我正在从事一个涉及计算日出/日落数据的宠物项目。我正在努力在 Swift 中实现以下公式:

Equation of Center:

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

这是我应该得到的给定纬度/经度的答案:

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 center = -1.015867439183884 而正确的 center = 0.61344892821988

我对方程式进行了双重和三次检查，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，但如果是这样的话我会很尴尬......

我正在根据提供的等式和答案工作 here .

编辑这是完整的代码:

//: 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)

最佳答案

正如@kennytm 所建议的，平均异常(太阳或其他任何东西)是一个角度。 Swift 中的角度(和 C，数学库的来源)都是弧度，而天文学家则用度数来表示。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

关于swift - 在 Swift 中计算中心方程，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34954208/