Techniques > Variations of Existing Techniques

Sunrise, Sunset and Solar Azimuth

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captain paranoia:
Some time ago, whilst looking at solar relief shading, I thought about where the sun's azimuth, and decided that the sun never appears above the W-E meridian in the UK.  Of course, that's completely wrong, as I realised when looking out of the bathroom window early one morning, to see the sun rising well to the north of east...

So, I started to think a little more clearly, and looked at the maths for calculating sunrise and sunset times.  Whilst searching to see if I'd got it right (I hadn't...), I found the following NOAA website, which has a nice integration with GoogleMaps, and will calculate sunrise, sunset & local solar noon times,  and draw sunrise/sunset and current solar azimuth lines over GoogleMaps, calculating for any specified time and date.  It's nice to check that the solar azimuth shown for the local solar noon is actually due south...

NOAA Solar Calculator

MoonMan:
The answer is found in Spherical Trigonometry:

sine Amplitude equals sine Declination times secant Latitude.
{Amplitude: the complement of the azimuth}.

Likewise, for the hours of daylight,
cosine Diurnal Arc equals -tangent Latitude times tangent Declination
Hours of daylight either side of Noon, & the complement of the hours from Sunset to midnight.

captain paranoia:
Thanks for that.  I'll have to go and convert the astronomical terms into things I understand...

However, the simple maths isn't really the problem, I think, when it comes to trying to replicate the NOAA website.  The problems are the approximations made in the simple maths.

Here's my simple maths:

Draw a circle, representing the Earth (approximation 1: the Earth is a perfect sphere).
Draw a vertical line through the centre, representing the suns' shadow (approximation 2: the Sun is at infinity).
Now draw two angled lines through the centre of the circle, showing the inclined Equator and polar axis.
Now draw a radial line from the centre, subtending an angle equal to the latitude of interest from the inclined equatorial line.
Now project a line, parallel to the equatorial line, from the point at which the radial line meets the circle.  This line will cross both the vertical, solar shadow line, and the inclined polar axis.

The point at which the parallel line crosses the inclined polar axis is r.sin(latitude) from the centre of the circle.
Drawing a right-angled triangle with points at the centre of the circle, and the intersections of the parallel line with the axial and shadow lines, the distance, d, from the intersection of the axial line to the intersection with the shadow line is r.sin(latitude).tan(inclination), using the tangent function, tan(theta)=opposite/adjacent, so opposite=adjacent.tan(theta)

Now draw the latitude circle, with radius r.cos(latitude), and draw a vertical line at distance d from the centre.  Where the line crosses the circle is the point on the horizon where the sun rises.
Draw a radial line from the centre to this point.  We know that the length of this radial line is r.cos(latitude), the latitude radius.
We can calculate the angle at which the sun rises using the simple trigonometry function: cos(angle)=adjacent/hypotenuse, where the adjacent is the distance d, and the hypotenuse is the radius.  So:

cos(angle) = r.sin(latitude).tan(inclination)/r.cos(latitude)
thus, angle = acos(r.sin(latitude).tan(inclination)/r.cos(latitude))
angle = acos(tan(latitude).tan(inclination))

We can find the inclination angle for a given day:

inclination=obliquity*sin(days_from_equinox/days_in_solar_year)

Thus, we can find the sunrise and sunset angles.

Which brings us to the third approximation, which is that the Earth hasn't moved around the sun between sunrise and sunset.

The simple maths gets us to within a few minutes of the NOAA calculations.  If we removed the approximations, we ought to get the same result as the NOAA...

captain paranoia:
Oh, and approximation 4: the sun has zero size, so subtends an angle of zero degrees.  With a real sun, sunrise (the first appearance of the sun's disk above the horizon) occurs earlier, and sunset later.

captain paranoia:
> cosine Diurnal Arc equals -tangent Latitude times tangent Declination

i.e.

cos(Diurnal Arc) = -tan(Latitude).tan(Declination)

thus, Diurnal Arc = acos(-tan(Latitude).tan(Declination))

> angle = acos(tan(latitude).tan(inclination))

Barring terminology and sign (which will depend on the sense of Declination/inclination), then, it looks like my 'simple maths' tallies with your definition.  In which case, neither matches the NOAA model...

I've attached a PDF of my drawing, showing the side view of a sunlit Earth, and the latitude circle with sunrise/sunset

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