OOPS. Mistake…
It also needs weighting, so there’s SEVEN (six) values.
Cumulative chance, chance, TL-long, TL-lat, BR-long, BR-lat, M-value.
0.011466566, 0.001483667, 129.5385591, -13.55964, 131.4940545, -17.00178, 977406
0.013444788, 0.001978222, 131.4940545, -13.55964, 132.6673518, -15.28071, 488942
Meaning that if the random is between .0114 and .0134 (0.197% of the time) you need a random long between 131.49 and 132.66, a random lat between -13.55 and -15.28, and position at m-value 488942.
Say https://www.google.com/maps/@-14.095484,132.330762,488942m
Which is a big chunk of Northern Territory.
Another, using
0.47400176, 0.00427296, 150.2668109, -33.524052, 151.0490091, -34.21248, 15316
0.48096510, 0.006963342, 150.625585, -33.588065, 151.2850431, -34.12051, 959
means a random position (which happens to cover Sydney) may give a much closer view of a suburb
Say https://www.google.com/maps/@-33.897852,150.943233,959m
And a POI
0.917910448, 0.001492537, 149.132023, -35.185223, 149.132023, -35.185223, 141
0.919402985, 0.001492537, 149.094029, -35.318151, 149.094029, -35.318151, 141
Say https://www.google.com/maps/@-35.318151,149.094029,141m
which is not-very-interesting except if you are a Numismatist!
Having thought more thoroughly, it should (will) be provided as EIGHT columns, so you only have to rerfer to a single row Low-Chance, High-Chance, chance, TL-long, TL-lat, BR-long, BR-lat, M-value. Needing Choose a random number Find row where number is between Col-1 and Col-2 Calculate Longitude as random between TL-long and BR-long Calculate Latitude as random between TL-lat and BR-lat Construct the html
Today’s co-ordinates on a USA map: 39.385338793006° N -114.768764088915° W Nevada. There’s even towns and industry out there. Australia is just barren land where you could hide a body that’ll NEVER be found