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Re: Basics of computing sunrise/sunset
From: Christian Scheele
Date: 2009 Jun 18, 00:15 +0200
From: Christian Scheele
Date: 2009 Jun 18, 00:15 +0200
Gentlemen,
found the mistake, I am acomplete idiot. I
converted 15,75 degrees from arc into time as 1h 45m instead of 1h
3m .... I must have seen the .75 and subconsciously associated it
with 3/4 hour, having immediately before recognised 15 deg as 1 hour.
Thanks for your help John Huth and Brad
Morris.
Best and good night for now.
CS
----- Original Message -----From: Apache RunnerTo: NavList@fer3.comSent: Wednesday, June 17, 2009 11:39 PMSubject: [NavList 8679] Re: Basics of computing sunrise/sunsetThe equation looks right
On Wed, Jun 17, 2009 at 5:35 PM, Christian Scheele <scheele@telkomsa.net> wrote:
I've considered EoT. I'm using the upper limb on horizon definition of
sunset and am therefore applying the semi-diameter of 16 minutes. I've also
corrected my result for error within time zone. Am still checking, it must
be a trig error.
Thanks for your time.
The very best regards,
Christian Scheele
----- Original Message -----
From: "Brad Morris" <bmorris@tactronics.com>
To: <NavList@fer3.com>
Sent: Wednesday, June 17, 2009 10:58 PM
Subject: [NavList 8676] Re: Basics of computing sunrise/sunset
Please note that you must consider the Equation of Time to convert from
Apparent Time to Mean Time.
Be careful about what you mean by sunset, there are many
1) Upper limb of sun touches the horizon
2) sun 6 degrees below horizon
3) sun 12 degrees below horizon
4) sun 18 degrees below horizon
You must consider your offset in longitude from the center longitude of your
time zone. That is, sunset occurs earlier in the eastern section of a given
time zone than a sunset in the same time zone, but on the western edge!
That should pretty much do it!
Best Regards
Brad
-----Original Message-----
From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf
Of Christian Scheele
Sent: Wednesday, June 17, 2009 4:41 PM
To: NavList@fer3.com
Subject: [NavList 8673] Basics of computing sunrise/sunset
I'm trying to solve the following sunrise/sunset astronomical triangle and
am encountering unexpected pitfalls.
cos(LHA) = [sin (a) - sin (lat) sin(dec)]/ cos(lat)cos(dec)
where I am assuming a = -0.83 degrees
It is of course something very basic and I'm almost embarassed to say that
although I keep rechecking what I'm doing, I remain with an error of about
15 minutes by the time I get to the final result in mean zone time. Before I
drag everybody into it, could somebody please tell me whether there are any
snags that I should look out for?
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