Hi Frank, Thanks for closing the loop. What goes around, comes around.
I remember the original question posed by Frank Evans, my awkward iterative solution, Gianni's reference to John Good and your brilliant solution. And now you have returned to the lesson that we never learned and used this example to teach us again this elegant solution. Well done. You now have my attention. When I first checked Gianni's reference to John Good's book, I found the description on pages 87-88 quite familiar. Good outlined the usual procedure for designing a vertical declining sundial. Knowing the latitude of the place and the declination of the vertical dial from south facing, you calculate the "substile's distance from the meridian, the height of the stile above the Plane, and the Planes difference in longitude". His "Cannon for Calculation" is identical to Alberta Waugh's technique, even the use of logs, outlined in Waugh's much more familiar and accessible book "Sundials: Their Theory and Construction". It seems that Waugh was following the old masters. Once I saw that Plane's Longitude was just the Waugh's familiar Difference in Longitude (Tan DL = Tan Dec / Sin Lat), I stopped thinking about it. SD, SH and the Equatorial line were all I was usually looking for. I did not make the require leap into mathematical abstraction to see this plane creating a great circle arc with a definable latitude and longitude, Plane's Latitude and Longitude. When I looked at Mahalah's drawings, my first interest was to calculate the curve on his map, a "line of position" of places where the Mecca was due east or west. This I was pleased to resolve to a simple equation, Tan Lat = Tan Mecca / Cos (Long - Long Mecca). Repeating this for Jerusalem and finding the common location as the fix seemed to be the simplest method to solve the puzzle. Brain damaged? No, just a simple Chemical Engineer with a small mathematical tool box. I thought of your solution to London Paris sunset question as relevant for defining the great circle of equal declination but missed how the shortest distance to the pole defined Plane's latitude and longitude. Now I can stop wondering who Mr. Plane, Esq. was. Thanks for a great lesson in plane and spherical geometry. Regards, Roger Bailey ps. yes, there was a typo in the spreadsheet. The Latitude of Mecca is 21.4225 according to the great pictures on Google Earth ----- Original Message ----- From: "Frank King" <[EMAIL PROTECTED]> To: "Roger Bailey" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]> Cc: <[email protected]>; <[EMAIL PROTECTED]> Sent: Thursday, July 19, 2007 12:04 PM Subject: Re: Solar Qiblah > 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 > > --------------------------------------------------- https://lists.uni-koeln.de/mailman/listinfo/sundial
