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winter solstice sunrise
Could anyone tell me how to calculate the position of the sunrise at
the winter sostice for a particular latitude. Thanks |
#2
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winter solstice sunrise
Teilo wrote:
Could anyone tell me how to calculate the position of the sunrise at the winter sostice for a particular latitude. Thanks Just use one of the popular planetarium programs readily available. If you need further help, let me know your latitude and I will get you the information from Chris Marriott's SkyMap Pro (an excellent program!). Anthony. |
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winter solstice sunrise
Thanks Anthony
My latitude is 52 degrees Regards Teilo |
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winter solstice sunrise
Could anyone tell me how to calculate the position of the sunrise at
the winter sostice for a particular latitude. Thanks Teilo, this page will calculate it for you... http://aa.usno.navy.mil/data/docs/AltAz.html -Florian |
#5
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winter solstice sunrise
Teilo wrote:
Could anyone tell me how to calculate the position of the sunrise at the winter sostice for a particular latitude. Thanks Assuming this isn't homework, use this formula: azimuth = 90° +/- arcsin(sin(23.4°)/cos(latitude)) where +/- 23.4 degrees is the declination of the Sun at the solstices. (It isn't that, exactly, but close enough for our purposes.) I use the north = 0, east = 90 convention. Thus, for instance, for your latitude of 52 degrees (north, I assume), the formula yields azimuth = 90° +/- arcsin(0.397/0.616) = 90° +/- arcsin(0.645) = 90° +/- 40° If you're in the northern hemisphere, azimuth of the sunrise is thus 50 degrees at the summer solstice and 130 degrees at the winter solstice; if you're in the southern hemisphere, it's vice versa. Note that at latitudes above 66.6 degrees in either hemisphere (in the Arctic and Antarctic circles, in other words), the argument of the arcsin function is above 1, and the formula correctly indicates that the Sun does not rise at all at the winter solstice. Some calculators/computers use radians instead of degrees; if so, be sure to take care with your units. Pi radians equal 180 degrees; or to put it another way, one radian equals about 57.3 degrees. This formula neglects two effects of some importance: refraction, and the fact that in many places, sunrise is defined as the moment at which the Sun's disc first peeks over the ideal horizon, not when its center does. At temperate latitudes, the combined impact is to decrease the azimuth of sunrise by a bit less than a degree in the northern hemisphere, and to increase it by the same amount in the southern hemisphere. -- Brian Tung The Astronomy Corner at http://astro.isi.edu/ Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html |
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winter solstice sunrise
To Brian
The real astronomical relevence of the solstices is in the asymmetrical relationship between the length of a day equidistant to the Equinoxes. The length of a day on Dec 21st in both hemispheres is asymmetrical from the length of a day on June 21st,unfortunately you insist on referencing the Sun's position off the Earth's Equator/axis allied to the calendar system and dispense with the asymmetry provided by variations in orbital motion. Hemispherical or quasi-geocentric astronomy is great if you wish to turn the great celestial motions into a personal sideshow but oh,what an undignified thing to do. There is an astronomical language waiting for genuine people who accept that the great motions cannot be bottled into a convenient calendar/celestial sphere bottle.There is also exciting avenues open for those who detest insincerity, pretensiousness and laziness for none of these things can have a part in astronomy and its noble tradition,regardless of the many imposters who deign otherwise. |
#7
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winter solstice sunrise
Thanks all for your help.
John |
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