<p>Mnyah… perhaps, depending on your definition of observable, and the lengths to which you are willing to go.</p>
<p>tl;dr You need refraction and altitude.</p>
<h3>Refraction</h3>
<p>The 33'25&quot; is, coincidentally, roughly equal to the diameter of the full Moon, meaning that, when α Cen culminates, it is a Moon diameter below the horizon. It is, also coincidentally, roughly equal to the angle of atmospheric refraction toward the horizon, as seen in this parametrization:</p>
<p><a href="https://i.sstatic.net/pGXxn.png" rel="noreferrer"><img src="https://i.sstatic.net/pGXxn.png" alt="refraction" /></a>
<sup><a href="https://en.wikipedia.org/wiki/Atmospheric_refraction#/media/File:BennettAtmRefractVsAlt.png" rel="noreferrer">Plot of atmospheric refraction vs. apparent altitude, using G.G. Bennett’s 1982 formula. Author: Jeff Conrad.</a></sup></p>
<p>That means that α Cen should <em>juuust</em> graze the horizon. According to Stellarium, tonight it will happen at 5:30 in the morning:</p>
<p><a href="https://i.sstatic.net/dR8f2.png" rel="noreferrer"><img src="https://i.sstatic.net/dR8f2.png" alt="stellarium" /></a>
<sup>Position of α Cen — here called &quot;Rigel Kent&quot; — at culmination on 19 January 2024, as seen from Eilat. It will reach roughly 30' below the horizon, but refraction should take it roughly 30' closer to the horizon. Created using the free software Stellarium.</sup></p>
<p>Refraction may vary quite a lot. Especially in the spring time, if you have a view over the sea, the heating air can increase quite a lot.</p>
<h3>Altitude</h3>
<p>Rather than relying on refraction, you could increase your altitude. Stellarium in principle allows you to type in altitude, but (my version of) this feature seems to be broken. So let's calculate:</p>
<p>Using <a href="https://en.wikibooks.org/wiki/Trigonometry/The_distance_and_dip_of_the_horizon" rel="noreferrer">standard trigonometry</a>, you will find that the distance <span class="math-container">$d$</span> to the horizon seen from an altitude <span class="math-container">$h$</span> is <span class="math-container">$d = \sqrt{2hR}$</span>, where <span class="math-container">$R$</span> is the radius of the Earth. The angle <span class="math-container">$\theta$</span> below the horizon that you can see is given by <span class="math-container">$\tan\theta = d/R$</span>. I've plotted this relation below, from which you can see that you need to increase your height to 200–300 meters (not taking into account refraction).</p>
<p><a href="https://i.sstatic.net/6JE8W.png" rel="noreferrer"><img src="https://i.sstatic.net/6JE8W.png" alt="dip vs. h" /></a></p>
<p>Google tells me you don't have many tall buildings in Eilat, but the nearby Mt. Shahmon may help you, if you can get a clear view toward the horizon.</p>
<p>Good luck!</p>

CC BY-SA 4.0
2024-01-18T17:28:18.607Z
The declination is -60 deg 50', so you can be as far north as 29.17 degrees latitude and have alpha Centauri graze the horizon (in theory).
JohnHoltz
CC BY-SA 4.0
2024-01-18T21:20:28.110Z
Wow great answers...still WHICH DAY of the year would one maximize one's chances of getting a glimpse?
C. Burns
CC BY-SA 4.0
2024-01-19T11:02:21.217Z
@C.Burns That depends on your preferred time of the night. With each month, the star culminates two hours earlier (because there's 24 hours/day, while there's 12 months/year). At the moment, you'd have to get up early (5:30). Going forward in time from now, α Cen will culminate earlier and earlier, so e.g. on 21 March, it culminates around 1:30 at night, while on 21 June it culminates at 20:30 (not 19:30 because summertime). I'm in Denmark, so I can't see any stars at 20:30 in the summer, but I suppose you can in Eilat.
pela
CC BY-SA 4.0
2024-01-19T15:22:06.780Z
@C.Burns But I can really recommend Stellarium. It's easy to use (once you've googled your way to a keyboard shortcut cheat sheet), it's beautiful, and you learn a lot about how the night sky works :)
pela
CC BY-SA 4.0
2024-01-18T16:30:56.563Z
2024-01-18T16:30:56.563Z
pela
<p>Wikipedia states that Alpha Centauri is the third brightest star, outshone by only Sirius and Canopus, and can be observed from latitudes south of ABOUT 29°N. With Eilat listed at 29°33′25″N, are the 33′25″ within the range covered by the word 'ABOUT' such that Alpha Centauri would be observable for a short time during the year?</p>

CC BY-SA 4.0
2024-01-18T23:24:30.890Z
This would seem to be a trivial question as the maximum altitude (ignoring refraction (but that is also easily known)) is easily calculated by hand. Isn't there a canonical question for this?
Peter Mortensen
CC BY-SA 4.0
2024-01-20T13:12:57.650Z
@PeterMortensen based on the answer, it seems this is quite an edge case, and so if there was a canonical answer for it one would have to work hard to find it.
uhoh
<observational-astronomy>
Can Alpha Centauri be observed from Eilat Israel, and if so when and for how long?
C. Burns
# Can Alpha Centauri be observed from Eilat Israel, and if so when and for how long?

(Asked by: username_0 on 2024-01-18)

Wikipedia states that Alpha Centauri is the third brightest star, outshone by only Sirius and Canopus, and can be observed from latitudes south of ABOUT 29°N. With Eilat listed at 29°33′25″N, are the 33′25″ within the range covered by the word 'ABOUT' such that Alpha Centauri would be observable for a short time during the year?

**Tags:** observational-astronomy

> **Comments:**
> **username_1**: This would seem to be a trivial question as the maximum altitude (ignoring refraction (but that is also easily known)) is easily calculated by hand. Isn't there a canonical question for this?  
> **username_2**: @PeterMortensen based on the answer, it seems this is quite an edge case, and so if there was a canonical answer for it one would have to work hard to find it.  

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## 1 answers

### Answer by username_3 on 2024-01-18. Score: 15

Mnyah… perhaps, depending on your definition of observable, and the lengths to which you are willing to go.

tl;dr You need refraction and altitude.

### Refraction

The 33'25" is, coincidentally, roughly equal to the diameter of the full Moon, meaning that, when α Cen culminates, it is a Moon diameter below the horizon. It is, also coincidentally, roughly equal to the angle of atmospheric refraction toward the horizon, as seen in this parametrization:

[![refraction](https://i.sstatic.net/pGXxn.png)](https://i.sstatic.net/pGXxn.png)
[Plot of atmospheric refraction vs. apparent altitude, using G.G. Bennett’s 1982 formula. Author: Jeff Conrad.](https://en.wikipedia.org/wiki/Atmospheric_refraction#/media/File:BennettAtmRefractVsAlt.png)

That means that α Cen should *juuust* graze the horizon. According to Stellarium, tonight it will happen at 5:30 in the morning:

[![stellarium](https://i.sstatic.net/dR8f2.png)](https://i.sstatic.net/dR8f2.png)
Position of α Cen — here called "Rigel Kent" — at culmination on 19 January 2024, as seen from Eilat. It will reach roughly 30' below the horizon, but refraction should take it roughly 30' closer to the horizon. Created using the free software Stellarium.

Refraction may vary quite a lot. Especially in the spring time, if you have a view over the sea, the heating air can increase quite a lot.

### Altitude

Rather than relying on refraction, you could increase your altitude. Stellarium in principle allows you to type in altitude, but (my version of) this feature seems to be broken. So let's calculate:

Using [standard trigonometry](https://en.wikibooks.org/wiki/Trigonometry/The_distance_and_dip_of_the_horizon), you will find that the distance $d$ to the horizon seen from an altitude $h$ is $d = \sqrt{2hR}$, where $R$ is the radius of the Earth. The angle $\theta$ below the horizon that you can see is given by $\tan\theta = d/R$. I've plotted this relation below, from which you can see that you need to increase your height to 200–300 meters (not taking into account refraction).

[![dip vs. h](https://i.sstatic.net/6JE8W.png)](https://i.sstatic.net/6JE8W.png)

Google tells me you don't have many tall buildings in Eilat, but the nearby Mt. Shahmon may help you, if you can get a clear view toward the horizon.

Good luck!

> **Comments:**
> **username_4**: The declination is -60 deg 50', so you can be as far north as 29.17 degrees latitude and have alpha Centauri graze the horizon (in theory).  
> **username_0**: Wow great answers...still WHICH DAY of the year would one maximize one's chances of getting a glimpse?  
> **username_3**: @C.Burns That depends on your preferred time of the night. With each month, the star culminates two hours earlier (because there's 24 hours/day, while there's 12 months/year). At the moment, you'd have to get up early (5:30). Going forward in time from now, α Cen will culminate earlier and earlier, so e.g. on 21 March, it culminates around 1:30 at night, while on 21 June it culminates at 20:30 (not 19:30 because summertime). I'm in Denmark, so I can't see any stars at 20:30 in the summer, but I suppose you can in Eilat.  
> **username_3**: @C.Burns But I can really recommend Stellarium. It's easy to use (once you've googled your way to a keyboard shortcut cheat sheet), it's beautiful, and you learn a lot about how the night sky works :)  

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