Using GIS to determine Average Land Distance From Equator?

How far away is the average piece of land on Earth from the equator? Put more precisely, what is the median absolute value of the latitude of landmass on Earth?

From spherical geometry, the median absolute value of latitude on the Earth's surface is 30 degrees. However, land might tend to be further or closer to the equator than this.

Note: I originally posted this in the Earth Science, but this question is a better fit here instead.

Answer

DISCLAIMER: Please see important caveats at end of message.

SHORT ANSWER: The average absolute latitude of land on Earth is 33.2924 degrees, or about 3764 km from the equator.

Longer Answer

The plot above shows the kilometers of land intersected by line of latitude sampled every 1.8' (6000 samples total). Some notes:

• The large bump on the left is Antarctica. As far north as $75 S {}^{\circ}$, Antarctica encircles 71.43% of the South Pole, a total of:

$(71.43 \%) \cos (-75 {}^{\circ}) (40700 km)\approx 7525 km$

• 
  

  If the equator were covered with land, the graph would peak at 40700km at the equator. However, only about 21.60% of the equator is covered with land.

  
  



• 
  

  The actual maximum occurs at about $30 {}^{\circ} 28' N$, where 15846 km (45.17% of this 35081 km circle) is covered by land.

  
  

Of course, you're interested in the absolute latitude:

And since you're interested in the total accumulated land measured from the equator, we "integrate" to get:

which shows the answer (the precise answer earlier was computed from functions, not approximated by looking at the graph).

Methodology and Caveats

I used Mathematica's 5743 world polygons and 232 additional Antarctica polygons to create a virtual 12000x6000 monochrome equiangular image of the world.

I never actually used the image directly, and it turns out my machine has insufficient memory to export it in any supported image format.

Here's a 1/4 size (6000x3000) version:

Note that I added a black bar near the north pole to help with scaling. My calculations ignore this bar.

I then simply counted the number of pixels per line of latitude, multiplied by the cosine of the latitude, and then by the Earth's circumference. I assumed a spherical Earth with a circumference of 40,700 km.

Since Mathematica gives world polygons to a precision of 0.0001 degrees (although they're not necessarily always accurate to the last digit), using a rasterization size of 1.8 minutes of arc (0.03 degrees) introduces rounding/pixellation errors.

My work is here:

https://github.com/barrycarter/bcapps/blob/master/STACK/bc-equ-dist.m

As always, it's possible I made an error, so please doublecheck my work before using it for anything important.

As noted above, Mathematica doesn't consider Antarctica to be part of the world. There are also several other known inaccuracies in Mathematica's model of the world:

https://mathematica.stackexchange.com/questions/10229

Improvements

There are several possible improvements to this answer if anyone's interested:

• 
  

  Use a more accurate set of world polygons.

  
  


• 
  

  Instead of rasterizing, use the polygons themselves to determine land length per line of latitude.

  
  
  


• 
  

  See how the results change if Antarctica is excluded.

  
  


• 
  

  I was hoping to provide a map $30 {}^{\circ} 28' N$ (the latitude line with the most land), but I couldn't find a good way to create a map that was 360 degrees wide and only a few degrees high.