<p>Given the angular diameter $a$ in <strong>radians</strong> and the distance $d$ in Mpc, you can get the actual diameter $D$ from:
$$D = d\tan{a}$$</p>

<p>Using the small <a href="http://en.wikipedia.org/wiki/Small-angle_approximation" rel="nofollow">angle approximation</a>, you get:
$$D = da$$</p>

<p>$a$ is in radians, so to get the distance in Mpc from the angular diameter in arc seconds you'd need to convert the angle in arc seconds to the angle in radians: $a = \frac{2\pi A}{360\times3600}$, where the factor $3600$ is used to convert arc seconds to degrees, and $\frac{2\pi}{360}$ to convert from degrees to radians:</p>

<p>To get the diameter in kpc:
$$D = 1000\times d \frac{2\pi A}{360\times3600}$$<br>
$$D = d \frac{\pi A}{648}$$
$$D \approx \frac{dA}{206} $$</p>

<p>where $d$ is in $\textrm{Mpc}$, $D$ is in $\textrm{kpc}$, and $A$ is in $\textrm{arcsec}$.</p>

<p>Here $D$ is the diameter for round objects (even for disk galaxies seen at an angle). If the object is not round then this would normally be the maximum diameter. </p>

<p>[EDIT (see answer and comments from HDE 226868)] For irregular galaxies you may also need to have more information (viewing angle) to find the <strong>real</strong> maximum diameter of the galaxy. But that information is (I think) only available for galaxies in the local group.</p>

CC BY-SA 3.0
2014-10-19T14:40:41.590Z
2014-10-19T15:19:14.720Z
2014-10-19T15:19:14.720Z
Dieudonné
<p>I'm not sure I understand the second formula (could you explain the variables?), but I <em>do</em> know that you can use <a href="https://en.wikipedia.org/wiki/Parallax" rel="nofollow">parallax</a> in certain situations. Parallax works by using the idea that objects appear to be in different places when you (the observer) are at different places - i.e. at different points in the Earth's orbit. This graphic gives a good description of the process.</p>

<p><img src="https://upload.wikimedia.org/wikipedia/commons/e/e3/Stellarparallax2.svg" alt="enter image description here">
Credit: <a href="https://upload.wikimedia.org/wikipedia/commons/e/e3/Stellarparallax2.svg" rel="nofollow">Wikipedia</a>.</p>

<p>What I think your friend is trying to get at is that if a galaxy is tilted at a certain angle, it will appear <a href="https://en.wikipedia.org/wiki/Perspective_(graphical)#Foreshortening" rel="nofollow">foreshortened</a>, and it will seem smaller than it should at that distance. This could lead you to make an incorrect measurement of its <em>angular</em> diameter and therefore an incorrect measurement of its actual diameter. You could correct this using trigonometry if you knew just at what angle the galaxy was tilted, but it would be hard to measure that angle. You <em>could</em> avoid this whole dilemma if the galaxy was perfectly circular, because you would always see the longer side even if it was foreshortened, but that could be an invalid assumption.</p>

<p>By the way, the reason this isn't a problem with ideal, perfectly spherical objects is that they are perfectly spherical and cannot appear foreshortened.</p>

CC BY-SA 3.0
2014-10-19T14:18:03.367Z
But is foreshortening not 'just' an optical illusion? If you use instrumentation to measure the angular diameter, this should not be a problem?
Dieudonné
CC BY-SA 3.0
2014-10-19T14:19:23.640Z
@Dieudonne I thought it *was* just an optical illusion. It depends on the viewer's perspective, right?
HDE 226868
CC BY-SA 3.0
2014-10-19T15:15:11.323Z
You're right, it depends on the user's perspective. I guess foreshortening might be an issue for irregular galaxies although I'm not sure how this is related to the equation in the question?
Dieudonné
CC BY-SA 3.0
2014-10-19T15:16:26.317Z
@Dieudonne It's not related to the equation; I was trying to explain why the original formula would not work in all cases. Your answer better addresses the question, though.
HDE 226868
CC BY-SA 3.0
2014-10-19T13:57:21.393Z
2014-10-19T13:57:21.393Z
HDE 226868
<p>I've been trying to figure out how to find the diameter in kpc of any galaxy given that one knows the distance to it in Mpc (d) and also the angular diameter/size (A) in arcsec. I had a formula before to find the diameter of a planet: $D=\frac{2\pi dA}{360degrees}$ but this is only if the celestial object is round. Then I asked my friend who gave me another formula $D=d*10^6 *sin (\frac{A}{3600})$</p>

<p>My question is, is the second formula the correct one if I want to find the diameter of any galaxy no matter what shape it is or is there another one which I do not know of?</p>

<galaxy><distances>
Diameter of any galaxy
Kennan
# Diameter of any galaxy

(Asked by: username_0 on 2014-10-19)

I've been trying to figure out how to find the diameter in kpc of any galaxy given that one knows the distance to it in Mpc (d) and also the angular diameter/size (A) in arcsec. I had a formula before to find the diameter of a planet: $D=\frac{2\pi dA}{360degrees}$ but this is only if the celestial object is round. Then I asked my friend who gave me another formula $D=d\*10^6 \*sin (\frac{A}{3600})$

My question is, is the second formula the correct one if I want to find the diameter of any galaxy no matter what shape it is or is there another one which I do not know of?

**Tags:** galaxy, distances

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## 2 answers

### Answer by username_1 on 2014-10-19. Score: 5

Given the angular diameter $a$ in **radians** and the distance $d$ in Mpc, you can get the actual diameter $D$ from:
$$D = d\tan{a}$$

Using the small [angle approximation](http://en.wikipedia.org/wiki/Small-angle_approximation), you get:
$$D = da$$

$a$ is in radians, so to get the distance in Mpc from the angular diameter in arc seconds you'd need to convert the angle in arc seconds to the angle in radians: $a = \frac{2\pi A}{360\times3600}$, where the factor $3600$ is used to convert arc seconds to degrees, and $\frac{2\pi}{360}$ to convert from degrees to radians:

To get the diameter in kpc:
$$D = 1000\times d \frac{2\pi A}{360\times3600}$$  
$$D = d \frac{\pi A}{648}$$
$$D \approx \frac{dA}{206} $$

where $d$ is in $\textrm{Mpc}$, $D$ is in $\textrm{kpc}$, and $A$ is in $\textrm{arcsec}$.

Here $D$ is the diameter for round objects (even for disk galaxies seen at an angle). If the object is not round then this would normally be the maximum diameter.

[EDIT (see answer and comments from HDE 226868)] For irregular galaxies you may also need to have more information (viewing angle) to find the **real** maximum diameter of the galaxy. But that information is (I think) only available for galaxies in the local group.


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### Answer by username_2 on 2014-10-19. Score: 1

I'm not sure I understand the second formula (could you explain the variables?), but I *do* know that you can use [parallax](https://en.wikipedia.org/wiki/Parallax) in certain situations. Parallax works by using the idea that objects appear to be in different places when you (the observer) are at different places - i.e. at different points in the Earth's orbit. This graphic gives a good description of the process.

![enter image description here](https://upload.wikimedia.org/wikipedia/commons/e/e3/Stellarparallax2.svg)
Credit: [Wikipedia](https://upload.wikimedia.org/wikipedia/commons/e/e3/Stellarparallax2.svg).

What I think your friend is trying to get at is that if a galaxy is tilted at a certain angle, it will appear [foreshortened](https://en.wikipedia.org/wiki/Perspective_(graphical)#Foreshortening), and it will seem smaller than it should at that distance. This could lead you to make an incorrect measurement of its *angular* diameter and therefore an incorrect measurement of its actual diameter. You could correct this using trigonometry if you knew just at what angle the galaxy was tilted, but it would be hard to measure that angle. You *could* avoid this whole dilemma if the galaxy was perfectly circular, because you would always see the longer side even if it was foreshortened, but that could be an invalid assumption.

By the way, the reason this isn't a problem with ideal, perfectly spherical objects is that they are perfectly spherical and cannot appear foreshortened.

> **Comments:**
> **username_1**: But is foreshortening not 'just' an optical illusion? If you use instrumentation to measure the angular diameter, this should not be a problem?  
> **username_2**: @Dieudonne I thought it *was* just an optical illusion. It depends on the viewer's perspective, right?  
> **username_1**: You're right, it depends on the user's perspective. I guess foreshortening might be an issue for irregular galaxies although I'm not sure how this is related to the equation in the question?  
> **username_2**: @Dieudonne It's not related to the equation; I was trying to explain why the original formula would not work in all cases. Your answer better addresses the question, though.  

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