Dear Frank,
It is not really possible to tidy up your efforts, because I think they
are as clean as they can possibly be - but I can try to mess up your
efforts by another concept:
CONCEPT
Sunset in Paris means that the rays of sunlight are in a plane
tangential to the earth in Paris.
Sunset in London means that the rays of sunlight are in a plane
tangential to the earth in London.
Sunset at the same time in London and Paris means that both of the above
conditions have to be fulfilled, which means that I have to find the
intersection of the two tangential planes, which gives a straight line.
The angle between this straight line and the earth's axis is equal to
90-declination of the sun.
MATH
la1 = latitude of place 1
la2 = latitude of place 2
d = their difference in longitude
The equations for the two tangential planes are (assuming the earth's
radius to be 1):
1) x.cos(la1)+z.sin(la1)=1
2) x.cos(la2).cos(d)+y.cos(la2).sin(d)+z.sin(la2)=1
If I find the intersection of these two planes and calculate the angle
between earth's axis and this line I find the result
sin(dec)=-sin(d)/sqrt(t1^2 - 2.t1.t2.cos(d) + t2^2 + sin^2(d))
Expressing tan(dec)=sin(dec)/sqrt(1-sin^2(dec)) I find exactly your
formula ! So I guess this is proof that you haven't goofed.
Best Wishes
Werner Riegler
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Frank King
Sent: Monday, June 18, 2007 7:20 PM
To: Frank Evans
Cc: Sundial
Subject: Re: simultaneous sunset
Dear Frank,
I do enjoy your puzzles!!
> ... a question appeared of the form:
>
> "Find a day on which the sun sets (altitude 0 deg.) at
> the same moment in London and Paris (positions given)."
Conceptually this is trivial. Mathematically it gets a
little messy but I think I can get a closed form of the
solution.
CONCEPT
I assume that at any instant half the Earth is in sunlight
and half is in darkness. A great circle separates the two
halves. Any place on this great circle is experiencing
the moment of (mathematical) sunset or sunrise.
The solution to your problem is to draw a great circle
from London to Paris and extend it until it reaches the
Equator. The angle this great circle makes to the plane
of the Equator is the complement of the solar declination
(subject to a minus sign).
MATHEMATICS
Let t1 = tangent of the latitude of place 1
Let t2 = tangent of the latitude of place 2
Let d = their difference in longitude
We then have:
tan(dec) = -sin(d)/sqrt(t1^2 - 2.t1.t2.cos(d) + t2^2)
Where dec is the required solar declination.
EXAMPLE
You cite London and Paris. I take the latitudes as being
51 deg 30' and 48 deg 52' and the difference in longitude
as being 2 deg 23'.
This gives the solar declination as -18.7 degrees.
I expect I have goofed. Some bright youngster can now
tidy up my efforts!!!
Best wishes
Frank King
Cambridge, U.K.
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