<p>Based on the math in the Wikipedia article about the <a href="https://en.wikipedia.org/wiki/Airy_disk#Mathematical_details" rel="nofollow noreferrer">Airy Disk</a>:</p>

<p>The intensity of the starlight as a function of angular distance $\theta$ from the center is reasonably approximated by a Gaussian $$ I(\theta) = I_0 \exp(-\theta^2/(2\sigma^2)) $$ where $\sigma$ is a measure for the width of the intensity profile, which for an ideal optical system and diffraction-limited images is $$ \sigma \approx 0.42 \lambda N $$ (where $\lambda$ is the wavelength and $N$ is the f-number) and which is independent of the peak intensity $I_0$.  For a non-ideal or not-diffraction-limited system, the Gaussian should still provide a reasonable approximation, but then $\sigma$ is (much) larger than for the ideal system.</p>

<p>The apparent edge of a star in an image occurs where the intensity $I(\theta)$ drops below a certain minimum $I_\text{limit}$, which (from the first equation) corresponds to an angular distance $$ \theta_\text{edge} = \sigma\sqrt{2\log(I_0/I_\text{limit})}$$</p>

<p>We can rewrite this in terms of visual magnitude $V$.  For fixed $\sigma$, $I_0$ is a measure for the brightness of the star.  Brightness is related to magnitude $V$ through $$ V = -2.5 \log_{10}(I/I_\text{ref}) $$ (with $I_\text{ref}$ the reference intensity for the magnitude scale) so $$\log(I_0/I_\text{limit}) = 0.4 \log(10) (V_\text{limit} - V_0)$$ and
\begin{eqnarray} 
   \theta_\text{edge} &amp; = &amp; \sigma \sqrt{0.8 \log(10) (V_\text{limit} - V_0)} \approx 1.36 \sigma \sqrt{V_\text{limit} - V_0}
\\ V_\text{limit} - V_0 &amp; = &amp; \frac{\theta_\text{edge}^2}{0.8 \log(10) \sigma^2} \approx 0.543 \left( \frac{\theta_\text{edge}}{\sigma} \right)^2
\end{eqnarray}</p>

<p>So, the difference between the limiting magnitude and the star's magnitude is proportional to the apparent angular area of the star in the image.</p>

CC BY-SA 3.0
2017-09-05T14:09:04.523Z
This prescription is reasonable, but has nothing to do with diffraction really. The angular size of stellar images is actually the same because of atmospheric "seeing", which would normally be much bigger than the Airy disk $\sigma$ you define. With CCDs there are other effects like charge transfer efficiency and saturation to consider; and there are serious non-linearities in the response of photographic plates.
ProfRob
CC BY-SA 3.0
2017-09-05T15:00:34.080Z
My description does not apply only to diffraction-limited cases.  Both diffraction-limited and seeing-limited cases are reasonably described using the Gaussian.  Saturation and non-linearities in the response are only important if they noticeably change the size of the stellar image defined by the $I_\text{limit}$ contour, which implies significant transfer by the recording device or medium of signal across where the star image's edge would have otherwise been.  It would be interesting to see a test of the final relationship based on the picture.
Louis Strous
CC BY-SA 3.0
2017-09-05T13:34:46.120Z
2017-09-05T13:34:46.120Z
Louis Strous
<p>This is the problem faced by all star map makers and has been "solved" by each map-maker. The solution has some aesthetic considerations if you use purely area Sirius becomes huge ... So instead of using a formula measure up the symbol sizes used in your favorite star atlas for each magnitude range. 
Check copy-right for your use.</p>

CC BY-SA 3.0
2017-09-10T23:02:00.977Z
I am not referring to any particular star chart or set of software. The point is that it's not a straight-forward problem but has been solved in slightly different ways by different star map makers.  So why not use what other people have done rather than re-invent the wheel.
TazAstroSpacial
CC BY-SA 3.0
2017-09-06T07:07:02.637Z
2017-09-06T07:07:02.637Z
TazAstroSpacial
<p>The sizes of the stars in the picture below have more to do with their brigtness than the actual size of the star.  I'm looking for a formula that relates the viewing size of a star with its brightness.</p>

<p><a href="https://i.sstatic.net/bxPjR.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bxPjR.jpg" alt="The Central Stars of Orion"></a></p>

<p>This is what I've worked on so far:</p>

<p>From Wikipedia on <a href="https://en.wikipedia.org/wiki/Apparent_magnitude#History" rel="nofollow noreferrer">Apparent Magnitude, History</a></p>

<p>The viewing brightness can be related to the apparent magnitude by</p>

<p>$$B=0.4^A$$</p>

<p>Before I sit down and count pixels and try to do a curve fit to find the  the relationship, I'm guessing this work has been done before.  I wonder if anyone has any leads that would get me to a good equation that I can use to generate the view sizes of stars.
<hr>
Edit:</p>

<p>My main goal is to make a reasonable 'cartoon' of the various constellations for the astronomy classes I'm teaching.  My rough work so far looks like this.  (I'm fairly certain the vertical scale is off.)</p>

<p><a href="https://i.sstatic.net/Pp3gO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Pp3gO.png" alt="This is a cartoon of some of the stars of Orion."></a></p>

CC BY-SA 3.0
2017-09-02T21:45:27.393Z
This is a function of the medium on which the image is recorded. There will be no simple relationship between the size of the stellar image and brightness.
adrianmcmenamin
CC BY-SA 3.0
2017-09-03T16:27:51.643Z
@barrycarter, Thanks!  There's a good lead there, the Airy disk might answer this, I'll look at the details today.
David Elm
CC BY-SA 3.0
2017-09-03T16:40:05.667Z
It looks like the intensity of the Airy disk goes like a Bessel Function, and in the center it approximates a Gaussian.  So does it make sense that the brighter stars would have bigger disks because the intensity curve is within the sensitivity range of the camera for longer?
David Elm
CC BY-SA 3.0
2017-09-05T13:52:42.160Z
The Airy disk has nothing to do with it (at least for larger telescopes).
ProfRob
<star><size>
How to Find the Viewing Size of a Star
David Elm
# How to Find the Viewing Size of a Star

(Asked by: username_0 on 2017-09-02)

The sizes of the stars in the picture below have more to do with their brigtness than the actual size of the star. I'm looking for a formula that relates the viewing size of a star with its brightness.

[![The Central Stars of Orion](https://i.sstatic.net/bxPjR.jpg)](https://i.sstatic.net/bxPjR.jpg)

This is what I've worked on so far:

From Wikipedia on [Apparent Magnitude, History](https://en.wikipedia.org/wiki/Apparent_magnitude#History)

The viewing brightness can be related to the apparent magnitude by

$$B=0.4^A$$

Before I sit down and count pixels and try to do a curve fit to find the the relationship, I'm guessing this work has been done before. I wonder if anyone has any leads that would get me to a good equation that I can use to generate the view sizes of stars.

---

Edit:

My main goal is to make a reasonable 'cartoon' of the various constellations for the astronomy classes I'm teaching. My rough work so far looks like this. (I'm fairly certain the vertical scale is off.)

[![This is a cartoon of some of the stars of Orion.](https://i.sstatic.net/Pp3gO.png)](https://i.sstatic.net/Pp3gO.png)

**Tags:** star, size
> **Comments:**
> **username_1**: This is a function of the medium on which the image is recorded. There will be no simple relationship between the size of the stellar image and brightness.  
> **username_0**: @barrycarter, Thanks!  There's a good lead there, the Airy disk might answer this, I'll look at the details today.  
> **username_0**: It looks like the intensity of the Airy disk goes like a Bessel Function, and in the center it approximates a Gaussian.  So does it make sense that the brighter stars would have bigger disks because the intensity curve is within the sensitivity range of the camera for longer?  
> **username_2**: The Airy disk has nothing to do with it (at least for larger telescopes).  

---

## 2 answers

### Answer by username_3 on 2017-09-05. Score: 2

Based on the math in the Wikipedia article about the [Airy Disk](https://en.wikipedia.org/wiki/Airy_disk#Mathematical_details):

The intensity of the starlight as a function of angular distance $\theta$ from the center is reasonably approximated by a Gaussian $$ I(\theta) = I\_0 \exp(-\theta^2/(2\sigma^2)) $$ where $\sigma$ is a measure for the width of the intensity profile, which for an ideal optical system and diffraction-limited images is $$ \sigma \approx 0.42 \lambda N $$ (where $\lambda$ is the wavelength and $N$ is the f-number) and which is independent of the peak intensity $I\_0$. For a non-ideal or not-diffraction-limited system, the Gaussian should still provide a reasonable approximation, but then $\sigma$ is (much) larger than for the ideal system.

The apparent edge of a star in an image occurs where the intensity $I(\theta)$ drops below a certain minimum $I\_\text{limit}$, which (from the first equation) corresponds to an angular distance $$ \theta\_\text{edge} = \sigma\sqrt{2\log(I\_0/I\_\text{limit})}$$

We can rewrite this in terms of visual magnitude $V$. For fixed $\sigma$, $I\_0$ is a measure for the brightness of the star. Brightness is related to magnitude $V$ through $$ V = -2.5 \log\_{10}(I/I\_\text{ref}) $$ (with $I\_\text{ref}$ the reference intensity for the magnitude scale) so $$\log(I\_0/I\_\text{limit}) = 0.4 \log(10) (V\_\text{limit} - V\_0)$$ and
\begin{eqnarray}
\theta\_\text{edge} & = & \sigma \sqrt{0.8 \log(10) (V\_\text{limit} - V\_0)} \approx 1.36 \sigma \sqrt{V\_\text{limit} - V\_0}
\\ V\_\text{limit} - V\_0 & = & \frac{\theta\_\text{edge}^2}{0.8 \log(10) \sigma^2} \approx 0.543 \left( \frac{\theta\_\text{edge}}{\sigma} \right)^2
\end{eqnarray}

So, the difference between the limiting magnitude and the star's magnitude is proportional to the apparent angular area of the star in the image.

> **Comments:**
> **username_2**: This prescription is reasonable, but has nothing to do with diffraction really. The angular size of stellar images is actually the same because of atmospheric "seeing", which would normally be much bigger than the Airy disk $\sigma$ you define. With CCDs there are other effects like charge transfer efficiency and saturation to consider; and there are serious non-linearities in the response of photographic plates.  
> **username_3**: My description does not apply only to diffraction-limited cases.  Both diffraction-limited and seeing-limited cases are reasonably described using the Gaussian.  Saturation and non-linearities in the response are only important if they noticeably change the size of the stellar image defined by the $I_\text{limit}$ contour, which implies significant transfer by the recording device or medium of signal across where the star image's edge would have otherwise been.  It would be interesting to see a test of the final relationship based on the picture.  

---

### Answer by username_4 on 2017-09-06. Score: 1

This is the problem faced by all star map makers and has been "solved" by each map-maker. The solution has some aesthetic considerations if you use purely area Sirius becomes huge ... So instead of using a formula measure up the symbol sizes used in your favorite star atlas for each magnitude range.
Check copy-right for your use.

> **Comments:**
> **username_4**: I am not referring to any particular star chart or set of software. The point is that it's not a straight-forward problem but has been solved in slightly different ways by different star map makers.  So why not use what other people have done rather than re-invent the wheel.  

---

