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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?

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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).

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