Re: Fwd: Where it wil be equinox, at noon
Also using Time Zone Master 1) Clicking any clock and going to the equinox page, you then click on the equinox (5:05 EDT for my case - Ottawa) 2) Then go to the sidereal page it shows 9:12:28 GST (sidereal time) 3) The time difference to noon is 2h 57m 31s Multiply x 15 - you get 44d 22m 45s E It shows a delta-T of 67 seconds On 2011-09-22 23:59, James E. Morrison wrote: My logic for answering the question of where on Earth will it be apparent noon at the instant of the autumnal equinox is: 1. The Greenwich sidereal time at the time of the equinox (Julian day 2455827.87793) is 09:10:36.1 2. We want the longitude where the local sidereal time = 12:00:00 at this instant. 3. The sidereal time difference is 02:49:24. 4. Times 15 for degrees = 42d 20m 59s E. This corresponds to a meridian passing through Georgia, east Turkey and bit west of Baghdad. I used The Electric Astrolabe with a Delta-T of 66 sec to get the GST value. Other programs may improve on the result. Best regards, Jim James E. Morrison janus.astrol...@verizon.net mailto:janus.astrol...@verizon.net Astrolabe web site at http://astrolabes.org From: axel törnvall gonzalez Date: Sep 22, 2011 8:58:05 PM Subject: Where it wil be equinox, at noon To: sundial@uni-koeln.de This is my Subject; Finding the position, Longitud, where, the sprig equinox, will ocurr at noon, I found some diference between the results of The Dialist´s Companion, and Sun v.5.6 Of R.Cernic, both programs I work for some time. In Longitud 42°12,80 E at 13:03:42 PM it will be Noon, for my studies Then I placed both programs in Latitud and longitud 0°, and in Sun v5.6 of R.Cernic I got UT 09:03:59 AM and in The Dialist´s Companion, I found the nearest cero declination at 08:44:47AM the altitude measure have a difference of almost 5°, I Know The Dialist Companion I Use vers 1.1.b is old. Sorry my english, it´s not my first language Best regards for all of you Axel 32°39'59S 70°42'41W --- https://lists.uni-koeln.de/mailman/listinfo/sundial --- https://lists.uni-koeln.de/mailman/listinfo/sundial --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: Fwd: Where it wil be equinox, at noon
Don't you hate it when you publish numbers with errors I manually subtracted from noon incorrectly 3) The time difference to noon is actually 2h 47m 31s Multiply by 15 gives 41d 52m 45s E Open a new clock and enter the longitude, and equator The result is near a town called Chisimayu in Somalia, near the Kenya border, where the sun will be overhead at solar noon, at the equinox On 2011-09-23 2:12, David Patte wrote: Also using Time Zone Master 1) Clicking any clock and going to the equinox page, you then click on the equinox (5:05 EDT for my case - Ottawa) 2) Then go to the sidereal page it shows 9:12:28 GST (sidereal time) 3) The time difference to noon is 2h 57m 31s Multiply x 15 - you get 44d 22m 45s E It shows a delta-T of 67 seconds On 2011-09-22 23:59, James E. Morrison wrote: My logic for answering the question of where on Earth will it be apparent noon at the instant of the autumnal equinox is: 1. The Greenwich sidereal time at the time of the equinox (Julian day 2455827.87793) is 09:10:36.1 2. We want the longitude where the local sidereal time = 12:00:00 at this instant. 3. The sidereal time difference is 02:49:24. 4. Times 15 for degrees = 42d 20m 59s E. This corresponds to a meridian passing through Georgia, east Turkey and bit west of Baghdad. I used The Electric Astrolabe with a Delta-T of 66 sec to get the GST value. Other programs may improve on the result. Best regards, Jim --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: Terminator
hi Doug Try www.daylightmap.com/ as an alternative regards Ian Maddocks Chester, UK -- From: Douglas Bateman douglas.bate...@btinternet.com Sent: Wednesday, September 14, 2011 10:25 PM To: Sundial sund...@rrz.uni-koeln.de Subject: Terminator Does anyone know of a link or an app that shows the terminator in graphic form over the world. I used to use a 'mapmaker' website but that seems to have disappeared. I have foundwww.worldtimezone.com/datetime12.php andhttp://24timezones.com/ but neither are as clear or as uncluttered and the site of a few years ago. The terminator has this fascinating shape as we approach the equinox. Regards, Doug --- https://lists.uni-koeln.de/mailman/listinfo/sundial --- https://lists.uni-koeln.de/mailman/listinfo/sundial
EOT
I’m trying to understand the equation of time All diagrams that I have seen (like the attached) are only two dimensional. For the sake of simplicity, the earth tilt is ignored. According to a two dimensional drawing, it will always require more than a 360 degree revolution for a point on the earth to face the sun again. However the earth is tilted. Does this mean that sometimes it takes more than 360 to make a revolution and sometimes it takes less? Or does this mean that it always takes more than a 360 degree revolution. However because of the tilt, the additional spin required to face the sun again is greater on some days than on others. I think its the later answer but I'm not sure. -- Cheers Donald 0423 102 090 This e-mail is privileged and confidential. If you are not the intended recipient please delete the message and notify the sender. Un-authorized use of this email is subject to penalty of law. So there! attachment: Sidereal_Time_en.PNG--- https://lists.uni-koeln.de/mailman/listinfo/sundial
