Dear Roger and Mashallah,

Many thanks to you both for a most interesting dialogue.

May I suggest three sundialling heroes who, you will see,
are very relevant your discussion:

  1.  Frank Evans - The Inspirer

  2.  Gianni Ferrari - who pointed us at...

  3.  John Good - who can laugh at us all!

Now let me give you some chronology...

  21 April 2007

  Frank Evans asked about the Hawkshead sundial and,
  in particular, asked what is meant by PL.

  29 April 2007

  Gianni Ferrari (correctly) told us that we were
  mistaken in our collective musings and that the
  answer was given a very long time ago in...

  1731

  This is the date of "The Art of Shadows" by
  John Good who tells us that any plane has a
  latitude and a longitude which we can call
  the Plane's Latitude and the Plane's Longitude.

  18 June 2007

  Frank Evans posed another puzzle which I express
  as: At what solar declination will there be
  simultaneous sunset in London and Paris?

  18 June 2007 (later!)

  I note that the required declination is the Plane's
  Latitude of the plane of the great circle through
  London and Paris.

  16 July 2007

  I read about Mashallah's puzzle and the Bailey
  answer to it.  The puzzle can be expressed as:
  At what point on the Earth's surface are both
  Mecca and Jerusalem due east?

Now for the exciting bit...

  Frank Evans's question and Mashallah's question
  ARE EXACTLY THE SAME.

If only you could put away your spreadsheets (these
things cause brain damage) and study John Good, you
would see why...

To answer Frank Evans's question you proceed thus:

  1. Take the great circle through London and Paris.

  2. Note that this great circle defines a plane.

  3. Consider that Plane's Latitude.

  4. Job done.

To answer Mashallah's question you proceed thus:

  1. Take the great circle through Mecca and Jerusalem.

  2. Note that this great circle defines a plane.

  3. Consider that Plane's Latitude AND Longitude.

  4. Job done.

OK, there is a tiny extra bit at step 3 but that is
a minor matter.  Let me repeat a few lines from my
message of 18 June...


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)


INTERPRETATION I

John Good would say that dec is the Plane's Latitude.
In the context of London and Paris this is the required
declination (which is why I called it dec).


INTERPRETATION II

In the context of Mecca and Jerusalem dec is STILL
the Plane's Latitude and comes out (using the figures
in your spreadsheet) at -20.8 degrees.  This can be
interpreted in two ways:

  1. It is the declination at which there will be
     simultaneous sunset in Mecca and Jerusalem.
     [Frank Evans hasn't asked that question yet
     but I expect he will!!!]

  2. Its absolute value (+20.8 degrees) is the
     distance of the required point from the North
     Pole which accords with your latitude of
     69.2 degrees north.


THE TINY EXTRA BIT

We still need the Plane's Longitude and this is given
as:

  cos(PLo) = tan(-PLa)*tan(Mecca's Latitude)

where:

  PLo is the Plane's Longitude (relative to Mecca)

  PLa is the Plane's Latitude and the minus sign is
         needed to make it +20.8 degrees.

You already gave this formula in another guise.

The value of PLo is relative to Mecca and you need 
to subtract the Longitude of Mecca from PLo to get
the Longitude relative to the Greenwich meridian.
Using your raw data you get 41.2 degrees east which
again accords with your result.


RAW DATA

I used the data from your spreadsheet:

  Mecca      39.82583 degrees East
             22.42254 degrees North

  Jerusalem  35.22011 degrees East
             31.80114 degrees North

A health warning is needed here!  In your message (as
distinct from your spreadsheet) you give the latitude
of Mecca as 21.422 degrees North which is actually
nearer the correct figure!  [Spreadsheets cause brain
damage!]

I am not certain what the correct figure is but the
various values I have found are around 21.5 degrees.


COMMENT

John Good would have been able to solve all these
puzzles nearly 300 years ago.  I salute him!!

Almost certainly, the ancient Arab astronomers would
also have been able to solve these puzzles and I
salute them too!

Many thanks again for entertaining me!

Best wishes

Frank

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