<p>You are forgetting parallax. </p>

<p>Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator.</p>

<p>From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear to be in the southern sky, and your declination coordinate will likely be south of 0°.</p>

CC BY-SA 3.0
2018-01-24T17:04:43.350Z
Thank you, you helped me finally pinpoint the essence of the problem. However, I knew there would have to be some corrections for locating the satellite in observer's frame (altitude and azimuth coordinates) because of the fact that satellite isn't a very distant object, but I thought that right ascension and declination were in earth-centered-inertial frame, spherical equivalent of cartesian position vector, independent of the observer's location. "Declination in astronomy is comparable to geographic latitude, projected onto the celestial sphere" https://en.wikipedia.org/wiki/Declination
Vekszor
CC BY-SA 3.0
2018-01-24T17:34:28.640Z
The software routines by D. Vallado may come in useful to cross-check the parallax correction described by @Mick.
https://celestrak.com/software/vallado-sw.asp
Ela
CC BY-SA 3.0
2018-01-25T00:48:29.023Z
@Vekszor For distant objects that is certainly true but on Solar System scales corrections do need to be made for the observer location, and for Earth-orbit position calculations even more so.
Mick
CC BY-SA 3.0
2018-01-24T01:26:12.827Z
2018-01-24T01:26:12.827Z
Mick
<p>My task is to calculate current declination and right ascension for some artificial Earth satellite. For example ISS. </p>

<p>I am using latest TLE data for extracting keplerian orbital elements and then calculate current cartesian orbital state vectors in ECI frame. 
From the position vector I directly calculate declination and right ascension, using cartesian to spherical coordinates conversion.</p>

<p>Right ascension gets calculated correctly but I am having problems getting the same result for declination as the one displayed by satellite trackers.</p>

<p>Speaking of satellite trackers, I don't understand how is their declination value even correct.</p>

<p>Declination is an angle measured north or south of the celestial equator.
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/9/98/Ra_and_dec_on_celestial_sphere.png/800px-Ra_and_dec_on_celestial_sphere.png" alt="declination visually"></p>

<p>Celestial equator of course is on the same plane as the equator of Earth.</p>

<p>Then how is it possible that, when a satellite (ISS in this example) is almost directly above Earth's equator, satellite trackers calculate declination to be around -29° ?</p>

<p><img src="https://i.sstatic.net/54jYl.png" alt="tracker 1">
<img src="https://i.sstatic.net/aElXu.png" alt="tracker 2">
<img src="https://i.sstatic.net/Q7YRf.png" alt="tracker 3"></p>

<p>Shouldn't the declination angle be zero? 
My calculations for this "ISS above equator" example did give declination to be close to 0°.</p>

<p>Who is doing the calculations wrong and why?</p>

<orbit><satellite><declination><right-ascension>
Declination calculation
Vekszor
# Declination calculation

(Asked by: username_0 on 2018-01-24)

My task is to calculate current declination and right ascension for some artificial Earth satellite. For example ISS.

I am using latest TLE data for extracting keplerian orbital elements and then calculate current cartesian orbital state vectors in ECI frame.
From the position vector I directly calculate declination and right ascension, using cartesian to spherical coordinates conversion.

Right ascension gets calculated correctly but I am having problems getting the same result for declination as the one displayed by satellite trackers.

Speaking of satellite trackers, I don't understand how is their declination value even correct.

Declination is an angle measured north or south of the celestial equator.
![declination visually](https://upload.wikimedia.org/wikipedia/commons/thumb/9/98/Ra_and_dec_on_celestial_sphere.png/800px-Ra_and_dec_on_celestial_sphere.png)

Celestial equator of course is on the same plane as the equator of Earth.

Then how is it possible that, when a satellite (ISS in this example) is almost directly above Earth's equator, satellite trackers calculate declination to be around -29° ?

![tracker 1](https://i.sstatic.net/54jYl.png)
![tracker 2](https://i.sstatic.net/aElXu.png)
![tracker 3](https://i.sstatic.net/Q7YRf.png)

Shouldn't the declination angle be zero?
My calculations for this "ISS above equator" example did give declination to be close to 0°.

Who is doing the calculations wrong and why?

**Tags:** orbit, satellite, declination, right-ascension


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## 1 answers

### Answer by username_1 on 2018-01-24. Score: 2 | Accepted answer

You are forgetting parallax.

Something which is very distant and lies on the celestial equator will have a declination of 0°, but if it is nearby then its declination will only be 0° to an observer located along Earth's equator.

From your location in Zagreb, something like the ISS, even if it is crossing Earth's equator at a longitude of ~16° will appear to be in the southern sky, and your declination coordinate will likely be south of 0°.

> **Comments:**
> **username_0**: Thank you, you helped me finally pinpoint the essence of the problem. However, I knew there would have to be some corrections for locating the satellite in observer's frame (altitude and azimuth coordinates) because of the fact that satellite isn't a very distant object, but I thought that right ascension and declination were in earth-centered-inertial frame, spherical equivalent of cartesian position vector, independent of the observer's location. "Declination in astronomy is comparable to geographic latitude, projected onto the celestial sphere" https://en.wikipedia.org/wiki/Declination  
> **username_2**: The software routines by D. Vallado may come in useful to cross-check the parallax correction described by @username_1. https://celestrak.com/software/vallado-sw.asp  
> **username_1**: @username_0 For distant objects that is certainly true but on Solar System scales corrections do need to be made for the observer location, and for Earth-orbit position calculations even more so.  

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