<p>As barrycarter mentions in the comments, if you take the average of the two altitudes, you'll get the altitude of the celestial pole—north if you're in the northern hemisphere, south if you're in the southern hemisphere.  In this case, the average is just the average in degrees plus the average in minutes:</p>

<p>$$
\frac{38+12}{2} = 25 \text{ degrees}
$$</p>

<p>and</p>

<p>$$
\frac{52+24}{2} = 38 \text{ minutes}
$$</p>

<p>This is equal to the latitude of the observation point.  One can also obtain the declination of the star in question, by halving the <em>difference</em> between the two altitudes, rather than their sum, and then subtracting the result from $90$ degrees:</p>

<p>$$
\frac{38-12}{2} = 13 \text{ degrees}
$$</p>

<p>and</p>

<p>$$
\frac{52-24}{2} = 14 \text{ minutes}
$$</p>

<p>If you subtract that from $90$ degrees, you get a declination of $76$ degrees, $46$ minutes.  I know of no bright star particularly near that declination, but perhaps this is just an academic exercise.</p>

<p>The foregoing neglects atmospheric refraction.  This turns out to be not entirely negligible at the level of precision provided in the question.  At an altitude of $39$ degrees, atmospheric refraction produces an additional elevation of about $1.2$ minutes (barring local and daily effects), meaning that if a star is observed at $38$ degrees $52$ minutes, it's "really" at $38$ degrees $51$ minutes or so.  At an altitude of $12$ degrees, the additional elevation is about $4.5$ minutes, meaning that the star observed at $12$ degrees $24$ minutes is "really" at $12$ degrees $19–20$ minutes.  If the altitude figures given are as observed, then the corrected figures should be used instead.</p>

<hr>

<p>ETA: Here's a page with information on the extent of <a href="http://www.mysundial.ca/tsp/refraction_of_light.html" rel="nofollow noreferrer">atmospheric refraction</a>.</p>

CC BY-SA 3.0
2017-02-07T18:26:08.633Z
2017-02-09T18:15:13.493Z
2017-02-09T18:15:13.493Z
Brian Tung
<p>All we know that a star's altitude at its upper culmination is 38 degrees and 52 arcminutes and that its altitude at its lower culmination is 12 degrees and 24 arcminutes (so it never really sets at that place).
How can we, through this information, find that place's latitude?</p>

CC BY-SA 3.0
2017-02-02T11:02:48.020Z
Is that homework?
chirlu
CC BY-SA 3.0
2017-02-02T11:15:22.857Z
It's not homework. My school doesn't teach that kind of stuff. I found it on the internet and I can't find the solution to it.
Potato Comet
CC BY-SA 3.0
2017-02-02T14:08:13.840Z
All the same, what have you tried, such as drawing some geometric diagrams?  And are you sure (just asking :-) ) it doesn't depend on the star's position in the heavens?
Carl Witthoft
<star><coordinate>
How to find a place's latitude when you know a star's minimum and maximum altitude?
Potato Comet
# How to find a place's latitude when you know a star's minimum and maximum altitude?

(Asked by: username_0 on 2017-02-01)

All we know that a star's altitude at its upper culmination is 38 degrees and 52 arcminutes and that its altitude at its lower culmination is 12 degrees and 24 arcminutes (so it never really sets at that place).
How can we, through this information, find that place's latitude?

**Tags:** star, coordinate

> **Comments:**
> **username_1**: Is that homework?  
> **username_0**: It's not homework. My school doesn't teach that kind of stuff. I found it on the internet and I can't find the solution to it.  
> **username_2**: All the same, what have you tried, such as drawing some geometric diagrams?  And are you sure (just asking :-) ) it doesn't depend on the star's position in the heavens?  

---

## 1 answers

### Answer by username_3 on 2017-02-07. Score: 1

As barrycarter mentions in the comments, if you take the average of the two altitudes, you'll get the altitude of the celestial pole—north if you're in the northern hemisphere, south if you're in the southern hemisphere. In this case, the average is just the average in degrees plus the average in minutes:

$$
\frac{38+12}{2} = 25 \text{ degrees}
$$

and

$$
\frac{52+24}{2} = 38 \text{ minutes}
$$

This is equal to the latitude of the observation point. One can also obtain the declination of the star in question, by halving the *difference* between the two altitudes, rather than their sum, and then subtracting the result from $90$ degrees:

$$
\frac{38-12}{2} = 13 \text{ degrees}
$$

and

$$
\frac{52-24}{2} = 14 \text{ minutes}
$$

If you subtract that from $90$ degrees, you get a declination of $76$ degrees, $46$ minutes. I know of no bright star particularly near that declination, but perhaps this is just an academic exercise.

The foregoing neglects atmospheric refraction. This turns out to be not entirely negligible at the level of precision provided in the question. At an altitude of $39$ degrees, atmospheric refraction produces an additional elevation of about $1.2$ minutes (barring local and daily effects), meaning that if a star is observed at $38$ degrees $52$ minutes, it's "really" at $38$ degrees $51$ minutes or so. At an altitude of $12$ degrees, the additional elevation is about $4.5$ minutes, meaning that the star observed at $12$ degrees $24$ minutes is "really" at $12$ degrees $19–20$ minutes. If the altitude figures given are as observed, then the corrected figures should be used instead.

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ETA: Here's a page with information on the extent of [atmospheric refraction](http://www.mysundial.ca/tsp/refraction_of_light.html).


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