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At high altitudes, the sun disappears earlier and appears later than at sea level. This question asks for how does light differs in such case but does not address the when.

How can I calculate the time at which the sun will appear and disappear based a date, location and altitude?

Low altitude do not make much difference but as I'll be between 3000m and 4500m above sea-level started next week, there may be a significant offset. The linked question mentions about one hour but does not mention at which altitude.

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    \$\begingroup\$ I'd guess it would depend on your latitude too; further north or south, the sun rises and sets at a shallow angle, meaning it would stay visible for longer a higher altitudes, but at the equator where it effectively moves vertically, it would disappear much more quickly. \$\endgroup\$
    – user456
    Dec 5, 2012 at 16:40
  • \$\begingroup\$ Yes, that is why I mentioned location. \$\endgroup\$
    – Itai
    Dec 5, 2012 at 16:48
  • \$\begingroup\$ Sorry, just thinking aloud, or whatever the equivalent is. Hopefully my answer is more helpful :) \$\endgroup\$
    – user456
    Dec 5, 2012 at 16:56
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    \$\begingroup\$ You have it backwards. The sun rises earlier and sets later at higher altitude, although this is a rather small effect relative to the length of day. \$\endgroup\$
    – Olin Lathrop
    Dec 5, 2012 at 17:02
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    \$\begingroup\$ @damned: No, sunrise will be earlier and sunset later as you go higher up above the same point on the ground. Think of a extreme case where you are in space way above the earth so that it apparent disk is the same size as the sun's. In that case full night is only a instance. Further up and you never have full night at all. \$\endgroup\$
    – Olin Lathrop
    Dec 6, 2012 at 15:00

4 Answers 4

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The Photographer's Ephemeris is a great software package to get started with these sorts of calculations. There is a free desktop version that you can use at home before your trip, and if it turns out to be really helpful, there are paid versions available for iOS and Android.

This tool lets you mark a spot on a map and then calculate sunrise, sunset, moonrise, and moonset for any day you choose. The vectors of these events are shown as colored rays on the map, and the times and moon-phase are shown, as well. As the question indicates, sunrise and sunset calculations depend, among other things, on elevation, and TPE can help with this. The mapping tools are aware of elevation, and based on various entries in the release notes, it appears that altitude is accounted for in its calculations.

Per some of the comments on this and others' answers, though, another factor is at work in practical instances -- the obscured horizon. In the simplest case, you'll have an unobscured view of the horizon:

Flat Horizon

In practice, however, very few of the really interesting landscapes feature perfectly flat horizons, so we'd ideally like to take into account the elevation and distance of those obstructions, as well:

enter image description here

Although TPE (to the best of my knowledge) won't account for all of the objects that might obscure the horizon for you, it does have some features that might help to an extent. There's an ability to set a secondary location on the map and compute the distance, bearing, apparent altitude change, and elevation change to that location. So long as you know that this is where your apparent horizon really is, you can plug this elevation back into TPE, which will calculate its effect on events.

enter image description here

As you can see, there's a bit of a knack to really getting the most out of TPE, but there are some great tutorials that walk you through scenarios like this.

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  • \$\begingroup\$ Looks good. Does it work without being on location? Meaning can I calculate today from where I am now, the sunset and sunrise times for next week in a different location? \$\endgroup\$
    – Itai
    Dec 5, 2012 at 17:35
  • \$\begingroup\$ The desktop version is free, and you can navigate to where you're going to be -- that's the purpose of the program, in fact. Sort of a neat tool; I should use it more often, in fact. \$\endgroup\$
    – D. Lambert
    Dec 5, 2012 at 17:59
  • \$\begingroup\$ After using this I understand better the complexity of the problem. The horizon is not always strait and rarely so at high altitude, so there is no way to simply adjust the times for a location. While in theory a program could compute it, it is too computationally intensive and there this application lets it as a manual calculation to the user. This explained in their Tutorial #4. \$\endgroup\$
    – Itai
    Dec 6, 2012 at 4:18
  • \$\begingroup\$ Note that the altitude taken into account is for atmospheric effects and not topology. In order to account for altitude one has to find the horizon elevation which can be done in this software by trial and error. \$\endgroup\$
    – Itai
    Dec 6, 2012 at 4:19
  • \$\begingroup\$ Since you found the software, I am inclined to accept this answer but I would like it to contain a detailed explanation. Could you? Or, if that is OK, I'll edit in what I understand now. \$\endgroup\$
    – Itai
    Dec 6, 2012 at 4:22
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At the equator, you would get 1 minute extra sun at either end of the day per 1.5km of altitude, according to this page.

Using trigonometry, for every degree north or south you travel, the extra time the sun would stay above the horizon (per 1.5km altitude) would be (1/cos (latitude)) * 1 minute per 1.5km, giving the following values:

  • 10° : 1.02 min = 1 min 1s
  • 20° : 1.06 min = 1 min 4s
  • 30° : 1.15 min = 1 min 9s
  • 40° : 1.31 min = 1 min 19s
  • 50° : 1.56 min = 1 min 35s
  • 60° : 2 min
  • 70° : 2.92 min = 2 min 55s
  • 80° : 5.76 min = 5 min 46s
  • 90° : infinity (i.e. never sets) - effectively at the poles

Note these values are approximate, as they do not take into account the tilt of the earth's axis, and the relationship between extra daylight and altitude approximates to a linear one given the small values relative to the Earth's circumference, but should be enough to give you a rough idea, if you cross-reference with a site like http://gaisma.com which gives sunrise and sunset for various locations.

EDIT: note the accuracy of these figures falls off the closer you get to the poles; but I'm assuming you won't be too far from the equator for this to make much difference.

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  • \$\begingroup\$ Interesting. That is the opposite described in the question I linked to. I'd guess a difference in assumptions. The article you reference talks about being on a plane while the question linked is about being on elevated ground, which will be my case too. \$\endgroup\$
    – Itai
    Dec 5, 2012 at 17:32
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    \$\begingroup\$ Well whether you're flying or climbing a mountain, the effect should be the same. The difference in the question you linked to is related to the presence of mountains, which would happen regardless of altitude, but if you define sunset to be the point where the sun passes the sea-level horizon, then it's purely a function of altitude (and latitude). There is a small village in north west Iceland (Siglufjörður) which has seen the last of the sun until the new year. This is because, even though the village is at sea-level, the nearby mountains cast a shadow. \$\endgroup\$
    – user456
    Dec 5, 2012 at 17:38
  • \$\begingroup\$ Without flying, at high altitude one is generally on a mountain! I am interested in when I see the sun appear and disappear (as said in the bold question in my question), not when it will go beyond or out of a point I cannot see. \$\endgroup\$
    – Itai
    Dec 5, 2012 at 17:53
  • \$\begingroup\$ Well then I'd have to say there's not enough information, as it also depends on the specific local geography of your location. If there are high mountains to your east, for example, then the sunrise will be later, but by how much depends on the height distribution of the mountains. You'd need a global database of terrain to calculate this precisely. \$\endgroup\$
    – user456
    Dec 5, 2012 at 17:56
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You can also use a free tool that I've developed with a colleague. It computes the actual sunrise and sunset times for any location worldwide, accounting for terrain. The example in the image is for Chamonix in France. Go to suncurves.com to find your own location. Hope you like it! I'm using it for all my outdoor shoots.

The sun curves for Chamonix, France

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The sun sets/rises later on a higher point than that point directly under it. The perfect ideal example is a person watching Sunset in the sea on a cliff while the other is down the cliff , the exact time difference can be calculated by analysing the triangle r , r+h where h is the height and r is Earth radius

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  • \$\begingroup\$ The sun rise would indeed be (a bit) later, but it should set (a bit) earlier. \$\endgroup\$
    – user20509
    Jun 19, 2013 at 23:28

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