<blockquote>
<p>This might be a dumb question.</p>
</blockquote>
<p>This is an <em>excellent question!</em></p>
<p>When we watch a solar eclipse, or the occultation of a star (or planet) by the Moon, and we want accurate timing information (or we want to record it) we must know our 3D position (lat, lon, elev) in order to either get an accurate time prediction or use measured timing to say something about the relative positions of objects in 3D space.</p>
<p>That's because of exactly what you point out; the two coordinate systems have different centers, and also because one (the celestial sphere) is at infinity.</p>
<blockquote>
<p>Why do people use this diagram to derive transformation equations?</p>
</blockquote>
<p>Because for say 99% of the time the things we look at are far enough away (e.g. light years) that the ~6370 (give or take) kilometers between the two centers make no difference at the level that we can measure.</p>
<p><em>When it does matter</em>, astronomers do in fact go to more rigorous calculations in 3D cartesian space rather than exclusively in spherical coordinates, and besides the distance to the object even take into account things like the speed of light and <a href="https://en.wikipedia.org/wiki/Aberration_(astronomy)" rel="nofollow noreferrer">astronomical aberration</a>.</p>
<p>So to</p>
<blockquote>
<p>Shouldn't the equatorial system and the horizontal system have different centers?</p>
</blockquote>
<p>the answer is a resounding <em><strong>Yes!</strong></em></p>

CC BY-SA 4.0
2022-06-12T16:08:18.280Z
Thank you so much! This makes a lot of sense and I definitely understand it now :)
CS8479
CC BY-SA 4.0
2022-06-12T21:24:49.160Z
@CS8479 glad to hear it! Like most things in science we are first exposed to basic principles then step by step we're told "Oh, you thought that you silly goose? No it's much more complicated." For example we learn $F=ma$ and think it's right, then someone discovers [impossible propulsion](https://space.stackexchange.com/a/39509/12102) because that's not true relativistically and $F$ really equals $dp/dt$. "Curiouser and curiouser." :-)
uhoh
CC BY-SA 4.0
2022-06-12T21:38:53.360Z
@CS8479 but please note that I have not answered "How can you derive equations between the Horizontal and Equatorial system...?" part beyond saying that these days with computers astronomers often work in 3D cartesian for solar system bodies at least. If you are asking for the derivation itself which certainly exists because astronomers needed to do this centuries before computers, I can add a bounty as soon as the question is 48 hours old.
uhoh
CC BY-SA 4.0
2022-06-13T16:57:04.113Z
Haha you don't have to worry about adding a bounty! My question was slightly misleading, and I was just wondering about the diagram and the center of each coordinate system.
CS8479
CC BY-SA 4.0
2022-06-12T02:55:09.290Z
2022-06-12T02:55:09.290Z
uhoh
<p>This might be a dumb question.</p>
<p>If the center of the celestial sphere is the center of the earth, and the center of the horizontal system is the observer, why do people use this diagram to derive transformation equations? Shouldn't the equatorial system and the horizontal system have different centers?<a href="https://i.sstatic.net/hrpaT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hrpaT.png" alt="Diagram from Fundamental Astronomy by H. Karttunen, et al." /></a></p>

CC BY-SA 4.0
2022-06-12T03:55:06.347Z
The center of the celestial sphere isn't always the geocenter.  RA/Dec coordinates as viewed from the geocenter are called $Geocentric$ coordinates, and RA/Dec coords based on an observer's location on the Earth's surface are called $Topocentric$ coordinates.  For stars and the planets, the two won't differ that much, but will differ considerably for anything in Earth orbit like the moon or a satellite.  For an example of this, go to JPL Horizons and compute the position for the Moon from two different locations.  You'll notice the RA/Dec likely differ considerably.
Greg Miller
<coordinate><positional-astronomy>
How can you derive equations between the Horizontal and Equatorial system if the two systems have different centers?
CS8479
# How can you derive equations between the Horizontal and Equatorial system if the two systems have different centers?

(Asked by: username_0 on 2022-06-12)

This might be a dumb question.

If the center of the celestial sphere is the center of the earth, and the center of the horizontal system is the observer, why do people use this diagram to derive transformation equations? Shouldn't the equatorial system and the horizontal system have different centers?[![Diagram from Fundamental Astronomy by H. Karttunen, et al.](https://i.sstatic.net/hrpaT.png)](https://i.sstatic.net/hrpaT.png)

**Tags:** coordinate, positional-astronomy
> **Comments:**
> **username_1**: The center of the celestial sphere isn't always the geocenter.  RA/Dec coordinates as viewed from the geocenter are called $Geocentric$ coordinates, and RA/Dec coords based on an observer's location on the Earth's surface are called $Topocentric$ coordinates.  For stars and the planets, the two won't differ that much, but will differ considerably for anything in Earth orbit like the moon or a satellite.  For an example of this, go to JPL Horizons and compute the position for the Moon from two different locations.  You'll notice the RA/Dec likely differ considerably.  

---

## 1 answers

### Answer by username_2 on 2022-06-12. Score: 2

> This might be a dumb question.

This is an *excellent question!*

When we watch a solar eclipse, or the occultation of a star (or planet) by the Moon, and we want accurate timing information (or we want to record it) we must know our 3D position (lat, lon, elev) in order to either get an accurate time prediction or use measured timing to say something about the relative positions of objects in 3D space.

That's because of exactly what you point out; the two coordinate systems have different centers, and also because one (the celestial sphere) is at infinity.

> Why do people use this diagram to derive transformation equations?

Because for say 99% of the time the things we look at are far enough away (e.g. light years) that the ~6370 (give or take) kilometers between the two centers make no difference at the level that we can measure.

*When it does matter*, astronomers do in fact go to more rigorous calculations in 3D cartesian space rather than exclusively in spherical coordinates, and besides the distance to the object even take into account things like the speed of light and [astronomical aberration](https://en.wikipedia.org/wiki/Aberration_(astronomy)).

So to

> Shouldn't the equatorial system and the horizontal system have different centers?

the answer is a resounding ***Yes!***

> **Comments:**
> **username_0**: Thank you so much! This makes a lot of sense and I definitely understand it now :)  
> **username_2**: @CS8479 glad to hear it! Like most things in science we are first exposed to basic principles then step by step we're told "Oh, you thought that you silly goose? No it's much more complicated." For example we learn $F=ma$ and think it's right, then someone discovers [impossible propulsion](https://space.stackexchange.com/a/39509/12102) because that's not true relativistically and $F$ really equals $dp/dt$. "Curiouser and curiouser." :-)  
> **username_2**: @CS8479 but please note that I have not answered "How can you derive equations between the Horizontal and Equatorial system...?" part beyond saying that these days with computers astronomers often work in 3D cartesian for solar system bodies at least. If you are asking for the derivation itself which certainly exists because astronomers needed to do this centuries before computers, I can add a bounty as soon as the question is 48 hours old.  
> **username_0**: Haha you don't have to worry about adding a bounty! My question was slightly misleading, and I was just wondering about the diagram and the center of each coordinate system.  

---

