<p>I do not have Stellarium, so I cannot confirm what is causing it with 100% certainty. But by using JPL's <a href="https://ssd.jpl.nasa.gov/horizons.cgi" rel="nofollow noreferrer">Horizons</a> website, I can confirm that the position on the Earth's surface changes the apparent Right Ascension (R.A.) and Declination (DEC) of Venus. The following output is for an observer at 80 degrees W longitude and the latitude listed. The time was chosen so that Venus was approximately 7 hours east of the meridian; that is, the hour angle was approximately 7 hours.</p>

<p><code>Date__(UT)__HR:MN     R.A.__(a-apparent)__DEC
2020-May-17 01:00     05 22 07.26 +27 00 24.9 Equator
2020-May-17 01:00     05 22 09.23 +27 00 01.6 North Pole
2020-May-17 01:00     05 22 09.23 +27 00 48.4 South Pole</code></p>

<p>These results show exactly what you were noticing in Stellarium:</p>

<ul>
<li>The R.A. of Venus changes by approximately 2 seconds from the equator to the poles.</li>
<li>The change is the same going to the North or South pole. I thought the values would not be the same because Venus is north of the equator, but the shift is identical in R.A. for those two locations. (Note that the shift in DEC is not the same.)</li>
</ul>

<p>Observing from the Earth's surface (topocentric position) is important for the Moon if you are looking for occultations or close conjunctions or precise hour angle, but not as important for other objects which are much farther away. Stellarium probably has a setting to change the calculation from topocentric to geocentric, in which case the hour angle discrepancy will disappear.</p>

CC BY-SA 4.0
2020-05-17T19:02:20.403Z
That's great, thanks. I found the setting on Stellarium to set topocentric/planetocentric coordinates, and the discrepancy has gone. Now I need to study the concept of HA a bit more because I thought it was independent of point of view; I thought it was just to do with the meridian of the observer and that of the body.
richTourist
CC BY-SA 4.0
2020-08-07T11:13:34.573Z
Actually, an observer at the south pole wouldn't see Venus, at all, or any celestial object with a significant north declination: they're all below the horizon. What the Horizons model seems to be doing is giving a result for a transparent Earth. Venus is below the horizon, but its parallax is less. An observer where Venus is in the nadir would see zero parallax looking through the Earth.
stretch
CC BY-SA 4.0
2020-05-17T17:56:44.493Z
2020-05-17T17:56:44.493Z
JohnHoltz
<p>Suppose we could look at Earth from Venus at that time:</p>
<p><img src="https://i.sstatic.net/3xhTl.png" alt="Earth as seen from Venus" />
<br/><em><a href="https://space.jpl.nasa.gov/" rel="nofollow noreferrer">Simulated image</a> courtesy NASA/JPL-Caltech</em></p>
<p>The central meridian is 101°56' W.
If you observe Venus from that longitude, its hour angle is 0h 0m 0s whether your latitude is 89° N, 27° N, or 63° S; there is no east-west parallax.</p>
<p>The 0° meridian is just beyond the right limb in this view.
Relative to a geocentric observer, someone observing Venus from the equator at this longitude should see an east-west parallax of
<span class="math-container">$$\frac{52.6''}{2} ~ \sin{102^\circ} = 25.7''$$</span>
of arc, which is
<span class="math-container">$$\frac{25.7''}{\cos{27^\circ}} ~ \frac{1\mathrm{s}}{15''} = 1.92\mathrm{s}$$</span>
in right ascension or hour angle, which agrees with your Stellarium result.</p>

CC BY-SA 4.0
2020-09-01T01:34:11.643Z
2020-09-01T01:34:11.643Z
Mike G
<p>It's not a correction for parallax or refraction. It's a bug. Your understanding that HA is not affected by observer location is correct. (Navigators commonly refer to it as LHA, for local hour angle. The other HA is SHA foe sidereal hour angle.)</p>
<p>Stellarium seems to be mainly for users with an interest in astronomy, users who input their location and what they want to see and get where to look in return. Navigators want to input time and what they saw and get location in return.
That's not the reason for the bug, but it's probably why no one has found it significant. It's a great program, but it seems to have some rough spots as an almanac for celestial navigation. I'm guessing that it throws in corrections for parallax and refraction, which redefines GHA. I also noticed that it uses local standard time, rather than GMT.</p>
<p>The anomaly in the question not a parallax correction because there's no location
on the meridian of the Venus geographical position (GP) that would have a parallax error affecting the apparent GHA of Venus. (The poles are the meeting points of all the meridians.) There would be a significant shift in the apparent declination, though. At the north pole it would be further south. Something like 0.25 minutes of arc.</p>
<p>And it's not a refraction correction because most of the points you tried can't see Venus: it's below the horizon. The two points that can see Venus (prime meridian, 80 and 90 deg N declination) are both about 63 degrees away from Venus' GP. That means the observed altitude is 1.9 arc minutes higher than actual altitude. (Table in the bookmark page in Increments and Corrections in the Nautical Almanac)</p>
<p>You'd be much better off using the Nautical Almanac, rather than Stellarium, for navigation. It's an official, joint publication of the US Naval Observatory and the UK Hydrographics Office. How to use it is in books I know of as Dutton and Bowditch. Both are updated frequently. There are free versions of the current Nautical Almanac available online and many versions of Bowditch. A lot of them are historic: it was first published around 1800.</p>
<p>Wikipedia also has good explanations. About everything, actually.</p>

CC BY-SA 4.0
2020-08-30T23:29:13.483Z
Only correct in the HA = 0h case. If HA ≈ ±6h, the observers are near the Earth's limb as seen from Venus.
Mike G
CC BY-SA 4.0
2020-09-01T00:26:11.840Z
To elaborate on Mike G's comment, the difference in the hour angle is caused be parallax. The OP confirmed that changing from a topocentric view to a geocentric view has made the LHA independent of the latitude.  I do not know navigation but your answer sounds true only if the object (Venus) is on the meridian. It is not in this example.
JohnHoltz
CC BY-SA 4.0
2020-09-02T16:03:03.580Z
@JohnHoltz Meridians of observer and Venus intersect at the poles. Observer at a pole is on both meridians. An observer standing on a pole and looking at some object would see any  correction applied make its apparent altitude and declination go up or down, but none. would make its azimuth go left or right. Your example from the Horizons website has values for apparent RA at the poles exactly equal to the Nautical Almanac derived values with no corrections applied. I still see differences in the Stellarium values but I might have made an error someplace.
stretch
CC BY-SA 4.0
2020-08-30T22:50:57.170Z
2020-08-31T22:30:23.347Z
2020-08-31T22:30:23.347Z
stretch
<p>I am using Stellarium for astronomical data to help in studying celestial navigation.
While trying to track down a source of errors in my sextant reading or sight reductions I noticed that the local hour angle (shown HA on Stellarium's ephemera) of Venus varied if I changed the latitude of my location whilst keeping time constant.</p>

<p>Can anyone explain this. I thought the HA would only be a function of time, longitude, and the position of Venus (RA, Dec). Why would my moving north along my meridian affect the HA of a body? Aren't both my meridian and Venus' fixed (time is stopped), so the angle between them wouldn't change?</p>

<p>I noticed that it changed systematically. For the date and meridian in  question (16/5/2020; 21:30:00 0° Long.) it varied from 6h 47m 41.86s at both the N and S poles, to 6h 47m 43.78s at the Equator. delta=1.92s</p>

<p>I know it's only a tiny amount, but I would like to understand what the principle behind it is. Everything I found on Google seemed to suggest it shouldn't change. I wondered if it was something to do with parallax changing as I changed position.</p>

<p>Thanks for any help.</p>

CC BY-SA 4.0
2020-05-17T15:33:45.663Z
If you literally tried 90 degrees latitude, the hour angle is undefined in reality, but perhaps it can be calculated mathematically. To avoid a possible issue, what are the results at 80 latitude?
JohnHoltz
CC BY-SA 4.0
2020-05-17T15:41:34.063Z
Hi, thanks for answering. At 80° both North and South HA = 6h 47m 42.19s
richTourist
CC BY-SA 4.0
2020-05-17T15:52:59.387Z
You may be correct that the calculation is using the topocentric position (parallax) instead of the geocentric position.  However,  Venus is currently near 27 degrees declination, so I would expect a slight difference between N and S pole. I would need to pull out a book to confirm that. Someone may provide the "real" answer before I get to that.
JohnHoltz
CC BY-SA 4.0
2020-05-17T16:09:22.853Z
I can't see why it should make a difference, though. I thought the hour angle was measured between two meridians, not from the observer's point of view. wikipedia: The hour angle of a point is the angle between two planes: one containing Earth's axis and the zenith (the meridian plane), and the other containing Earth's axis and the given point (the hour circle passing through the point).
richTourist
CC BY-SA 4.0
2020-05-17T17:11:58.410Z
The hour angle is the difference between two right ascension, and the right ascension of Venus changes based on your location due to parallax. The difference would be small, but 2 seconds HA given by Stellarium is also small!
JohnHoltz
<celestial-mechanics><navigation>
Why does the latitude of an observer affect the hour angle of a body in Stellarium?
richTourist
# Why does the latitude of an observer affect the hour angle of a body in Stellarium?

(Asked by: username_0 on 2020-05-17)

I am using Stellarium for astronomical data to help in studying celestial navigation.
While trying to track down a source of errors in my sextant reading or sight reductions I noticed that the local hour angle (shown HA on Stellarium's ephemera) of Venus varied if I changed the latitude of my location whilst keeping time constant.

Can anyone explain this. I thought the HA would only be a function of time, longitude, and the position of Venus (RA, Dec). Why would my moving north along my meridian affect the HA of a body? Aren't both my meridian and Venus' fixed (time is stopped), so the angle between them wouldn't change?

I noticed that it changed systematically. For the date and meridian in question (16/5/2020; 21:30:00 0° Long.) it varied from 6h 47m 41.86s at both the N and S poles, to 6h 47m 43.78s at the Equator. delta=1.92s

I know it's only a tiny amount, but I would like to understand what the principle behind it is. Everything I found on Google seemed to suggest it shouldn't change. I wondered if it was something to do with parallax changing as I changed position.

Thanks for any help.

**Tags:** celestial-mechanics, navigation

> **Comments:**
> **username_1**: If you literally tried 90 degrees latitude, the hour angle is undefined in reality, but perhaps it can be calculated mathematically. To avoid a possible issue, what are the results at 80 latitude?  
> **username_0**: Hi, thanks for answering. At 80° both North and South HA = 6h 47m 42.19s  
> **username_1**: You may be correct that the calculation is using the topocentric position (parallax) instead of the geocentric position.  However,  Venus is currently near 27 degrees declination, so I would expect a slight difference between N and S pole. I would need to pull out a book to confirm that. Someone may provide the "real" answer before I get to that.  
> **username_0**: I can't see why it should make a difference, though. I thought the hour angle was measured between two meridians, not from the observer's point of view. wikipedia: The hour angle of a point is the angle between two planes: one containing Earth's axis and the zenith (the meridian plane), and the other containing Earth's axis and the given point (the hour circle passing through the point).  
> **username_1**: The hour angle is the difference between two right ascension, and the right ascension of Venus changes based on your location due to parallax. The difference would be small, but 2 seconds HA given by Stellarium is also small!  

---

## 3 answers

### Answer by username_1 on 2020-05-17. Score: 2 | Accepted answer

I do not have Stellarium, so I cannot confirm what is causing it with 100% certainty. But by using JPL's [Horizons](https://ssd.jpl.nasa.gov/horizons.cgi) website, I can confirm that the position on the Earth's surface changes the apparent Right Ascension (R.A.) and Declination (DEC) of Venus. The following output is for an observer at 80 degrees W longitude and the latitude listed. The time was chosen so that Venus was approximately 7 hours east of the meridian; that is, the hour angle was approximately 7 hours.

`Date__(UT)__HR:MN R.A.__(a-apparent)__DEC
2020-May-17 01:00 05 22 07.26 +27 00 24.9 Equator
2020-May-17 01:00 05 22 09.23 +27 00 01.6 North Pole
2020-May-17 01:00 05 22 09.23 +27 00 48.4 South Pole`

These results show exactly what you were noticing in Stellarium:

* The R.A. of Venus changes by approximately 2 seconds from the equator to the poles.
* The change is the same going to the North or South pole. I thought the values would not be the same because Venus is north of the equator, but the shift is identical in R.A. for those two locations. (Note that the shift in DEC is not the same.)

Observing from the Earth's surface (topocentric position) is important for the Moon if you are looking for occultations or close conjunctions or precise hour angle, but not as important for other objects which are much farther away. Stellarium probably has a setting to change the calculation from topocentric to geocentric, in which case the hour angle discrepancy will disappear.

> **Comments:**
> **username_0**: That's great, thanks. I found the setting on Stellarium to set topocentric/planetocentric coordinates, and the discrepancy has gone. Now I need to study the concept of HA a bit more because I thought it was independent of point of view; I thought it was just to do with the meridian of the observer and that of the body.  
> **username_2**: Actually, an observer at the south pole wouldn't see Venus, at all, or any celestial object with a significant north declination: they're all below the horizon. What the Horizons model seems to be doing is giving a result for a transparent Earth. Venus is below the horizon, but its parallax is less. An observer where Venus is in the nadir would see zero parallax looking through the Earth.  

---

### Answer by username_3 on 2020-09-01. Score: 2

Suppose we could look at Earth from Venus at that time:

![Earth as seen from Venus](https://i.sstatic.net/3xhTl.png)
  
*[Simulated image](https://space.jpl.nasa.gov/) courtesy NASA/JPL-Caltech*

The central meridian is 101°56' W.
If you observe Venus from that longitude, its hour angle is 0h 0m 0s whether your latitude is 89° N, 27° N, or 63° S; there is no east-west parallax.

The 0° meridian is just beyond the right limb in this view.
Relative to a geocentric observer, someone observing Venus from the equator at this longitude should see an east-west parallax of
$$\frac{52.6''}{2} ~ \sin{102^\circ} = 25.7''$$
of arc, which is
$$\frac{25.7''}{\cos{27^\circ}} ~ \frac{1\mathrm{s}}{15''} = 1.92\mathrm{s}$$
in right ascension or hour angle, which agrees with your Stellarium result.


---

### Answer by username_2 on 2020-08-30. Score: 1

It's not a correction for parallax or refraction. It's a bug. Your understanding that HA is not affected by observer location is correct. (Navigators commonly refer to it as LHA, for local hour angle. The other HA is SHA foe sidereal hour angle.)

Stellarium seems to be mainly for users with an interest in astronomy, users who input their location and what they want to see and get where to look in return. Navigators want to input time and what they saw and get location in return.
That's not the reason for the bug, but it's probably why no one has found it significant. It's a great program, but it seems to have some rough spots as an almanac for celestial navigation. I'm guessing that it throws in corrections for parallax and refraction, which redefines GHA. I also noticed that it uses local standard time, rather than GMT.

The anomaly in the question not a parallax correction because there's no location
on the meridian of the Venus geographical position (GP) that would have a parallax error affecting the apparent GHA of Venus. (The poles are the meeting points of all the meridians.) There would be a significant shift in the apparent declination, though. At the north pole it would be further south. Something like 0.25 minutes of arc.

And it's not a refraction correction because most of the points you tried can't see Venus: it's below the horizon. The two points that can see Venus (prime meridian, 80 and 90 deg N declination) are both about 63 degrees away from Venus' GP. That means the observed altitude is 1.9 arc minutes higher than actual altitude. (Table in the bookmark page in Increments and Corrections in the Nautical Almanac)

You'd be much better off using the Nautical Almanac, rather than Stellarium, for navigation. It's an official, joint publication of the US Naval Observatory and the UK Hydrographics Office. How to use it is in books I know of as Dutton and Bowditch. Both are updated frequently. There are free versions of the current Nautical Almanac available online and many versions of Bowditch. A lot of them are historic: it was first published around 1800.

Wikipedia also has good explanations. About everything, actually.

> **Comments:**
> **username_3**: Only correct in the HA = 0h case. If HA ≈ ±6h, the observers are near the Earth's limb as seen from Venus.  
> **username_1**: To elaborate on username_3 G's comment, the difference in the hour angle is caused be parallax. The OP confirmed that changing from a topocentric view to a geocentric view has made the LHA independent of the latitude.  I do not know navigation but your answer sounds true only if the object (Venus) is on the meridian. It is not in this example.  
> **username_2**: @username_1 Meridians of observer and Venus intersect at the poles. Observer at a pole is on both meridians. An observer standing on a pole and looking at some object would see any  correction applied make its apparent altitude and declination go up or down, but none. would make its azimuth go left or right. Your example from the Horizons website has values for apparent RA at the poles exactly equal to the Nautical Almanac derived values with no corrections applied. I still see differences in the Stellarium values but I might have made an error someplace.  

---

