# Equations for converting axial tilt to hours of daylight

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.

• "24 hour rotation": Earth has a rotation period of 23 hours and 56 minutes. Why is a mean solar day 24 hours, four minutes longer than the rotation period of the planet? Jun 4, 2023 at 16:45
• @AlexP Your comments have really gotten far too aggressive lately. The value of the information you're adding is being outweighed by your manner, IMO. You're coming across as both pedantic and nasty. Rhetorical (or alternatively, very patronising) questions are rude and should nor be used where someone's mistakes are either small, or rooted in simple lack of knowledge. You have knowledge? Great. Share it in a friendly or completely neutral manner. "Remember, use 23:56, not 24:00 hours" was all that was needed here. Or even "Earth's day is 23 hours 56 minutes, not 24 hours".
– user86462
Jun 4, 2023 at 21:08
• @JBH Understood. I do note that a) I am not the only one to have said so in the last couple of weeks, and b) No one likes opening formal dispute resolution or flagging a well respected (as Alex is, and deserves to be) contributor.
– user86462
Jun 5, 2023 at 1:15
• @AncientGiantPottedPlant Yeah, I get it, but flagging and/or using the Contact link are the methods for guaranteeing the matter is brought to the attention of the Mods, who have far more options for judging and dealing with the situation than an open dispute. Just sayin'. Thanks for your understanding.
– JBH
Jun 5, 2023 at 3:01
• I think the issue was more that he was pointing to a slight simplification rather than asking me to do additional research. That being said, Alex gave another response which gives me a lot of what I think I need to understand the problem. Jun 6, 2023 at 1:32

Given:

• The planet completes a rotation in 24 Earth-hours.
• The planet completes a revolution around its sun in 154 Earth-days.
• The orbit of the planet is just about circular.

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.

• I'm not sure I'm following the equations you gave correctly. Unless I'm misunderstanding it, it gives a result of, for example, 27 hours of daylight at the summer solstice at 40 degrees latitude. Shifting it up to 45 degrees gives me -10. I'm assuming the units here are hours? Jun 6, 2023 at 2:02
• @AHerrera: d = duration of a solar day = 24 h 9 min 21 s = 84961 s. λ = 40°, θ = 18°. tan 40° = 0.84. tan 18° = 0.32. Duration of daytime at summer solstice at 40° = ½ × d × (1 + 0.84 * 0.32) = ½ × 84961 × (1 + 0.27) = 55335 s, or 15 h 22 min 15 s. Jun 6, 2023 at 5:40
• ok. i feel very silly now - I just realized that my excel sheet I was using to do the calculations uses radians instead of degrees for the trigonometric functions. Jun 6, 2023 at 14:57

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.

• After reading the article, I don't think this is really the same - this seems to be looking at the disparity between our standardized 24 hour clock. What I'm looking for is hours of literal daylight - sunrise to sunset hours. Jun 4, 2023 at 21:36
• @AHerrera L.Dutch isn't wrong. The Equation of Time is where you begin the process. For example, the Sunrise equation depends on understanding the equation of time. But even if you apply those two fairly complex sets of equations, the number of hours between any two locations on the planet, even when nearby, can be very, very different due to mountains, climate, etc. You're looking for a simple equation that depends on only one varible. There isn't one.
– JBH
Jun 5, 2023 at 0:52