Web search found how to calculate shadow length and others but not my specific question. I have a photo. I found location on a map (so I know directions to objects: north/west/... and latitude). There is a shadow visible with clear direction, I want to know time of day as correctly as plausible from the photo.

I've found out for https://en.wikipedia.org/wiki/Gnomon the Sun's shadow "travel" by 15 degrees each hour (and to North during actual midday in Northern hemisphere, but later understood Gnomon has to be parallel to Earth's axis (is it correct BTW?).

I think there is already written formula to use shadow on Earth, not on parallel to axis sundial. TIA

On this site there is similar QA: Is it possible to tell the time of day by shadows on the photo. Well, I'm almost sure it is possible (that QA's answers do not explain how; there is link to some app but I want simpler thing - a formula), it is as I see it some geometry to convert to plane perpendicular to Earth's axis. Just hoping formula is known to somebody here already, if not maybe worth asking on Math...


Now I realize shadow on equatorial subdial indeed might be different for same shadow on the ground. I know date approximately (in the comments it is mentioned that by date one can find azimuth). Alternatively and more generally, I also noticed length of the shadow is about same as height of vertical object - I can get azimuth from that, can't I?

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    $\begingroup$ You really need at least one more piece of information, otherwise there are too many solutions. You could solve the Alt/Az equations at the link below for the hour angle, but you still need either the Alt of the sun, or the declination of the Sun to produce an unique solution. celestialprogramming.com/convert_ra_dec_to_alt_az.html $\endgroup$ Sep 15 at 2:02
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    $\begingroup$ To answer your question about sundials, note there are a lot of different types of sundials. On an equatorial sundial (gnomon parallel to Earth's axis, plate parallel to the plane of the equator), you do get 15deg per hour. A horizontal sundial (gnomon parallel to Earth's axis, plate level to the ground), that does not hold true anymore. It isn't possible to make a sundial with a vertical gnomon and a horizontal plate. $\endgroup$ Sep 15 at 2:14
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    $\begingroup$ @GregMiller, I've written I know latitude. So I think knowing it one can convert one sundial's angles to the other. If not, why? $\endgroup$ Sep 15 at 7:19
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    $\begingroup$ I saw that, you still need either the Sun's declination or it's Alt. If you look at the equations mentioned in my first comment, you'll see they are needed. $\endgroup$ Sep 15 at 12:00
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    $\begingroup$ What are you trying to do? If you also know the date you can use the azimuth of the Sun to find the time. What are your constraints? You indicate you don't want to use an app, but you must know that you will need to do trig lookups with any formula. $\endgroup$
    – stretch
    Sep 15 at 12:53

1 Answer 1


The edge of the shadow does give you the azimuth of the Sun. Also, there is only one time of day when the Sun will be at that azimuth for your location on a given date. So you certainly can find time of day with azimuth of the Sun and the date as the starting points.

How to do it depends on how much accuracy you need. Over the course of a year the position of the Sun at a given time of day varies. You could try search terms analemma, equation of time, and declination of the Sun for explanations. There are convenient formlas on Wikipedia for good approximations to the equation of time and declination. There are other sources with more accuracy than you need considering how much accuracy you'll get from a shadow.

There must be an easier way, but successive approximation would work. Start by calculating the Greenwich hour angle (GHA) of the Sun at your best guess time. The Sun passes the 180 degree meridian at 0000 UTC corrected for the equation of time. It moves westward at 15 degrees per hour. So, for example, after 17 hours and 20 minutes it would be somewhere over the east coast of the United states, and its GHA would be 80 degrees. Knowing the Sun's GHA and declination you know all that's needed to solve the spherical triangle with vertices at the North Pole, the geographical position (GP: the lat/long of the spot on the Earth the Sun's directly over) of the Sun, and the lat/long of the position where the picture of the shadow was taken. The angle at the pole is the difference between GHA of the Sun and the observer's west longitude. The known sides are 90 degrees minus the observer's latitude, and 90 degrees minus the Sun's declination. Using the laws of sines and cosines for spherical triangles you can determine what the azimuth of the Sun would have been at your best guess time. If it's smaller or larger than the azimuth in the photograph, try again with a guess that's earlier or later.


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