sorry for taking so long, but I had to find and decipher my notes.
this is based on the equation in book 6 for temps. That equation is a simple
blackbody simulation where the value 374.025 appears to be various constants
and convertion stuff lumped into a single value.
T= K*G*(1-A)*(L^.25)/(D^.5)
while its use in book 6 is for an entire surface, it can be used to find
average temp for different parts of the surface as well. All thats needed
is the angle of sun to determine how much energy is falling on the surface.
Taking axial tilt into consideration, I came up with this;
*someone who is actually good at math should double check this to see if its
right.*
Q=cos((x-1)*8.57+4.3 plus/minus axial tilt )
where x is equal to the number of hew row on a standard trav worldmap (
unfolded icosohedron ). the 8.57 and the 4.3 simple convert hex row into degree
value of center of hex row in question. 'Q' would have to be figured for each
season.
changing 'K' in the original book6 equatation would then look like this;
K= (((1-F)*Q*425)+(F*374.025))
Its been a while, but I think the '425' would give the avergae temp given the
average sunangle over the enitre planet to give 288Kelvin for earth, and 'F" is
a mixing factor for how atmosphere/winds and ocean currents distribute the heat
across the globe. An experiment in that would be (hyd%+atm*.1)/2. More oceans
make temps more even across world...thicker atmosphere spread heat more evenly
around globe. I guessed at that and it'd most likely need to be changed. In a
program that works it out for 'cells', a simple smoothing function using
adjacent 'cells' would probably work just as well.
Albedo is interesting because it cannot be figured on planet as whole, but
taking into consideration variations of terrain. Each 'cell' would have its own
albedo number modified by cloud coverand even sunangle on water. Basicly, this
figures energy getting past clouds and then figures energy reflected back off
surface. Cloud cover% is proportional to hyd%----- hyd%*.7 is a good place to
start. Water albedo changes with angle light hits it. I came up with
((lat^3)/34200)+6 to fit one list of water albedo with relationship to sun angle
Albedo.cloud = ( hyd%*.7)*.7 the second.7 is around avergae for actual cloud
albdeo numbers
Albedo.ground = ((1-(hyd%*.7))*ground.albedo)*(1-hyd%)
this is energy past clouds hitting the surface and not water
ground.albedo varies a bit based on ground cover..ice/snow is high and
forest/darksoil is low
Albedo.water = ((1-(hyd%*.7)*(((lat^3)/34200)+6)*hyd%))
this is energy getting past clouds and hitting water and not land.
you're on your own for greenhouse gasses for now. I do have an idea and found
a method for figuring water vapor in air, which might be fun seeing as water is
a major greenhouse gas. You might also be anal enough to figure atm composition
and figure it from that...but thats a lot of work.
thats pretty much what I have
if you figure how cold makes ice which raises albedo, the feedback can
'snowball' and cause ice ages.
I wonder if heating would cause more clouds and increase albedo to cool world
off....just curious.
this eems to work for me, but I could be way off base, thus I refuse to
guarantee this is correct. I'm only a high school grad and that was nearly 30
years ago, so take it all with a grain of salt.
well, back to work gutting basement and filling dumpster