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Sunset time calculation

 
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carvell



Joined: 20 Sep 2016
Posts: 4

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Sunset time calculation
PostPosted: Tue Sep 20, 2016 1:42 pm     Reply with quote

There was a sunrise/sunset time calculator on this forum, but it was intertwined with a lot of other functions and was fairly huge, so I implemented the following.

I followed this method, implementing it step by step:
http://williams.best.vwh.net/sunrise_sunset_algorithm.htm

Confirmed it against a variety of sources.

Parameters:
int8 offset - Sunset/sunrise times are calculated in UTC. The offset given here will be applied to the UTC result before returning a value.
int1 riseorset - 1 to calculate rise time, 0 to calculate set time
int16 day - Day of the month, e.g. 31.
int16 month - Month, e.g. 12.
int16 year - Year, e.g. 2016.
float lat - Latitude position, e.g. 51.5074 for London.
float lon - Longitude position, e.g. -0.1278 for London.

The time is returned in decimal format. I.e., for 18:30 it will return 18.5, for 09:45 it'll return 9.75, etc.

Code:

#pragma device ANSI //this is required in order that variables default to signed instead of unsigned

#include <math.h>

#define PI 3.141592654
#define ZENITH 90.833333333

float calc_suntime(int8 offset, int1 riseorset, int16 day, int16 month, int16 year, float lat, float lon)
{
   //offset:       offset to apply from UTC (in hours)
   //riseorset:   1 for rise time, 0 for set time

   float lngHour, t, M, L, RA, Lquadrant, RAquadrant, sinDec;
   float cosDec, cosH, H, T, UT, localT;
   int16 N1, N2, N3, N;
   
   N1 = floor(275 * month / 9);
   N2 = floor((month + 9) / 12);
   N3 = (1 + floor((year - 4 * floor(year / 4) + 2) / 3));
   N = N1 - (N2 * N3) + day - 30;
   
   lngHour = lon / 15;
   
   if (riseorset)
   {
      //sunrise time
      t = N + ((6 - lngHour) / 24);
   }
   else
   {
      //sunset time
      t = N + ((18 - lngHour) / 24);
   }
   
   M = (0.9856 * t) - 3.289;
   
   L = M + (1.916 * sin((PI/180) * M)) + (0.020 * sin((PI/180) * 2 * M)) + 282.634;
   //need to adjust L to be in range (0,360)
   while ((L > 360) || (L < 0))
   {
      if (L > 360) L -= 360;
      if (L < 0) L += 360;
   }
   
   RA = (180/PI) * atan(0.91764 * tan((PI/180) * L));
   //need to adjust RA to be in range (0,360)
   while ((RA > 360) || (RA < 0))
   {
      if (RA > 360) RA -= 360;
      if (RA < 0) RA += 360;
   }
   
   Lquadrant  = (floor( L/90)) * 90;
   RAquadrant = (floor(RA/90)) * 90;
   RA = RA + (Lquadrant - RAquadrant);
   
   RA = RA / 15;
   
   sinDec = 0.39782 * sin((PI/180) * L);
   cosDec = cos((PI/180) * ((180/PI) * asin(sinDec)));
   
   cosH = (cos((PI/180) * ZENITH) - (sinDec * sin((PI/180) * lat))) / (cosDec * cos((PI/180) * lat));
   
   /*
   if (cosH >  1)
     the sun never rises on this location (on the specified date)
   if (cosH < -1)
     the sun never sets on this location (on the specified date)
   */
   
   if (riseorset)
   {
      //sunrise time
      H = 360 - ((180/PI) * acos(cosH));
   }
   else
   {
      //sunset time
      H = (180/PI) * acos(cosH);
   }
   
   H = H / 15;
   
   T = H + RA - (0.06571 * t) - 6.622;
   
   UT = T - lngHour;
   while ((UT > 24) || (UT < 0))
   {
      if (UT > 360) UT -= 24;
      if (UT < 0) UT += 24;
   }
   
   localT = UT + offset;

   return localT;
}
Ttelmah



Joined: 11 Mar 2010
Posts: 13054

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PostPosted: Mon Oct 10, 2016 1:44 am     Reply with quote

One comment.

You need to think in terms of using a smaller unit for 'offset'. Quite a lot of countries/places use 0.5 hour timezones. (Adelaide in Australia, India, and Newfoundland for example).
So I'd suggest changing 'offset' to be in integer 0.5hour steps, then the offset line just becomes:

localT = UT + (offset*0.5);

Which can then handle these locations. Smile

Best Wishes
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