<p>For angles of magnitude 1/50 arcsecond (of the order <span class="math-container">$10^{-7}$</span>, the difference between <span class="math-container">$\tan\alpha$</span> and <span class="math-container">$\alpha$</span> is truly negligible (if alpha is in radians) and only a matter of a choice of units if you don't. It is quite correct to say: distance (in AU) = 1/angle (in radians)  to get the distance from the angle in radians.</p>

<p>Using parsecs a means that we don't need to use a different unit for the angle, and the point of parsecs is that we can use "arcseconds".
So it is correct to use distance (in parsec) = 1/parallax angle (in seconds)</p>

CC BY-SA 4.0
2020-05-15T23:07:11.850Z
2020-05-16T03:44:47.747Z
2020-05-16T03:44:47.747Z
James K
<p>For small angles the <a href="https://en.wikipedia.org/wiki/Laurent_series" rel="nofollow noreferrer">Laurent series</a> of <span class="math-container">$1/ \tan(x)$</span> about zero <a href="https://www.wolframalpha.com/input/?i=taylor+series+1%2Ftan%28x%29" rel="nofollow noreferrer">begins</a>:</p>

<p><span class="math-container">$$\frac{1}{\tan(x)} = \frac{1}{x} - \frac{x}{3} - ... = \frac{1}{x} \left( 1 - \frac{x^2}{3} -...\right)$$</span></p>

<p>so for small <span class="math-container">$\alpha$</span> the approximation <span class="math-container">$1/ \alpha$</span> will be high by about the fraction <span class="math-container">$\alpha^2/3$</span>.</p>

<p>This means that the <em>error in the distance</em> determined by parallax will be only 1 part per million if <span class="math-container">$\alpha$</span> is 1.73 mrad or about 0.1° or 357 arcseconds.</p>

<p>The distance error will be 1 part per thousand with a parallax of 3.1° or 11,300 arcseconds.</p>

<p>The only way for such large parallax values to occur is if the object is so close that it is under the gravitational influence of the Sun and therefore this method of distance determination is no longer valid, it becomes an celestial mechanics question instead.</p>

<p>Your teacher is demonstrating that a feel for the problem and understanding the phenomenon are just as important as strict adherence to rote equations. </p>

<p>When using pencil and paper astronomers will use <span class="math-container">$1/ \alpha$</span> and when writing programs meant for general use and scrutinized, astronomers <em>might</em> then use <span class="math-container">$ 1 / \tan(\alpha)$</span>.</p>

CC BY-SA 4.0
2020-05-16T00:08:10.937Z
2020-05-16T00:17:34.670Z
2020-05-16T00:17:34.670Z
uhoh
<p>high school student here. 
In astronomy my teacher says that if the parsec angle (<span class="math-container">$\alpha$</span>) of a star is 1/50 arcsecond, then that star is 50 parsecs away.
I'm not really convinced - calculation of distances from parsecs requires using the formula <span class="math-container">$$\text{distance in AU}=\frac{1 AU}{\tan(\alpha)}.$$</span>
This means that distance is inversely proportional to <span class="math-container">$tan{\alpha}$</span>. Distance is not inversely proportional to <span class="math-container">$\alpha$</span>.</p>

<p>I know you can use a small angle approximation, but that feels to me as very dodgy - you can get fairly large discrepancies in the calculated distance depending on whether you use <span class="math-container">$tan(\alpha)$</span> or <span class="math-container">$\alpha$</span> to find the distance.</p>

<p>So, what is the best practice when calculating distances in parsecs? Is my teacher's approximation acceptable in "real life" astronomy?</p>

CC BY-SA 4.0
2020-05-16T20:30:06.747Z
The error in using $\tan\alpha$ instead of $\alpha$ is less than the error of making the Kessel run in a "time" of 12 parsecs
Hagen von Eitzen
CC BY-SA 4.0
2020-10-18T00:03:55.727Z
if you feel that one of the answers sufficiently answers your question, feel free to click "accept" check mark. If you need more, feel free cot comment.
uhoh
<distances><parallax><parsec>
Do parsecs scale as an inverse proportion with parallax angle?
Hypatia of Alexandria
# Do parsecs scale as an inverse proportion with parallax angle?

(Asked by: username_0 on 2020-05-15)

high school student here.
In astronomy my teacher says that if the parsec angle ($\alpha$) of a star is 1/50 arcsecond, then that star is 50 parsecs away.
I'm not really convinced - calculation of distances from parsecs requires using the formula $$\text{distance in AU}=\frac{1 AU}{\tan(\alpha)}.$$
This means that distance is inversely proportional to $tan{\alpha}$. Distance is not inversely proportional to $\alpha$.

I know you can use a small angle approximation, but that feels to me as very dodgy - you can get fairly large discrepancies in the calculated distance depending on whether you use $tan(\alpha)$ or $\alpha$ to find the distance.

So, what is the best practice when calculating distances in parsecs? Is my teacher's approximation acceptable in "real life" astronomy?

**Tags:** distances, parallax, parsec

> **Comments:**
> **username_1**: The error in using $\tan\alpha$ instead of $\alpha$ is less than the error of making the Kessel run in a "time" of 12 parsecs  
> **username_2**: if you feel that one of the answers sufficiently answers your question, feel free to click "accept" check mark. If you need more, feel free cot comment.  

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## 2 answers

### Answer by username_2 on 2020-05-16. Score: 3 | Accepted answer

For small angles the [Laurent series](https://en.wikipedia.org/wiki/Laurent_series) of $1/ \tan(x)$ about zero [begins](https://www.wolframalpha.com/input/?i=taylor+series+1%2Ftan%28x%29):

$$\frac{1}{\tan(x)} = \frac{1}{x} - \frac{x}{3} - ... = \frac{1}{x} \left( 1 - \frac{x^2}{3} -...\right)$$

so for small $\alpha$ the approximation $1/ \alpha$ will be high by about the fraction $\alpha^2/3$.

This means that the *error in the distance* determined by parallax will be only 1 part per million if $\alpha$ is 1.73 mrad or about 0.1° or 357 arcseconds.

The distance error will be 1 part per thousand with a parallax of 3.1° or 11,300 arcseconds.

The only way for such large parallax values to occur is if the object is so close that it is under the gravitational influence of the Sun and therefore this method of distance determination is no longer valid, it becomes an celestial mechanics question instead.

Your teacher is demonstrating that a feel for the problem and understanding the phenomenon are just as important as strict adherence to rote equations.

When using pencil and paper astronomers will use $1/ \alpha$ and when writing programs meant for general use and scrutinized, astronomers *might* then use $ 1 / \tan(\alpha)$.


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### Answer by username_3 on 2020-05-15. Score: 3

For angles of magnitude 1/50 arcsecond (of the order $10^{-7}$, the difference between $\tan\alpha$ and $\alpha$ is truly negligible (if alpha is in radians) and only a matter of a choice of units if you don't. It is quite correct to say: distance (in AU) = 1/angle (in radians) to get the distance from the angle in radians.

Using parsecs a means that we don't need to use a different unit for the angle, and the point of parsecs is that we can use "arcseconds".
So it is correct to use distance (in parsec) = 1/parallax angle (in seconds)


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