Position of the sun given time of day, latitude and longitude

Question

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.

Code below is in R. I\'ve

Answer 1

This is a suggested update to Josh's excellent answer.

Much of the start of the function is boilerplate code for calculating the number of days since midday on 1st Jan 2000. This is much better dealt with using R's existing date and time function.

I also think that rather than having six different variables to specify the date and time, it's easier (and more consistent with other R functions) to specify an existing date object or a date strings + format strings.

Here are two helper functions

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)
}

And the start of the function now simplifies to

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)
  #...

---

The other oddity is in the lines like

mnlong[mnlong < 0] <- mnlong[mnlong < 0] + 360

Since mnlong has had %% called on its values, they should all be non-negative already, so this line is superfluous.