<p>Hard to follow your query. Maybe if you labelled <span class="math-container">$i_{\rm star}$</span> and <span class="math-container">$i_{\rm LOS}$</span>. The inclination angle is defined in exactly the way you show - but the diagram is not unique.</p>
<p>Perhaps the source of confusion is that just knowing the inclination angle does <em>not</em> uniquely identify<span class="math-container">$^{\dagger}$</span> the orientation of a binary system in space. i.e. The vector that you have labelled <span class="math-container">$\vec{N}$</span>, which could be the orbital angular momentum vector, can be rotated around the line of sight without changing observables like the measured radial velocity of the binary components. Furthermore, most observables are also agnostic to whether the inclination is <span class="math-container">$i$</span> or <span class="math-container">$180^\circ - i$</span>.</p>
<p><span class="math-container">$\dagger$</span> Except that an inclination of zero or 180 degrees <em>does</em> uniquely define the orientation.</p>

CC BY-SA 4.0
2024-04-16T16:09:53.707Z
2024-04-16T16:20:54.227Z
2024-04-16T16:20:54.227Z
ProfRob
<p>I'm encountering some difficulty understanding the concept of line of sight inclination. Typically, we assume that the observer is positioned perpendicular to the system being observed. Consequently, the orbital inclination (which is crucial for calculating the radial velocity of the system) is represented by the angle 'i,' as illustrated in the figure below.
<a href="https://i.sstatic.net/4awufALj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4awufALj.png" alt="enter image description here" /></a></p>
<p>My question is this: What happens if the observer's line of sight (LOS) is tilted at an angle? Moreover, what would occur if both the stellar system and the observer's line of sight are tilted?</p>
<p>I would deduce that the orbital inclination (<span class="math-container">$i$</span>) can be calculated using the equation <span class="math-container">$$i = \arccos\left(\cos\left(i_{\text{LOS}} \right) \cos\left(i_{\text{star}} \right)  \right) \,.$$</span>
Here, <span class="math-container">$i_{\text{LOS}}$</span> represents the inclination on the line of sight plane, and <span class="math-container">$i_{\text{star}}$</span> denotes the inclination on the star's plane.  Or you just sum the inclinations?</p>

<observational-astronomy><astrophysics><inclination>
Relative inclination with respect to line of sight
RKerr
# Relative inclination with respect to line of sight

(Asked by: username_0 on 2024-04-16)

I'm encountering some difficulty understanding the concept of line of sight inclination. Typically, we assume that the observer is positioned perpendicular to the system being observed. Consequently, the orbital inclination (which is crucial for calculating the radial velocity of the system) is represented by the angle 'i,' as illustrated in the figure below.
[![enter image description here](https://i.sstatic.net/4awufALj.png)](https://i.sstatic.net/4awufALj.png)

My question is this: What happens if the observer's line of sight (LOS) is tilted at an angle? Moreover, what would occur if both the stellar system and the observer's line of sight are tilted?

I would deduce that the orbital inclination ($i$) can be calculated using the equation $$i = \arccos\left(\cos\left(i\_{\text{LOS}} \right) \cos\left(i\_{\text{star}} \right) \right) \,.$$
Here, $i\_{\text{LOS}}$ represents the inclination on the line of sight plane, and $i\_{\text{star}}$ denotes the inclination on the star's plane. Or you just sum the inclinations?

**Tags:** observational-astronomy, astrophysics, inclination

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## 1 answers

### Answer by username_1 on 2024-04-16. Score: 5 | Accepted answer

Hard to follow your query. Maybe if you labelled $i\_{\rm star}$ and $i\_{\rm LOS}$. The inclination angle is defined in exactly the way you show - but the diagram is not unique.

Perhaps the source of confusion is that just knowing the inclination angle does *not* uniquely identify$^{\dagger}$ the orientation of a binary system in space. i.e. The vector that you have labelled $\vec{N}$, which could be the orbital angular momentum vector, can be rotated around the line of sight without changing observables like the measured radial velocity of the binary components. Furthermore, most observables are also agnostic to whether the inclination is $i$ or $180^\circ - i$.

$\dagger$ Except that an inclination of zero or 180 degrees *does* uniquely define the orientation.


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