<p>It's not hours, minutes, and seconds, its degrees, arcminutes and arcseconds.</p>

<p>There are 360 degrees in a circle, 60 arcminutes in a degree and 60 arcseconds in an arcminute.</p>

<p>Since the planets lie close to the ecliptic, and the ecliptic tilted at 23 degrees relative to the equator, the declension of the planets rarely exceeds +-23 degrees, giving an appearance that their dec. is like a time.</p>

CC BY-SA 3.0
2017-08-24T06:42:27.680Z
So I was right! In that I was "lacking something incredibly fundamental". Thank you, that seems to have done it.
Jonathan Pearl
CC BY-SA 3.0
2017-08-24T06:09:44.280Z
2017-08-24T06:09:44.280Z
James K
<p>I'm running the program found at <a href="https://www.projectpluto.com/source.htm" rel="nofollow noreferrer">https://www.projectpluto.com/source.htm</a> in order to find the Right ascensions and declinations of celestial bodies at various points in time. An odd thing I've noticed is that the declination is reported in hours, minutes, seconds as opposed to degrees, minutes, seconds and I'm not sure how to use that. Especially considering the hours seem to lie on [-24,24] which doesn't make sense for the [-90,90] range for declination measured in degrees.</p>

<p>An example:</p>

<blockquote>
  <p>Sun RA=10:11:50.1 Dec= 11:08:50.8</p>
</blockquote>

<p><a href="https://i.sstatic.net/I7IFH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/I7IFH.png" alt="image showing right ascensions and declinations of various planets"></a></p>

<p>This is sampled from Los Angeles, California at 8 PM PDT (3:00 UTC) so I would expect when I ran these numbers to calculate elevation and azimuth that the elevation would be a good bit under the horizon.</p>

<p>I tried scaling the hours by 90/24 so that it would be back in the [-90,90] range but when I tried running those numbers through <a href="http://www.convertalot.com/celestial_horizon_co-ordinates_calculator.html" rel="nofollow noreferrer">this tool</a>, it ended up reporting that the sun would actually be just above the horizon which is definitely wrong.</p>

<p><a href="https://i.sstatic.net/9iudr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9iudr.png" alt="wrong calculation"></a></p>

<p>So with that said, how should I be interpreting this data to get the result I expect? I'm pretty new to this, so I may be lacking something incredibly fundamental.</p>

<positional-astronomy><declination>
Measuring declination in negative hours of arc?
Jonathan Pearl
# Measuring declination in negative hours of arc?

(Asked by: username_0 on 2017-08-24)

I'm running the program found at <https://www.projectpluto.com/source.htm> in order to find the Right ascensions and declinations of celestial bodies at various points in time. An odd thing I've noticed is that the declination is reported in hours, minutes, seconds as opposed to degrees, minutes, seconds and I'm not sure how to use that. Especially considering the hours seem to lie on [-24,24] which doesn't make sense for the [-90,90] range for declination measured in degrees.

An example:

> Sun RA=10:11:50.1 Dec= 11:08:50.8

[![image showing right ascensions and declinations of various planets](https://i.sstatic.net/I7IFH.png)](https://i.sstatic.net/I7IFH.png)

This is sampled from Los Angeles, California at 8 PM PDT (3:00 UTC) so I would expect when I ran these numbers to calculate elevation and azimuth that the elevation would be a good bit under the horizon.

I tried scaling the hours by 90/24 so that it would be back in the [-90,90] range but when I tried running those numbers through [this tool](http://www.convertalot.com/celestial_horizon_co-ordinates_calculator.html), it ended up reporting that the sun would actually be just above the horizon which is definitely wrong.

[![wrong calculation](https://i.sstatic.net/9iudr.png)](https://i.sstatic.net/9iudr.png)

So with that said, how should I be interpreting this data to get the result I expect? I'm pretty new to this, so I may be lacking something incredibly fundamental.

**Tags:** positional-astronomy, declination


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## 1 answers

### Answer by username_1 on 2017-08-24. Score: 2 | Accepted answer

It's not hours, minutes, and seconds, its degrees, arcminutes and arcseconds.

There are 360 degrees in a circle, 60 arcminutes in a degree and 60 arcseconds in an arcminute.

Since the planets lie close to the ecliptic, and the ecliptic tilted at 23 degrees relative to the equator, the declension of the planets rarely exceeds +-23 degrees, giving an appearance that their dec. is like a time.

> **Comments:**
> **username_0**: So I was right! In that I was "lacking something incredibly fundamental". Thank you, that seems to have done it.  

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