35

C# 中有没有办法计算给定一天的太阳落下和升起时的纬度和经度？

4

9 回答 9

15

我知道这篇文章很旧，但万一有人还在寻找......

CoordinateSharp作为 Nuget 包提供。它是一个独立的包，可以处理太阳时间和月亮时间。

Celestial cel = Celestial.CalculateCelestialTimes(85.57682, -70.75678, new DateTime(2017,8,21));
Console.WriteLine(cel.SunRise.Value.ToString());

笔记：

它假定 DateTimes 始终采用 UTC。

.Condition最后，如果日期返回 null ，您可能需要引用 Celestial 对象 Sun/Moon 。这发生在太阳全天升起/落下时。

编辑 2019 年 1 月 9 日

自这篇文章以来，图书馆发生了巨大的变化。它现在也可以处理当地时间。

于 2017-04-14T13:27:35.940 回答
5

这个 API 似乎对我有用：

http://sunrise-sunset.org/api

于 2015-12-07T13:47:23.700 回答
5

我已经在 UWP 中测试了这个 nuget 包。

https://www.nuget.org/packages/SolarCalculator/

文档有点粗略，在这里：

https://github.com/porrey/Solar-Calculator

你可以用它来获得日出，给定

la = 纬度；和 lo = 经度；对于您所在的地区：

            SolarTimes solarTimes = new SolarTimes(DateTime.Now, la, lo);
            DateTime sr = solarTimes.Sunrise;
            DateTime dt = Convert.ToDateTime(sr);
            textblockb.Text = dt.ToString("h:mm:ss");

您可以使用 PM 管理器在 Visual Studio 中安装它

Install-Package SolarCalculator -Version 2.0.2

或通过在“管理 NuGet 包”Visual Studio 库中查找 SolarCalculator。

于 2017-11-06T13:01:37.323 回答
4

对此接受的答案是 JavaScript 实现，它不适合我的应用程序，因为我需要在 C# 中进行计算。

我已经使用了这个 C# 代码：http ://wiki.crowe.co.nz/Calculate%20Sunrise%2fSunset.ashx ，我已经验证了这里的日出/日落时间：http: //www.timeanddate.com/天文学/ .

如果我将秒四舍五入到最接近的分钟，C# 实现的日出和日落时间与 timeanddate.com 上显示的相应值匹配，包括夏令时的情况。虽然代码有点压倒性（除非您也想要月相数据），所以我将对其进行重构以专门执行我现在需要的数字是正确的。

于 2015-01-28T16:57:44.213 回答
3

dotsa 答案的 VB.Net 版本，它也可以自动确定时区。

输出（通过今晚看日落检查）：

输出

主.VB：

Module Main

Sub Main()

    ' http://www.timeanddate.com/sun/usa/seattle
    ' http://www.esrl.noaa.gov/gmd/grad/solcalc/

    ' Vessy, Switzerland
    Dim latitude As Double = 46.17062
    Dim longitude As Double = 6.161667
    Dim dst As Boolean = True
    Dim timehere As DateTime = DateTime.Now

    Console.WriteLine("It is currently {0:HH:mm:ss} UTC", DateTime.UtcNow)
    Console.WriteLine("The time here, at {0}°,{1}° is {2:HH:mm:ss}", latitude, longitude, timehere)
    Dim local As TimeZoneInfo = TimeZoneInfo.Local
    Dim zone As Integer = local.BaseUtcOffset().TotalHours

    If local.SupportsDaylightSavingTime Then
        Dim standard As String = local.StandardName
        Dim daylight As String = local.DaylightName
        dst = local.IsDaylightSavingTime(timehere)
        Dim current As String = IIf(dst, daylight, standard)
        Console.WriteLine("Daylight-saving time is supported here. Current offset {0:+0} hours, {1}", zone, current)
    Else
        Console.WriteLine("Daylight-saving time is not supported here")
    End If

    System.Console.WriteLine("Sunrise today {0}", Sunrises(latitude, longitude))
    System.Console.WriteLine("Sunset  today {0}", Sunsets(latitude, longitude))
    System.Console.ReadLine()
End Sub

End Module

太阳.vb：

Public Module Sun
' Get sunrise time at latitude, longitude using local system timezone
Function Sunrises(latitude As Double, longitude As Double) As DateTime
    Dim julian As Double = JulianDay(DateTime.Now)
    Dim rises As Double = SunRiseUTC(julian, latitude, longitude)
    Dim timehere As DateTime = DateTime.Now
    Dim local As TimeZoneInfo = TimeZoneInfo.Local
    Dim dst As Boolean = local.IsDaylightSavingTime(timehere)
    Dim zone As Integer = local.BaseUtcOffset().TotalHours
    Dim result As DateTime = getDateTime(rises, zone, timehere, dst)
    Return result
End Function
' Get sunset time at latitude, longitude using local system timezone
Function Sunsets(latitude As Double, longitude As Double) As DateTime
    Dim julian As Double = JulianDay(DateTime.Now)
    Dim rises As Double = SunSetUTC(julian, latitude, longitude)
    Dim timehere As DateTime = DateTime.Now
    Dim local As TimeZoneInfo = TimeZoneInfo.Local
    Dim dst As Boolean = local.IsDaylightSavingTime(timehere)
    Dim zone As Integer = local.BaseUtcOffset().TotalHours
    Dim result As DateTime = getDateTime(rises, zone, timehere, dst)
    Return result
End Function
' Convert radian angle to degrees
Public Function Degrees(angleRad As Double) As Double
    Return (180.0 * angleRad / Math.PI)
End Function
' Convert degree angle to radians
Public Function Radians(angleDeg As Double) As Double
    Return (Math.PI * angleDeg / 180.0)
End Function
'* Name: JulianDay  
'* Type: Function   
'* Purpose: Julian day from calendar day    
'* Arguments:   
'* year : 4 digit year  
'* month: January = 1   
'* day : 1 - 31 
'* Return value:    
'* The Julian day corresponding to the date 
'* Note:    
'* Number is returned for start of day. Fractional days should be   
'* added later. 
Public Function JulianDay(year As Integer, month As Integer, day As Integer) As Double
    If month <= 2 Then
        year -= 1
        month += 12
    End If
    Dim A As Double = Math.Floor(year / 100.0)
    Dim B As Double = 2 - A + Math.Floor(A / 4)

    Dim julian As Double = Math.Floor(365.25 * (year + 4716)) + Math.Floor(30.6001 * (month + 1)) + day + B - 1524.5
    Return julian
End Function

Public Function JulianDay([date] As DateTime) As Double
    Return JulianDay([date].Year, [date].Month, [date].Day)
End Function

'***********************************************************************/
'* Name: JulianCenturies    
'* Type: Function   
'* Purpose: convert Julian Day to centuries since J2000.0.  
'* Arguments:   
'* julian : the Julian Day to convert   
'* Return value:    
'* the T value corresponding to the Julian Day  
'***********************************************************************/

Public Function JulianCenturies(julian As Double) As Double
    Dim T As Double = (julian - 2451545.0) / 36525.0
    Return T
End Function


'***********************************************************************/
'* Name: JulianDayFromJulianCentury 
'* Type: Function   
'* Purpose: convert centuries since J2000.0 to Julian Day.  
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* the Julian Day corresponding to the t value  
'***********************************************************************/

Public Function JulianDayFromJulianCentury(t As Double) As Double
    Dim julian As Double = t * 36525.0 + 2451545.0
    Return julian
End Function


'***********************************************************************/
'* Name: calGeomMeanLongSun 
'* Type: Function   
'* Purpose: calculate the Geometric Mean Longitude of the Sun   
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* the Geometric Mean Longitude of the Sun in degrees   
'***********************************************************************/

Public Function GemoetricMeanLongitude(t As Double) As Double
    Dim L0 As Double = 280.46646 + t * (36000.76983 + 0.0003032 * t)
    While L0 > 360.0
        L0 -= 360.0
    End While
    While L0 < 0.0
        L0 += 360.0
    End While
    Return L0
    ' in degrees
End Function


'***********************************************************************/
'* Name: calGeomAnomalySun  
'* Type: Function   
'* Purpose: calculate the Geometric Mean Anomaly of the Sun 
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* the Geometric Mean Anomaly of the Sun in degrees 
'***********************************************************************/

Public Function GemoetricMeanAnomaly(t As Double) As Double
    Dim M As Double = 357.52911 + t * (35999.05029 - 0.0001537 * t)
    Return M
    ' in degrees
End Function

'***********************************************************************/
'* Name: EarthOrbitEccentricity 
'* Type: Function   
'* Purpose: calculate the eccentricity of earth's orbit 
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* the unitless eccentricity    
'***********************************************************************/


Public Function EarthOrbitEccentricity(t As Double) As Double
    Dim e As Double = 0.016708634 - t * (0.000042037 + 0.0000001267 * t)
    Return e
    ' unitless
End Function

'***********************************************************************/
'* Name: SunCentre  
'* Type: Function   
'* Purpose: calculate the equation of center for the sun    
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* in degrees   
'***********************************************************************/


Public Function SunCentre(t As Double) As Double
    Dim m As Double = GemoetricMeanAnomaly(t)

    Dim mrad As Double = Radians(m)
    Dim sinm As Double = Math.Sin(mrad)
    Dim sin2m As Double = Math.Sin(mrad + mrad)
    Dim sin3m As Double = Math.Sin(mrad + mrad + mrad)

    Dim C As Double = sinm * (1.914602 - t * (0.004817 + 0.000014 * t)) + sin2m * (0.019993 - 0.000101 * t) + sin3m * 0.000289
    Return C
    ' in degrees
End Function

'***********************************************************************/
'* Name: SunTrueLongitude   
'* Type: Function   
'* Purpose: calculate the true longitude of the sun 
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* sun's true longitude in degrees  
'***********************************************************************/


Public Function SunTrueLongitude(t As Double) As Double
    Dim l0 As Double = GemoetricMeanLongitude(t)
    Dim c As Double = SunCentre(t)

    Dim O As Double = l0 + c
    Return O
    ' in degrees
End Function

'***********************************************************************/
'* Name: SunTrueAnomaly 
'* Type: Function   
'* Purpose: calculate the true anamoly of the sun   
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* sun's true anamoly in degrees    
'***********************************************************************/

Public Function SunTrueAnomaly(t As Double) As Double
    Dim m As Double = GemoetricMeanAnomaly(t)
    Dim c As Double = SunCentre(t)

    Dim v As Double = m + c
    Return v
    ' in degrees
End Function

'***********************************************************************/
'* Name: SunDistanceAU  
'* Type: Function   
'* Purpose: calculate the distance to the sun in AU 
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* sun radius vector in AUs 
'***********************************************************************/

Public Function SunDistanceAU(t As Double) As Double
    Dim v As Double = SunTrueAnomaly(t)
    Dim e As Double = EarthOrbitEccentricity(t)

    Dim R As Double = (1.000001018 * (1 - e * e)) / (1 + e * Math.Cos(Radians(v)))
    Return R
    ' in AUs
End Function

'***********************************************************************/
'* Name: SunApparentLongitude   
'* Type: Function   
'* Purpose: calculate the apparent longitude of the sun 
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* sun's apparent longitude in degrees  
'***********************************************************************/

Public Function SunApparentLongitude(t As Double) As Double
    Dim o As Double = SunTrueLongitude(t)

    Dim omega As Double = 125.04 - 1934.136 * t
    Dim lambda As Double = o - 0.00569 - 0.00478 * Math.Sin(Radians(omega))
    Return lambda
    ' in degrees
End Function

'***********************************************************************/
'* Name: MeanObliquityOfEcliptic    
'* Type: Function   
'* Purpose: calculate the mean obliquity of the ecliptic    
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* mean obliquity in degrees    
'***********************************************************************/

Public Function MeanObliquityOfEcliptic(t As Double) As Double
    Dim seconds As Double = 21.448 - t * (46.815 + t * (0.00059 - t * (0.001813)))
    Dim e0 As Double = 23.0 + (26.0 + (seconds / 60.0)) / 60.0
    Return e0
    ' in degrees
End Function

'***********************************************************************/
'* Name: calcObliquityCorrection    
'* Type: Function   
'* Purpose: calculate the corrected obliquity of the ecliptic   
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* corrected obliquity in degrees   
'***********************************************************************/

Public Function calcObliquityCorrection(t As Double) As Double
    Dim e0 As Double = MeanObliquityOfEcliptic(t)

    Dim omega As Double = 125.04 - 1934.136 * t
    Dim e As Double = e0 + 0.00256 * Math.Cos(Radians(omega))
    Return e
    ' in degrees
End Function

'***********************************************************************/
'* Name: SunRightAscension  
'* Type: Function   
'* Purpose: calculate the right ascension of the sun    
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* sun's right ascension in degrees 
'***********************************************************************/

Public Function SunRightAscension(t As Double) As Double
    Dim e As Double = calcObliquityCorrection(t)
    Dim lambda As Double = SunApparentLongitude(t)

    Dim tananum As Double = (Math.Cos(Radians(e)) * Math.Sin(Radians(lambda)))
    Dim tanadenom As Double = (Math.Cos(Radians(lambda)))
    Dim alpha As Double = Degrees(Math.Atan2(tananum, tanadenom))
    Return alpha
    ' in degrees
End Function

'***********************************************************************/
'* Name: SunDeclination 
'* Type: Function   
'* Purpose: calculate the declination of the sun    
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* sun's declination in degrees 
'***********************************************************************/

Public Function SunDeclination(t As Double) As Double
    Dim e As Double = calcObliquityCorrection(t)
    Dim lambda As Double = SunApparentLongitude(t)

    Dim sint As Double = Math.Sin(Radians(e)) * Math.Sin(Radians(lambda))
    Dim theta As Double = Degrees(Math.Asin(sint))
    Return theta
    ' in degrees
End Function

'***********************************************************************/
'* Name: TrueSolarToMeanSolar   
'* Type: Function   
'* Purpose: calculate the difference between true solar time and mean   
'*   solar time 
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* Return value:    
'* equation of time in minutes of time  
'***********************************************************************/

Public Function TrueSolarToMeanSolar(t As Double) As Double
    Dim epsilon As Double = calcObliquityCorrection(t)
    Dim l0 As Double = GemoetricMeanLongitude(t)
    Dim e As Double = EarthOrbitEccentricity(t)
    Dim m As Double = GemoetricMeanAnomaly(t)

    Dim y As Double = Math.Tan(Radians(epsilon) / 2.0)
    y *= y

    Dim sin2l0 As Double = Math.Sin(2.0 * Radians(l0))
    Dim sinm As Double = Math.Sin(Radians(m))
    Dim cos2l0 As Double = Math.Cos(2.0 * Radians(l0))
    Dim sin4l0 As Double = Math.Sin(4.0 * Radians(l0))
    Dim sin2m As Double = Math.Sin(2.0 * Radians(m))

    Dim Etime As Double = y * sin2l0 - 2.0 * e * sinm + 4.0 * e * y * sinm * cos2l0 - 0.5 * y * y * sin4l0 - 1.25 * e * e * sin2m

    Return Degrees(Etime) * 4.0
    ' in minutes of time
End Function

'***********************************************************************/
'* Name: SunriseHourAngle   
'* Type: Function   
'* Purpose: calculate the hour angle of the sun at sunrise for the  
'*   latitude   
'* Arguments:   
'* lat : latitude of observer in degrees    
'*  solarDec : declination angle of sun in degrees  
'* Return value:    
'* hour angle of sunrise in radians 
'***********************************************************************/

Public Function SunriseHourAngle(lat As Double, solarDec As Double) As Double
    Dim latRad As Double = Radians(lat)
    Dim sdRad As Double = Radians(solarDec)

    Dim HAarg As Double = (Math.Cos(Radians(90.833)) / (Math.Cos(latRad) * Math.Cos(sdRad)) - Math.Tan(latRad) * Math.Tan(sdRad))

    Dim HA As Double = (Math.Acos(Math.Cos(Radians(90.833)) / (Math.Cos(latRad) * Math.Cos(sdRad)) - Math.Tan(latRad) * Math.Tan(sdRad)))

    Return HA
    ' in radians
End Function

'***********************************************************************/
'* Name: SunsetHourAngle    
'* Type: Function   
'* Purpose: calculate the hour angle of the sun at sunset for the   
'*   latitude   
'* Arguments:   
'* lat : latitude of observer in degrees    
'*  solarDec : declination angle of sun in degrees  
'* Return value:    
'* hour angle of sunset in radians  
'***********************************************************************/

Public Function SunsetHourAngle(lat As Double, solarDec As Double) As Double
    Dim latRad As Double = Radians(lat)
    Dim sdRad As Double = Radians(solarDec)

    Dim HAarg As Double = (Math.Cos(Radians(90.833)) / (Math.Cos(latRad) * Math.Cos(sdRad)) - Math.Tan(latRad) * Math.Tan(sdRad))

    Dim HA As Double = (Math.Acos(Math.Cos(Radians(90.833)) / (Math.Cos(latRad) * Math.Cos(sdRad)) - Math.Tan(latRad) * Math.Tan(sdRad)))

    Return -HA
    ' in radians
End Function


'***********************************************************************/
'* Name: SunRiseUTC 
'* Type: Function   
'* Purpose: calculate the Universal Coordinated Time (UTC) of sunrise   
'*   for the given day at the given location on earth   
'* Arguments:   
'* julian : julian day  
'* latitude : latitude of observer in degrees   
'* longitude : longitude of observer in degrees 
'* Return value:    
'* time in minutes from zero Z  
'***********************************************************************/

'Public  Function SunRiseUTC(julian As Double, latitude As Double, longitude As Double) As Double
'    Dim t As Double = JulianCenturies(julian)

'    ' *** Find the time of solar noon at the location, and use
'    ' that declination. This is better than start of the 
'    ' Julian day

'    Dim noonmin As Double = SolarNoonUTC(t, longitude)
'    Dim tnoon As Double = JulianCenturies(julian + noonmin / 1440.0)

'    ' *** First pass to approximate sunrise (using solar noon)

'    Dim eqTime As Double = TrueSolarToMeanSolar(tnoon)
'    Dim solarDec As Double = SunDeclination(tnoon)
'    Dim hourAngle As Double = SunriseHourAngle(latitude, solarDec)

'    Dim delta As Double = longitude - Degrees(hourAngle)
'    Dim timeDiff As Double = 4 * delta
'    ' in minutes of time
'    Dim timeUTC As Double = 720 + timeDiff - eqTime
'    ' in minutes
'    ' alert("eqTime = " + eqTime + "\nsolarDec = " + solarDec + "\ntimeUTC = " + timeUTC);

'    ' *** Second pass includes fractional julianay in gamma calc

'    Dim newt As Double = JulianCenturies(JulianDayFromJulianCentury(t) + timeUTC / 1440.0)
'    eqTime = TrueSolarToMeanSolar(newt)
'    solarDec = SunDeclination(newt)
'    hourAngle = SunriseHourAngle(latitude, solarDec)
'    delta = longitude - Degrees(hourAngle)
'    timeDiff = 4 * delta
'    timeUTC = 720 + timeDiff - eqTime
'    ' in minutes
'    ' alert("eqTime = " + eqTime + "\nsolarDec = " + solarDec + "\ntimeUTC = " + timeUTC);

'    Return timeUTC
'End Function

'***********************************************************************/
'* Name: SolarNoonUTC   
'* Type: Function   
'* Purpose: calculate the Universal Coordinated Time (UTC) of solar 
'*   noon for the given day at the given location on earth  
'* Arguments:   
'* t : number of Julian centuries since J2000.0 
'* longitude : longitude of observer in degrees 
'* Return value:    
'* time in minutes from zero Z  
'***********************************************************************/

Public Function SolarNoonUTC(t As Double, longitude As Double) As Double
    ' First pass uses approximate solar noon to calculate eqtime
    Dim tnoon As Double = JulianCenturies(JulianDayFromJulianCentury(t) + longitude / 360.0)
    Dim eqTime As Double = TrueSolarToMeanSolar(tnoon)
    Dim solNoonUTC As Double = 720 + (longitude * 4) - eqTime
    ' min
    Dim newt As Double = JulianCenturies(JulianDayFromJulianCentury(t) - 0.5 + solNoonUTC / 1440.0)

    eqTime = TrueSolarToMeanSolar(newt)
    ' double solarNoonDec = SunDeclination(newt);
    solNoonUTC = 720 + (longitude * 4) - eqTime
    ' min
    Return solNoonUTC
End Function

'***********************************************************************/
'* Name: SunSetUTC  
'* Type: Function   
'* Purpose: calculate the Universal Coordinated Time (UTC) of sunset    
'*   for the given day at the given location on earth   
'* Arguments:   
'* julian : julian day  
'* latitude : latitude of observer in degrees   
'* longitude : longitude of observer in degrees 
'* Return value:    
'* time in minutes from zero Z  
'***********************************************************************/

Public Function SunSetUTC(julian As Double, latitude As Double, longitude As Double) As Double
    Dim t = JulianCenturies(julian)
    Dim eqTime = TrueSolarToMeanSolar(t)
    Dim solarDec = SunDeclination(t)
    Dim hourAngle = SunriseHourAngle(latitude, solarDec)
    hourAngle = -hourAngle
    Dim delta = longitude + Degrees(hourAngle)
    Dim timeUTC = 720 - (4.0 * delta) - eqTime
    ' in minutes
    Return timeUTC
End Function

Public Function SunRiseUTC(julian As Double, latitude As Double, longitude As Double) As Double
    Dim t = JulianCenturies(julian)
    Dim eqTime = TrueSolarToMeanSolar(t)
    Dim solarDec = SunDeclination(t)
    Dim hourAngle = SunriseHourAngle(latitude, solarDec)
    Dim delta = longitude + Degrees(hourAngle)
    Dim timeUTC = 720 - (4.0 * delta) - eqTime
    ' in minutes
    Return timeUTC
End Function

Public Function getTimeString(time As Double, timezone As Integer, julian As Double, dst As Boolean) As String
    Dim timeLocal = time + (timezone * 60.0)
    Dim riseT = JulianCenturies(julian + time / 1440.0)
    timeLocal += (If((dst), 60.0, 0.0))
    Return getTimeString(timeLocal)
End Function

Public Function getDateTime(time As Double, timezone As Integer, [date] As DateTime, dst As Boolean) As System.Nullable(Of DateTime)
    Dim julian As Double = JulianDay([date])
    Dim timeLocal = time + (timezone * 60.0)
    Dim riseT = JulianCenturies(julian + time / 1440.0)
    timeLocal += (If((dst), 60.0, 0.0))
    Return getDateTime(timeLocal, [date])
End Function

Private Function getTimeString(minutes As Double) As String

    Dim output As String = ""

    If (minutes >= 0) AndAlso (minutes < 1440) Then
        Dim floatHour = minutes / 60.0
        Dim hour = Math.Floor(floatHour)
        Dim floatMinute = 60.0 * (floatHour - Math.Floor(floatHour))
        Dim minute = Math.Floor(floatMinute)
        Dim floatSec = 60.0 * (floatMinute - Math.Floor(floatMinute))
        Dim second = Math.Floor(floatSec + 0.5)
        If second > 59 Then
            second = 0
            minute += 1
        End If
        If (second >= 30) Then
            minute += 1
        End If
        If minute > 59 Then
            minute = 0
            hour += 1
        End If
        output = [String].Format("{0:00}:{1:00}", hour, minute)
    Else
        Return "error"
    End If

    Return output
End Function

Private Function getDateTime(minutes As Double, [date] As DateTime) As System.Nullable(Of DateTime)

    Dim retVal As System.Nullable(Of DateTime) = Nothing

    If (minutes >= 0) AndAlso (minutes < 1440) Then
        Dim floatHour = minutes / 60.0
        Dim hour = Math.Floor(floatHour)
        Dim floatMinute = 60.0 * (floatHour - Math.Floor(floatHour))
        Dim minute = Math.Floor(floatMinute)
        Dim floatSec = 60.0 * (floatMinute - Math.Floor(floatMinute))
        Dim second = Math.Floor(floatSec + 0.5)
        If second > 59 Then
            second = 0
            minute += 1
        End If
        If (second >= 30) Then
            minute += 1
        End If
        If minute > 59 Then
            minute = 0
            hour += 1
        End If
        Return New DateTime([date].Year, [date].Month, [date].Day, CInt(hour), CInt(minute), CInt(second))
    Else
        Return retVal
    End If
End Function
End Module
于 2016-08-23T20:11:31.237 回答
2

从这个信息开始：

日出方程式

我正在使用它来编写仍在制作中的 ruby​​ 脚本。我无法理解多部分的朱利安日期。

很清楚的一件事是，您应该选择准确的太阳过境时间。然后根据您的纬度和太阳赤纬减去并添加 semi_diurnal_arc = acos(cos_omega)。哦！并且一定要包括太阳中心和地球折射。看来这个地球真是个魔术师。

于 2011-02-23T22:10:15.060 回答
1

您需要一个包含时间等式的公式，以允许地球月球系统围绕太阳的偏心轨道。您需要使用具有适当基准点的坐标，例如 WGS84 或 NAD27 或类似的东西。您需要使用 JULIAN 日历，而不是我们每天使用的日历，才能正确获得 5 这些时间。在一秒钟之内猜测出来并不是一件容易的事情。我想在我的位置有时间，阴影长度等于任何高度。这应该每天发生两次，当太阳在中午前后升高到地平线以上 60 度时。此外，据我了解，您只需要每年增加一天即可获得恒星时间，所以如果您想将时钟频率提高 X 366.25/365.25，您现在可能有一个恒星时钟而不是民用时钟？？？"

于 2013-08-10T14:39:47.527 回答
1

是的，放弃一些。

一些模式链接。

http://williams.best.vwh.net/sunrise_sunset_example.htm

http://www.codeproject.com/Articles/29306/C-Class-for-Calculating-Sunrise-and-Sunset-Times

https://social.msdn.microsoft.com/Forums/vstudio/en-US/a4fad4c3-6d18-41fc-82b7-1f3031349837/get-sunrise-and-sunset-time-based-on-latitude-and-longitude？论坛=csharpgeneral

https://gist.github.com/cstrahan/767532

http://pointofint.blogspot.com/2014/06/sunrise-and-sunset-in-c.html

http://yaddb.blogspot.com/2013/01/how-to-calculate-sunrise-and-sunset.html

https://forums.asp.net/t/1810934.aspx?Sunrise+and+Sunset+timings+Calculation+

http://www.ip2location.com/tutorials/display-sunrise-sunset-time-using-csharp-and-mysql-database

http://en.pudn.com/downloads270/sourcecode/windows/csharp/detail1235934_en.html

http://regator.com/p/25716249/c_class_for_calculating_sunrise_and_sunset_times

http://forums.xkcd.com/viewtopic.php?t=102253

http://www.redrok.com/solar_position_algorithm.pdf

http://sidstation.loudet.org/sunazimuth-en.xhtml

https://sourceforge.net/directory/os:windows/?q=sunrise/set%20times

https://www.nuget.org/packages/SolarCalculator/

http://www.grasshopper3d.com/forum/topics/solar-calculation-plugin

这是我很久以前为 Planet Source Code 所做的一个项目，但幸运的是我将它保存在其他地方，因为该站点丢失了数据。

https://github.com/DouglasAllen/SunTimes.VSCS.Net

使用这个要点加

https://gist.github.com/DouglasAllen/c682e4c412a0b9d8f536b014c1766f20

现在简要解释一下实现这一点的技术。

首先，您需要真正的太阳正午或您所在位置的过境。

这考虑到您当地的经度。只需将其除以 15 即可转换为时间。

那就是您距祖鲁地区时间或零经度的时间。

从中午 12:00 或中午开始。

并根据经度计算您的时间。

现在是困难的部分。你需要一种计算时间方程的方法。

这是由地球倾斜和围绕太阳运行的轨道造成的时差。

这会给你一个想法...... https://en.wikipedia.org/wiki/Equation_of_time

但他们有一个更容易的公式...... https://en.wikipedia.org/wiki/Sunrise_equation

这家伙有一些书，很多人都会去或买。:-D https://en.wikipedia.org/wiki/Jean_Meeus

使用您的第一个计算来计算平均太阳过境并计算 JDN... https://en.wikipedia.org/wiki/Julian_day

这被所有角度公式用作儒略世纪的时间 https://en.wikipedia.org/wiki/Julian_year_(astronomy)

https://en.wikipedia.org/wiki/Epoch_（天文学）

它基本上是您的 JDN 减去诸如 J2000 或 2451545.0 之类的纪元，全部除以 36525.0 得到儒略世纪或 t，它用于大多数以 t 作为参数的公式。有时使用儒略千年。在这种情况下，它是 3652500.0

诀窍是找到那些可以帮助您解决时间方程的角度公式。

然后你得到你真正的太阳过境并减去半天或为你的位置增加半天的阳光。您会在答案和软件中找到这些内容。

一旦你得到一些东西，你可以通过搜索时间或在线计算器来检查它。

我希望这足以让你继续前进。到处都有图书馆，但制作自己的图书馆并不难。我做了，但它是在 Ruby 中的。它可能很有用.... https://github.com/DouglasAllen/gem-equationoftime

祝你好运！

于 2016-10-14T01:08:53.227 回答
1

如果您更喜欢外部服务，您可以使用这个漂亮且免费的日出和日落时间 API：http ://sunrise-sunset.org/api

我已经在几个项目中使用它并且效果很好，数据似乎非常准确。只需向http://api.sunrise-sunset.org/json发出 HTTP GET 请求

接受的参数：

  • lat：十进制度的纬度。必需的。
  • lng：十进制度的经度。必需的。
  • 日期：YYYY-MM-DD 格式的日期。还接受其他日期格式，甚至是相对日期格式。如果不存在，日期默认为当前日期。可选的。
  • callback：JSONP 响应的回调函数名。可选的。
  • 格式化：0 或 1（默认为 1）。响应的时间值将按照 ISO 8601 表示，day_length 将以秒表示。可选的。

响应包括日出和日落时间以及黄昏时间。

于 2015-09-26T05:30:26.423 回答