## Exercise: Find a location from Sun data, date, and time

An exercise in navigation, finding your own position? I flew away but don't know where I landed. To do this, I measured the times of sunrise and sunset and the maximum height of the Sun above the horizon. Thank God the sky was cloudless and clear yesterday and I was able to take the measurements directly over bodies of water. Here are the readings I took yesterday on October 28th. Sunrise was at 9:20:25 and sunset was at 20:29:55. My watch was set to CEST (Central European Summer Time) on October 26th before departure and the time deviation of the watch can be kept as 0 seconds. The solar altitude measured with the Quadrant was somewhere around 48° +/-0.5°. Find me!

This table can help you:

Day | EOT | Declination |
---|---|---|

October 26 | 16.0 | -12.5 |

October 27 | 16.1 | -12.84 |

October 28 | 16.2 | -13.17 |

October 29 | 16.3 | -13.51 |

October 30 | 16.3 | -13.83 |

Would you like to try to solve this? Everything you need for the solution can be found in GQT5^{[1]}. Imagine that it is an emergency that has happened to you or that while walking on the beach you find a bottle with this message and this shipwrecked Central European is sitting on the same beach a few kilometers or hundreds of meters away and is hoping for your help. The easiest way to find the latitude is from the altitude of the Sun, as described in the chapter on the star compass. Once you know the latitude, the time of sunrise and sunset (true local solar time) can be calculated (chapter on the Sine Quadrant). We then correct this calculated time with EOT and compare it with the measured time, convert it into angles and then calculate the longitude as the difference. Another little note, since I was lazy about using my compass I couldn't measure directions, but I did notice that the Sun moved from left to right over time (before you look for me on a completely different beach). Also, think what a true local lunchtime is (not necessarily 12:00). OK, as long as you're looking for me, I'll go catch some fish so I can set a table when you get there to save me.

I'm reassured that if you find my coordinates with an accuracy of 15 km, you'll find me much faster. And if you are familiar with David Burch's methods^{[2]}, you can find the longitude directly using the difference between the true local solar noon and the measured one.

Sorry, I can't give you my GPS coordinates either because my GPS receiver went bad and didn't come back.

### Note on EOT (Equation of time)

When EOT in the table is positive, the Sun is traveling faster than average, when negative, then slower. If the Sun is fast, it reaches the local meridian earlier than average, if slower, then later. In other words, if you are at the Greenwich Prime Meridian and your watch is set to this Greenwich time, you will see the Sun earlier than 12:00 over the local meridian. When EOT is 16.2 minutes, the Sun crosses the Greenwich Meridian at 11:43:59 (time). To get the true local solar time (12:00), you had to add these 16.2 minutes to your time.

Unfortunately, there are two different definitions of EOT used in different countries:

- EOT = Average time - time on a specific day (like in the table above, in Germany)
- EOT = Time on a specific day - average time

If you are in a different location, you also have to add the difference in longitude between the local meridian and Greenwich Meridian (0°) multiplied by 4 minutes/degree. This applies to both sunrise and sunset time measurements. The formulas for calculating sunrise time provide the true local solar time. If you leave out the minus sign in front of the tangent in the formula from the sunrise time angle calculation book and divide the result by 15, you get the time from midnight instead of from noon.

## Solution

Ok, let's solve the problem. What is given:

- Date of observation and measurement: October 28, 2023
- Time of sunrise:
`τ`

_{r}= 09:20:25 - Time of sunset:
`τ`

_{s}= 20:28:55 - Altitude of the Sun above the horizon at noon:
`H`

_{s}= 48° +/- 0.5° - EOT on Oct-28: 16.2 minutes
- Declination of the Sun on Oct-28:
`δ = -13.17°`

- The Sun moves over the horizon from left to right in front of the observer

We are looking for the latitude `φ`

and the longitude `λ`

of the observation location.

### The latitude

First, we find the latitude of the observation using the formula from the book. There are two formulas, one for a star (the Sun is a star) above the southern horizon and one for a star above the northern horizon. From the observation we know that the Sun moves from east to west from left to right. Therefore, the Sun moves along the southern horizon and we use the formula:

```
φ = δ + z
```

z is the zenith distance, the distance from the Sun to the zenith above the observer's head.

`z = 90° - H`_{s}

Putting them together we get:

`φ = δ + (90° - H`_{s})
φ = -13.17° + (90° - 48°)
= 28.83°
=> 28°49'48.00"

The *actual location* we are looking for is at a latitude of 28.4541160°N.

Because the degree scale on the Quadrant has a 1° scale, we can measure the angle with an accuracy of 1°. That's why the measured angle is in the range of +/- 0.5°. Therefore, the location you are looking for is within +/-0.5° or 30 NM (nautical miles) around the calculated latitude. The location we are actually looking for is within the measurement error range, which is very good, even though 30 NM (approx. 55.560 km) is quite a large distance.

### The longitude

Now we calculate our longitude using the measured times for sunrise and sunset. The first method would be to compare the measured time of sunrise and sunset with the sunrise and sunset table or local tides and calculate the longitude from this. Since we don't have such a table for this location, we will use the formula for calculating the sunrise hour angle from the book:

```
cos(τ) = - tan(δ) · tan(φ)
```

to calculate the hour angle of sunrise from noon. We will omit the minus sign before the first tan() to get the hour angle of midnight to save further calculation. For this, we also take the previously calculated latitude into account:

`τ`_{ra} = arccos(tan(δ) · tan(φ))
= arccos(tan(-13.17°) · tan(28°49'48")
= 97.400239
=> 97°24'0.86"

We divide the result by 15 to convert the hour angle into time (true solar time):

`τ`_{rc} = τ_{ra} / 15
= 6.493273
=> 6h 29m 36s

Now we take the measured sunrise time, add the EOT to adjust it to the true solar time and subtract the calculated (target) sunrise time to find the time difference between the true local solar time and the true solar time of the time meridian (CEST), to which my watch is set.

Multiply the result by 15 to convert this time difference into the angle and take the result from the length of the time meridian (30°) of my clock. The result is the observer's local longitude.

```
λ = 30° - (9:20:25 + 0:16.2:00 - 6:29:36) · 15
= -16°45'14.10"
=> -16.753817
```

As we can see, the result is almost one degree farther east than expected. Why is that?

- The latitude we calculated differs from the actual position by 0.3°.
- The formula uses an ideal sphere, but the Earth is an ellipsoid.
- The formula assumes that the observer is infinitely small and lies directly on the surface of the ideal sphere. I am on the other hand slightly higher than 0 mm and standing on a cliff that is about 10 m or more above the ocean surface. From this I see the horizon a little further away (approx. 12 km or 7 NM) and therefore sunrise is a little earlier and sunset is a little later than calculated.

If we do the same calculations but instead with the sunset times, we get a longitude about 1° west.

```
λ = 30° - (20:29:55 + 0:16.2:00 - (24:00:00 - 6:29:36)) · 15
= -18°55'45.90"
=> -18.929417
```

Let's take the average.

```
((-16°45'14.10") + (-18°55'45.90")) / 2
= -17°50'30.00"
=> -17.841666
```

## Result

The calculated location is thus at:

**28°49'48.00" latitude** and **-17°50'30.00" longitude** or **28.83° N 17.841666° W**

We are almost getting the exact point. The error is in arcseconds, less than 1 NM! The longitude we are looking for is:

```
-17.8382748 => -17°50'17.79"
```

This small error is caused by the fact that over the course of the day the solar declination and the EOT changed slightly. See the values in the table above for the following day.

## Calculated location

The calculated location is at:

28.83°N 17.841666°W (compare on Google Maps)

## Exact location

The exact location we are looking for is at:

28.4541160°N 17.8382748°W (compare on Google Maps)

See the discussion above about the measurement precision of +/-0.5° or 30 NM for the latitude.

### Don't forget the minus sign!

We have seen that using the sunrise and sunset times together for longitude calculations helps to increase the accuracy of the position, and minimize the errors caused by the calculation of the sunrise/sunset time due to inaccurate latitude, altitude of the observer, etc.

In the second method we will use the times for sunrise and sunset to determine the time of noon and calculate the longitude from this. To do this, we take the average to get the “measured” time of noon:

`τ`_{n} = (τ_{r} + τ_{s}) / 2
= (9:20:25 + 20:29:55) / 2
= 29:50:20 / 2
=> 14:55:10

Let's correct this time with EOT to get the measured solar time.

```
14:55:10 + 0:16.2 = 15:11:22
```

The true local solar noon is at 12:00:00. Let's subtract the true local solar noon from our measured corrected noon to get the time difference to the time zone meridian of our clock:

`τ`_{d} = 15:11:22 - 12:00:00
= 3:11:22

If this time difference is positive, our position is west of the time zone meridian, if negative, then east. Let's convert this time difference into the angle difference by multiplying by 15.

```
3:11:22 · 15 = 47°50'30"
```

To get the longitude of our position, we subtract our difference angle from the longitude of the time zone meridian. Pay attention to the sign!

```
30° - 47°50'30" = -17°50'30"
```

For the same longitude as the calculation above. This method is called the “equal solar altitude method”.

The "method of equal solar altitudes" means that the times of the same altitudes of the Sun are determined before and after the culmination, which are used to determine the true local solar noon and, if necessary, the longitude. The sunrise and sunset have the same elevation of 0° when measured over a horizon.

### Other methods

Often there are low hanging clouds or other obstacles, so observing the sunrise or sunset is not possible. In this case, the Telemeter Quadrant can be used as a Kamal to create an artificial horizon by aligning it in parallel to the horizon, thereby elevating the horizon to where the Sun becomes visible. Both altitude measurements will be made above that artificial horizon.

Since we don't know what the horizon looks like in the morning, it is recommended to take several measurements at different altitudes each in pairs for sunrise and sunset.

Over land, it is rarely possible to measure the sunrise and sunset above the horizon, since it is rare that the height of the horizon in the east is equal to the height in the west. In this case, measuring the Sun's altitude as a zenith distance is the best solution, as is done with the Quadrant. The Indian circle method is useful here.

I will end my story with the words of one song: “so that you are not lost on earth, try not to lose yourself!”

(Thanks to Telemeter user Andrej for this tutorial!)

**Resources:**

**Footnotes:**