我想根据给定的纬度和经度以及日期时间计算夏令时 我的意思是根据地理坐标计算日出时间和日落时间，并根据地理坐标。
答案 0 :(得分：6)
选中此Latitude and Longitude and Daylight Hours
D = daylength
L = latitude
J = day of the year
P = asin[.39795*cos(.2163108 + 2*atan{.9671396*tan[.00860(J-186)]})]
_ _
/ sin(0.8333*pi/180) + sin(L*pi/180)*sin(P) \
D = 24 - (24/pi)*acos{ ----------------------------------------- }
\_ cos(L*pi/180)*cos(P) _/
答案 1 :(得分：5)
这是一个python函数，它返回带有纬度和一年中某一天的参数的日光小时数（1-356之间的数字）：
import math
def Daylight(latitude,day):
P = math.asin(0.39795 * math.cos(0.2163108 + 2 * math.atan(0.9671396 * math.tan(.00860 * (day - 186)))))
pi = math.pi
daylightamount = 24 - (24 / pi) * math.acos(
(math.sin((0.8333 * pi / 180) + math.sin(latitude * pi / 180) * math.sin(P)) / (math.cos(latitude * pi / 180) * math.cos(P))))
return daylightamount
答案 2 :(得分：0)
我刚刚回答了另一个问题，并认为我的解决方案也适用于此。这是一个Javascript解决方案，因此如果需要，您应该能够轻松转换为其他语言。
我在GitHub Sundial下创建了一个存储库，它在许可的修改后的BSD许可下获得许可，因此您可以在自己的项目中自由使用它。
它应精确到0.0001分钟，并考虑到地球的axial tilt和equation of time。
/* Credit and References */
// http://lexikon.astronomie.info/zeitgleichung/ EOT
// http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1989MNRAS.238.1529H&db_key=AST&page_ind=2&plate_select=NO&data_type=GIF&type=SCREEN_GIF&classic=YES
// http://code.google.com/p/eesim/source/browse/trunk/EnergySim/src/sim/_environment.py?spec=svn6&r=6
// http://mathforum.org/library/drmath/view/56478.html
// http://www.jgiesen.de/elevaz/basics/meeus.htm
// http://www.ehow.com/how_8495097_calculate-sunrise-latitude.html
// http://www.jgiesen.de/javascript/Beispiele/TN_Applet/DayNight125d.java
// http://astro.unl.edu/classaction/animations/coordsmotion/daylighthoursexplorer.html
// http://www.neoprogrammics.com/nutations/Nutation_In_Longitude_And_RA.php
(function (factory) {
if (typeof define === 'function' && define.amd ) {
// AMD. Register as module
if(typeof dojo === 'object') {
define(["dojo/_base/declare"], function(declare){
return declare( "my.calc.Sun", null, factory());
});
} else {
define( 'Sundial', null, factory());
}
} else {
Sun = new factory();
}
}(function () {
return {
date : new Date(),
getDate : function(){
return this.date;
},
setDate : function(d){
this.date = d;
return this;
},
getJulianDays: function(){
this._julianDays = Math.floor(( this.date / 86400000) - ( this.date.getTimezoneOffset() / 1440) + 2440587.5);
return this._julianDays;
},
// Calculate the Equation of Time
// The equation of time is the difference between apparent solar time and mean solar time.
// At any given instant, this difference will be the same for every observer on Earth.
getEquationOfTime : function (){
var K = Math.PI/180.0;
var T = (this.getJulianDays() - 2451545.0) / 36525.0;
var eps = this._getObliquity(T); // Calculate the Obliquity (axial tilt of earth)
var RA = this._getRightAscension(T);
var LS = this._getSunsMeanLongitude(T);
var deltaPsi = this._getDeltaPSI(T);
var E = LS - 0.0057183 - RA + deltaPsi*Math.cos(K*eps);
if (E>5) {
E = E - 360.0;
}
E = E*4; // deg. to min
E = Math.round(1000*E)/1000;
return E;
},
getTotalDaylightHoursInYear : function(lat){
// We can just use the current Date Object, and incrementally
// Add 1 Day 365 times...
var totalDaylightHours = 0 ;
for (var d = new Date(this.date.getFullYear(), 0, 1); d <= new Date(this.date.getFullYear(), 11, 30); d.setDate(d.getDate() + 1)) {
this.date = d;
// console.log( this.getDaylightHours(lat) );
totalDaylightHours += this.getDaylightHours(lat);
}
return totalDaylightHours;
},
getDaylightHours : function (lat) {
var K = Math.PI/180.0;
var C, Nenner, C2, dlh;
var T = (this.getJulianDays() - 2451545.0) / 36525.0;
this._getRightAscension(T); // Need to get the Suns Declination
Nenner = Math.cos(K*lat)*Math.cos(K*this._sunDeclination);
C = -Math.sin(K*this._sunDeclination)*Math.sin(K*lat)/Nenner;
C2=C*C;
// console.log( T, C2, C, Nenner, lat, K, Math.cos(K*lat) );
if ((C>-1) && (C<1)) {
dlh=90.0 - Math.atan(C / Math.sqrt(1 - C2)) / K;
dlh=2.0*dlh/15.0;
dlh=Math.round(dlh*100)/100;
}
if (C>1) {
dlh=0.0;
}
if (C<-1) {
dlh=24.0;
}
return dlh;
},
_getRightAscension : function(T) {
var K = Math.PI/180.0;
var L, M, C, lambda, RA, eps, delta, theta;
L = this._getSunsMeanLongitude(T); // Calculate the mean longitude of the Sun
M = 357.52910 + 35999.05030*T - 0.0001559*T*T - 0.00000048*T*T*T; // Mean anomoly of the Sun
M = M % 360;
if (M<0) {
M = M + 360;
}
C = (1.914600 - 0.004817*T - 0.000014*T*T)*Math.sin(K*M);
C = C + (0.019993 - 0.000101*T)*Math.sin(K*2*M);
C = C + 0.000290*Math.sin(K*3*M);
theta = L + C; // get true longitude of the Sun
eps = this._getObliquity(T);
eps = eps + 0.00256*Math.cos(K*(125.04 - 1934.136*T));
lambda = theta - 0.00569 - 0.00478*Math.sin(K*(125.04 - 1934.136*T)); // get apparent longitude of the Sun
RA = Math.atan2(Math.cos(K*eps)*Math.sin(K*lambda), Math.cos(K*lambda));
RA = RA/K;
if (RA<0) {
RA = RA + 360.0;
}
delta = Math.asin(Math.sin(K*eps)*Math.sin(K*lambda));
delta = delta/K;
this._sunDeclination = delta;
return RA;
},
// Calculate the Mean Longitude of the Sun
_getSunsMeanLongitude : function(T){
var L = 280.46645 + 36000.76983*T + 0.0003032*T*T;
L = L % 360;
if (L<0) {
L = L + 360;
}
return L;
},
// Nutation in ecliptical longitude expressed in degrees.
_getDeltaPSI : function(T){
var K = Math.PI/180.0;
var deltaPsi, omega, LS, LM;
LS = this._getSunsMeanLongitude(T);
LM = 218.3165 + 481267.8813*T;
LM = LM % 360;
if (LM<0) {
LM = LM + 360;
}
// Longitude of ascending node of lunar orbit on the ecliptic as measured from the mean equinox of date.
omega = 125.04452 - 1934.136261*T + 0.0020708*T*T + T*T*T/450000;
deltaPsi = -17.2*Math.sin(K*omega) - 1.32*Math.sin(K*2*LS) - 0.23*Math.sin(K*2*LM) + 0.21*Math.sin(K*2*omega);
deltaPsi = deltaPsi/3600.0;
return deltaPsi;
},
// T = Time Factor Time factor in Julian centuries reckoned from J2000.0, corresponding to JD
// Calculate Earths Obliquity Nutation
_getObliquity : function (T) {
var K = Math.PI/180.0;
var LS = this._getSunsMeanLongitude(T);
var LM = 218.3165 + 481267.8813*T;
var eps0 = 23.0 + 26.0/60.0 + 21.448/3600.0 - (46.8150*T + 0.00059*T*T - 0.001813*T*T*T)/3600;
var omega = 125.04452 - 1934.136261*T + 0.0020708*T*T + T*T*T/450000;
var deltaEps = (9.20*Math.cos(K*omega) + 0.57*Math.cos(K*2*LS) + 0.10*Math.cos(K*2*LM) - 0.09*Math.cos(K*2*omega))/3600;
return eps0 + deltaEps;
}
};
}));
您可以查看一下如何在jsfiddle上使用它的演示。
然后当我开始讨论它时，请查看存储库中的示例用例。 GitHub Sundial
答案 3 :(得分：0)
sin24 +（24cos-18 ^ 12）^（一年中的天数）+（纬度）^ 24 = #of daylight hours