>From a purely analytical approach, we can look at the simultaneous equations:
cos t = - tan dec tan 48.8667 cos (t - 2.3833) = - tan dec tan 51.5 These can be solved for t to obtain: tan t = ((tan 51.5 / tan 48.8667) - cos 2.3833) / sin 2.3833 so t = 67.1851. This is the hour angle at sunset in Paris. The hour angle at sunset in London is 67.1851 - 2.383 = 64.8018 Now we can retrieve the value for dec from: tan dec = - cos t / tan 48.8667 so dec = -18.7 Fred Sawyer ----- Original Message ----- From: "Werner Riegler" <[EMAIL PROTECTED]> To: "Frank King" <[EMAIL PROTECTED]>; "Frank Evans" <[EMAIL PROTECTED]> Cc: "Sundial" <[EMAIL PROTECTED]> Sent: Monday, June 18, 2007 6:59 PM Subject: RE: simultaneous sunset > 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. > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial > > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial > > --------------------------------------------------- https://lists.uni-koeln.de/mailman/listinfo/sundial
