Sunrise and sunset at different elevations

[bjarkeandersen]

I'm trying to take the elevation of a coordinate into consideration when calculating the sunrise and sunset. It's quite simple - you just add elevation as parameter to the creation of the observer object:

var observer_8850 = new Observer(27.987, 86.925, 8850);
var observer_0 = new Observer(27.987, 86.925, 0);
var astroTime = new AstroTime(new DateTime(2025,4,23).ToUniversalTime());

// 23:41:14
var nextEvent_8850 = Astronomy.SearchRiseSet(Body.Sun, observer_8850, Direction.Rise, astroTime, 1);

// 23:39:36
var nextEvent_0 = Astronomy.SearchRiseSet(Body.Sun, observer_0, Direction.Rise, astroTime, 1);

Here I try to calculate the sunrise of the 23rd of april 2025 just around where Mount Everest is located. I do this for 8850 meters and 0 meters. I would expect that the sun should rise earlier at 8850 meters, but it seams that the opposite is happening? Could somebody please explain this behaviour to me. I also read a couple of places on the internet, that the sun should rise around 1 minute earlier for every 1000 meters you go upwards, but that doesn't seem to be the case when I look at the timestamps I get from the SearchRiseSet-function.

[cosinekitty]

In general, sunrises will occur earlier at higher elevations, but only when surrounding terrain makes the horizon altitude angle negative, that is, lower than the observer. This is not always the case, for example on a plateau.

Try taking a look at the more generic function SearchAltitude. You will have to make a couple of corrections: subtract 0.25 degrees for the angular radius of the Sun, and also subtract whatever angle below horizontal to drop the horizon due to elevation. This depends on surrounding terrain at the observation site.

[bjarkeandersen]

I found out that I forgot to send the metersAboveGround parameter to the SearchRiseSet function (it defaults to 0). Doing that the rise time at 8850 meters is 23:24:18, which is more what I would expect.