给定一天中的时间、纬度和经度的太阳位置 [英] Position of the sun given time of day, latitude and longitude
问题描述
已提出此问题 在三年多以前.给出了答案，但是我发现解决方案中有一个小问题.
This question has been asked before a little over three years ago. There was an answer given, however I've found a glitch in the solution.
下面的代码是用 R 编写的.我已将其移植到另一种语言，但已直接在 R 中测试了原始代码，以确保问题与我的移植无关.
Code below is in R. I've ported it to another language, however have tested the original code directly in R to ensure the issue wasn't with my porting.
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(南)夏至 12:00 对位置 0ºE 和 41ºS、3ºS、3ºN 和 41ºN 运行该函数:
The problem I'm hitting is that the azimuth it returns seems wrong. For example, if I run the function on the (southern) summer solstice at 12:00 for locations 0ºE and 41ºS, 3ºS, 3ºN and 41ºN:
> 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
这些数字似乎不正确.我对海拔高度感到满意——前两个应该大致相同，第三个稍微低一点，第四个要低得多.然而，第一个方位角应该大致是正北，而它给出的数字是完全相反的.其余三个应该大致指向正南，但只有最后一个指向.中间的两个就在北边，再次向外 180 度.
These numbers just don't seem right. The elevation I'm happy with - the first two should be roughly the same, the third a touch lower, and the fourth much lower. However the first azimuth should be roughly due North, whereas the number it gives is the complete opposite. The remaining three should point roughly due South, however only the last one does. The two in the middle point just off North, again 180º out.
如您所见，低纬度地区(靠近赤道)也触发了一些错误
As you can see there are also a couple of errors triggered with the low latitudes (close the equator)
我认为错误在本节，错误在第三行触发(以 elc 开头).
I believe the fault is in this section, with the error being triggered at the third line (starting with 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
我搜索了一下，发现了一段类似的 C 代码，转换为 R 用于计算方位角的代码类似于
I googled around and found a similar chunk of code in C, converted to R the line it uses to calculate the azimuth would be something like
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，但是当它转换回度数时，我无法让它一直给我正确的答案.
The output here seems to be heading in the right direction, but I just can't get it to give me the right answer all the time when it's converted back to degrees.
更正代码(怀疑只是上面的几行)以使其计算出正确的方位角会很棒.
A correction of the code (suspect it's just the few lines above) to make it calculate the correct azimuth would be fantastic.
推荐答案
这似乎是一个重要的话题，所以我发布了一个比典型答案更长的答案:如果这个算法将来被其他人使用，我想重要的是要附上参考文献.
This seems like an important topic, so I've posted a longer than typical answer: if this algorithm is to be used by others in the future, I think it's important that it be accompanied by references to the literature from which it has been derived.
正如您所指出的，您发布的代码不适用于赤道附近或南半球的位置.
As you've noted, your posted code does not work properly for locations near the equator, or in the southern hemisphere.
要修复它，只需在原始代码中替换这些行:
To fix it, simply replace these lines in your original code:
elc <- asin(sin(dec) / sin(lat))
az[el >= elc] <- pi - az[el >= elc]
az[el <= elc & ha > 0] <- az[el <= elc & ha > 0] + twopi
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.
It should now work for any location on the globe.
您示例中的代码几乎逐字改编自 J.J. 1988 年的一篇文章.Michalsky(太阳能.40:227-235).那篇文章反过来改进了 R. Walraven 1978 年文章中提出的算法(太阳能.20:393-397).Walraven 报告说，该方法已成功使用数年，用于在加利福尼亚州戴维斯市(38° 33' 14" N，121° 44' 17" W)精确定位偏振辐射计.
The code in your example is adapted almost verbatim from a 1988 article by J.J. Michalsky (Solar Energy. 40:227-235). That article in turn refined an algorithm presented in a 1978 article by R. Walraven (Solar Energy. 20:393-397). Walraven reported that the method had been used successfully for several years to precisely position a polarizing radiometer in Davis, CA (38° 33' 14" N, 121° 44' 17" W).
Michalsky 和 Walraven 的代码都包含重要/致命错误. 特别是，虽然 Michalsky 的算法在美国大部分地区运行良好，但在靠近赤道，或在南半球.1989 年，J.W.澳大利亚维多利亚的斯宾塞也指出了同样的事情(Solar Energy. 42(4):353):
Both Michalsky's and Walraven's code contains important/fatal errors. In particular, while Michalsky's algorithm works just fine in most of the United States, it fails (as you've found) for areas near the equator, or in the southern hemisphere. In 1989, J.W. Spencer of Victoria, Australia, noted the same thing (Solar Energy. 42(4):353):
尊敬的先生:
Michalsky 将计算出的方位角分配给正确象限的方法源自 Walraven，但在应用于南(负)纬度时不会给出正确的值.此外，由于除以零，临界海拔 (elc) 的计算对于零纬度将失败.只需考虑 cos(azimuth) 的符号，将方位角分配到正确的象限，就可以避免这两种反对意见.
Michalsky's method for assigning the calculated azimuth to the correct quadrant, derived from Walraven, does not give correct values when applied for Southern (negative) latitudes. Further the calculation of the critical elevation (elc) will fail for a latitude of zero because of division by zero. Both these objections can be avoided simply by assigning the azimuth to the correct quadrant by considering the sign of cos(azimuth).
我对您的代码所做的修改是基于 Spencer 在发表的评论中建议的更正.我只是稍微改变了它们以确保 R 函数 sunPosition() 保持矢量化"(即在点位置的矢量上正常工作，而不是需要一次通过一个点).
My edits to your code are based on the corrections suggested by Spencer in that published Comment. I have simply altered them somewhat to ensure that the R function sunPosition() remains 'vectorized' (i.e. working properly on vectors of point locations, rather than needing to be passed one point at a time).
为了测试 sunPosition() 是否正常工作，我将其结果与美国国家海洋和大气管理局 太阳能计算器.在这两种情况下，太阳位置都是在 2012 年南夏至(12 月 22 日)中午(下午 12:00)计算的.所有结果均在 0.02 度以内.
To test that sunPosition() works correctly, I've compared its results with those calculated by the National Oceanic and Atmospheric Administration's Solar Calculator. In both cases, sun positions were calculated for midday (12:00 PM) on the southern summer solstice (December 22nd), 2012. All results were in agreement to within 0.02 degrees.
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其他(相当轻微的)错误.第一个导致闰年的 2 月 29 日和 3 月 1 日都被记录为一年中的第 61 天.第二个错误源于原始文章中的拼写错误，Michalsky 在 1989 年的注释(Solar Energy. 43(5):323)中对其进行了更正.
Other errors in the code
There are at least two other (quite minor) errors in the posted code. The first causes February 29th and March 1st of leap years to both be tallied as day 61 of the year. The second error derives from a typo in the original article, which was corrected by Michalsky in a 1989 note (Solar Energy. 43(5):323).
这个代码块显示了有问题的行，被注释掉并紧跟更正的版本:
This code block shows the offending lines, commented out and followed immediately by corrected versions:
# 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
sunPosition()
的更正版本这是上面验证的更正代码:
Corrected version of sunPosition()
Here is the corrected code that was verified above:
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))
}
参考资料:
米哈尔斯基，J.J.1988. 天文年历的近似太阳位置算法(1950-2050).太阳能.40(3):227-235.
References:
Michalsky, J.J. 1988. The Astronomical Almanac's algorithm for approximate solar position (1950-2050). Solar Energy. 40(3):227-235.
米哈尔斯基，J.J.1989. 勘误表.太阳能.43(5):323.
Michalsky, J.J. 1989. Errata. Solar Energy. 43(5):323.
斯宾塞，J.W.1989.评论天文历书的近似太阳位置算法(1950-2050)".太阳能.42(4):353.
Spencer, J.W. 1989. Comments on "The Astronomical Almanac's Algorithm for Approximate Solar Position (1950-2050)". Solar Energy. 42(4):353.
Walraven, R. 1978.计算太阳的位置.太阳能.20:393-397.
Walraven, R. 1978. Calculating the position of the sun. Solar Energy. 20:393-397.
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