Question

我正在开展一项涉及计算日出/日落数据的宠物项目。我正在努力在Swift中实现以下公式：

中心公式：

C =（1.9148 * sin（meanSolarAnomaly））+（0.0200 *   sin（2 * meanSolarAnomaly））+（0.0003 * sin（3 * meanSolarAnomaly））

以下是我给出的Lat / Lon应该得到的答案：

C = 1.9148 * sin（18.30143135945）+ 0.0200 * sin（2 * 18.30143135945）+ 0.0003 * sin（3 * 18.30143135945）= 0.61344892821988

这是我的代码，它没有给出正确的值作为最终值：

// meanSolarAnomaly has a value of 18.30143135945036 at this point
let center = (1.9148 * sin(meanSolarAnomaly)) + (0.0200 * sin(2*meanSolarAnomaly)) + (0.0003 * sin(3*meanSolarAnomaly))

我的代码说中心= -1.015867439183884而正确的中心= 0.61344892821988

我对方程进行了双重和三重检查，但我似乎无法发现我的错误。我希望这是一个简单的语法错误，但如果它是......将会很尴尬。

我正在使用等式和答案提供here。

编辑以下是完整的代码：

//: Playground - noun: a place where people can play

import UIKit
//calculator to determine what time of day Sunset / Sunrise will occur
func jdFromDate(date : NSDate) -> Double {
    let JD_JAN_1_1970_0000GMT = 2440587.5
    return JD_JAN_1_1970_0000GMT + date.timeIntervalSince1970 / 86400
}

func dateFromJd(jd : Double) -> NSDate {
    let JD_JAN_1_1970_0000GMT = 2440587.5
    return  NSDate(timeIntervalSince1970: (jd - JD_JAN_1_1970_0000GMT) * 86400)
}
let julianOffset = 2451545 as Double
let refDateFormatter = NSDateFormatter()
refDateFormatter.dateFormat = "MM-dd-yyyy"

let today = NSDate()

let julianDaysToToday = round(jdFromDate(today))

//get the lat/lon variables set (Tampa in example)
let lat = 27.9681
let lon = 82.4764

//now we need to calculate julian cycle
let nRaw = (julianDaysToToday - julianOffset - 0.0009) - (lon / 360)
let n = round(nRaw)

//n now contains the julian cycle
//next we must calculate the julian date of solar noon (approximately)
//J* = 2451545 + 0.0009 + (lw/360) + n

let jSolarNoon = julianOffset + 0.0009 + (lon/360) + n

//next calculate the mean solar anomaly
//M = [357.5291 + 0.98560028 * (J* - 2451545)] mod 360
let meanSolarAnomaly = (357.5291 + 0.98560028 * (jSolarNoon - julianOffset)) % 360

//next calculate the equation of center

let center = (1.9148 * sin(meanSolarAnomaly)) + (0.0200 * sin(2*meanSolarAnomaly)) + (0.0003 * sin(3*meanSolarAnomaly))

//Now, using Center  and Mean, calculate the ecliptical longitude of the sun.
//λ = (M + 102.9372 + C + 180) mod 360
let eclLonOfSun = (meanSolarAnomaly + 102.9372 + center + 180) % 360

//now we can finally get an accurate julian date for solar noon
let jTransit = jSolarNoon + (0.0053 * sin(meanSolarAnomaly)) - (0.0069 * sin(2 * eclLonOfSun))

//To calculate the hour angle we need to find the declination of the sun
//δ = arcsin( sin(λ) * sin(23.45) )
let declinationOfSun = asin(sin(eclLonOfSun) * sin(23.45))

//now calculate the hour angle
//H = arccos( [sin(-0.83) - sin(ln) * sin(δ)] / [cos(ln) * cos(δ)] )
let hourCosNum = sin(-0.83) - sin(lat) * sin(declinationOfSun)
let hourDenom = cos(lat)*cos(declinationOfSun)
let hourAngle = acos(hourCosNum)/hourDenom

//time to go back through the approximation again using the hour angle
let finalJulianApproximation = 2451545 + 0.0009 + ((hourAngle + lon)/360) + n

//The values of M and λ from above don't really change from solar noon to sunset, so there is no need to recalculate them before calculating sunset.
let jSet = finalJulianApproximation + (0.0053 * sin(meanSolarAnomaly)) - (0.0069 * sin(2*eclLonOfSun))

let sunset = dateFromJd(jSet)

Answer 1

正如@kennytm所暗示的那样，（太阳或其他任何东西）的异常是一个角度。 Swift中的角度（以及数学图书馆来自的C）都是弧度，而天文学家则用度数说话。这是您在Playground中的代码：

var meanSolarAnomaly = 18.30143135945036
var c = (1.9148 * sin(meanSolarAnomaly)) + (0.0200 * sin(2 * meanSolarAnomaly)) + (0.0003 * sin(3 * meanSolarAnomaly)) 
// = -1.01586743918389 - wrong answer

meanSolarAnomaly = meanSolarAnomaly * M_PI / 180.0 
// Convert it to radians

c = (1.9148 * sin(meanSolarAnomaly)) + (0.0200 * sin(2 * meanSolarAnomaly)) + (0.0003 * sin(3 * meanSolarAnomaly)) 
// = 0.6134489282198807 - right answer

在Swift中计算中心方程

1 个答案: