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Declination
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==Relation to latitude== When an object is directly overhead its declination is almost always within 0.01 degrees of the observer's latitude; it would be exactly equal except for two complications.<ref>{{Cite web|url=http://www.austincc.edu/jheath/Stellar/Hand/ccord.htm|title=Celestial Coordinates|website=www.austincc.edu|access-date=2017-03-24|archive-date=2016-11-11|archive-url=https://web.archive.org/web/20161111100549/http://www.austincc.edu/jheath/Stellar/Hand/ccord.htm|url-status=dead}}</ref><ref>{{cite web| url = https://web.ecs.baylor.edu/faculty/grady/EGR1301_FALL2015_Masters_1stEd_Chapter7_The_Solar_Resource.pdf| title = ''baylor.edu''}}</ref> The first complication applies to all celestial objects: the object's declination equals the observer's astronomical latitude, but the term "latitude" ordinarily means geodetic latitude, which is the latitude on maps and GPS devices. In the continental United States and surrounding area, the difference (the [[vertical deflection]]) is typically a few [[Minute of arc|arcseconds]] (1 arcsecond = {{sfrac|3600}} of a degree) but can be as great as 41 arcseconds.<ref>{{cite web |url = http://www.ngs.noaa.gov/GEOID/USDOV2009/ |title = USDOV2009 |date = 2011 |publisher = [[U.S. National Geodetic Survey]] |location = Silver Spring, Maryland }}</ref> The second complication is that, assuming no deflection of the vertical, "overhead" means perpendicular to the ellipsoid at observer's location, but the perpendicular line does not pass through the center of the Earth; almanacs provide declinations measured at the center of the Earth. (An ellipsoid is an approximation to [[sea level]] that is mathematically manageable).<ref>{{cite book |editor = P. Kenneth Seidelmann |title = Explanatory Supplement to the Astronomical Almanac |publisher = University Science Books |location = Sausalito, CA |date = 1992 |pages = 200β5}}</ref>
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