The plate scale, $S = \delta s / \delta \theta$, is the displacement, $\delta s$, at the telescope focal plane for a given difference in incoming light angle, $\delta \theta$. For a given telescope focal length:
$$\sin (\delta \theta) = \frac{\delta s}{f}$$
Using the small-angle approximation:
$$\sin(\delta \theta) \approx \delta \theta$$
so
$$\frac{\delta s}{f} \approx \delta \theta$$
$$\frac{\delta s}{\delta \theta} \approx \frac{1}{f}$$
$S \approx \frac{1}{f}$ with S radians / length
or, converting radians to arcseconds
$S \approx 206265/f$ arcsec / length
The pixel scale, $P$, for a given telescope focal length, $f$, and sensor pixel length or width, $w$, is the angular distance on the sky between sensor pixels.
$P = Sw \approx \frac{w}{f}$ in radians / pixel assuming w and f are the same units
converting to arcsec:
$P = Sw \approx 206265 \times \frac{w}{f}$ in arcsec / pixel assuming w and f are the same units
and adding factors to make w in μm and f in mm:
$P = Sw \approx 206.265 \times \frac{w}{f}$ in arcsec / pixel assuming w in μm and f in mm
The AT65EDQ has a nominal 420 mm focal length, while the ASI533MC Pro has 3.76 μm square pixels.
According to PixInsight plate solve:
The PDF for the observed camera pixel value, $O$, is:
$$f(O) = G \times \big[\mathrm{Poisson}(S+B)+\mathrm{Gaussian}(0,r)\big]$$
where $O$ is the camera readout value in ADU, $G$ is the electron to ADU gain, $S$ is the true number of signal electrons, $B$ is the thermal background in electrons, and $r$ is the readout noise in electrons.
So the mean of $O$:
$$\overline{O} = G(\overline{S+B})$$
and variance of $O$:
$$\sigma^2_{O} = G^2(\sigma^2_{S+B} + r^2)$$
Because the variance and mean of the Poisson distribution are equal, $\sigma^2_{S+B} = \overline{S+B}$, so:
$$\sigma^2_{O} = G^2(\overline{S+B} + r^2) = G\overline{O} + G^2r^2$$
To estimate $G$ and $r$, take many light frames at each of various $\overline{O}$. That can be accomplished using various brightnesses or exposure times. For each pixel and $\overline{O}$, fit a line to the $\sigma^2_O$ vs. the trimmed mean or median of $O$. The slope of the line is $G$, and the intercept is $G^2r^2$ as in the equation:
$$\sigma^2_{O} = G\overline{O} + G^2r^2$$
To estimate $B$, take many dark frames. They have no signal. The median or trimmed mean of each pixel is the thermal background in ADU:
$$\overline{O} = G\overline{B}$$
You can check your estimate of $G$ and $r$ with many bias frames. There is no signal or thermal background noise in a bias frame so:
$$\sigma^2_{O} = G^2r^2$$
Photon flux can be calculated from Jy by:
1 Jy = $1.51\times 10^7$ photons s$^{-1}$m$^{-2}$ $\left(\frac{\Delta\lambda}{\lambda}\right)^{-1}$
| Band | Central Wavelength (nm) | $\frac{\Delta \lambda}{\lambda}$ | Flux at m=0 (Jy) |
|---|---|---|---|
| U | 360 | 0.15 | 1810 |
| B | 440 | 0.22 | 4260 |
| V | 550 | 0.16 | 3640 |
| R$_c$ | 640 | 0.23 | 3080 |
| I$_c$ | 790 | 0.19 | 2550 |
From Bessel Astronomical Society of the Pacific, Publications, vol. 91, Oct.-Nov. 1979, p. 589-607.