[OC] I took the mean of all forest pixels on earth from 20 years of satellite data. Here’s actual statistical Forest Green:

12 comments

1. Yeah that’s about where I’d have put it.

   Cool idea btw. 

2. What’s the hour range you used? Does this include all night-time shots?

3. Too impacted by shadow, I’d guess.

4. I can see this being used to make camouflage.

5. It’d be quirky if there were a website where questions like “what’s the average colour of forests on earth” or “what’s the average colour of all oceans” is asked and the user is presented a colour pallet to drag around, then see how close their answers are, could collate all that information and see if a load of users answers averaged together themselves is even close to actual.

6. How on Earth did you crunch that much data??? All forests on earth? Would take you years just to identify them surely. 

7. That’s neat. What other colors or findings were you able to identify as a result of your research? Or were you just focused on finding that one color?

8. Are you planning on doing the other surfaces as well? Water, snow, desert, etc.

9. If it’s possible, you could make an array of the greens (at whatever temporal resolution you’d like), it would be interesting to see the variation/lack of variation over time.

   Just a thought, great work!

10. That’s a dark and spooky forest.

11. Neat idea! How did you handle deciduous forests in winter?

12. I’m interested in whether this average shows any seasonality or latitude dependence, could be an idea for a future post.

    What approach did you use for averaging the colours? Are you summing the red, green and blue values and dividing by the number of samples? i.e. (R1 + R2 + R3)/3,(G1 + G2 + G3)/3,(B1 + B2 + B3)/3

    This is the most obvious approach but will give inaccurate results, because brightness isn’t linear with RGB value. RGB goes from zero to 100% brightness (0 to 255), but for [complicated reasons](https://www.youtube.com/watch?v=LKnqECcg6Gw) a G value of 127 isn’t half as bright as 255 – it’s one fifth as bright. Half-of-full-brightness is actually a value around 180.

    What you need to do [as explained in this article](https://kevinsimper.medium.com/how-to-average-rgb-colors-together-6cd3ef1ff1e5) is square all of your RGB components, sum them, get the average of *that* number, then take *the square root of the average*.

    edit: I’ll give a short example. This table defines 21 variants of green ([taken from here](https://www.rapidtables.com/web/color/RGB_Color.html)):

    | R | G | B |
    |——|—–|—–|
    | 154 | 205 | 50 |
    | 85 | 107 | 47 |
    | 107 | 142 | 35 |
    | 124 | 252 | 0 |
    | 127 | 255 | 0 |
    | 173 | 255 | 47 |
    | 0 | 100 | 0 |
    | 0 | 128 | 0 |
    | 34 | 139 | 34 |
    | 0 | 255 | 0 |
    | 50 | 205 | 50 |
    | -144 | 238 | 144 |
    | -152 | 251 | 152 |
    | -143 | 188 | 143 |
    | 0 | 250 | 154 |
    | 0 | 255 | 127 |
    | 46 | 139 | 87 |
    | -102 | 205 | 170 |
    | -60 | 179 | 113 |
    | -32 | 178 | 170 |
    | 47 | 79 | 79 |

    If we simply average each column the result is [#0fbf4c](https://imagecolorpicker.com/color-code/0fbf4c) but if we use the squares method, it’s [#5fc762](https://imagecolorpicker.com/color-code/5fc762). The difference is significant!

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