astronomysun
Knowing that astronomical twilight (i.e. astronomical dawn) is when the sun is 18 degrees below the horizon, I am calculating the astronomical twilight time this way:
Sunrise of day1 - [ (Sunrise of day1 - Sunset of day0) / 180° ] * 18°
Is that correct?
It was really difficult to find any mentioning of the "lunar twilight" in astronomical literature. But I managed to find one in an old book by G.V.Rosenberg called "General picture of twilight phenomena". The book is in russian, but there is not too much about the moon. So I'll just translate all the relevant stuff.
First of all there is this plot for the solar twilight (I've translated it):
Where E is the illuminance of a horizontal surface in luxes. And lg -- is a logarithm base 10.
And then there is a small chapter:
Lunar twilight
Lunar twilight is similar to solar twilight, but it is much more bleak. The illumination from the moon varies from $10^{-9}$ (new moon) to $2\cdot 10^{-8}$ (full moon) from the illumination from the sun at the same point in the sky. Therefore the illumination from the full moon high in the sky corresponds approximately to the middle of nautical twilight. And lunar twilight ends virtually when the moon gets under the horizon.
Formulas on that page are about as simple as you'll find and should meet your needs. If you're confused by something specific, just ask!
For shortening the period of night, you should go by the distance (in degrees, say) the Sun is below the horizion, rather than any fixed time. That is the setting of the "zenith" variable. You might like http://en.wikipedia.org/wiki/Twilight for background, then the values "official", "civil", "nautical", and "astronomical" on the page you referenced should make sense. Without even knowing the use for your sensor, I would suggest starting with the nautical twilight value, that is set the variable zenith to 102. (Nautical twilight is pretty dark!)
UPD: The math on the page looks good. I tried coding it and it gave the right answer (for my location today anyway) to within about a minute.
Best Answer
No, that's not correct. Or, rather, it would only be correct on the equator at the equinox, when the sun rises and sets straight up and traces a full great circle across the sky.
Instead, start with the formula for the solar elevation angle $\theta_\mathrm{s}$,
$$\sin \theta_\mathrm{s} = \cos h \cos \delta \cos \Phi + \sin \delta \sin \Phi,$$
and solve for the hour angle $h$ (in the local solar time):
$$\cos h = \frac{\sin \theta_\mathrm{s} - \sin \delta \sin \Phi}{\cos \delta \cos \Phi}$$
where $\delta$ is the current sun declination, $\Phi$ is the local latitude and $\theta_\mathrm{s} = -18^\circ$ (plus a correction for atmospheric refraction, if that's not already included in the 18° figure).