Dear Mr. Evans and Mr. King,
I liked the problem. I have not worked out the math yet. I am not very
good at quadratics in trig. I was going to work on "iterations" to close in
on the answer.
I tried a few examples with your math. They seem to calculate reasonable
and interesting answers:
--lat of 50° and 35.8° need a difference of longitude of 11° to get
solar dec at solstice of around 23.5°
--two locations at different lat need a difference of long of zero to
set on the equinox (solar dec of 0°).
--two locations with too great of a difference of longitude, would be
too far apart to ever have the same moment of sunset.
-- here are a few situations - all at the winter solstice
Lat1 Lat2 Diff in Long
42°N 33°N 6.7°
42°N 10°N 18.6° (one hour)
42°N -20°S 32° (two hours)
42°N -51°S 56° (three or four hours)
Upon the first reading of the problem, I thought it was a continuation of
the "rotating platform on a polar axis with a pointer of EOT correction"
thread. IF you had a pointer on a scale, and you didn't use it for
corrections for EOT or longitude, then it could point to the name of cities
that have the same sunset time. It might require a condition given on the
horizontal "dial" surface that would have to relate to solar declination. I
don't know if I have the correct picture in my head for set up -- but I will
think more about it. Any other ideas on this?
Warren
----- Original Message -----
From: "Frank King" <[EMAIL PROTECTED]>
To: "Frank Evans" <[EMAIL PROTECTED]>
Cc: "Sundial" <[EMAIL PROTECTED]>
Sent: Monday, June 18, 2007 11:20 AM
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.
>
> ---------------------------------------------------
> https://lists.uni-koeln.de/mailman/listinfo/sundial
>
---------------------------------------------------
https://lists.uni-koeln.de/mailman/listinfo/sundial
