<p>My original perception of the Declination was wrong, that's why I was confusing Declination with Amplitude.</p>
<p>Here's a correct way to imagine Declination and Amplitude. And in this picture you can also see why they're NOT the same.</p>
<p><a href="https://i.sstatic.net/OkdkV.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OkdkV.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-03-16T14:12:10.820Z
2023-03-16T14:12:10.820Z
Harbard
<p>An easy way to see this is to note that the &quot;Amplitude&quot; (the azimuth of a body when it is on the horizon) will depend on both the position of the body in the sky (the declination) and the position of the observer on the ground (the observer's latitude).</p>
<p>In particular <span class="math-container">$\sin(\delta) = \cos(\lambda)\sin(A)$</span> where <span class="math-container">$\delta,\ \lambda,\ A$</span> are declination, latitude and Amplitude resp.</p>
<p>So A isn't the same as declination.</p>

CC BY-SA 4.0
2023-04-15T18:47:14.860Z
2023-04-16T15:32:01.000Z
2023-04-16T15:32:01.000Z
James K
<p><a href="https://i.sstatic.net/lVvb7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lVvb7.png" alt="enter image description here" /></a>
From the way I understand <a href="https://maritimesa.org/nautical-science-grade-12/2020/09/07/the-amplitude/" rel="nofollow noreferrer">Amplitude</a> and Declination, it seems to me that they should always be the same, yet they are not, why?</p>
<p>P.S. In the picture I am imagining that the declination remains constant (25 degrees N)</p>

CC BY-SA 4.0
2022-10-08T12:31:02.120Z
What do you mean by 'amplitude' in this context?
Barry Carter
CC BY-SA 4.0
2022-10-08T12:48:15.550Z
Yes, it's definitely with respect to the rational horizon  https://cultofsea.com/navigation/amplitude/
PM 2Ring
CC BY-SA 4.0
2022-10-08T16:12:36.123Z
@Harbard Feel free to write your own answer. This site could do with more celestial navigation content...
PM 2Ring
CC BY-SA 4.0
2023-03-17T00:34:16.520Z
@PM2Ring: It seems "celestial navigation content..." uses different vocabulary than standard observational astronomy, so maybe some sort of translation almanach first would be in place...
AtmosphericPrisonEscape
<declination>
Why isn't Amplitude the same as Declination?
Harbard
# Why isn't Amplitude the same as Declination?

(Asked by: username_0 on 2022-10-08)

[![enter image description here](https://i.sstatic.net/lVvb7.png)](https://i.sstatic.net/lVvb7.png)
From the way I understand [Amplitude](https://maritimesa.org/nautical-science-grade-12/2020/09/07/the-amplitude/) and Declination, it seems to me that they should always be the same, yet they are not, why?

P.S. In the picture I am imagining that the declination remains constant (25 degrees N)

**Tags:** declination
> **Comments:**
> **username_1**: What do you mean by 'amplitude' in this context?  
> **username_2**: Yes, it's definitely with respect to the rational horizon  https://cultofsea.com/navigation/amplitude/  
> **username_2**: @Harbard Feel free to write your own answer. This site could do with more celestial navigation content...  
> **username_3**: @PM2Ring: It seems "celestial navigation content..." uses different vocabulary than standard observational astronomy, so maybe some sort of translation almanach first would be in place...  

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## 2 answers

### Answer by username_0 on 2023-03-16. Score: 2

My original perception of the Declination was wrong, that's why I was confusing Declination with Amplitude.

Here's a correct way to imagine Declination and Amplitude. And in this picture you can also see why they're NOT the same.

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


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### Answer by username_4 on 2023-04-15. Score: 1

An easy way to see this is to note that the "Amplitude" (the azimuth of a body when it is on the horizon) will depend on both the position of the body in the sky (the declination) and the position of the observer on the ground (the observer's latitude).

In particular $\sin(\delta) = \cos(\lambda)\sin(A)$ where $\delta,\ \lambda,\ A$ are declination, latitude and Amplitude resp.

So A isn't the same as declination.


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