Dear Frank et al,

An intriguing side issue to your puzzle is that
it relates to the discussion about the Hawkeshead
dial and the notion of a Plane's Longitude and,
implicitly, the notion of a Plane's Latitude.

Once you have taken on board these notions, the
simplest way of expressing the solution to your
new puzzle is to say:

 1.  Consider the plane defined by the great
     circle through London and Paris.

 2.  The required declination is that
     Plane's Latitude.

I am fairly confident that John Good of "The Art
of Shadows" could have come up with a solution
200 years ago.

Since it was a 1950s problem I took a 1950s approach
rather than an 18th century one!

I have enjoyed reading the comments by Roger Bailey,
Warren Thom, Werner Riegler and Fred Sawyer.

I have a comment on one of Warren's notes:

> two locations with too great of a difference of
> longitude, would be too far apart to ever have
> the same moment of sunset.

This is not quite the whole story...

  The great circle which separates light from dark
  necessarily encompasses EVERY longitude, so there
  will be points almost 180 degrees apart which
  have the same moment of sunset.

  Half the points on this circle will correspond to
  sunset and half to sunrise.  The most northerly
  and most southerly points will, respectively, be
  points where sunset is immediately followed by
  sunrise and sunrise is immediately followed by
  sunset.

My formula doesn't distinguish between sunrise and
sunset and the leading minus sign could equally be
a plus sign:

tan(dec) = [-]sin(d)/sqrt(t1^2 - 2.t1.t2.cos(d) + t2^2)

In the case of London and Paris a declination of
-18.7 degrees is the condition for common sunset
and +18.7 degrees is the condition for common sunrise.

I usually like to sleep on any Mathematics before
writing about it and I worried a little about
whether the argument of the square root function
could be negative.

Happily it cannot be negative for real t1, t2
and d.  I'll leave it as an exercise for you all
to demonstrate this!

It can be zero and this corresponds to a declination
of 90 degrees which doesn't apply to the real sun
but I am not bothered about trifles like that!

Best wishes

Frank

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