I have determined the average global temperature of my world based on the luminosity of the star, distance to the star, and the bond albedo and greenhouse factor of the world.

What I'm having a devil of a time figuring out is how to determine how much hotter / colder than the average the planet is at a given latitude, expressed as a percentage + / - of the average temperature.


  1. Assume sea level
  2. Ignoring prevailing winds, ocean currents, land / sea, distance from coasts, atmospheric refraction, and everything else -- those things can be adjusted on top of the base temp at latitude.
  3. Axial tilt of the world is 28 degrees
  4. Year is 321.7181 local days
  5. Day is 25.6382 Earth hours (sidereal) or 25.71814 Earth hours (solar)
  6. For simplicity, I want to figure temps at the solstices and equinoxes

I know this much: the length of day at a given latitude on the solstices and equinoxes will play into the equation, because that will determine literally how long the surface receives heat from the sun.

What I don't possess... and I curse my disinterest in school thirty five + years ago... is a good grasp of calculus / trigonometry.

Ideally, I would love to create a spreadsheet with pages that show:

  1. daylight as a percentage at a given latitude / calendar day
  2. Base temperature at a given latitude on the summer and winter solstices and the spring and autumnal equinoxes

... once I plug in the variables appropriate to my planet.

Has anyone figured this out? I turn to all of you after a whole lot of Googling, much of which went over my head. THANK YOU.


1 Answer 1


The easiest way to understand this for a maths novice is to leave axial tilt to the end, and just think of this using an abstract sphere for as long as possible.

Step 1 - Angle of incidence to power ratio.

If we ignore axial tilt, and my crude mspaint, the amount of energy absorbed from the sun for any point on the planet at any given time looks roughly like this:

enter image description here

When it's hitting square on - it's 100%. If the sun is parallel to the surface, it's 0%, and hitting somewhere in between it's half as a strong. We can define the power ratio of any point as $P$.

$$ P = \cos\theta $$ Where $\theta$ is the effective latitude, 0 for on the equator, $+\pi\over2$ for the north pole,$-\pi\over2$ for the south pole. $P$ will be 1.0 (100%) at the equator, and 0.0 (0%) at the poles.

Step 2 - day length

How hot it gets also depends on how long the sun is up. You've given total rotation time, but how can we get day length? (once again, ignoring axial tilt for the moment).

Same formula. When the midday sun is perfectly vertical, the day length is 100%, when it's perfectly parallel to the horizon, the day will be 0% (the sun doesn't rise).

$$ D = \cos \theta $$ Where $\theta$ is the effective latitude, 0 for on the equator, $+\pi\over2$ for the north pole,$-\pi\over2$ for the south pole. $P$ will be 1.0 (100% day length) at the equator, and 0.0 (0% day length) at the poles.

Step 3 - Both together

We can take some shortcuts here because we're not calculating actual values, only the proportion of the maximum sun energy transfer. Doing the calculus in full adds a scale and constant factor here, but we then remove it to keep within the range 0 to 1. So defining total sun contribution $S$ as the unsatisfying simple $S = D \cdot P$

Step 4 - Latitude and date to effective latitude

Now we start considering axial tilt. 28 degrees is 0.488 radians.

$$ \theta = L + 0.488 \sin ({Y \cdot 2\pi}) $$

Where $L$ is the actual latitude, so 0 for the true equator, $\pi\over2$ for the true north pole, $-\pi\over2$ for the true south pole, and Y is the date of the year relative to the equinox. So 0.0 and 0.5 for the equinoxes, 0.25 and 0.75 for the solstices.

If $\theta$ is greater than $\pi\over2$ (or less than $-\pi\over2$) then the sun is below the horizon all day (ie winter at poles), so if you see this state, it's pitch black 24/7.

Step 5 - Putting it all together

Lat   Solstice1 Power    Solstice2 Power        Equinox Power
-75     46%                 0% (dark)             6.6%
-50     85%                 4.3%                  41%
-25     99%                 36%                   82%
-10     90%                 62%                   96%
 0      78%                 78%                   100%
 10     62%                 90%                   96%
 25     36%                 99%                   82%
 50     4.3%                85%                   41%
 75     0 (dark)            46%                   6.5%

Step 6 - To temperature.

This is beyond what I can answer, this maths says that the poles in winter should be 0 Kelvin (-273 Degrees C), but we know that's not true, it's a property of residual heat and greenhouse effect and convection through oceans and things like that, which you said to ignore.

But, since you've calculated the average temperature, we can use some trig and calculate that the average power input for the entire sphere occurs on the equinox at the 45 degree latitude band, it's, helpfully, 0.5 at that exact point in time.

This creates a very nice range, where the planet is coldest when the power is 0, it's the average temperature where the power is 0.5, and the hottest when the power is 1.0.

This seems too simple to be true, but the values are within a few degrees of earths so we can use it for reference to be sure we're on the right track.

(Earths orbit is eccentric by just under 2%, slightly favouring a hotter northern hemisphere and a colder southern hemisphere, but the figures should be close enough.)

Your poles are a bit below the average planet temperature in middle of summer. Earths average temperature is currently 14 degrees C. North pole has hit 13 degrees. (South pole has -12 record, but global warming is effecting the arctic more than Antarctica). I'd say this is a loose fit.

We'd also expect peak December temperatures at about 28 degrees latitude south. Hottest city on Earth in December is Brisbane (27 degrees latitude). Peak July temperatures would be at about 28 degrees north. Las vegas (at 36N) is the record holder, Death Valey is also at 36N. I'd say this is also a loose fit.

  • $\begingroup$ FANTASTIC, @Ash. Really broke it down for me. I can't wait to dig in and plug all this into my suddenly do-able spreadsheet. Many, many thanks. $\endgroup$ Commented Sep 17, 2020 at 16:49
  • $\begingroup$ Running into a snag, @Ash. I don't understand how to convert the formula into something a spreadsheet will understand. I've tried various combinations to replicate your results (for example, for 75 degrees north on the Equinox, you show 6.5% power). So for the autumnal Equinox (day zero) cell for 75 degrees north, I have: =cos((1.308996939+.488)*(sin(0*(2*pi())))) 1.308996939 being 75 degrees in radians. But no the latitude, the result = 100%. I'm sure something's getting lost in the translation, but... what? Thank you. $\endgroup$ Commented Sep 18, 2020 at 4:40
  • $\begingroup$ @MattSelznick docs.google.com/spreadsheets/d/… $\endgroup$
    – Ash
    Commented Sep 18, 2020 at 8:44
  • $\begingroup$ Thanks, @Ash, for the spreadsheet... I did not realize (not well versed in equation symbols) the dot between Y and 2pi was "divided by." However... in your description, you write that Y is the date of the year relative to the equinox. But in your spreadsheet, you have nothing for Y for the first solstice and 3 for Y for the second solstice. This is confusing. $\endgroup$ Commented Sep 20, 2020 at 18:50
  • $\begingroup$ Okay, I think I've sussed out what you were trying to do, @Ash, but one thing I don't understand on your sample spreadsheet is why, on the solstice, the % begins to rise again after around +66 / -66 degrees. (If anyone else would care to chime in, too, all input is welcome!) $\endgroup$ Commented Sep 30, 2020 at 16:32

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