<p><s>I found an equation on <a href="http://www2.naic.edu/~phil/mbeam/mbeamrotation.html" rel="nofollow noreferrer">this website</a>,</p>

<p><span class="math-container">$\require{cancel} \cancel{\sin{\theta_p}=\sin{\theta_{az}} \cos{\theta_{lat}}/ \cos{\delta}}$</span>,</p>

<p>with <span class="math-container">$\theta_p$</span> being the parllactic angle, <span class="math-container">$\theta_{az}$</span> being the azimuth, <span class="math-container">$\theta_{lat}$</span> being the latitude of the observing stations, and <span class="math-container">$\delta$</span> being the declination.  The website cites this equation as being from "Spherical astronomy, Small pg. 49."  I am unable to find such a book (based on my searches, I'm wondering if the correct reference might actually be "<em>Textbook on</em> Spherical Astronomy, <em>Smart</em>", which I do not have ready access to), but in my experience so far this equation has been correct.</s></p>

<p>The above equation, my original answer, seems to provide something similar to the parallactic angle (something like the complement in each quadrant, with an extra negative sign, but I haven't investigated in detail).  However, when testing against values gotten from astropy, it did not match the astropy values.</p>

<p>Instead, I found the following, taken from <em>Astronomical Algorithms</em> by Jean Meeus, which does match the astropy values:</p>

<p><span class="math-container">$\theta_p = \text{atan2}( \sin{\theta_H}, \tan{\theta_{lat}}\cos{\delta} - \sin{\delta}\cos{\theta_H}),$</span></p>

<p>with <span class="math-container">$\theta_H$</span> being the hour angle, calculated via</p>

<p><span class="math-container">$\theta_H = \text{atan2}( \sin{\theta_{az,mod}}, \cos{\theta_{az,mod}} \sin{\theta_{lat}} + \tan{\theta_{el}}  \cos{ \theta_{lat} } ),$</span></p>

<p>Where <span class="math-container">$\theta_{el}$</span> is the elevation and <span class="math-container">$\theta_{az,mod}$</span> is the azimuth, but defined differently from <span class="math-container">$\theta_{az}$</span> as used originally and elsewhere in my answer.  The zero-point <span class="math-container">$\theta_{az}$</span> is due north, increasing clockwise, and the zero-point <span class="math-container">$\theta_{az,mod}$</span> is due south, increasing clockwise.</p>

<p>Declination can be gotten from latitude, elevation, and azimuth:</p>

<p><span class="math-container">$ \sin{\delta} = \sin{\theta_{el}} \sin{ \theta_{lat} } + \cos{ \theta_{el} } \cos{ \theta_{lat} }  \cos{\theta_{az}} $</span>,</p>

<p>Thus indeed, given an alt-az (or az-el depending on your preferred terminology) position and the latitude of the observing station, it is unnecessary to use RA or an independently determined hour angle in a calculation of parallactic angle. I say "independently determined hour angle" meaning that although hour angle is used in the calculation, it can be determined entirely from az/el and latitude, so one does not need to provide an hour angle for this calculation to work.</p>

CC BY-SA 4.0
2020-04-21T23:13:16.610Z
You wrote "it is unnecessary to use RA or hour angle in a calculation of parallactic angle". However, your solution is based on calculating the hour angle! P.S. Thanks for the update and corrected equations.
JohnHoltz
CC BY-SA 4.0
2020-04-22T18:00:24.250Z
@JohnHoltz, thanks, I've updated accordingly.  While hour angle is used in the calculation, it can be determined entirely from az/el and latitude, and technically could be removed from the final formula (but keeping it around as an intermediate value cleans up the formula a lot).
NeutronStar
CC BY-SA 4.0
2019-11-27T16:37:00.963Z
2020-04-22T17:59:00.597Z
2020-04-22T17:59:00.597Z
NeutronStar
<p>Say we have an observer at a specified latitude and longitude on Earth who would like to observer a specific fixed alt-az sky position (rather than RA and dec), for example to look at geosynchronous satellites.  Since the alt-az position of the observation is not changing, it seems to me that <a href="https://en.wikipedia.org/wiki/Parallactic_angle" rel="nofollow noreferrer">parallactic angle</a> also would not be changing and so should be derivable from just alt-az and lat-long, but I haven't been able to figure out how to do it.  </p>

<p>All the calculations I've seen online use RA or hour angles in calculating parallactic angle, which intermediary values I would like to avoid if mathematically possible.</p>

CC BY-SA 4.0
2019-11-26T16:00:57.963Z
@barrycarter, it wouldn't be 90 degrees minus the elevation.  As one specific example, objects on the meridian, independent of elevation, all have a parallactic angle of zero.
NeutronStar
CC BY-SA 4.0
2019-11-26T17:44:27.717Z
I do not know with 100% certainty, but I think that it will not be possible. The parallactic angle (q) is based on the lines of RA, so that is needed to calculate q (or the hour angle which is related to RA). For the geosat belt, the declination is almost constant (right?). If you only need a good approximation for q, you could assume the declination is constant for all hour angles. Then using hour angles from -90 to +90, the known declination, and latitude, you could calculate the altitude, azimuth, and q for the entire geosat belt.
JohnHoltz
CC BY-SA 4.0
2019-11-26T18:37:02.187Z
@JohnHoltz, I need a bit more general that just GEO.  I have updated my question to indicate "geosynchronous orbit", i.e. even geosynchronous satellites with nonzero inclinations.

Parallactic angle depends on an object's hour circle, which is related to RA, but for objects with hour circles that are fixed in the sky for a given observer (like GEO satellites), the hour circle does not track with RA.  This is what leads me to think parallactic angle can be expressed independent of RA.
NeutronStar
<observational-astronomy><coordinate><mathematics>
How to calculate parallactic angle from a fixed alt-az position?
NeutronStar
# How to calculate parallactic angle from a fixed alt-az position?

(Asked by: username_0 on 2019-11-25)

Say we have an observer at a specified latitude and longitude on Earth who would like to observer a specific fixed alt-az sky position (rather than RA and dec), for example to look at geosynchronous satellites. Since the alt-az position of the observation is not changing, it seems to me that [parallactic angle](https://en.wikipedia.org/wiki/Parallactic_angle) also would not be changing and so should be derivable from just alt-az and lat-long, but I haven't been able to figure out how to do it.

All the calculations I've seen online use RA or hour angles in calculating parallactic angle, which intermediary values I would like to avoid if mathematically possible.

**Tags:** observational-astronomy, coordinate, mathematics
> **Comments:**
> **username_0**: @barrycarter, it wouldn't be 90 degrees minus the elevation.  As one specific example, objects on the meridian, independent of elevation, all have a parallactic angle of zero.  
> **username_1**: I do not know with 100% certainty, but I think that it will not be possible. The parallactic angle (q) is based on the lines of RA, so that is needed to calculate q (or the hour angle which is related to RA). For the geosat belt, the declination is almost constant (right?). If you only need a good approximation for q, you could assume the declination is constant for all hour angles. Then using hour angles from -90 to +90, the known declination, and latitude, you could calculate the altitude, azimuth, and q for the entire geosat belt.  
> **username_0**: @JohnHoltz, I need a bit more general that just GEO.  I have updated my question to indicate "geosynchronous orbit", i.e. even geosynchronous satellites with nonzero inclinations.  Parallactic angle depends on an object's hour circle, which is related to RA, but for objects with hour circles that are fixed in the sky for a given observer (like GEO satellites), the hour circle does not track with RA.  This is what leads me to think parallactic angle can be expressed independent of RA.  

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## 1 answers

### Answer by username_0 on 2019-11-27. Score: 2 | Accepted answer

~~I found an equation on [this website](http://www2.naic.edu/~phil/mbeam/mbeamrotation.html),~~

$\require{cancel} \cancel{\sin{\theta\_p}=\sin{\theta\_{az}} \cos{\theta\_{lat}}/ \cos{\delta}}$,

with $\theta\_p$ being the parllactic angle, $\theta\_{az}$ being the azimuth, $\theta\_{lat}$ being the latitude of the observing stations, and $\delta$ being the declination. The website cites this equation as being from "Spherical astronomy, Small pg. 49." I am unable to find such a book (based on my searches, I'm wondering if the correct reference might actually be "*Textbook on* Spherical Astronomy, *Smart*", which I do not have ready access to), but in my experience so far this equation has been correct.

The above equation, my original answer, seems to provide something similar to the parallactic angle (something like the complement in each quadrant, with an extra negative sign, but I haven't investigated in detail). However, when testing against values gotten from astropy, it did not match the astropy values.

Instead, I found the following, taken from *Astronomical Algorithms* by Jean Meeus, which does match the astropy values:

$\theta\_p = \text{atan2}( \sin{\theta\_H}, \tan{\theta\_{lat}}\cos{\delta} - \sin{\delta}\cos{\theta\_H}),$

with $\theta\_H$ being the hour angle, calculated via

$\theta\_H = \text{atan2}( \sin{\theta\_{az,mod}}, \cos{\theta\_{az,mod}} \sin{\theta\_{lat}} + \tan{\theta\_{el}} \cos{ \theta\_{lat} } ),$

Where $\theta\_{el}$ is the elevation and $\theta\_{az,mod}$ is the azimuth, but defined differently from $\theta\_{az}$ as used originally and elsewhere in my answer. The zero-point $\theta\_{az}$ is due north, increasing clockwise, and the zero-point $\theta\_{az,mod}$ is due south, increasing clockwise.

Declination can be gotten from latitude, elevation, and azimuth:

$ \sin{\delta} = \sin{\theta\_{el}} \sin{ \theta\_{lat} } + \cos{ \theta\_{el} } \cos{ \theta\_{lat} } \cos{\theta\_{az}} $,

Thus indeed, given an alt-az (or az-el depending on your preferred terminology) position and the latitude of the observing station, it is unnecessary to use RA or an independently determined hour angle in a calculation of parallactic angle. I say "independently determined hour angle" meaning that although hour angle is used in the calculation, it can be determined entirely from az/el and latitude, so one does not need to provide an hour angle for this calculation to work.

> **Comments:**
> **username_1**: You wrote "it is unnecessary to use RA or hour angle in a calculation of parallactic angle". However, your solution is based on calculating the hour angle! P.S. Thanks for the update and corrected equations.  
> **username_0**: @JohnHoltz, thanks, I've updated accordingly.  While hour angle is used in the calculation, it can be determined entirely from az/el and latitude, and technically could be removed from the final formula (but keeping it around as an intermediate value cleans up the formula a lot).  

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