<p>Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude.
Neglecting atmospheric refraction, the altitude at culmination is
$$\mathsf{alt_{max}} = 90^\circ - |\phi - \delta| $$</p>

<p>Here are some special cases:</p>

<ul>
<li><p>A star whose declination equals the observer's latitude ($\delta = \phi$) culminates at the zenith (altitude 90$^\circ$).</p></li>
<li><p>A star on the celestial equator ($\delta$ = 0) culminates at altitude 90$^\circ$ - $|\phi|$.</p></li>
</ul>

<p>For the example in the question, if we solve for $\phi$,
$$\phi = \delta \pm (90^\circ - \mathsf{alt_{max}}) $$</p>

<p>Then the observer's latitude is 20$^\circ$ $\pm$ 40$^\circ$, either 20$^\circ$S or 60$^\circ$N.
GCSE Astronomy's answer is incorrect.</p>

CC BY-SA 4.0
2018-05-08T18:44:48.433Z
2018-05-08T18:44:48.433Z
Mike G
<p>From gcseastronomy.co.uk:
<em>A star culminates at 50°. It has a declination of +20°. What is the latitude from where it is observed?</em> </p>

<p>What is the relationship between a culmination, declination, and latitude? How would one work this out?</p>

<p>This is the answer:
50 - 30 = 20°
Latitude = 20°</p>

CC BY-SA 4.0
2018-05-08T16:09:59.053Z
There is something wrong, either in the website's answer or the way you typed it. A star at +20 declination would culminate at the zenith (90 degrees) for an observer at +20 latitude. No wonder you are confused!
JohnHoltz
<star>
How to calculate observer's latitude from declination and culmination of a star?
Ben D
# How to calculate observer's latitude from declination and culmination of a star?

(Asked by: username_0 on 2018-05-08)

From gcseastronomy.co.uk:
*A star culminates at 50°. It has a declination of +20°. What is the latitude from where it is observed?*

What is the relationship between a culmination, declination, and latitude? How would one work this out?

This is the answer:
50 - 30 = 20°
Latitude = 20°

**Tags:** star

> **Comments:**
> **username_1**: There is something wrong, either in the website's answer or the way you typed it. A star at +20 declination would culminate at the zenith (90 degrees) for an observer at +20 latitude. No wonder you are confused!  

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## 1 answers

### Answer by username_2 on 2018-05-08. Score: 5 | Accepted answer

Let $\delta$ be the star's declination and $\phi$ be the observer's terrestrial latitude.
Neglecting atmospheric refraction, the altitude at culmination is
$$\mathsf{alt\_{max}} = 90^\circ - |\phi - \delta| $$

Here are some special cases:

* A star whose declination equals the observer's latitude ($\delta = \phi$) culminates at the zenith (altitude 90$^\circ$).
* A star on the celestial equator ($\delta$ = 0) culminates at altitude 90$^\circ$ - $|\phi|$.

For the example in the question, if we solve for $\phi$,
$$\phi = \delta \pm (90^\circ - \mathsf{alt\_{max}}) $$

Then the observer's latitude is 20$^\circ$ $\pm$ 40$^\circ$, either 20$^\circ$S or 60$^\circ$N.
GCSE Astronomy's answer is incorrect.


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