line instead of as a link.

Answer 1

See picture below: a star is circumpolar if it lies within the grey cone of half-width $L$, the same as the latitude. Hence declination of such a star must be greater than $\pi/2-L$.

Answer 2

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

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