Geographical Point of the Sun

Question

I want to draw a Sun in a 3d scene, so I figured the easiest way would be to calculate the latitude and longitude of the point perpendicular to Sun position and then just move it "up".

As a starting point, I want to calculate it's equatorial coordinates: declination and right ascension.

I followed instructions from this page, also cross-referencing with this one

Here's my code (in Swift):

import Foundation

class Astronomy {

  static func julianDays(since: Date) -> Double {
    return since.timeIntervalSince1970 / 86400 + 2440587.5
  }

  static func daysSinceJ2000(for date: Date) -> Double {
    return julianDays(since: date) - 2451545
  }

  static func sunAxialTilt(for daysSinceJ200: Double) -> Double {
    return 23.4393 - 3.563e-7 * daysSinceJ200
  }

  static func sunEquatorialPosition(on date: Date) -> (declination: Double, RA: Double, distance: Double) {
    let daysSinceGreenwichNoon = daysSinceJ2000(for: date)
    var meanLongitude = 280.460 + 0.9856474*daysSinceGreenwichNoon
    var meanAnomaly = 357.528 + 0.9856003*daysSinceGreenwichNoon
    meanLongitude = meanLongitude.rangedUpTo360
    meanAnomaly = meanAnomaly.rangedUpTo360
    let eclipticLongitude = meanLongitude + 1.915*sin(meanAnomaly.radians) + 0.020*sin(2*meanAnomaly.radians)

    let distanceToEarth = 1.00014 - 0.01671*cos(meanAnomaly.radians) - 0.00014*cos(2*meanAnomaly.radians)

    let axialTilt = sunAxialTilt(for: daysSinceGreenwichNoon)
    let rightAscension = atan2(cos(axialTilt.radians)*sin(eclipticLongitude.radians), cos(eclipticLongitude.radians))
    let declination = asin(sin(axialTilt.radians)*sin(eclipticLongitude.radians))

    return (declination: declination.degrees, RA: rightAscension.degrees, distance: distanceToEarth)
  }
}

extension Double {
  var radians: Double {
    return self * .pi / 180
  }

  var degrees: Double {
    return self  * 180 / .pi
  }

  var rangedUpTo360: Double {
    let multiples = (self/360).rounded(.down)
    return self - 360*multiples
  }
}

However, it does not seem to be working correctly. As of the time I write this article, the result is declination -16.98772292697088, RA -44.80674952684117. Compared with https://theskylive.com/sun-info the declination is about right (-17° 03’ 38”), but right ascension is way off: 20h 59m 43s, which is 314.93 degrees according to this site

What went wrong?

Edit: As it turned out nothing was wrong, I can get to the second part: calculating the geographical point. (the title suggests it after all)

AFAIK for latitude, I can use declination as is and for longitude, I need to calculate Greenwich Hour Angle(GHA)

After examining several articles, I found these equations:

SHA[°] = 360 – 15 · RA[h] 
GHAAries[°] = 15 · GST[h] 
GHA = SHA + GHAAries 

For my case, I suppose I may use RA as is, without multiplying its hour component by 15. What keeps me a bit puzzled is GST:

1. May I use Greenwich Mean Sidereal Time instead if I don't need the calculations to be super precise?
2. Is there any "GST for dummies"? All the calculations I found rely on some floating second offset, I found this table but have no idea how to use it.

Answer 1

If we draw a circle and draw a 315 degree angle on that circle, in the counter clockwise direction, we would roughly get the same point as when we draw an angle of 44.8 degrees in the opposite direction (the equivalent of drawing -44.8 in the same direction) because 315+45=360. There therefore does not seem to be an error in your results... Visually the two angles are the same. As for the slight inaccuracy, it may be due to floating point precision errors or measurement errors.
If you want positive right ascension, all you need to do is add 360 if it is negative. Before doing that (this is general advice, it may not be important to your code), you ought to subtract the decimal part, get the remainder on dividing by 360, and re-add the decimal part.
Edit: The difference between GST and GMST is called the equation of time (for Grinich), and it does not grow beyond 16 minutes, so I think you can use it for estimates, but then there would be no point of rounding your results to the tenth decimal or so. I honestly don't know if there is a "GST for dummies" (normally I wouldn't post an incomplete answer, but the question was edited), but using GMST should suffice if the 16 minute error margin is acceptable.

Answer 2

For any who might be looking for the same sort of thing, my own solution to my question does roughly what this question is asking, though the calculations are shown in TypeScript.