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PaulMurrayCbr
  • Member for 10 years
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Equal-area projections of the hyperbolic plane
It's ok - I think I know what I want to do. The mapping will be: For a point A, find the point B on the horizontal axis such that A is above or below it (i.e., AB is at right angles to OB). The x coordinate is simply the hyperbolic distance between OB, the ycoordinate the hyperbolic distance BA. This will not be equal-area, but will line p the octagons along the axes nicely. So delete the question, I'm good. Ta.
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Equal-area projections of the hyperbolic plane
You know - with the octagons-along-a-horizonatl-line model, it's clear that each octagon will have a column of octagons above and below it (with a large nuber of other octagons fitted into the gaps). We could treat this as a repeating patter of (say) 360 octagons along that base, and glue the whole structure into a cylinder. That might be nice for game - each octagon is a day, the Sun travels along that circuit once a year.
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