rhu: (Default)
[personal profile] rhu
So a few weeks back I queried: What's the median latitude of land area on the planet Earth? Not having found an answer online, I finally resorted to solving it myself. The answer and how I found it are :

I started by downloading a reasonably hi-res image of the world in an equirectangular projection. I used ImageMagick to convert this TIFF into a BMP, and I wrote a Python script to iterate over the rows in this image. For each row, I computed the latitude and what the proportional area of that latitude's band is. Then I counted the ratio of white pixels to the length of the row, which when multiplied by the latitudinal band's area gave me the ratio of land in this latitudinal band to the total surface area of the Earth. (For purposes of this exercise, I treated the Earth as a sphere. Close enough.)

Storing each band's contribution in an array, I also tracked the total. A second pass over the array quickly identifies the point where the total-so-far is 50% of the total. And the answer I got is:

23°12′ North, which by coincidence is just 14 minutes south of the Tropic of Cancer.

(no subject)

Date: 2010-06-30 03:42 pm (UTC)
cnoocy: green a-e ligature (Default)
From: [personal profile] cnoocy
Does your map consider icecap to be land?

(no subject)

Date: 2010-06-30 11:18 pm (UTC)
cellio: (Default)
From: [personal profile] cellio
Nicely done!

I was guessing about 10 degrees farther south. Boy was I off!

(no subject)

Date: 2010-07-30 02:36 am (UTC)
ext_12246: (Default)
From: [identity profile] thnidu.livejournal.com
OK, that is geeky.

(no subject)

Date: 2010-07-30 02:37 am (UTC)
ext_12246: (Default)
From: [identity profile] thnidu.livejournal.com
But what I wanted to ask is... What are you doing to that poor E. coli at 530nm?

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rhu: (Default)
Andrew M. Greene

January 2013

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