这是Josh r code的php实现,用于计算给定日期和时间的太阳位置:
这是MvG帮助后的更正代码:
function getSunPosition($lat, $long, $year, $month, $day, $hour, $min) {
// From https://stackoverflow.com/questions/8708048/position-of-the-sun-given-time-of-day-latitude-and-longitude?rq=1
// Get Julian date for date at noon
$jd = gregoriantojd($month,$day,$year);
//correct for half-day offset
$dayfrac = $hour / 24 - .5;
//now set the fraction of a day
$frac = $dayfrac + $min / 60 / 24;
$jd = $jd + $frac;
// The input to the Atronomer's almanach is the difference between
// the Julian date and JD 2451545.0 (noon, 1 January 2000)
$time = ($jd - 2451545);
// Ecliptic coordinates
// Mean longitude
$mnlong = (280.460 + 0.9856474 * $time);
$mnlong = fmod($mnlong,360);
if ($mnlong < 0) $mnlong = ($mnlong + 360);
// Mean anomaly
$mnanom = (357.528 + 0.9856003 * $time);
$mnanom = fmod($mnanom,360);
if ($mnanom < 0) $mnanom = ($mnanom + 360);
$mnanom = deg2rad($mnanom);
// Ecliptic longitude and obliquity of ecliptic
$eclong = ($mnlong + 1.915 * sin($mnanom) + 0.020 * sin(2 * $mnanom));
$eclong = fmod($eclong,360);
if ($eclong < 0) $eclong = ($eclong + 360);
$oblqec = (23.439 - 0.0000004 * $time);
$eclong = deg2rad($eclong);
$oblqec = deg2rad($oblqec);
// Celestial coordinates
// Right ascension and declination
$num = (cos($oblqec) * sin($eclong));
$den = (cos($eclong));
$ra = (atan($num / $den));
if ($den < 0) $ra = ($ra + pi());
if ($den >= 0 && $num <0) $ra = ($ra + 2*pi());
$dec = (asin(sin($oblqec) * sin($eclong)));
// Local coordinates
// Greenwich mean sidereal time
//$h = $hour + $min / 60 + $sec / 3600;
$h = $hour + $min / 60;
$gmst = (6.697375 + .0657098242 * $time + $h);
$gmst = fmod($gmst,24);
if ($gmst < 0) $gmst = ($gmst + 24);
// Local mean sidereal time
$lmst = ($gmst + $long / 15);
$lmst = fmod($lmst,24);
if ($lmst < 0) $lmst = ($lmst + 24);
$lmst = deg2rad($lmst * 15);
// Hour angle
$ha = ($lmst - $ra);
if ($ha < pi()) $ha = ($ha + 2*pi());
if ($ha > pi()) $ha = ($ha - 2*pi());
// Latitude to radians
$lat = deg2rad($lat);
// Azimuth and elevation
$el = (asin(sin($dec) * sin($lat) + cos($dec) * cos($lat) * cos($ha)));
$az = (asin(-cos($dec) * sin($ha) / cos($el)));
// For logic and names, see Spencer, J.W. 1989. Solar Energy. 42(4):353
if ((sin($dec) - sin($el) * sin($lat)) >00) {
if(sin($az) < 0) $az = ($az + 2*pi());
} else {
$az = (pi() - $az);
}
$el = rad2deg($el);
$az = rad2deg($az);
$lat = rad2deg($lat);
return array(number_format($el,2),number_format($az,2));
}
已经与刚果(近赤道)纬度/长度进行了测试:2013年9月1日10点00分,测试时间为-4.77867 / 11.86364。在这种情况下,正确答案是: 海拔= 67.77503 方位角= 54.51532
感谢您帮助调试此PHP代码!
Greg Fabre。
答案 0 :(得分:2)
我相信这一行
if ($dayfrac < 0) $dayfrac += 1;
出错了。如果你在中午之前,你不想在一天后的同一时间参考,而是你想在中午之前指定一个时间,即减去代表中午的朱利安日期。
Removing that line,您的示例日期对应于使用http://www.imcce.fr/en/grandpublic/temps/jour_julien.php计算的日期,即2456536.9166666665。结果
$el = 67.775028608168
$az = 54.515316112281
对我来说很好看。特别是,它同意R run
elevation = 67.77503
azimuth = 54.51532
以及Stellarium所说的内容(虽然我在上面的评论中错误地引用了这一点):
Alt = 67°46'30" = 67.775
Az = 54°30'60" = 45.5167
它(几乎)同意sunearthtools.com,所以我猜你在第一次输入数据时犯了错误:
所以我说这可以解决问题。