On Mon, 2005-08-08 at 06:31, Durk Talsma wrote:
On Sunday 07 August 2005 15:54, Erik Hofman wrote:
Erik Hofman wrote:
Is this correct or am I missing something?
I just realized that you also need to adjust for day-of-year to
compensate for the out-of-center rotation that causes long
Steve Hosgood wrote:
Durk is quite right. It's non-trivial, but dealing with all these issues
is not really a problem. I'm in the middle of doing it myself, but have
just only just got back from a week on vacation.
My main interest is in removing the use of Xearth's find the point on
the planet
Erik Hofman wrote:
But if you don't mind I'm still working on my own solution, I already
have the code to compensate for lon and daylight saving time.
Which is silly since daylight saving time isn't something nature
provides for us, but which is man made. So that code is commented out again.
Hi,
I just tried an idea I had which was the following:
* Unix time represents the current time in seconds since 00:00:00 UTC,
January 1, 1970.
* There are 86400 seconds in a day
By combining these tow I can get the normalized UTC time-of-day (ranging
from 0.0 .. 1.0) at lat/lon (0.0 , 0.0)
Erik Hofman wrote:
Given that 12.00 UTC is high noon and 0.00 is midnight it is now easy to
determine the sun azimuth for (0,0) by using the following formula:
double azimuth = -90.0 + 180 * daytime.
This should be:
double azimuth = -90 + 180.0*(0.5 + 0.5*daytime);
Erik
Erik Hofman wrote:
Is this correct or am I missing something?
I just realized that you also need to adjust for day-of-year to
compensate for the out-of-center rotation that causes long days (and
nights) for both polar areas.
Erik
___
On Sunday 07 August 2005 15:54, Erik Hofman wrote:
Erik Hofman wrote:
Is this correct or am I missing something?
I just realized that you also need to adjust for day-of-year to
compensate for the out-of-center rotation that causes long days (and
nights) for both polar areas.
Erik
Hi
