Calculate Sun Position in Lat/Lon from Date/time

Posted on 2008-06-20
Medium Priority
Last Modified: 2010-05-18
I am developing an application that automates the 3D Globe control provided by ESRI.

ESRI use this globe themselves in there ArcGlobe application, and within this app it is possible to specify a date/time and plot the sun against the globe.

I wish to replicate this functionality in my application, however ESRI do not appear to have exposed this functionality in the API.

As far as I am aware it is only possible to set the sun position directly as a Lat/Lon

What I need is either :
a) a pointer to the ESRI API to get Sun Lat/Lon from Date/Time


b) code to derive the Sun Lat/Lon from Date/Time.

Note: I do not wish the Sun Azimuth/Altitute I have found these calculations many times. I wont the Lat/Lon position that the sun is directly above. As shown on this site : http://www.fourmilab.ch/cgi-bin/Earth

Thanks in Advance

Question by:Gordonp
  • 2
LVL 53

Accepted Solution

Infinity08 earned 2000 total points
ID: 21829728
I've quickly written the following code. I assume that's what you intended ? The precision is relatively good (using Spencer's algorithm), but could be better with more accurate formula's (like Michalsky's almanac algorithm for example, from his "The Astronomical Almanac's algorithm for approximate solar position" paper).
#include <stdio.h>
#include <time.h>
#include <math.h>
#define PI 3.14159265358979323846
double solar_declination(struct tm *t) {
  /* day angle : */
  double J = 1.0 + ((double) t->tm_yday);
  double tau_d = 2.0 * PI * (J - 1.0) / 365.0;
  /* solar declination */
  double delta_s = 0.006981 - 0.399912 * cos(      tau_d) + 0.070257 * sin(      tau_d)
                            - 0.006758 * cos(2.0 * tau_d) + 0.000907 * sin(2.0 * tau_d)
                            - 0.002697 * cos(3.0 * tau_d) + 0.001480 * sin(3.0 * tau_d);
  return delta_s;
int main(void) {
  time_t now = time(0);
  struct tm *now_tm = gmtime(&now);
  double hour = now_tm->tm_hour + (now_tm->tm_min / 60.0) + (now_tm->tm_sec / 3600.0);
  double latitude = solar_declination(now_tm);
  double longitude = PI - (2.0 * PI * hour / 24.0);
  fprintf(stdout, "latitude  : %10.5f radians (%10.5f degrees)\n", latitude, latitude * 180.0 / PI);
  fprintf(stdout, "longitude : %10.5f radians (%10.5f degrees)\n", longitude, longitude * 180.0 / PI);
  return 0;

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Author Comment

ID: 21829780
This appears to be exactly what I'm after. Just need to check it out, and I'll be back to Accept if formally.

Author Closing Comment

ID: 31469075
Thanks for your help

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