Procedures in brief for setting uyp the isis terrain in step by step fashion.

Determine scale of current BMP based map to the pixel=Miles level.

i.e. if world is 10,00 miles at the equator and there are 2048 pixels then each pixel is 4.88281 Miles across. It might be good to adjust this to a even number so 2048x5=10240 miles in diameter at the equator.

Next, (for my own info) calculate the scale of the existing map. If the map is 44" across then the scale becomes 10240/44=miles per inch or 232.72 MPI. Or checking to see aberation on a more normalized system, @250MPI diameter difference would be 10240-11000 = 760 miles for an aberation % of less than 1 which is acceptable. So the scale of the map is 250 miles per inch? If that is so then one inch would contain 50 pixels?

Anyway, some important points, The map projection is only accurate at the equator. The land masses represented are accurate, but the distances between the contenants are not, but bays and inlets are accurate. If this is so, the land masses must be checked to see that they will fit into the reduced space of a tapering sphere!

To continue with procedures. We have the scale, and now we need to determine the elevation express in values from 1 to 256. We need to go back to the original GIF map to see the pallate setting for the first land mass area which is set to yellow. So if yellow pixels have a value of 100, then the undersea elevation scale would be Max depth of the ocean/number of pixel colors allocated to blue. If we stipulate that the lowest part of the ocean is a 5 miles and sea blue pixels start at 0 and go to 100, then 5mi/100=distance down per change in pixel color. The reverse calculation would be true for land masses. If the highest mountain is at 6.5 miles and 156 pixels (256-100 for blue sea) are used, then each pixel elevation would represent .041 miles. (at 2000 ft. per mile thats about 83 feet per elevation increment (pixel value).

The next thing that enters into the calculation is the distance that seperates the given pixel from water. There is also a relation to the area of the most adjacent water mass. In otherwords, a pixel close to the ocean is going to be somewhat different than a pixel close to a mountain lake both in terms of elevation and in terms of ambient moisture.

Next, the prevailing winds will have much to do with transporting available moisture to a location. In general, winds will be blowing from the opposite direction of the planet’s spin. However, it would be more interesting and accurate to manualy plot air and water currents than to have that calculated. Any such manual plot should have a chance of being modified by a local "MicoClimate" variable.

Remember the idea of all this is to calculate seasonal tempritures and annual rainfall. Some additional factors to compensate for are obsticles to the transport of moisture. A high range of moisture will tend to reduce moisture on the other side. Another factor is the geology of the ground in specific area. This affects the areas albedo which has a lot to due with the temp and rainfall in an area. The higher the albedo, the higher the temp and the lower the rainfall.

Another factor is the distance of the point from the equator; the further away, the lower the average temp and in general the higher the rainfall.

Additional info can be easily obtained from the random weather generators available in the Outdoor adventures and greyhawk rule sets. It would also be a good idea to look in an atlas for currents info.

FOR calculating temp gradient expressed as distance from the equatior towared either pole determine max and min temps at the pose and at the equator. Then sbutract the max and max and the min from the min and then devide them by the distance from equator to pole. The result is the temp/distance gradient.

POLE MinTemp -50deg MaxTemp 80deg

Equator MinTemp 50deg MaxTemp 120deg

Means:

Min gradient = -50-50=100/5600Miles=.018 Deg per mile or 1 degree per 100 miles.

Max gradient= 120-80=40/5600miles=.008 or 1 deg per 800 miles Now we modify these numbers based on air moisture/proximity to water. The closer to a body of water, the lower the temp.

This should be calculate as follows: -10 degrees at the coast and reduce that number by 1 degree for 3miles-2x(total delta of elevation)

flat jungle min temp average = 80 at coast 70, 6 miles from coast 72, 15 miles from coast 75 etc.

Coastal Hills min temp 50 at coast 40, at 9 miles 43 but at 10 miles there is a ridge of hills at 1000 feet which drops back down to roughly sea level after 5 miles. Total delta= 1 mile elevation (1000up and 1000 down). Then at 12 miles we get:

12/3=-4 + 2x1 for -4 or a temp of 44

Costal Mountain range min temp 60 range is 10000 feet and is 3 miles broad, we get:

• 1+3x(10) for 21