## Thursday, October 18, 2007

### Approximate sunrise and sunset times

This function is not perfect, but it does a reasonable job estimating sunrise and sunset times for my field site. If more accurate data are required, try here. Note, the command to calculate day of year is: strptime(x, "%m/%d/%Y")\$yday+1
``````suncalc<-function(d,Lat=48.1442,Long=-122.7551){
## d is the day of year
## Lat is latitude in decimal degrees
## Long is longitude in decimal degrees (negative == West)

##This method is copied from:
##Teets, D.A. 2003. Predicting sunrise and sunset times.
##  The College Mathematics Journal 34(4):317-321.

## At the default location the estimates of sunrise and sunset are within
## seven minutes of the correct times (http://aa.usno.navy.mil/data/docs/RS_OneYear.php)
## with a mean of 2.4 minutes error.

## Function to convert degrees to radians

##Radius of the earth (km)
R=6378

##Radians between the xy-plane and the ecliptic plane

##Convert observer's latitude to radians

## Calculate offset of sunrise based on longitude (min)
## If Long is negative, then the mod represents degrees West of
## a standard time meridian, so timing of sunrise and sunset should
## be made later.
timezone = -4*(abs(Long)%%15)*sign(Long)

## The earth's mean distance from the sun (km)
r = 149598000

theta = 2*pi/365.25*(d-80)

z.s = r*sin(theta)*sin(epsilon)
r.p = sqrt(r^2-z.s^2)

t0 = 1440/(2*pi)*acos((R-z.s*sin(L))/(r.p*cos(L)))

##a kludge adjustment for the radius of the sun
that = t0+5

## Adjust "noon" for the fact that the earth's orbit is not circular:
n = 720-10*sin(4*pi*(d-80)/365.25)+8*sin(2*pi*d/365.25)

## now sunrise and sunset are:
sunrise = (n-that+timezone)/60
sunset = (n+that+timezone)/60

return(list("sunrise" = sunrise,"sunset" = sunset))
}

``````

Anonymous said...

thanks for providing this function, and all these tips!

Anne Ghisla

John (aka Bruno) said...

I'm using this algorithm for a variety of cities and I get NaN errors for cities which are very close to the North Pole:

city=Eureka, Nunavut, country=Canada, latitude=79.9833, longtude=-85.95
city=Nord, country=Greenland, latitude=81.6, longtude=-16.6667

For instance, if I try to calculate for Alert for today (the 52nd day of the year), I have these values in R:

> z.s
[1] -27578819
> r.p
[1] 147033909
> L
[1] 1.395972
> acos(R-z.s*sin(L))/(r.p*cos(L))
[1] NaN
Warning message:
In acos(R - z.s * sin(L)) : NaNs produced
> R-z.s*sin(L)
[1] 27164815
> acos(R-z.s*sin(L))
[1] NaN
Warning message:
In acos(R - z.s * sin(L)) : NaNs produced
>

Forester said...

@John I believe that is because the sun is down all day for those latitudes this time of the year. There is twilight, but no sunrise/set.