How to Find the Viewing Size of a Star 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.



This is what I've worked on so far:

From Wikipedia on 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.)
Based on the math in the Wikipedia article about the Airy Disk:

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} &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}

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.