Re: simultaneous sunset
Dear Warren, Happy Summer Solstice! You ask an astute question: Are you saying below that ANY two locations MUST have a moment of mutual sunrise/sunset? Well, I am ALMOST saying that and the mathematics IS saying that... The declination of the sun is, of course, constrained to be between -23.5 degrees and +23.5 degrees. If it could do the full range -90 to +90 then the answer WOULD be yes. The explanation is simple. If you take ANY two locations then there is guaranteed to be a great circle joining them. If the locations are different and not diametrically opposite, this great circle is unique and defines a plane. The perpendicular to that plane through the centre of the Earth cuts the surface of the Earth at two points, one on the sunny side and one on the dark side. The point on the sunny side will be the instantaneous position of the sub-solar point. John Good would call the location of this point the Plane's Latitude and Longitude. The Plane's Latitude is the latitude of the sub-solar point which, of course, is the declination of the sun. If this calculation leads to a declination is outside the range +/- 23.5 degrees then the mathematics hasn't failed but the sun won't cooperate!! Best wishes Frank --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: simultaneous sunset
Dear Frank et al, After a second night sleeping on your nice puzzle I realised that I DID make a small goof in one of my assertions and no one has picked me up on it!! In the formula: tan(dec) = [-]sin(d)/sqrt(t1^2 - 2.t1.t2.cos(d) + t2^2) I asserted (correctly) that the argument of the square root function couldn't be negative. Dave Bell kindly confirmed this. I (also correctly) asserted that the argument could be zero. I then (incorrectly) asserted that when the argument is zero, the associated declination would be 90 degrees. My (false) reasoning was that with zero as the value of the square root, the right-hand side would have a zero denominator and hence an infinite value overall. In fact, for the argument of the square root function to be zero we require... Either t1 = t2 and d = 0 or t1 = -t2 and d = 180 In both cases the numerator, sin(d), is also zero and we have nought-over-nought which is well known to give bad vibrations to mathematicians! These two sets of requirements correspond to: Either two places at the same latitude and with zero separation of longitude (meaning that the two places are coincident). or two places of opposite latitude and 180 degrees of longitude apart (meaning the two places are diametrically opposite on the Earth's surface). In both sets of circumstances there is no unique great circle through the pairs of points and, in consequence, no single associated solar declination. The second case is interesting in demonstrating something fairly obvious but perhaps not widely known: If, standing in a particular place, you note the instant of sunset (or sunrise) you can ponder the thought that someone at the other end of the Earth's diameter from you is simultaneously experiencing sunrise (or sunset). This is independent of where you are or the time of year. One of the many nice features of this puzzle is that there is no need to know the time of sunset (or sunrise) and, in consequence, you can lay any understanding of hour-angles on one side. Hey, isn't that just what you wanted to know! Best wishes Frank --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: simultaneous sunset
Dear Mr. King, I have really enjoyed this problem. In one sense it is simple because it not about hours, but about events. In a recent post you said: From: Frank King Subject: Re: simultaneous sunset One of the many nice features of this puzzle is that there is no need to know the time of sunset (or sunrise) and, in consequence, you can lay any understanding of hour-angles on one side. But I am having a problem understanding something. Are you saying below that ANY two locations MUST have a moment of mutual sunrise/sunset? Or are you saying, for any location there exists a second location at EVERY longitude that will have one mutual moment sunrise/sunset? I can understand the second statement, but not the first. At first glance I rejected the second statement, but clearly it must be true if the sun covers half the Earth every instant. For example, if I calculate the solar declination for the two latitudes of 42°N and 10°N and 90° difference of longitude -- I get a declination of -47°. Certainly this is outside the range of -23.5° to +23.5° for our current situation on Earth. This would be two locations that never share a sunrise/sunset event. Yet 42°N and -64.7S and 90° longitude difference share a common event on winter solstice. Location 2 is determined given location 1, difference in longitude and solstice solar declination of 23.5° . Warren 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) --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: simultaneous sunset
Warren and others, After installing Sun Clock which can be found on www.mapmaker.com you can easily see that there is a whole range of locations on the earth with different longitudes where the sun sets simultaneously. A picture tells us sometimes much more then a story. Thibaud Chabot At 14:28 20-06-2007, Warren Thom wrote: But I am having a problem understanding something. Are you saying below that ANY two locations MUST have a moment of mutual sunrise/sunset? Or are you saying, for any location there exists a second location at EVERY longitude that will have one mutual moment sunrise/sunset? I can understand the second statement, but not the first. At first glance I rejected the second statement, but clearly it must be true if the sun covers half the Earth every instant. -- Th. Taudin Chabot, . [EMAIL PROTECTED] --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: simultaneous sunset
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 --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: simultaneous sunset
Hello Franks and all, I assume we are to neglect altitude differences and use the fictional spherical earth model, but just what do we mean by 'the same time'? I would assume that when two or more persons look at their watches, corrected for standard zone time, that they read the same numbers? If so, this would give a larger number of answers as the sun sets in each time zone, it will probably occur some place within it at the same clock time as in fixed places in other time zones. Something like the New Year always starts at midnight, so each time zone has it's own New Year, 24 in total. Since the time zones are so non-uniform, the problem could be quite difficult to solve. But you probably mean at the same GMT or Universal time. Then we have our troubles with synchronizing. Given that light travels at about 11.8 inches in a nanosecond, someone 11.8 feet away would be off by 12 nanoseconds. Still, the problem is a neat one! Having fun with Sundials! Edley. Greetings, fellow dialists, A little while ago there was a dialling discussion about pairs of places where the sun rose or set at the same moment. I recall that to solve the problem a terminator programme was called into play. There are earlier examples of this question. Around the year 1950 airline pilots (who still used sextants) sat two successive navigation examinations, the higher of which was called the First N. In the written paper 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). This problem caused the pilots much head scratching until it was realised that this type of question occurred in every paper that was set. The solution is presumably a simultaneous equation in spherical trigonometry to discover the sun's declination. Would any dialist care to try it? Frank 55N 1W -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.5.472 / Virus Database: 269.9.0/852 - Release Date: 17/06/2007 08:23 --- https://lists.uni-koeln.de/mailman/listinfo/sundial --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: simultaneous sunset
Hello Edley and listeners everywhere, Edley, I don't quite know how to answer you. I explained in an earlier message that this was a real question set for airline pilots but it was fifty eight years ago and the exact wording unfortunately eludes me. Did they have nanoseconds then? And what about refraction, parallax and height of eye? Is Paris higher than London? It looks lower on the map. Which reminds me that the south east of England is running out of water. We have plenty here in the north and the good folks of London are under the impression that if they built a pipe the water would just run down to them. After all, we are at the top of our English maps and they are at the bottom. Frank 55N 1W Edley McKnight wrote: Hello Franks and all, I assume we are to neglect altitude differences and use the fictional spherical earth model, but just what do we mean by 'the same time'? I would assume that when two or more persons look at their watches, corrected for standard zone time, that they read the same numbers? If so, this would give a larger number of answers as the sun sets in each time zone, it will probably occur some place within it at the same clock time as in fixed places in other time zones. Something like the New Year always starts at midnight, so each time zone has it's own New Year, 24 in total. Since the time zones are so non-uniform, the problem could be quite difficult to solve. But you probably mean at the same GMT or Universal time. Then we have our troubles with synchronizing. Given that light travels at about 11.8 inches in a nanosecond, someone 11.8 feet away would be off by 12 nanoseconds. Still, the problem is a neat one! Having fun with Sundials! Edley. Greetings, fellow dialists, A little while ago there was a dialling discussion about pairs of places where the sun rose or set at the same moment. I recall that to solve the problem a terminator programme was called into play. There are earlier examples of this question. Around the year 1950 airline pilots (who still used sextants) sat two successive navigation examinations, the higher of which was called the First N. In the written paper 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). This problem caused the pilots much head scratching until it was realised that this type of question occurred in every paper that was set. The solution is presumably a simultaneous equation in spherical trigonometry to discover the sun's declination. Would any dialist care to try it? Frank 55N 1W -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.5.472 / Virus Database: 269.9.0/852 - Release Date: 17/06/2007 08:23 --- https://lists.uni-koeln.de/mailman/listinfo/sundial No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.5.472 / Virus Database: 269.9.1/854 - Release Date: 19/06/2007 13:12 -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.5.472 / Virus Database: 269.9.1/854 - Release Date: 19/06/2007 13:12 --- https://lists.uni-koeln.de/mailman/listinfo/sundial
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
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
Re: simultaneous sunset
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°N6.7° 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
Re: simultaneous sunset
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
