<p>The method is generally reliable between 9 am and 3 pm.</p>
<p>The other consideration is when one is using the method in either of the tropics: Cancer or Capricorn.</p>
<p>It is generally accepted that in the northern hemisphere, the sun is located in the southern part of the sky and the opposite in the southern hemisphere. This applies to latitudes greater than the latitudes of the Tropics. Between the equator and the Tropics, depending on the time of year, the position of the sun in the sky alternates between north and south.</p>

CC BY-SA 4.0
2020-10-21T23:02:12.510Z
Is it possible to support "The method is generally reliable between 9 am and 3 pm" somehow? Is it also possible to address the actual question more directly? "...what further correction could they apply to estimate the north/south direction more accurately?"
uhoh
CC BY-SA 4.0
2020-10-20T21:03:39.273Z
2020-10-20T21:03:39.273Z
Fred
<p>I wrote a test program to compute the deviation between the wristwatch estimation and the real sun position, based on a <a href="https://gist.github.com/dabasler/298cd9da1c4c7675e9c7a91073e23009" rel="nofollow noreferrer">program by dabasler</a>.</p>
<p>The results quality depends on latitude, month and time of day.</p>
<ul>
<li>Between latitudes 45 south and 45 north, the results are barely usable (can be off by 45 degrees or more).</li>
<li>Typically, you get the worst results around 09:00 or 15:00.</li>
<li>In winter, you get better results than in summer, with the best ones in between, around February and November.</li>
</ul>
<p>I've assembled a table for the northern hemisphere.</p>
<p>How to read: e.g. the cell for latitude 55N in June reads <code>19°/15</code>, meaning that on June 1st, at 15:00 you have the worst results for that day, being 19 degrees off the true north.</p>
<pre><code>      JAN    FEB    MAR    APR    MAY    JUN    JUL    AUG    SEP    OCT    NOV    DEC
25N: 12°/17 10°/10 19°/10 30°/10 47°/14 66°/13 70°/11 53°/11 34°/10 22°/14 13°/14 12°/07
30N: 12°/17  9°/17 15°/10 25°/10 39°/14 51°/14 53°/10 44°/10 28°/14 18°/14 11°/14 12°/07
35N:  6°/16  9°/17 12°/10 20°/10 32°/14 42°/14 44°/10 36°/10 22°/09 15°/14  9°/14 12°/07
40N:  7°/16  9°/17 10°/10 16°/09 26°/15 34°/14 36°/10 30°/10 18°/09 12°/14  8°/07  7°/08
45N:  7°/16  9°/17  8°/10 13°/09 22°/15 28°/15 30°/09 25°/09 15°/15 10°/14  8°/07  8°/08
50N:  7°/16  6°/16  6°/10 11°/09 18°/15 23°/15 25°/09 21°/09 12°/15  8°/14  8°/07  8°/08
55N:  5°/15  6°/16  5°/10  9°/09 14°/15 19°/15 20°/09 17°/09  9°/15  6°/14  6°/08  6°/09
60N:  5°/15  7°/16  4°/10  7°/09 11°/15 15°/16 16°/08 14°/09  7°/15  5°/14  6°/08  6°/09
65N:  1°/13  5°/15  4°/17  5°/09  9°/16 12°/16 13°/08 11°/08  5°/15  4°/14  6°/08  4°/10
70N:  -----  4°/14  4°/17  3°/09  7°/16  9°/16 10°/08  9°/08  4°/16  3°/14  5°/09  -----
75N:  -----  -----  3°/16  2°/08  5°/16  6°/17  7°/07  6°/08  2°/16  3°/14  4°/11  -----
80N:  -----  -----  3°/15  2°/08  3°/17  4°/17  5°/07  4°/07  1°/16  2°/07  -----  -----
85N:  -----  -----  -----  1°/08  2°/17  2°/17  3°/07  3°/07  0°/17  2°/08  -----  -----
</code></pre>
<p>(The entries given as dashes mean that on these days, the sun isn't visible at all.)</p>
<p>Results for the southern hemisphere are similar, just shift the months by half a year.</p>
<p>Acknowledgment: Thanks to the people who made the computation for sun azimuth and elevation freely available.</p>

CC BY-SA 4.0
2020-10-24T20:42:22.813Z
2020-10-24T20:42:22.813Z
Ralf Kleberhoff
<p>As discussed in a 2015 <a href="/q/11719">question here</a>
and shown in a 2013 <a href="https://www.primermagazine.com/2013/learn/spy-school-how-to-use-an-analog-watch-as-a-compass-to-find-direction" rel="nofollow noreferrer">illustration in Primer Magazine</a>,
it's possible to estimate the north/south direction on the ground using a correlation between the hour hand of an analog watch and the azimuth of the Sun.</p>
<p><img src="https://i.sstatic.net/wbghT.png" alt="Illustration &quot;How to use an analog watch...to find direction&quot;" /></p>
<p>As <a href="https://en.wikipedia.org/wiki/Analog_watch" rel="nofollow noreferrer">Wikipedia:Analog watch</a> suggests, the relation between hour angle and azimuth depends on latitude:</p>
<blockquote>
<p>The method functions less well as one gets closer to the equator.</p>
</blockquote>
<p>Also the Sun's declination affects its azimuth around 6 AM or PM.
Assuming that the observer has already determined the local apparent solar time,
what further correction could they apply to estimate the north/south direction more accurately?
I expect that this would be a function of the observer's latitude <em>φ</em>, the Sun's declination <em>δ</em>, and the Sun's hour angle <em>h</em>.</p>

CC BY-SA 4.0
2020-10-21T13:34:51.027Z
Is this just for an instantaneous measurement? How about using a matchstick (or similar) as a gnomon?  If the observer has a few hours to spare, and can make a couple of measurements of the shadow of a vertical post, before & after noon, they can determine the meridian fairly accurately. That doesn't even need a watch.
PM 2Ring
CC BY-SA 4.0
2020-10-21T15:13:09.530Z
@PM2Ring I want to quantify the error in this particular method.
Mike G
CC BY-SA 4.0
2020-10-22T08:42:16.587Z
The method doesn't state how to orient the wristwatch: flat to the earth surface (meaning that the hour hand points to the horizon) or pointing up into the sky so the hour hand points upwards to the sun. And that makes a big difference.
Ralf Kleberhoff
CC BY-SA 4.0
2020-10-22T11:20:40.730Z
@RalfKleberhoff "use...as a compass" implies that the dial is parallel to the horizon.
Mike G
CC BY-SA 4.0
2020-10-22T11:34:36.440Z
The method (with a watch held flat) is exact if the sun is moving parallel to the horizon, which is true only for north and south pole. Closer to the equator, the method's result deviates from the correct direction. The method assumes that the sun azimuth changes linearly from 090 (sunrise, east) at 06:00 over 180 (noon, south) at 12:00 to 270 (sunset, west). E.g. suncalc.org can compute azimuth/elevation for a given location and time, allowing for (manual) comparison between the linear simplification and the real thing.
Ralf Kleberhoff
CC BY-SA 4.0
2020-10-22T11:35:57.710Z
Haven't got enough free time right now, otherwise I'd write up a program to calculate max. deviation for a range of situations.
Ralf Kleberhoff
<navigation><azimuth>
Refinement to sun/wristwatch estimate of N/S direction
Mike G
# Refinement to sun/wristwatch estimate of N/S direction

(Asked by: username_0 on 2020-10-20)

As discussed in a 2015 [question here](/q/11719)
and shown in a 2013 [illustration in Primer Magazine](https://www.primermagazine.com/2013/learn/spy-school-how-to-use-an-analog-watch-as-a-compass-to-find-direction),
it's possible to estimate the north/south direction on the ground using a correlation between the hour hand of an analog watch and the azimuth of the Sun.

![Illustration "How to use an analog watch...to find direction"](https://i.sstatic.net/wbghT.png)

As [Wikipedia:Analog watch](https://en.wikipedia.org/wiki/Analog_watch) suggests, the relation between hour angle and azimuth depends on latitude:

> The method functions less well as one gets closer to the equator.

Also the Sun's declination affects its azimuth around 6 AM or PM.
Assuming that the observer has already determined the local apparent solar time,
what further correction could they apply to estimate the north/south direction more accurately?
I expect that this would be a function of the observer's latitude *φ*, the Sun's declination *δ*, and the Sun's hour angle *h*.

**Tags:** navigation, azimuth

> **Comments:**
> **username_1**: Is this just for an instantaneous measurement? How about using a matchstick (or similar) as a gnomon?  If the observer has a few hours to spare, and can make a couple of measurements of the shadow of a vertical post, before & after noon, they can determine the meridian fairly accurately. That doesn't even need a watch.  
> **username_0**: @username_1 I want to quantify the error in this particular method.  
> **username_2**: The method doesn't state how to orient the wristwatch: flat to the earth surface (meaning that the hour hand points to the horizon) or pointing up into the sky so the hour hand points upwards to the sun. And that makes a big difference.  
> **username_0**: @username_2 "use...as a compass" implies that the dial is parallel to the horizon.  
> **username_2**: The method (with a watch held flat) is exact if the sun is moving parallel to the horizon, which is true only for north and south pole. Closer to the equator, the method's result deviates from the correct direction. The method assumes that the sun azimuth changes linearly from 090 (sunrise, east) at 06:00 over 180 (noon, south) at 12:00 to 270 (sunset, west). E.g. suncalc.org can compute azimuth/elevation for a given location and time, allowing for (manual) comparison between the linear simplification and the real thing.  
> **username_2**: Haven't got enough free time right now, otherwise I'd write up a program to calculate max. deviation for a range of situations.  

---

## 2 answers

### Answer by username_3 on 2020-10-20. Score: 1

The method is generally reliable between 9 am and 3 pm.

The other consideration is when one is using the method in either of the tropics: Cancer or Capricorn.

It is generally accepted that in the northern hemisphere, the sun is located in the southern part of the sky and the opposite in the southern hemisphere. This applies to latitudes greater than the latitudes of the Tropics. Between the equator and the Tropics, depending on the time of year, the position of the sun in the sky alternates between north and south.

> **Comments:**
> **username_4**: Is it possible to support "The method is generally reliable between 9 am and 3 pm" somehow? Is it also possible to address the actual question more directly? "...what further correction could they apply to estimate the north/south direction more accurately?"  

---

### Answer by username_2 on 2020-10-24. Score: 1

I wrote a test program to compute the deviation between the wristwatch estimation and the real sun position, based on a [program by dabasler](https://gist.github.com/dabasler/298cd9da1c4c7675e9c7a91073e23009).

The results quality depends on latitude, month and time of day.

* Between latitudes 45 south and 45 north, the results are barely usable (can be off by 45 degrees or more).
* Typically, you get the worst results around 09:00 or 15:00.
* In winter, you get better results than in summer, with the best ones in between, around February and November.

I've assembled a table for the northern hemisphere.

How to read: e.g. the cell for latitude 55N in June reads `19°/15`, meaning that on June 1st, at 15:00 you have the worst results for that day, being 19 degrees off the true north.

```
      JAN    FEB    MAR    APR    MAY    JUN    JUL    AUG    SEP    OCT    NOV    DEC
25N: 12°/17 10°/10 19°/10 30°/10 47°/14 66°/13 70°/11 53°/11 34°/10 22°/14 13°/14 12°/07
30N: 12°/17  9°/17 15°/10 25°/10 39°/14 51°/14 53°/10 44°/10 28°/14 18°/14 11°/14 12°/07
35N:  6°/16  9°/17 12°/10 20°/10 32°/14 42°/14 44°/10 36°/10 22°/09 15°/14  9°/14 12°/07
40N:  7°/16  9°/17 10°/10 16°/09 26°/15 34°/14 36°/10 30°/10 18°/09 12°/14  8°/07  7°/08
45N:  7°/16  9°/17  8°/10 13°/09 22°/15 28°/15 30°/09 25°/09 15°/15 10°/14  8°/07  8°/08
50N:  7°/16  6°/16  6°/10 11°/09 18°/15 23°/15 25°/09 21°/09 12°/15  8°/14  8°/07  8°/08
55N:  5°/15  6°/16  5°/10  9°/09 14°/15 19°/15 20°/09 17°/09  9°/15  6°/14  6°/08  6°/09
60N:  5°/15  7°/16  4°/10  7°/09 11°/15 15°/16 16°/08 14°/09  7°/15  5°/14  6°/08  6°/09
65N:  1°/13  5°/15  4°/17  5°/09  9°/16 12°/16 13°/08 11°/08  5°/15  4°/14  6°/08  4°/10
70N:  -----  4°/14  4°/17  3°/09  7°/16  9°/16 10°/08  9°/08  4°/16  3°/14  5°/09  -----
75N:  -----  -----  3°/16  2°/08  5°/16  6°/17  7°/07  6°/08  2°/16  3°/14  4°/11  -----
80N:  -----  -----  3°/15  2°/08  3°/17  4°/17  5°/07  4°/07  1°/16  2°/07  -----  -----
85N:  -----  -----  -----  1°/08  2°/17  2°/17  3°/07  3°/07  0°/17  2°/08  -----  -----

```

(The entries given as dashes mean that on these days, the sun isn't visible at all.)

Results for the southern hemisphere are similar, just shift the months by half a year.

Acknowledgment: Thanks to the people who made the computation for sun azimuth and elevation freely available.


---

