太阳的位置给定时间,经度和纬度
Spo*_*nNZ 82 math geometry r azimuth astronomy
这个问题在三年多前就被提出过了.给出了答案,但是我发现解决方案存在问题.
下面的代码在R.我已经将它移植到另一种语言,但是直接在R中测试了原始代码,以确保问题不在于我的移植.
sunPosition <- function(year, month, day, hour=12, min=0, sec=0,
lat=46.5, long=6.5) {
twopi <- 2 * pi
deg2rad <- pi / 180
# Get day of the year, e.g. Feb 1 = 32, Mar 1 = 61 on leap years
month.days <- c(0,31,28,31,30,31,30,31,31,30,31,30)
day <- day + cumsum(month.days)[month]
leapdays <- year %% 4 == 0 & (year %% 400 == 0 | year %% 100 != 0) & day >= 60
day[leapdays] <- day[leapdays] + 1
# Get Julian date - 2400000
hour <- hour + min / 60 + sec / 3600 # hour plus fraction
delta <- year - 1949
leap <- trunc(delta / 4) # former leapyears
jd <- 32916.5 + delta * 365 + leap + day + hour / 24
# The input to the Atronomer's almanach is the difference between
# the Julian date and JD 2451545.0 (noon, 1 January 2000)
time <- jd - 51545.
# Ecliptic coordinates
# Mean longitude
mnlong <- 280.460 + .9856474 * time
mnlong <- mnlong %% 360
mnlong[mnlong < 0] <- mnlong[mnlong < 0] + 360
# Mean anomaly
mnanom <- 357.528 + .9856003 * time
mnanom <- mnanom %% 360
mnanom[mnanom < 0] <- mnanom[mnanom < 0] + 360
mnanom <- mnanom * deg2rad
# Ecliptic longitude and obliquity of ecliptic
eclong <- mnlong + 1.915 * sin(mnanom) + 0.020 * sin(2 * mnanom)
eclong <- eclong %% 360
eclong[eclong < 0] <- eclong[eclong < 0] + 360
oblqec <- 23.429 - 0.0000004 * time
eclong <- eclong * deg2rad
oblqec <- oblqec * deg2rad
# Celestial coordinates
# Right ascension and declination
num <- cos(oblqec) * sin(eclong)
den <- cos(eclong)
ra <- atan(num / den)
ra[den < 0] <- ra[den < 0] + pi
ra[den >= 0 & num < 0] <- ra[den >= 0 & num < 0] + twopi
dec <- asin(sin(oblqec) * sin(eclong))
# Local coordinates
# Greenwich mean sidereal time
gmst <- 6.697375 + .0657098242 * time + hour
gmst <- gmst %% 24
gmst[gmst < 0] <- gmst[gmst < 0] + 24.
# Local mean sidereal time
lmst <- gmst + long / 15.
lmst <- lmst %% 24.
lmst[lmst < 0] <- lmst[lmst < 0] + 24.
lmst <- lmst * 15. * deg2rad
# Hour angle
ha <- lmst - ra
ha[ha < -pi] <- ha[ha < -pi] + twopi
ha[ha > pi] <- ha[ha > pi] - twopi
# Latitude to radians
lat <- lat * deg2rad
# Azimuth and elevation
el <- asin(sin(dec) * sin(lat) + cos(dec) * cos(lat) * cos(ha))
az <- asin(-cos(dec) * sin(ha) / cos(el))
elc <- asin(sin(dec) / sin(lat))
az[el >= elc] <- pi - az[el >= elc]
az[el <= elc & ha > 0] <- az[el <= elc & ha > 0] + twopi
el <- el / deg2rad
az <- az / deg2rad
lat <- lat / deg2rad
return(list(elevation=el, azimuth=az))
}
我遇到的问题是它返回的方位角似乎是错误的.例如,如果我在12:00(夏令时)和40ºS,3ºS,3ºN和41ºN的位置(夏令时)运行该功能:
> sunPosition(2012,12,22,12,0,0,-41,0)
$elevation
[1] 72.42113
$azimuth
[1] 180.9211
> sunPosition(2012,12,22,12,0,0,-3,0)
$elevation
[1] 69.57493
$azimuth
[1] -0.79713
Warning message:
In asin(sin(dec)/sin(lat)) : NaNs produced
> sunPosition(2012,12,22,12,0,0,3,0)
$elevation
[1] 63.57538
$azimuth
[1] -0.6250971
Warning message:
In asin(sin(dec)/sin(lat)) : NaNs produced
> sunPosition(2012,12,22,12,0,0,41,0)
$elevation
[1] 25.57642
$azimuth
[1] 180.3084
这些数字似乎不对.我很满意的高度 - 前两个应该大致相同,第三个接触较低,第四个低得多.然而,第一个方位角大致应该是北方,而它给出的数字完全相反.其余三个应该大致指向南方,但只有最后一个.中间的两个位于北方,再次180º.
正如你所看到的那样,低纬度也会引发一些错误(关闭赤道)
我认为故障在本节中,错误在第三行触发(从开始elc).
# Azimuth and elevation
el <- asin(sin(dec) * sin(lat) + cos(dec) * cos(lat) * cos(ha))
az <- asin(-cos(dec) * sin(ha) / cos(el))
elc <- asin(sin(dec) / sin(lat))
az[el >= elc] <- pi - az[el >= elc]
az[el <= elc & ha > 0] <- az[el <= elc & ha > 0] + twopi
我用Google搜索并在C中找到了类似的代码块,转换为R,它用来计算方位角的行就像是
az <- atan(sin(ha) / (cos(ha) * sin(lat) - tan(dec) * cos(lat)))
这里的输出似乎朝着正确的方向前进,但是当它被转换回度数时,我无法让它给我正确的答案.
对代码进行修正(怀疑它只是上面的几行)以使其计算出正确的方位角将是非常棒的.
Jos*_*ien 107
这似乎是一个重要的主题,所以我发布了一个比典型答案更长的时间:如果这个算法将来被其他人使用,我认为重要的是它必须附带参考文献的来源. .
简短的回答
如您所知,您发布的代码无法在赤道附近或南半球的位置正常工作.
要修复它,只需在原始代码中替换这些行:
elc <- asin(sin(dec) / sin(lat))
az[el >= elc] <- pi - az[el >= elc]
az[el <= elc & ha > 0] <- az[el <= elc & ha > 0] + twopi
用这些:
cosAzPos <- (0 <= sin(dec) - sin(el) * sin(lat))
sinAzNeg <- (sin(az) < 0)
az[cosAzPos & sinAzNeg] <- az[cosAzPos & sinAzNeg] + twopi
az[!cosAzPos] <- pi - az[!cosAzPos]
它现在应该适用于地球上的任何位置.
讨论
你的例子中的代码几乎逐字地改编自JJ Michalsky的1988年文章(Solar Energy.40:227-235).该文章反过来改进了R. Walraven在1978年的一篇文章中提出的算法(Solar Energy.20:393-397).Walraven报道说,该方法已经成功使用了几年,以便在加利福尼亚州戴维斯(38°33'14"N,121°44'17"W)精确定位偏振辐射计.
Michalsky和Walraven的代码都包含重要/致命的错误.特别是,虽然Michalsky的算法在美国的大部分地区运作得很好,但是对于赤道附近或南半球的区域来说,它失败了(正如你所发现的那样).1989年,澳大利亚维多利亚州的JW Spencer注意到同样的事情(太阳能.42(4):353):
亲爱的先生:
Michalsky将计算出的方位角分配给正确象限的方法(源自Walraven)在应用于南方(负)纬度时没有给出正确的值.此外,临界高程(elc)的计算将因纬度为零而失败,因为除以零.通过考虑cos(方位角)的符号,可以简单地通过将方位角分配给正确的象限来避免这两个异议.
我对您的代码的编辑是基于Spencer在已发布的评论中提出的更正.我只是稍微修改它们以确保R函数sunPosition()保持"矢量化"(即在点位置的矢量上正常工作,而不是一次需要传递一个点).
功能的准确性 sunPosition()
为了测试它sunPosition()是否正常,我将其结果与国家海洋和大气管理局的太阳能计算器计算结果进行了比较.在这两种情况下,太阳位置都计算在2012年南部夏至(12月22日)的正午(中午12:00).所有结果均在0.02度范围内.
testPts <- data.frame(lat = c(-41,-3,3, 41),
long = c(0, 0, 0, 0))
# Sun's position as returned by the NOAA Solar Calculator,
NOAA <- data.frame(elevNOAA = c(72.44, 69.57, 63.57, 25.6),
azNOAA = c(359.09, 180.79, 180.62, 180.3))
# Sun's position as returned by sunPosition()
sunPos <- sunPosition(year = 2012,
month = 12,
day = 22,
hour = 12,
min = 0,
sec = 0,
lat = testPts$lat,
long = testPts$long)
cbind(testPts, NOAA, sunPos)
# lat long elevNOAA azNOAA elevation azimuth
# 1 -41 0 72.44 359.09 72.43112 359.0787
# 2 -3 0 69.57 180.79 69.56493 180.7965
# 3 3 0 63.57 180.62 63.56539 180.6247
# 4 41 0 25.60 180.30 25.56642 180.3083
代码中的其他错误
发布的代码中至少还有两个(非常小的)错误.第一个原因是闰年的2月29日和3月1日两者都计入了第61天.第二个误差来源于原始文章中的拼写错误,Michalsky在1989年的一篇文章(太阳能.43(5):323)中对其进行了修正.
此代码块显示违规行,注释掉并立即跟踪更正的版本:
# leapdays <- year %% 4 == 0 & (year %% 400 == 0 | year %% 100 != 0) & day >= 60
leapdays <- year %% 4 == 0 & (year %% 400 == 0 | year %% 100 != 0) &
day >= 60 & !(month==2 & day==60)
# oblqec <- 23.429 - 0.0000004 * time
oblqec <- 23.439 - 0.0000004 * time
更正版本 sunPosition()
以下是经过验证的更正代码:
sunPosition <- function(year, month, day, hour=12, min=0, sec=0,
lat=46.5, long=6.5) {
twopi <- 2 * pi
deg2rad <- pi / 180
# Get day of the year, e.g. Feb 1 = 32, Mar 1 = 61 on leap years
month.days <- c(0,31,28,31,30,31,30,31,31,30,31,30)
day <- day + cumsum(month.days)[month]
leapdays <- year %% 4 == 0 & (year %% 400 == 0 | year %% 100 != 0) &
day >= 60 & !(month==2 & day==60)
day[leapdays] <- day[leapdays] + 1
# Get Julian date - 2400000
hour <- hour + min / 60 + sec / 3600 # hour plus fraction
delta <- year - 1949
leap <- trunc(delta / 4) # former leapyears
jd <- 32916.5 + delta * 365 + leap + day + hour / 24
# The input to the Atronomer's almanach is the difference between
# the Julian date and JD 2451545.0 (noon, 1 January 2000)
time <- jd - 51545.
# Ecliptic coordinates
# Mean longitude
mnlong <- 280.460 + .9856474 * time
mnlong <- mnlong %% 360
mnlong[mnlong < 0] <- mnlong[mnlong < 0] + 360
# Mean anomaly
mnanom <- 357.528 + .9856003 * time
mnanom <- mnanom %% 360
mnanom[mnanom < 0] <- mnanom[mnanom < 0] + 360
mnanom <- mnanom * deg2rad
# Ecliptic longitude and obliquity of ecliptic
eclong <- mnlong + 1.915 * sin(mnanom) + 0.020 * sin(2 * mnanom)
eclong <- eclong %% 360
eclong[eclong < 0] <- eclong[eclong < 0] + 360
oblqec <- 23.439 - 0.0000004 * time
eclong <- eclong * deg2rad
oblqec <- oblqec * deg2rad
# Celestial coordinates
# Right ascension and declination
num <- cos(oblqec) * sin(eclong)
den <- cos(eclong)
ra <- atan(num / den)
ra[den < 0] <- ra[den < 0] + pi
ra[den >= 0 & num < 0] <- ra[den >= 0 & num < 0] + twopi
dec <- asin(sin(oblqec) * sin(eclong))
# Local coordinates
# Greenwich mean sidereal time
gmst <- 6.697375 + .0657098242 * time + hour
gmst <- gmst %% 24
gmst[gmst < 0] <- gmst[gmst < 0] + 24.
# Local mean sidereal time
lmst <- gmst + long / 15.
lmst <- lmst %% 24.
lmst[lmst < 0] <- lmst[lmst < 0] + 24.
lmst <- lmst * 15. * deg2rad
# Hour angle
ha <- lmst - ra
ha[ha < -pi] <- ha[ha < -pi] + twopi
ha[ha > pi] <- ha[ha > pi] - twopi
# Latitude to radians
lat <- lat * deg2rad
# Azimuth and elevation
el <- asin(sin(dec) * sin(lat) + cos(dec) * cos(lat) * cos(ha))
az <- asin(-cos(dec) * sin(ha) / cos(el))
# For logic and names, see Spencer, J.W. 1989. Solar Energy. 42(4):353
cosAzPos <- (0 <= sin(dec) - sin(el) * sin(lat))
sinAzNeg <- (sin(az) < 0)
az[cosAzPos & sinAzNeg] <- az[cosAzPos & sinAzNeg] + twopi
az[!cosAzPos] <- pi - az[!cosAzPos]
# if (0 < sin(dec) - sin(el) * sin(lat)) {
# if(sin(az) < 0) az <- az + twopi
# } else {
# az <- pi - az
# }
el <- el / deg2rad
az <- az / deg2rad
lat <- lat / deg2rad
return(list(elevation=el, azimuth=az))
}
参考文献:
Michalsky,JJ 1988.天文年历的近似太阳位置算法(1950-2050).太阳能.40(3):227-235.
Michalsky,JJ 1989.勘误表.太阳能.43(5):323.
Spencer,JW 1989.评论"天文年历的近似太阳位置算法(1950-2050)".太阳能.42(4):353.
Walraven,R.1978.计算太阳的位置.太阳能.20:393-397.
- 非常好的答案 (18认同)
mba*_*ask 19
使用上面某个链接中的"NOAA Solar Calculations",我通过使用一种略有不同的算法改变了函数的最后部分,我希望这种算法能够无误地翻译.我已经注释掉了现在没用的代码,并在纬度到弧度转换之后添加了新算法:
# -----------------------------------------------
# New code
# Solar zenith angle
zenithAngle <- acos(sin(lat) * sin(dec) + cos(lat) * cos(dec) * cos(ha))
# Solar azimuth
az <- acos(((sin(lat) * cos(zenithAngle)) - sin(dec)) / (cos(lat) * sin(zenithAngle)))
rm(zenithAngle)
# -----------------------------------------------
# Azimuth and elevation
el <- asin(sin(dec) * sin(lat) + cos(dec) * cos(lat) * cos(ha))
#az <- asin(-cos(dec) * sin(ha) / cos(el))
#elc <- asin(sin(dec) / sin(lat))
#az[el >= elc] <- pi - az[el >= elc]
#az[el <= elc & ha > 0] <- az[el <= elc & ha > 0] + twopi
el <- el / deg2rad
az <- az / deg2rad
lat <- lat / deg2rad
# -----------------------------------------------
# New code
if (ha > 0) az <- az + 180 else az <- 540 - az
az <- az %% 360
# -----------------------------------------------
return(list(elevation=el, azimuth=az))
为了验证你提到的四种情况中的方位角趋势,我们将其与一天中的时间进行对比:
hour <- seq(from = 0, to = 23, by = 0.5)
azimuth <- data.frame(hour = hour)
az41S <- apply(azimuth, 1, function(x) sunPosition(2012,12,22,x,0,0,-41,0)$azimuth)
az03S <- apply(azimuth, 1, function(x) sunPosition(2012,12,22,x,0,0,-03,0)$azimuth)
az03N <- apply(azimuth, 1, function(x) sunPosition(2012,12,22,x,0,0,03,0)$azimuth)
az41N <- apply(azimuth, 1, function(x) sunPosition(2012,12,22,x,0,0,41,0)$azimuth)
azimuth <- cbind(azimuth, az41S, az03S, az41N, az03N)
rm(az41S, az03S, az41N, az03N)
library(ggplot2)
azimuth.plot <- melt(data = azimuth, id.vars = "hour")
ggplot(aes(x = hour, y = value, color = variable), data = azimuth.plot) +
geom_line(size = 2) +
geom_vline(xintercept = 12) +
facet_wrap(~ variable)
附图:
Ric*_*ton 12
这是一个重写,它更加惯用于R,更容易调试和维护.这基本上是Josh的答案,但是使用Josh和Charlie的算法计算方位角进行比较.我还从其他答案中包含了对日期代码的简化.基本原则是将代码分成许多较小的函数,您可以更轻松地编写单元测试.
astronomersAlmanacTime <- function(x)
{
# Astronomer's almanach time is the number of
# days since (noon, 1 January 2000)
origin <- as.POSIXct("2000-01-01 12:00:00")
as.numeric(difftime(x, origin, units = "days"))
}
hourOfDay <- function(x)
{
x <- as.POSIXlt(x)
with(x, hour + min / 60 + sec / 3600)
}
degreesToRadians <- function(degrees)
{
degrees * pi / 180
}
radiansToDegrees <- function(radians)
{
radians * 180 / pi
}
meanLongitudeDegrees <- function(time)
{
(280.460 + 0.9856474 * time) %% 360
}
meanAnomalyRadians <- function(time)
{
degreesToRadians((357.528 + 0.9856003 * time) %% 360)
}
eclipticLongitudeRadians <- function(mnlong, mnanom)
{
degreesToRadians(
(mnlong + 1.915 * sin(mnanom) + 0.020 * sin(2 * mnanom)) %% 360
)
}
eclipticObliquityRadians <- function(time)
{
degreesToRadians(23.439 - 0.0000004 * time)
}
rightAscensionRadians <- function(oblqec, eclong)
{
num <- cos(oblqec) * sin(eclong)
den <- cos(eclong)
ra <- atan(num / den)
ra[den < 0] <- ra[den < 0] + pi
ra[den >= 0 & num < 0] <- ra[den >= 0 & num < 0] + 2 * pi
ra
}
rightDeclinationRadians <- function(oblqec, eclong)
{
asin(sin(oblqec) * sin(eclong))
}
greenwichMeanSiderealTimeHours <- function(time, hour)
{
(6.697375 + 0.0657098242 * time + hour) %% 24
}
localMeanSiderealTimeRadians <- function(gmst, long)
{
degreesToRadians(15 * ((gmst + long / 15) %% 24))
}
hourAngleRadians <- function(lmst, ra)
{
((lmst - ra + pi) %% (2 * pi)) - pi
}
elevationRadians <- function(lat, dec, ha)
{
asin(sin(dec) * sin(lat) + cos(dec) * cos(lat) * cos(ha))
}
solarAzimuthRadiansJosh <- function(lat, dec, ha, el)
{
az <- asin(-cos(dec) * sin(ha) / cos(el))
cosAzPos <- (0 <= sin(dec) - sin(el) * sin(lat))
sinAzNeg <- (sin(az) < 0)
az[cosAzPos & sinAzNeg] <- az[cosAzPos & sinAzNeg] + 2 * pi
az[!cosAzPos] <- pi - az[!cosAzPos]
az
}
solarAzimuthRadiansCharlie <- function(lat, dec, ha)
{
zenithAngle <- acos(sin(lat) * sin(dec) + cos(lat) * cos(dec) * cos(ha))
az <- acos((sin(lat) * cos(zenithAngle) - sin(dec)) / (cos(lat) * sin(zenithAngle)))
ifelse(ha > 0, az + pi, 3 * pi - az) %% (2 * pi)
}
sunPosition <- function(when = Sys.time(), format, lat = 46.5, long = 6.5)
{
if(is.character(when)) when <- strptime(when, format)
when <- lubridate::with_tz(when, "UTC")
time <- astronomersAlmanacTime(when)
hour <- hourOfDay(when)
# Ecliptic coordinates
mnlong <- meanLongitudeDegrees(time)
mnanom <- meanAnomalyRadians(time)
eclong <- eclipticLongitudeRadians(mnlong, mnanom)
oblqec <- eclipticObliquityRadians(time)
# Celestial coordinates
ra <- rightAscensionRadians(oblqec, eclong)
dec <- rightDeclinationRadians(oblqec, eclong)
# Local coordinates
gmst <- greenwichMeanSiderealTimeHours(time, hour)
lmst <- localMeanSiderealTimeRadians(gmst, long)
# Hour angle
ha <- hourAngleRadians(lmst, ra)
# Latitude to radians
lat <- degreesToRadians(lat)
# Azimuth and elevation
el <- elevationRadians(lat, dec, ha)
azJ <- solarAzimuthRadiansJosh(lat, dec, ha, el)
azC <- solarAzimuthRadiansCharlie(lat, dec, ha)
data.frame(
elevation = radiansToDegrees(el),
azimuthJ = radiansToDegrees(azJ),
azimuthC = radiansToDegrees(azC)
)
}
Ric*_*ton 10
这是Josh优秀答案的建议更新.
该函数的大部分开始是用于计算自2000年1月1日中午以来的天数的样板代码.使用R的现有日期和时间函数可以更好地处理.
我还认为,不是指定六个不同的变量来指定日期和时间,而是指定现有日期对象或日期字符串+格式字符串更容易(并且与其他R函数更一致).
这是两个辅助函数
astronomers_almanac_time <- function(x)
{
origin <- as.POSIXct("2000-01-01 12:00:00")
as.numeric(difftime(x, origin, units = "days"))
}
hour_of_day <- function(x)
{
x <- as.POSIXlt(x)
with(x, hour + min / 60 + sec / 3600)
}
现在,该功能的开始简化为
sunPosition <- function(when = Sys.time(), format, lat=46.5, long=6.5) {
twopi <- 2 * pi
deg2rad <- pi / 180
if(is.character(when)) when <- strptime(when, format)
time <- astronomers_almanac_time(when)
hour <- hour_of_day(when)
#...
另一个奇怪的是像
mnlong[mnlong < 0] <- mnlong[mnlong < 0] + 360
由于mnlong已经%%调用了它的值,它们都应该都是非负的,所以这条线是多余的.
- 漂亮的变化@Richie Cotton.请注意,赋值小时< - hour_of_day实际上应该是小时< - hour_of_day(when),并且该变量时间应该包含天数,而不是类"difftime"的对象.第二行函数astronomers_almanac_time应该改为as.numeric(difftime(x,origin,units ="days"),units ="days"). (2认同)
我需要在 Python 项目中定位太阳。我改编了 Josh O'Brien 的算法。
谢谢乔希。
如果它对任何人有用，这是我的改编。
请注意，我的项目只需要即时太阳位置，因此时间不是参数。
def sunPosition(lat=46.5, long=6.5):
# Latitude [rad]
lat_rad = math.radians(lat)
# Get Julian date - 2400000
day = time.gmtime().tm_yday
hour = time.gmtime().tm_hour + \
time.gmtime().tm_min/60.0 + \
time.gmtime().tm_sec/3600.0
delta = time.gmtime().tm_year - 1949
leap = delta / 4
jd = 32916.5 + delta * 365 + leap + day + hour / 24
# The input to the Atronomer's almanach is the difference between
# the Julian date and JD 2451545.0 (noon, 1 January 2000)
t = jd - 51545
# Ecliptic coordinates
# Mean longitude
mnlong_deg = (280.460 + .9856474 * t) % 360
# Mean anomaly
mnanom_rad = math.radians((357.528 + .9856003 * t) % 360)
# Ecliptic longitude and obliquity of ecliptic
eclong = math.radians((mnlong_deg +
1.915 * math.sin(mnanom_rad) +
0.020 * math.sin(2 * mnanom_rad)
) % 360)
oblqec_rad = math.radians(23.439 - 0.0000004 * t)
# Celestial coordinates
# Right ascension and declination
num = math.cos(oblqec_rad) * math.sin(eclong)
den = math.cos(eclong)
ra_rad = math.atan(num / den)
if den < 0:
ra_rad = ra_rad + math.pi
elif num < 0:
ra_rad = ra_rad + 2 * math.pi
dec_rad = math.asin(math.sin(oblqec_rad) * math.sin(eclong))
# Local coordinates
# Greenwich mean sidereal time
gmst = (6.697375 + .0657098242 * t + hour) % 24
# Local mean sidereal time
lmst = (gmst + long / 15) % 24
lmst_rad = math.radians(15 * lmst)
# Hour angle (rad)
ha_rad = (lmst_rad - ra_rad) % (2 * math.pi)
# Elevation
el_rad = math.asin(
math.sin(dec_rad) * math.sin(lat_rad) + \
math.cos(dec_rad) * math.cos(lat_rad) * math.cos(ha_rad))
# Azimuth
az_rad = math.asin(
- math.cos(dec_rad) * math.sin(ha_rad) / math.cos(el_rad))
if (math.sin(dec_rad) - math.sin(el_rad) * math.sin(lat_rad) < 0):
az_rad = math.pi - az_rad
elif (math.sin(az_rad) < 0):
az_rad += 2 * math.pi
return el_rad, az_rad