Generating one's oen trig values would be extremely tedious, It essentially
involves solving a Pythagorean triangle, and it is trial and error!!
How computers generate trig values.
Copilot:
Trigonometry; what is the strategy used by calcula | Shared Copilot Chat
The role of the arccos function is to capture how long the sun is at
work on a particular day. One needs to appreciate the notion of azimuth
on the plane on which the observer stands.
Gemini:
The arccos term measures the solar hour angle at the exact moment of
sunrise or sunset.
Been working through the actual math of using a formula on a sphere
to solve a daylight hours problem. Using Paris, France as the example.
Paris sits at Latitude 48.9 degrees N and Longitude 2.2 degrees E.
https://share.gemini.google/6GfkuFKVc0r9
1.146 x 0.434 = 0.497
I want to calculate arccos(0.497) - (arccos is the angle for a given arc length) - but my
Windows machine will only do this if I am working in rads...
(There are 2 pi rads in 180 degrees.)
arccos(0.497) is 1.05 rads.
pi - 1.5 is 2.09 rads
Or more directly, I run arccos on -0.497:
(minus because the tilt is giving me more hours!... while the numbers run counter clockwise on the sphere!))
To get hours, I divide 24 hours by pi rads.
I then multiply by 2.09, which gives 15.9 hours.
I am free to add more minutes, as per the Gemini solution!
I was looking for somw insight into yesterday US elections from La Presse.
Stumbled on this fascinating piece about a deal between Hydro-Quebec and the Mohawk
people...
I was asked to explain the math in the daylight hours equation. Google
produced interesting visuals for me:
Could I see both tanϕ and tanδ on the same globe rendering... - Google Search
Latitude
Gemini ai on the Copilot equation:
https://share.gemini.google/5qZ0lWWl60ru
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Washington daylight hours from the Microsoft Weather App:
With respect of Montreal and Edmonton longitude calculations...
Copilot:
Does this formula assume the Earth is a sphere?
Yes. The standard formula I showed assumes the Earth is a perfect sphere.
In reality, the Earth is slightly flattened at the poles and bulges at the equator, making it an oblate spheroid. The equatorial radius is about 21 km larger than the polar radius.
For sunrise and sunset calculations, however, the spherical assumption works remarkably well because:
The Earth's radius is about 6,371 km.
The flattening is only about 0.3%.
The resulting error in sunrise or sunset times is typically much smaller than other effects, such as atmospheric refraction.
In fact, the largest source of uncertainty is often the atmosphere. Variations in temperature and pressure can shift the apparent sunrise or sunset by a minute or more because they cange how strongly light bends near the horizon.
Modern astronomical software therefore includes corrections for:
Earth's actual shape,
atmospheric refraction,
observer elevation,
the Sun's finite size,
and the precise Earth-Sun geometry on a given date.
An interesting consequence is that if you stand on a hill or in a tall building, you can see the Sun for longer than someone at sea level because your horizon is farther away. From the top of a 100 m hill, sunrise occurs about a minute earlier and sunset about a minute later than at sea level.
For a place like St-Jean-sur-Richelieu (about 45°N), treating the Earth as a sphere gives sunrise and sunset times that are extremely close to the values published by weather services. The atmosphere usually matters more than the Earth's slight flattening.
In a sense, the spherical model is to astronomers what a straight line is to engineers: not perfectly true, but often accurate enough to get very useful answers.
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Quick review of Latitude and Longitude:
