我正在编写几种计算太阳穿过特定点的路径所需的方法。我使用两个不同的源编写代码进行计算,但都没有产生所需的结果。来源是:http://www.pveducation.org/pvcdrom/properties-of-sunlight/suns-position and
http://www.esrl.noaa.gov/gmd/grad/solcalc/solareqns.PDF
注意:度数转弧分为 Deg * 60 分钟。
localSolartime:我已将经度转换为“分钟”,从 localStandardTimeMeridian 方法派生的当地标准时间子午线(lstm)返回一个以“分钟”为单位的值,而equationOfTime也以“分钟”为单位返回。使用 pveducation 中的方程,我计算了时间校正,该校正考虑了给定时区内的微小时间变化。当我将此结果和 localTime(均以分钟为单位)应用于当地太阳时 (lst) 方程时,结果为 676.515(此时),这对我来说没有任何意义。据我了解,当地太阳时代表相对于太阳的时间,当太阳在当地天空中处于最高点时,被认为是太阳正午。 676.515 没有意义。有谁明白可能是什么原因造成的。
HourAngle:我希望一旦修复了 localSolarTime 方法,就不需要纠正这个问题。
我选择华盛顿特区作为纬度和经度。天顶角和方位角读数都应该是正值,对于我所在的区域,此时分别为 66 和 201。
public class PathOfSun {
static LocalTime localTime = LocalTime.now();
static double dcLat = 38.83;
static double dcLong = -77.02;
static DecimalFormat df = new DecimalFormat("#.0");
public static void main(String [] args) {
int day = dayOfYear();
double equationOfTime = equationOfTime(day);
double lstm = localTimeMeridian();
double lst = localSolarTime(equationOfTime, dcLong, lstm);
double declination = declination(day);
double hourAngle = hourAngle(lst);
double zenith = zenith(dcLat, declination, hourAngle);
double azimuth = azimuth(dcLong, declination, zenith, hourAngle);
}
//Longitude of timezone meridian
public static double localTimeMeridian() {
TimeZone gmt = TimeZone.getTimeZone("GMT");
TimeZone est = TimeZone.getTimeZone("EST");
int td = gmt.getRawOffset() - est.getRawOffset();
double localStandardTimeMeridian = 15 * (td/(1000*60*60)); //convert td to hours
//System.out.println("Local Time Meridian: " + localStandardTimeMeridian);
return localStandardTimeMeridian;
}
//Get the number of days since Jan. 1
public static int dayOfYear() {
Calendar localCalendar = Calendar.getInstance(TimeZone.getDefault());
int dayOfYear = localCalendar.get(Calendar.DAY_OF_YEAR);
//System.out.println("Day: " + dayOfYear);
return dayOfYear;
}
//Emperical equation to correct the eccentricity of Earth's orbit and axial tilt
public static double equationOfTime (double day) {
double d =(360.0/365.0)*(day - 81);
d = Math.toRadians(d);
double equationTime = 9.87*sin(2*d)-7.53*cos(d)-1.54*sin(d);
//System.out.println("Equation Of Time: " + equationTime);
return equationTime;
}
//The angle between the equator and a line drawn from the center of the Sun(degrees)
public static double declination(int dayOfYear) {
double declination = 23.5*sin((Math.toRadians(360.0/365.0))*(dayOfYear - 81));
//System.out.println("Declination: " + df.format(declination));
return declination;
}
//Add the number of minutes past midnight localtime//
public static double hourAngle(double localSolarTime) {
double hourAngle = 15 * (localSolarTime - 13);
System.out.println("Hour Angle: " + df.format(hourAngle)); //(degrees)
return hourAngle;
}
//Account for the variation within timezone - increases accuracy
public static double localSolarTime(double equationOfTime, double longitude, double lstm) {
//LocalSolarTime = 4min * (longitude + localStandardTimeMeridian) + equationOfTime
//Time Correction is time variation within given time zone (minutes)
//longitude = longitude/60; //convert degrees to arcminutes
double localStandardTimeMeridian = lstm;
double timeCorrection = (4 * (longitude + localStandardTimeMeridian) + equationOfTime);
System.out.println("Time Correction: " + timeCorrection); //(in minutes)
//localSolarTime represents solar time where noon represents sun's is highest position
// in sky and the hour angle is 0 -- hour angle is negative in morning, and positive after solar noon.
double localSolarTime = (localTime.toSecondOfDay() + (timeCorrection*60)); //(seconds)
localSolarTime = localSolarTime/(60*60); //convert from seconds to hours
//Convert double to Time (HH:mm:ss) for console output
int hours = (int) Math.floor(localSolarTime);
int minutes = (int) ((localSolarTime - hours) * 60);
//-1 for the daylight savings
Time solarTime = new Time((hours-1), minutes, 0);
System.out.println("Local Solar Time: " + solarTime); //hours
return localSolarTime;
}
public static double azimuth(double lat, double declination, double zenith, double hourAngle) {
double azimuthDegree = 0;
double elevation = 90 - zenith;
elevation = Math.toRadians(elevation);
zenith = Math.toRadians(zenith);
lat = Math.toRadians(lat);
declination = Math.toRadians(declination);
hourAngle = Math.round(hourAngle);
hourAngle = Math.toRadians(hourAngle);
//double azimuthRadian = -sin(hourAngle)*cos(declination) / cos(elevation);
double azimuthRadian = ((sin(declination)*cos(lat)) - (cos(hourAngle)*cos(declination)*
sin(lat)))/cos(elevation);
//Account for time quadrants
Calendar cal = Calendar.getInstance();
int hour = cal.get(Calendar.HOUR_OF_DAY);
if(hour > 0 && hour < 6) {
azimuthDegree = Math.toDegrees(acos(azimuthRadian));
}
else if(hour >= 6 && hour < 12) {
azimuthDegree = Math.toDegrees(acos(azimuthRadian));
azimuthDegree = 180 - azimuthDegree;
} else if (hour >= 12 && hour < 18) {
azimuthDegree = Math.toDegrees(acos(azimuthRadian));
azimuthDegree = azimuthDegree - 180;
} else if (hour >= 18 && hour < 24) {
azimuthDegree = Math.toDegrees(acos(azimuthRadian));
azimuthDegree = 360 - azimuthDegree;
}
System.out.println("Azimuth: " + df.format(azimuthDegree));
return azimuthDegree;
}
public static double zenith(double lat, double declination, double hourAngle) {
lat = Math.toRadians(lat);
declination = Math.toRadians(declination);
hourAngle = Math.round(hourAngle);
hourAngle = Math.toRadians(hourAngle);
//Solar Zenith Angle
double zenith = Math.toDegrees(acos(sin(lat)*sin(declination) + (cos(lat)*cos(declination)*cos(hourAngle))));
//Solar Elevation Angle
double elevation = Math.toDegrees(asin(sin(lat)*sin(declination) + (cos(lat)*cos(declination)*cos(hourAngle))));
System.out.println("Elevation: " + df.format(elevation));
System.out.println("Zenith: " + df.format(zenith));
return zenith;
}
}
Just to reiterate, the day, local time meridian are exactly correct, and the equation of time and declination are accurate but not exact.
----UPDATE OUTPUT----
- - -更新 - - -
使用散点图显示全天太阳的高度/方位角。我仍然无法确定方位角输出。它在很长一段时间内都是正确的,但随后它会从增加变为减少(~270-->0)。一旦我最终得到正确的输出,我一定会更新代码。