su_liam wrote:Any chance of your exposing the source code?
If I can bear the embarrassment

su_liam wrote:Any chance of your exposing the source code?
Anguipes wrote:su_liam wrote:Any chance of your exposing the source code?
If I can bear the embarrassment
Anguipes wrote:su_liam wrote:Any chance of your exposing the source code?
If I can bear the embarrassment
Zompist's Markov generator wrote:it was labelled" orange marmalade," but that is unutterably hideous.
Nancy Blackett wrote:Anguipes wrote:su_liam wrote:Any chance of your exposing the source code?
If I can bear the embarrassment
As long as you can apply it to the Earth so that I'll know what the weather will be like when I'm on holiday, your code will be beyond any reasonable criticism
Daistallia wrote:Also, I assume you were planning on making it adjustable to various star type, yes?
Anguipes wrote:Only if your holiday is in Manchester.
Zompist's Markov generator wrote:it was labelled" orange marmalade," but that is unutterably hideous.
Anguipes wrote:If anyone plans on working with highly elliptical orbits (eccentricity over 0.3), they're going to have to find me some C++ code to solve Kepler's equation for those cases. I'm currently running the (C?) code from here, which runs the < .3 case fine but sulks at everything else.
su_liam wrote:P.S: Hey Nancy(are you ever going to change that back?), who's the angry lady in your avatar? Kinda makes me think of a very young Jodie Foster(yum).
Zompist's Markov generator wrote:it was labelled" orange marmalade," but that is unutterably hideous.
su_liam wrote:It's not because it's C. It actually looks like it should handle eccentricity up to 0.8 ...pretty... well. Its just that over about e=0.3 it takes a lot of iterations to find the root unless you make a really good initial guess. Doing it by hand usually bogs down scary bad at 0.1, so don't complain...
Two hints. One, unless they're darting around little class-M stars, planets with orbital eccentricities much over 0.3 are really unlikely to be remotely fit for humans. Without invoking magic... Two, assuming that your angular coordinates are zeroed at perihelion, a good guess for about nine tenths of the year would be right around pi(180º). With these two points in mind, a fastish computer and a widdle bit of patience, you got it made my man!
Anguipes wrote:Thanks. I'm aware that high eccentricities are unlikely in strictly realistic conworlds, but I would like this program to (eventually) cover as many crazy possibilities as possible.
XinuX wrote:I learned this language, but then I sneezed and now am in prison for high treason. 0/10 would not speak again.
Anguipes wrote:the latitude equal to the axial incline of the planet
the duke of nuke wrote:Yes - these latitudes are the tropics.
su_liam wrote:I remember reading... somewhere(which I can't find in my current state)... that, on planets with an axial inclination in excess of 45º, the annual mean insolation is greatest at the poles. I have yet to figure out the calculations for annual mean of insolation, so I don't know.
XinuX wrote:I learned this language, but then I sneezed and now am in prison for high treason. 0/10 would not speak again.
the duke of nuke wrote: out of interest, will we be able to use this for planets that are tidally locked, or that have ridiculously long days?
bulbaquil wrote:...refraction...