Modulated scan high cadence LAT

Introduction

This post refines the work shown in High_cadence_LAT_from_Chile. We show how to modulate the scan rate to achieve maximally uniform integration depth across a maximal sky area.

Geometrical considerations

Let us assume that,

• $\theta _{0}$ is the observatory latitude, measured in radians from the Equator
• $\alpha$ is the observing azimuth, measured in radians from North. East is at $\pi /2$
• $\beta$ is the observing elevation, measured in radians from the horizon.
• $\theta$ is the declination on the celestial sphere.

Then we can write the declination as a function of the observatory latitude and observing azimuth and elevation.

$(1)\quad \quad \quad \theta =\sin ^{-1}\left(\cos \theta _{0}\cos \beta \cos \alpha +\sin \theta _{0}\sin \beta \right)$

Moreover, we can write for the rate at which declination changes in terms of the azimuthal rate

$(2)\quad \quad \quad {\frac {\mathrm {d} \theta }{\mathrm {d} t}}={\frac {\mathrm {d} \theta }{\mathrm {d} \alpha }}{\frac {\mathrm {d} \alpha }{\mathrm {d} t}}={\frac {\cos \theta _{0}\cos \beta \sin \alpha }{\sqrt {1-\left(\cos \theta _{0}\cos \beta \cos \alpha +\sin \theta _{0}\sin \beta \right)^{2}}}}{\frac {\mathrm {d} \alpha }{\mathrm {d} t}}={\frac {\cos \theta _{0}\cos \beta \sin \alpha }{\cos \theta }}{\frac {\mathrm {d} \alpha }{\mathrm {d} t}}$

where the final form follows from application of $(1)$. The time, $T$, spent observing at a specific declination is inversely proportional to the rate at which the declination changes

$(3)\quad \quad \quad T(\theta )\propto \left({\frac {\mathrm {d} \theta }{\mathrm {d} t}}\right)^{-1}={\frac {\cos \theta }{\cos \theta _{0}\cos \beta \sin \alpha }}\left({\frac {\mathrm {d} \alpha }{\mathrm {d} t}}\right)^{-1}$

If we consider the fact that the sky area at each latitude is proportional to $\cos \theta$, the integration time per unit sky area, $T_{a}$, is

$(4)\quad \quad \quad T_{a}(\theta _{0},\beta ,\alpha )\propto {\frac {1}{\cos \beta \sin \alpha }}\left({\frac {\mathrm {d} \alpha }{\mathrm {d} t}}\right)^{-1}$

where we have dropped the constant factor $\cos \theta _{0}$

From $(4)$ it is evident that for the observing depth to be uniform ( $T_{a}$ to be constant), the azimuthal scanning rate must satisfy

$(5)\quad \quad \quad {\frac {\mathrm {d} \alpha }{\mathrm {d} t}}\propto {\frac {1}{\cos \beta \sin \alpha }}$

and explicitly, the azimuthal rate on the sky, $\omega$, will depend on some base azimuthal rate on the telescope, $\omega _{0}$, as

$(6)\quad \quad \quad \omega ={\frac {\omega _{0}}{\cos \beta \sin \alpha }}$

Hardware limits

Modulating the azimuthal scanning rate is limited by maximum azimuthal rate of the telescope and, to a lesser degree, the azimuthal acceleration of the telescope.

Here is a plot of the modulation factor as a function of the telescope azimuth:

It is obvious that sweeping close to 180 degrees is unfeasible as it would require the telescope to move very fast as it approaches the turnaround. A factor of 2 or 3 (throw of 160 or 170 degrees) is achievable, especially if the base scanning rate is low enough.

Pushing for extreme azimuth ranges is not even required for the sky area, as we quickly enter the domain of diminishing returns: