<h2>What can you tell from a picture of some stars?</h2>

<p><strong>At the very least, you need a recognisable asterism</strong></p>

<p>If the exposure is too low or the light pollution too high to identify unambiguously an asterism of three or more stars, you're simply not going to be able to tell much from the photo.</p>

<p><strong>An asterism on its own can only tell you where the picture wasn't shot</strong>.</p>

<p>If a star in a picture has a declination of <span class="math-container">$60^o$</span>, you can rule out all shooting locations south of <span class="math-container">$-30^o$</span> latitude. A star like Polaris can only really be seen in the northern hemisphere, while Sirius can be seen everywhere south of Svalbard.</p>

<p><strong>An asterism plus the horizon is sufficient for determining latitude</strong></p>

<p>If the edge of land in the picture is sufficiently close to being the actual horizon (and not a range of mountains) and it is present alongside a recognisable asterism in the photo, you can calculate the latitude. This is true for any magnification, as you can use the angular distance between stars (a known quantity) to measure the angle to the horizon. I can't think of any special locations where you wouldn't be able to determine the latitude from the stars + horizon.</p>

<p>I haven't really done the math for how <em>accurately</em> one would be able to determine the latitude, but here are some considerations: There are two sources of error, the position of the stars and the position of the horizon. The star image may spread out over several pixels, and if the picture was shot through a standard camera and not a telescope, there would be an error, at a wild guess, of around ten arcminutes. The horizon can appear lower than <span class="math-container">$0^o$</span> depending on your distance above ground, but the <a href="https://en.wikipedia.org/wiki/Horizon" rel="noreferrer">effect is small</a>. Looking out from Mt Everest, the horizon will appear a twentieth of a degree lower down than it ought to. I believe the horizon's exact position can also be obscured through atmospheric refraction.</p>

<p>All in all, the error in the angle between an identified star and the horizon equals the error in latitude of your identified position. For a 10 arcminute error, that corresponds to ten arcminutes of latitude, which is about 18 kilometres. </p>

<p><strong>For longitude, you need global, not local time</strong></p>

<p>If you only have local time, the photo could have been taken anywhere along that latitude. If you have some form of global time (UTC, GMT, whatever, it doesn't matter which as long as you know what it is) you can figure out longitude from there.</p>

CC BY-SA 4.0
2019-08-27T20:17:18.727Z
FWIW, there has been military R&D along these lines with a hope of creating partial backups for military systems if GPS fails.
ohwilleke
CC BY-SA 4.0
2019-08-27T01:37:36.867Z
2019-08-27T01:37:36.867Z
Ingolifs
<p>To my knowledge, this question can be related to celestial navigation on one hand and to attitude (or orientation) determination from stars direction on the other hand. However the combination of this two methods does not solve the problem, so I am wondering if there is a different approach.</p>

<p>In celestial navigation, the altitude of the stars (elevation wrt to the horizon) at a given time can be used to determine the observer's position. The limit of this method is that the horizon must be visible, which is not necessarly the case in a photo of the night sky.</p>

<p>An image of the night sky can also provide information about the attitude (orientation) of the camera wrt to an inertial frame. The measured direction of a few stars (from their position on the image) and their known direction in an inertial frame give an estimate of the attitude. This method is used in devices called star trackers onboard satellites for instance.</p>

<p>The combination of this 2 approaches fails to solve the problem because the information about the horizon is still missing. Indeed, the orientation is found with respect to the inertial frame, not the local frame, and thus the horizon is still unknown.</p>

CC BY-SA 4.0
2019-08-26T13:27:54.550Z
This is a *really cool question!* For clarification, I'm guessing by "time of the shot" you mean something like [UTC](https://en.wikipedia.org/wiki/Coordinated_Universal_Time), and a position accuracy of say a few kilometers would be okay, and a clear horizon such as an ocean or some plains known to be flat?
uhoh
CC BY-SA 4.0
2019-08-26T18:47:23.143Z
Yep - you need some kind of reference in the "other axis,"  since a narrow FOV image could be pointed to zenith at  1AM , then rotated and tilted to get the same image at 4 AM.  The "trick" here is that the planets & moon will have moved infinitesimally over those 3 hours, so with a high - enough angular accuracy and precision, you could retrieve the time of the image but not the location.
Carl Witthoft
CC BY-SA 4.0
2019-08-27T02:45:35.990Z
@CarlWitthoft If we are allowed to have the Moon in the image, won't it have a different apparent position relative to  stars depending on our latitude?
Keith McClary
CC BY-SA 4.0
2019-08-27T11:41:30.373Z
@KeithMcClary  sounds right -- dunno about precision.
Carl Witthoft
CC BY-SA 4.0
2019-10-17T20:06:08.657Z
@uhoh yes that is it :) However the horizon is not always visible on a picture, which is why it is necessary to somehow estimate the orientation of the camera with respect to the local horizontal plane.
LaTilz
CC BY-SA 4.0
2024-06-15T13:43:14.867Z
Instead of using the obscured horizon, a practical alternative in cities is the zenith, which can be found using tall buildings (which all intersect at a [vanishing point](https://en.wikipedia.org/wiki/Vanishing_point)). Furthermore, applying the [pinhole camera model](https://en.wikipedia.org/wiki/Pinhole_camera_model), one can find the distance from a photo to its virtual image plane and coordinate of its principal point. This allows one to make reasonably accurate angular measurements of stars in spite of lens distortions.
比尔盖子
CC BY-SA 4.0
2024-06-15T14:39:08.830Z
In late 2023, a Flat Earther in China posted a photo of the sky and challenged others. Soon their longitude and latitude was determined by an amateur astronomer within a few kilometers using the principles described above, and that post subsequently became viral. So, at least in some cases, celestial navigation using a *single timestamped photo without horizon* is possible... But it does require some (unidentified) city landmarks to establish the zenith with respect to ground, a pure sky won't work. I plan to see whether I can replicate it and write an answer if I succeed at it.
比尔盖子
<star><navigation>
Is it possible to determine an observer's position on Earth from a photo of the night sky and the time of the shot?
LaTilz
# Is it possible to determine an observer's position on Earth from a photo of the night sky and the time of the shot?

(Asked by: username_0 on 2019-08-26)

To my knowledge, this question can be related to celestial navigation on one hand and to attitude (or orientation) determination from stars direction on the other hand. However the combination of this two methods does not solve the problem, so I am wondering if there is a different approach.

In celestial navigation, the altitude of the stars (elevation wrt to the horizon) at a given time can be used to determine the observer's position. The limit of this method is that the horizon must be visible, which is not necessarly the case in a photo of the night sky.

An image of the night sky can also provide information about the attitude (orientation) of the camera wrt to an inertial frame. The measured direction of a few stars (from their position on the image) and their known direction in an inertial frame give an estimate of the attitude. This method is used in devices called star trackers onboard satellites for instance.

The combination of this 2 approaches fails to solve the problem because the information about the horizon is still missing. Indeed, the orientation is found with respect to the inertial frame, not the local frame, and thus the horizon is still unknown.

**Tags:** star, navigation

> **Comments:**
> **username_1**: This is a *really cool question!* For clarification, I'm guessing by "time of the shot" you mean something like [UTC](https://en.wikipedia.org/wiki/Coordinated_Universal_Time), and a position accuracy of say a few kilometers would be okay, and a clear horizon such as an ocean or some plains known to be flat?  
> **username_2**: Yep - you need some kind of reference in the "other axis,"  since a narrow FOV image could be pointed to zenith at  1AM , then rotated and tilted to get the same image at 4 AM.  The "trick" here is that the planets & moon will have moved infinitesimally over those 3 hours, so with a high - enough angular accuracy and precision, you could retrieve the time of the image but not the location.  
> **username_3**: @username_2 If we are allowed to have the Moon in the image, won't it have a different apparent position relative to  stars depending on our latitude?  
> **username_2**: @username_3  sounds right -- dunno about precision.  
> **username_0**: @username_1 yes that is it :) However the horizon is not always visible on a picture, which is why it is necessary to somehow estimate the orientation of the camera with respect to the local horizontal plane.  
> **username_4**: Instead of using the obscured horizon, a practical alternative in cities is the zenith, which can be found using tall buildings (which all intersect at a [vanishing point](https://en.wikipedia.org/wiki/Vanishing_point)). Furthermore, applying the [pinhole camera model](https://en.wikipedia.org/wiki/Pinhole_camera_model), one can find the distance from a photo to its virtual image plane and coordinate of its principal point. This allows one to make reasonably accurate angular measurements of stars in spite of lens distortions.  
> **username_4**: In late 2023, a Flat Earther in China posted a photo of the sky and challenged others. Soon their longitude and latitude was determined by an amateur astronomer within a few kilometers using the principles described above, and that post subsequently became viral. So, at least in some cases, celestial navigation using a *single timestamped photo without horizon* is possible... But it does require some (unidentified) city landmarks to establish the zenith with respect to ground, a pure sky won't work. I plan to see whether I can replicate it and write an answer if I succeed at it.  

---

## 1 answers

### Answer by username_5 on 2019-08-27. Score: 9

## What can you tell from a picture of some stars?

**At the very least, you need a recognisable asterism**

If the exposure is too low or the light pollution too high to identify unambiguously an asterism of three or more stars, you're simply not going to be able to tell much from the photo.

**An asterism on its own can only tell you where the picture wasn't shot**.

If a star in a picture has a declination of $60^o$, you can rule out all shooting locations south of $-30^o$ latitude. A star like Polaris can only really be seen in the northern hemisphere, while Sirius can be seen everywhere south of Svalbard.

**An asterism plus the horizon is sufficient for determining latitude**

If the edge of land in the picture is sufficiently close to being the actual horizon (and not a range of mountains) and it is present alongside a recognisable asterism in the photo, you can calculate the latitude. This is true for any magnification, as you can use the angular distance between stars (a known quantity) to measure the angle to the horizon. I can't think of any special locations where you wouldn't be able to determine the latitude from the stars + horizon.

I haven't really done the math for how *accurately* one would be able to determine the latitude, but here are some considerations: There are two sources of error, the position of the stars and the position of the horizon. The star image may spread out over several pixels, and if the picture was shot through a standard camera and not a telescope, there would be an error, at a wild guess, of around ten arcminutes. The horizon can appear lower than $0^o$ depending on your distance above ground, but the [effect is small](https://en.wikipedia.org/wiki/Horizon). Looking out from Mt Everest, the horizon will appear a twentieth of a degree lower down than it ought to. I believe the horizon's exact position can also be obscured through atmospheric refraction.

All in all, the error in the angle between an identified star and the horizon equals the error in latitude of your identified position. For a 10 arcminute error, that corresponds to ten arcminutes of latitude, which is about 18 kilometres.

**For longitude, you need global, not local time**

If you only have local time, the photo could have been taken anywhere along that latitude. If you have some form of global time (UTC, GMT, whatever, it doesn't matter which as long as you know what it is) you can figure out longitude from there.

> **Comments:**
> **username_6**: FWIW, there has been military R&D along these lines with a hope of creating partial backups for military systems if GPS fails.  

---

