I have half million km^2 triangles... still want to push out quads to triangles. Sigh.

Rows of quads to triangles...

These diamond quads...

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My next gigasecond?

My next megasecond

Quads in a Sinusoidal World Map

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Now that I can find the column and row, I can pick a point randomly or by clicking inside the sinusoid. I can walk to a random adjacent tile and make it land or sea, blue or green, or use position to generate a seed for a not so random number and average them with adjacent heights and subdivide quads into triangle fans with vertices above or below sea level.

Campaign?

I'm working on a solo game...

I should go back to the sinusoidal world map and then subdivide the quads and triangle pairs into adjacent parent triangle corner quads/triangle pairs.

Make the diamond quads 7 px wide and 12 px tall

These diamond quads...

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Equilateral diamond quads 700 km wide, 600 km tall

Area at each row.

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Equilateral diamond quads 900 kilometers wide

Area at each row.

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1 million km^2 equilateral diamond quads

Area at each row.

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I have an isometric grid notebook, so I want to set up a grid of equilateral triangles, each one half m. km^2, one million km^2 for a diamond quad of two equilateral triangles. So an equilateral diamond quad is the fourth root of 4/3, (4/3)^1/4, bigger than an equilateral triangle a thousand km on a side, so 1,074.57 km wide and 930.6 km tall. Earth is 37.224 quads wide and 21.49 quads high... The actual arrangement rounds down so that the actual areas are a little more; the idea is at least a million km^2, although the comparable area in Barnes' article is more like two thirds, so an Earth equivalent would be 40 diamond quads wide at 1,000 km and 100 km/pixel, and 23.09 rows 8.6 pixels tall.

Area by latitude

Area by latitude from the pole, an angle. Make rows out from the initial one million km^2, each subsequent row a whole number of million km^2; round up and find a new height.

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N Rows, 1000 km/row

Area at each row.

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 Area of a sphere = 

A

=

4

π

r

2

Area between the latitude, or 

A = 2*pi*r^2*(1-cos(theta))

1 - cos(theta) = 1 m. km^2 / (2*pi*r^2)

theta = arcos(1 - 10^6 km^2 / (2*pi*r^2) ), r = circ/(2*pi) = 40,000/2/pi = 6366.2 km

theta = arccos(1 - 10^6/2/pi/6366.2/6366.2) = arccos(1 - 0.003026988008686) = arccos(0.996073011991313) = 0.080877478519613

0.080877478519613/pi*20,000 km = 514.9 km

One million km^2

Points within each square.

Oops!