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This may be a bit of a noob question, but I am wondering how exactly to get the parallax angle in degrees when looking at a nearby star relative to background stars.

I have found the information online of how to do the calculations etc, but they all say to look at how much the star has moved and you have the angle. Really? Where has this angle number actually come from? Nowhere seems to give that info (that I can find).

The method I assume is the correct one is to measure how much the star has shifted (relative to background stars) in arcseconds, then convert this to degrees. Assuming 1" of movement, the parallax angle would be 0.000277778°.

Is this the correct way to get the value in degrees, or am I just being silly? This is something that has been bugging me for a while and if it really is that simple, why don't online sources just say, measure the distance in arcseconds and convert rather than just say "Look how much the star has moved and you have your angle", and assume you know how to get it?

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  • $\begingroup$ Probably because arcseconds are also angular measurements, and arcseconds or radians are more convenient or more customary angular units than degrees for most things you'd be calculating from parallax measurement. $\endgroup$
    – notovny
    Jun 7 at 10:24
  • $\begingroup$ It's not quite so simple. The proper motion of a star often exceeds the parallax movement, which must also be accounted for. $\endgroup$ Jun 7 at 12:03
  • $\begingroup$ @GregMiller I understand that. I was just asking if that is the way to do it. Converting the arcseconds to degrees. Everything else I'm fine with. I didn't want to overcomplicate it so just singled out the 1 thing I wanted to know $\endgroup$
    – MCG
    Jun 7 at 12:10
  • $\begingroup$ @notovny so are you saying yes that is the way to do it? or not? $\endgroup$
    – MCG
    Jun 7 at 12:11
  • $\begingroup$ That is correct. To convert degrees to arcseconds, you divide by 3600. So 1" = .00027777... degrees. $\endgroup$ Jun 7 at 12:15

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Your assertion that 1" = 0.000277778° is correct. An arcsecond is 1/60th of an arcminute, and an arcminute is 1/60th of a degree. So, dividing arcseconds by 3600, will produce a result in degrees.

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  • $\begingroup$ However, the result in arcseconds is much more convenient because you can easily find the distance of the star in parsecs: $d=\frac 1 \pi$, where $\pi$ is the parallax value in arcseconds. $\endgroup$
    – User123
    Jun 8 at 13:48
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    $\begingroup$ Actually, that's an approximation, the more precise equation is $$ d = \frac{1au}{\tan \theta} $$, which uses degrees. $\endgroup$ Jun 8 at 13:59
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    $\begingroup$ Its an extremely good approximation! For $\theta$ converted to radius, where $\theta=1"\cdot\frac{\pi}{180\cdot 3600}$ then $\frac{1}{\tan(\theta)}\approx\frac{1}{\theta}$ is accurate to better than 1 part in 10^10. $\endgroup$
    – Sheldon
    Jun 8 at 17:59
  • $\begingroup$ Ooh, I knew the d = 1/p (parallax) equation, but haven't seen the other one. I might re-do my stuff using this equation instead $\endgroup$
    – MCG
    Jun 9 at 7:04
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Suppose you are asked to "Find the distance between two points drawn on a piece of paper" Is this the correct procedure?

  1. Measure with a ruler and get a length in cm,
  2. Convert to miles (1 cm =0.000000621731).

The second part of that process seems unnecessary. There's no need to convert cm to miles when you are finding a distance. You can measure distance in cm or miles, (or km or light-years etc). You can choose whichever units are convenient.

Similarly there is no need to convert arcseconds to degrees. You can measure angles in arcseconds or degrees (or radians or mils etc). You can choose whichever units are convenient.

For parallax calculations, arcseconds are a convenient unit to describe an angle. Multiplying by 0.00027777... would convert from arcseconds to degrees. But this is not necessary and does, in fact, make the next stage of the calculation harder.

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    $\begingroup$ The cool kids use attoparsecs for small distances. ;) $\endgroup$
    – PM 2Ring
    Jun 8 at 19:38
  • $\begingroup$ It wasn't really the fact that I particularly wanted to use degrees, I just wanted to be sure I could do it correctly if I wanted to $\endgroup$
    – MCG
    Jun 9 at 7:07

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