<p>Jacob's staff is probably all you need.  It is a pole with a cross arm. By positioning the arm at a certain point on the pole and sighting down the pole one can get a position on the sky.  A skilled user can achieve sub-degree accuracy with such a tool.</p>
<p><a href="https://i.sstatic.net/TCEmx.png" rel="noreferrer"><img src="https://i.sstatic.net/TCEmx.png" alt="enter image description here" /></a></p>
<p>The position of the cross beam is easy enough to calculate by trigonometry, or by careful construction of triangles.</p>
<p><span class="math-container">$\tan(\theta/2)= c/2l$</span></p>
<p>Where <span class="math-container">$\theta$</span> is the angle above the horizon, <span class="math-container">$l$</span> is the position of the crossbar from the eye and <span class="math-container">$c$</span> is the length of the crossbar.</p>

CC BY-SA 4.0
2020-12-07T12:43:21.720Z
Thanks. The tricky bit of the tool is to keep it horizontal at the eye level but perhaps a table could help.
luchonacho
CC BY-SA 4.0
2020-12-07T12:52:01.440Z
Btw, woudln't something like a quadrant be much easier? Just select the angle which then you can project to the sky. Like the last photo [here](https://en.wikipedia.org/wiki/Quadrant_(instrument)). No need to do calculations about horizontal and vertical length.
luchonacho
CC BY-SA 4.0
2020-12-06T21:36:04.797Z
2020-12-06T21:36:04.797Z
James K
<p>Consider this example. I am standing in a flat surface, at latitude X, looking to the equator, in a random day of the year. (For simplicity, let's assume X &gt; 23.5º)</p>
<p>I want to find the position of the sun at noon on the winter solstice (that is, the highest point in the sky of the sun, the day which this reaches the lowest altitude), location which has a declination of 23.5º.</p>
<p>We know the sextant does exactly that. But say I don't have a sextant. <strong>I want to find an alternative method to do the same.</strong></p>
<p>If my latitude is X, then such position of the sun is X-23.5º above the horizon (looking North). I could then divide this result by 90, and get a proportion of the sky between the horizon and my zenit where that point is. But this is too rough.</p>
<p>Perhaps one way would be to have a long stick and a measuring tape. For a given value of X-23.5º there must be some relationship between my eye, the vertical height of the stick and my horizontal distance to the stick which allows me to find exactly the position in the sky.</p>
<p>Or perhaps another method?</p>

CC BY-SA 4.0
2020-12-06T19:41:57.167Z
Are you able to manufacture a tool to help with this?
James K
CC BY-SA 4.0
2020-12-06T21:20:38.540Z
@JamesK Absolutely. As long as it is rudimentary
luchonacho
CC BY-SA 4.0
2020-12-07T04:34:16.017Z
Seems like pretty much anything that works is effectively a sextant. If you can get something with a joint with enough friction to stay wherever you leave it, but still lets you move it easily, you can move it so it matches the declenation.
Acccumulation
<positional-astronomy><declination>
How to manually locate a certain declination in the sky for a given location? (no sextant available)
luchonacho
# How to manually locate a certain declination in the sky for a given location? (no sextant available)

(Asked by: username_0 on 2020-12-06)

Consider this example. I am standing in a flat surface, at latitude X, looking to the equator, in a random day of the year. (For simplicity, let's assume X > 23.5º)

I want to find the position of the sun at noon on the winter solstice (that is, the highest point in the sky of the sun, the day which this reaches the lowest altitude), location which has a declination of 23.5º.

We know the sextant does exactly that. But say I don't have a sextant. **I want to find an alternative method to do the same.**

If my latitude is X, then such position of the sun is X-23.5º above the horizon (looking North). I could then divide this result by 90, and get a proportion of the sky between the horizon and my zenit where that point is. But this is too rough.

Perhaps one way would be to have a long stick and a measuring tape. For a given value of X-23.5º there must be some relationship between my eye, the vertical height of the stick and my horizontal distance to the stick which allows me to find exactly the position in the sky.

Or perhaps another method?

**Tags:** positional-astronomy, declination

> **Comments:**
> **username_1**: Are you able to manufacture a tool to help with this?  
> **username_0**: @JamesK Absolutely. As long as it is rudimentary  
> **username_2**: Seems like pretty much anything that works is effectively a sextant. If you can get something with a joint with enough friction to stay wherever you leave it, but still lets you move it easily, you can move it so it matches the declenation.  

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## 1 answers

### Answer by username_1 on 2020-12-06. Score: 5

Jacob's staff is probably all you need. It is a pole with a cross arm. By positioning the arm at a certain point on the pole and sighting down the pole one can get a position on the sky. A skilled user can achieve sub-degree accuracy with such a tool.

[![enter image description here](https://i.sstatic.net/TCEmx.png)](https://i.sstatic.net/TCEmx.png)

The position of the cross beam is easy enough to calculate by trigonometry, or by careful construction of triangles.

$\tan(\theta/2)= c/2l$

Where $\theta$ is the angle above the horizon, $l$ is the position of the crossbar from the eye and $c$ is the length of the crossbar.

> **Comments:**
> **username_0**: Thanks. The tricky bit of the tool is to keep it horizontal at the eye level but perhaps a table could help.  
> **username_0**: Btw, woudln't something like a quadrant be much easier? Just select the angle which then you can project to the sky. Like the last photo [here](https://en.wikipedia.org/wiki/Quadrant_(instrument)). No need to do calculations about horizontal and vertical length.  

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