计算太阳的路径

Question

我正在编写几种必要的方法来计算特定点的太阳路径.我已经使用两个不同的来源编写代码用于我的计算,并且都没有产生期望的结果.来源是:http://www.pveducation.org/pvcdrom/properties-of-sunlight/suns-position和 http://www.esrl.noaa.gov/gmd/grad/solcalc/solareqns.PDF

注意:弧度为Deg*60分钟.

1. localSolartime:我已经将经度转换为'分钟',从localStandardTimeMeridian方法派生的本地标准时间子午线(lstm)返回一个以'分钟'为单位的值,以及在'分钟'中也返回的equationOfTime.使用pveducation中的等式,我计算了时间校正,它解释了给定时区内的小时间变化.当我将这个结果和localTime(每分钟几分钟)应用到当地的太阳时(lst)方程时,结果是676.515(此刻),这对我没有任何意义.据我所知,当地的太阳时代表示相对于太阳的时间,当它处于天空的最高点时,在当地被认为是太阳正午.676.515没有意义.有没有人理解可能导致这种情况的原因.

2. HourAngle:我希望一旦我修复了localSolarTime方法,就不需要纠正了.

我选择了华盛顿特区的纬度和经度.Zenith和Azimuth读数都应该是正值,而对于我这个时刻,它们分别是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;
    }
}

重申一下,当天,当地时间子午线是完全正确的,时间和赤纬方程是准确的但不准确.----更新输出----

-----更新-----使用散点图全天显示太阳的仰角/方位角.我仍然无法确定方位角输出.这是正确的很长一段时间,但它会从增加和开始变为减少(~270 - > 0).一旦我最终得到正确的输出,我一定会更新代码.

Answer 1

您将经度传递为localSolarTime()度数，然后将其除以 60，并附有注释，声称这是为了转换为弧分。这是错误的；你后面的计算需要度数，即使你需要弧分，你也会乘以 60，而不是除法。

这种错误的划分导致经度为 -1.3°，当您找到当地时间子午线与您的位置之间的角度时，您会得到一个很大的角度（大约 75°）。应该是一个小角度，一般为±7.5°。大角度会导致大量的时间校正，并且使一切都变得混乱。

更新：在该azimuth()方法的更新版本中，象限选择应基于太阳的小时角度，或者等效地，基于当地太阳时，而不是标准挂钟时间。并且，所有计算中使用的小时角不应四舍五入。该方法可以如下所示，而不是测试四个不同的象限：

public static double azimuth(double lat, double declination, double zenith, double hourAngle)
{
  double elevation = Math.toRadians(90 - zenith);
  lat = Math.toRadians(lat);
  declination = Math.toRadians(declination);
  hourAngle = Math.toRadians(hourAngle);
  double azimuthRadian = acos(((sin(declination) * cos(lat)) - (cos(hourAngle) * cos(declination) * sin(lat))) / cos(elevation));
  double azimuthDegree = Math.toDegrees(azimuthRadian);
  if (hourAngle > 0)
    azimuthDegree = 360 - azimuthDegree;
  System.out.println("Azimuth: " + df.format(azimuthDegree));
  return azimuthDegree;
}

最后，您dcLong作为方法lat的参数传入azimuth()；这应该是dcLat。

我建议在内部始终使用弧度，并且仅在输入和输出上进行度数转换。这将有助于防止错误，并减少舍入错误和不必要的混乱。