<ol>
<li>Yes it is just a step in calculating the local time</li>
<li>It is essentially the same as your simple formula. The fuss is about the local time - they did not have accurate clocks so they measured the time by getting accurate angles between horizon and a star or sun etc at the highest point. There were many tables that recorded what the angles and time should be for various objects in the sky at their highest point - azimuth. There was a very large reward offered for the development of an accurate clock.
Remember, a swaying ship with wind and waves made each determination difficult and you only got one chance with each star and often clouds, rain, storm. Get it wrong and reefs tear the bottom out of your ship or you miss the trade winds and have to tack for weeks. An accurate clock, which we take for granted, avoided all that fuss.</li>
</ol>

CC BY-SA 4.0
2022-04-08T00:22:40.820Z
You really should expand on point #1 a lot.
Greg Miller
CC BY-SA 4.0
2022-04-12T01:37:15.657Z
'how to find you position with a sextant - idea 21C' the yellow underlined sections are the practical bits - the modern sextant is a lot more developed than what they had but the idea is the same
Moggsy8
CC BY-SA 4.0
2022-04-12T01:52:32.097Z
'Step-by Step Sextant Users Guide by Andrew Evans' contains latitude tables. both this and 'how to find your' respond to Google entry. The Davis Quadrant aka backstaff is an earlier instrument, see Wiki. The angle between the horizon and 'whatever you can see and have tables for' is the key datum.
Moggsy8
CC BY-SA 4.0
2022-04-07T05:40:41.190Z
2022-04-07T05:40:41.190Z
Moggsy8
<p>I'm studying the history of longitude and Greenwich Time, and I'm currently confused about the "longitude by chronometer" technique.</p>

<p>Conceptually, the method seems extremely simple.  With a chronometer (ie. an accurate clock) set to the time of your home port, compare to your local time.  Multiply the difference by <span class="math-container">$\frac{15°}{hour}$</span> and that will give you the degrees of longitude you have traveled.</p>

<p>According to Derek Howse's <em>Greenwich Time and the Longitude</em> and Dava Sobel's <em>Longitude</em>, this method was first published by Gemma Frisius in 1530, then described again by Richard Eden in 1555.  Both of these authors say that you must first find your latitude.  Here is Frisius in 1530 (emphasis mine):</p>

<blockquote>
  <p>We must wait until the hand of our clock exactly touches the point of an hour and at the same moment by means of an astrolabe or by means of our globe, we must find out the time of the place at which we now find ourselves.  If this time agrees to the minute with the time shown on our watch, it is certain that we are still on the same meridian or in the same longitude, and our journey has been made towards the south.  But if it differs by one hour or by a number of minutes, then these should be turned into degrees, or minutes of degrees, by the method I set out in the previous chapter, and in this way the longitude is discovered.  In this way I would be able to find the longitude of places, even if I was dragged off unawares across a thousand miles, and even though the distance of my journey was unknown. <strong>But then first of all, as always, the latitude must be learnt.  I have already explained this before and also that it can be found out by various methods of finding out the tirne.</strong></p>
</blockquote>

<p>Here is <a href="https://books.google.com/books?id=VHoKAQAAIAAJ&amp;lpg=PA390&amp;ots=DzQNZhRtmk&amp;dq=and%20so%20shall%20the%20longitude%20bee%20founde.%20And%20by%20this%20arte%20can%20I%20fynde%20the%20longitude%20of%20regions&amp;pg=PA390#v=onepage&amp;q&amp;f=false" rel="noreferrer">Eden in 1555</a>:</p>

<pre><code>And so shall the longitude bee found.  And by
   this arte can I fynde the longitude of
       regions althowgh I were a thou-
        sand myles owt of my attempt- 
         ted course &amp; in an unkno-
           wen distance, but the
            latitude must fyrst
               bee perfectly
                  knowen.
</code></pre>

<p>I can't figure out why knowing the latitude would be necessary for calculating longitude by chronometer, unless you need it to calculate the local time.  But it seems to be well established that local time can be easily calculated with a noon sighting of the Sun, corrected for the equation of time.  This method is described <a href="https://en.wikipedia.org/wiki/Longitude_by_chronometer#Noon_sight_for_Longitude" rel="noreferrer">here</a>.</p>

<p>Most of the Wikipedia page for <a href="https://en.wikipedia.org/wiki/Longitude_by_chronometer" rel="noreferrer">Longitude by Chronometer</a> seems to describe a more complicated method that involves finding the intersection of a position line with an assumed latitude.  It also describes solving the navigational triangle.  I'm having trouble following the method described there.  I can't tell whether this is the same method described by Frisius or something else entirely.</p>

<p>So overall my questions are:</p>

<ol>
<li>Why do Frisius and Eden say that you need to calculate latitude before you can do longitude by chronometer?  Is it just a step in calculating the local time?</li>
<li>What is the method described in the <a href="https://en.wikipedia.org/wiki/Longitude_by_chronometer" rel="noreferrer">"Longitude by Chronometer" Wikipedia page</a>, and the navigational triangle stuff?  How does it work, and what benefit does it have over the simple <span class="math-container">$(t_{home} - t_{local})\frac{15°}{hour}$</span> method?</li>
</ol>

CC BY-SA 4.0
2019-09-03T17:06:25.560Z
I might be able to re-parse your quoted text as stating that you can't determine the entire distance covered unless you know both lat and long, which makes sense.  You know your new longitude but don't know yet how far South or North you also moved.
Carl Witthoft
CC BY-SA 4.0
2019-09-03T17:08:49.197Z
Hmmm.. reading that wikipedia page, I think their point is that you need to know your latitude to be able to determine the LocalTime as a function of sun elevation angle.  See the section where it explains why trying to identify noon - zenith -  has a large uncertainty.
Carl Witthoft
<time><navigation>
Why does "longitude by chronometer" require first calculating latitude?
Josh Haberman
# Why does "longitude by chronometer" require first calculating latitude?

(Asked by: username_0 on 2019-09-02)

I'm studying the history of longitude and Greenwich Time, and I'm currently confused about the "longitude by chronometer" technique.

Conceptually, the method seems extremely simple. With a chronometer (ie. an accurate clock) set to the time of your home port, compare to your local time. Multiply the difference by $\frac{15°}{hour}$ and that will give you the degrees of longitude you have traveled.

According to Derek Howse's *Greenwich Time and the Longitude* and Dava Sobel's *Longitude*, this method was first published by Gemma Frisius in 1530, then described again by Richard Eden in 1555. Both of these authors say that you must first find your latitude. Here is Frisius in 1530 (emphasis mine):

> We must wait until the hand of our clock exactly touches the point of an hour and at the same moment by means of an astrolabe or by means of our globe, we must find out the time of the place at which we now find ourselves. If this time agrees to the minute with the time shown on our watch, it is certain that we are still on the same meridian or in the same longitude, and our journey has been made towards the south. But if it differs by one hour or by a number of minutes, then these should be turned into degrees, or minutes of degrees, by the method I set out in the previous chapter, and in this way the longitude is discovered. In this way I would be able to find the longitude of places, even if I was dragged off unawares across a thousand miles, and even though the distance of my journey was unknown. **But then first of all, as always, the latitude must be learnt. I have already explained this before and also that it can be found out by various methods of finding out the tirne.**

Here is [Eden in 1555](https://books.google.com/books?id=VHoKAQAAIAAJ&lpg=PA390&ots=DzQNZhRtmk&dq=and%20so%20shall%20the%20longitude%20bee%20founde.%20And%20by%20this%20arte%20can%20I%20fynde%20the%20longitude%20of%20regions&pg=PA390#v=onepage&q&f=false):

```
And so shall the longitude bee found.  And by
   this arte can I fynde the longitude of
       regions althowgh I were a thou-
        sand myles owt of my attempt- 
         ted course & in an unkno-
           wen distance, but the
            latitude must fyrst
               bee perfectly
                  knowen.

```

I can't figure out why knowing the latitude would be necessary for calculating longitude by chronometer, unless you need it to calculate the local time. But it seems to be well established that local time can be easily calculated with a noon sighting of the Sun, corrected for the equation of time. This method is described [here](https://en.wikipedia.org/wiki/Longitude_by_chronometer#Noon_sight_for_Longitude).

Most of the Wikipedia page for [Longitude by Chronometer](https://en.wikipedia.org/wiki/Longitude_by_chronometer) seems to describe a more complicated method that involves finding the intersection of a position line with an assumed latitude. It also describes solving the navigational triangle. I'm having trouble following the method described there. I can't tell whether this is the same method described by Frisius or something else entirely.

So overall my questions are:

1. Why do Frisius and Eden say that you need to calculate latitude before you can do longitude by chronometer? Is it just a step in calculating the local time?
2. What is the method described in the ["Longitude by Chronometer" Wikipedia page](https://en.wikipedia.org/wiki/Longitude_by_chronometer), and the navigational triangle stuff? How does it work, and what benefit does it have over the simple $(t\_{home} - t\_{local})\frac{15°}{hour}$ method?

**Tags:** time, navigation

> **Comments:**
> **username_1**: I might be able to re-parse your quoted text as stating that you can't determine the entire distance covered unless you know both lat and long, which makes sense.  You know your new longitude but don't know yet how far South or North you also moved.  
> **username_1**: Hmmm.. reading that wikipedia page, I think their point is that you need to know your latitude to be able to determine the LocalTime as a function of sun elevation angle.  See the section where it explains why trying to identify noon - zenith -  has a large uncertainty.  

---

## 1 answers

### Answer by username_2 on 2022-04-07. Score: 1

1. Yes it is just a step in calculating the local time
2. It is essentially the same as your simple formula. The fuss is about the local time - they did not have accurate clocks so they measured the time by getting accurate angles between horizon and a star or sun etc at the highest point. There were many tables that recorded what the angles and time should be for various objects in the sky at their highest point - azimuth. There was a very large reward offered for the development of an accurate clock.
   Remember, a swaying ship with wind and waves made each determination difficult and you only got one chance with each star and often clouds, rain, storm. Get it wrong and reefs tear the bottom out of your ship or you miss the trade winds and have to tack for weeks. An accurate clock, which we take for granted, avoided all that fuss.

> **Comments:**
> **username_3**: You really should expand on point #1 a lot.  
> **username_2**: 'how to find you position with a sextant - idea 21C' the yellow underlined sections are the practical bits - the modern sextant is a lot more developed than what they had but the idea is the same  
> **username_2**: 'Step-by Step Sextant Users Guide by Andrew Evans' contains latitude tables. both this and 'how to find your' respond to Google entry. The Davis Quadrant aka backstaff is an earlier instrument, see Wiki. The angle between the horizon and 'whatever you can see and have tables for' is the key datum.  

---

