I have a world that I am intending to give about an 18° axial tilt. I know generally that this will reduce seasonal variation in the length of daylight (sunrise to sunset), however, what I'm unsure of is exactly how to go about calculating it precisely at any given latitude.
The planet also has a 154 day orbit around its sun, which I don't think matters but just throwing it out there in case it does. 24 hour rotation. About the same distance to the sun, again, not sure that matters.
Given:
Then
If the planet rotates in the same direction as it revolves around its sun, that is, the planet's rotation is direct, then an average solar day will have 24 × (155/154) = 24 hours 9 minutes 21 seconds.
If the planet rotates in the opposite direction of its revolution around its sun, that is, the planet's rotation is retrograde, then an average solar day will have 24 × (153/154) = 23 hours 50 minutes 39 seconds.
Going forward, I will assume that the planet's rotation is direct, which is the most common arrangement.
The planet's year will be 153 solar days plus 23 minutes 14 seconds. They can take their civil year to be 153 days and insert a leap day every 150 years, skipping every 50th leap year. (If they remember the rule after 7500 years...)
I will also ignore the effect of atmospheric refraction, which will lift the apparent position of the sun near the horizon, so that daytime is longer than what the naive geometric calculation will predict, and night-time is shorter.
On Earth, atmospheric refraction lifts the position of the sun by about 10 minutes of arc near the horizon, which makes day-time about 5 minutes of time longer than calculated, and night-time about 5 minutes shorter than calculated at 45° latitude. The daytime lengthening and night-time shortening varies with latitude, being maximum near the poles and minimum on the equator.
On the planet's equator all the time, and at all latitudes at equinoxes (twice per year), days and nights are equal, each having 12 hours 4 minutes and 41 seconds.
Let's put the duration of the average solar day as $d$, and the axial tilt as $\theta$. At a reasonable latitude $\lambda$, at the summer solstice the daytime will be about $\frac12 d(1 + \tan \lambda \tan \theta)$, and at winter solstice it will be about $\frac12 d(1 - \tan \lambda \tan \theta)$.
Note: The formula is only an approximation; deriving an exact formula would require actual work. It tends to overestimate the variation of daytime duration at mid-latitudes, but it is good enough for fiction. For example, on Earth the formula gives a daytime of 17 hours at 45° latitude at the summer solstice, instead of the correct 15 hours and a half.
The polar circle is at 90° − $\theta$ latitude; for the given 18° axial tilt, the polar circles will be at 72° latitude. On the polar circle, at the summer solstice the daytime extends to the entire solar day. (The approximative formula above calculates this correctly.)
(In reality, atmospheric refraction will make daytime equal to an entire solar day at the summer solstice a little bit south of the northern polar circle and a little bit north of the southern polar circle. By contrast, to make night time take an entire solar day at the winter solstice we need to move a little bit north of the northern polar circle, and a little bit south of the southern polar circle.)
As we move north of the northern polar circle or south of the southern polar circle, the sun will stay above the horizon more and more days around the summer solstice, and it will stay below the horizon more and more days around the winter solstice. At the poles, daytime will be half a year and night-time the other half of the year.
What you are looking for is called the equation of time:
The equation of time describes the discrepancy between two kinds of solar time. The two times that differ are the apparent solar time, which directly tracks the diurnal motion of the Sun, and mean solar time, which tracks a theoretical mean Sun with uniform motion along the celestial equator. Apparent solar time can be obtained by measurement of the current position (hour angle) of the Sun, as indicated (with limited accuracy) by a sundial. Mean solar time, for the same place, would be the time indicated by a steady clock set so that over the year its differences from apparent solar time would have a mean of zero.
The link above provide the equation of time for Earth.
