Original by @jagarikin here: https://twitter.com/jagarikin/status/1149492270250393600 …
It works when the outer points are in pairs opposite each other (any shape / speed ok). The relative influence of each outer point becomes evenly distributed for each inner point. Here's one with the center rigged.pic.twitter.com/geUNZ4B745
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#pico8 cart: https://lexaloffle.com/pico-8.php?play=midpoints … (press up to increase depth, x to toggle outer average display, ctrl-r to re-run w/ random speeds)Prikaži ovu nit
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huh, that's interesting. Is it because each midpoint is basically the average position of the surrounding two? So the orange line is average of two points, yellow is average of three, and by the time you past 6 you're actually averaging the position of all the spinners?
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Yes, almost -- the yellow one is the average of 4 points, with the center point counted twice. You can take the relative influence from the nth row of pascal's triangle. (But then the contributions also wrap around, so the distribution becomes flat quit quickly).
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It reminds me of the Napoleon's theorem. If you draw equilateral triangles in the three sides of any triangle, the triangle formed by their baricentres is equilateral as well.pic.twitter.com/iDhOo0f9Rl
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That's what I'd expect, since calculating a midpoint is basically a low-pass filter, and you're doing it recursively. That's always going to trend towards uniformity.
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Not to burst your bubble, but your observation is not entirely accurate. The polygon being approached will always have equal side lengths and will have vertices that lie on an ellipse, but in general it's not approaching a regular polygon. Try a wider initial polygon to see this.pic.twitter.com/Tefvpyv27A
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Initial conditions determine polygon. Non-symmetric points equals non-symmetric polygons
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Čini se da učitavanje traje već neko vrijeme.
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