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.

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

Moreover, we can write for the rate at which declination changes whilst scanning at constant azimuthal rate

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

The time, $T$, spent observing at a specific declination is inversely proportional to the rate at which the declination changes

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

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

$t(\theta _{0},\beta ,\alpha )={\frac {1}{\cos \theta _{0}\cos \beta \sin \alpha }}$