HERA has a derotator, which ensures that the pixels do not rotate on the
sky. The sky can thus be mapped by scanning along *e.g.* the right ascension
or the declination axis in equatorial coordinates. We aim at obtaining a
fully sampled map, implying a distance between the rows of
, where
is the beam full width at half maximum: At 1 mm,
this corresponds typically to . However, the pixels are typically
separated by
. We thus have to find the best
scanning strategy which will fill the hole of the instantaneous footprint
of the multi-pixel. To do this, we will use a property of the deroratator,
*i.e.* it can be configured so that one of the main axes of the multi-pixel
is rotated by an angle (
) from the scanning direction. Indeed, we
can ask what is the value of
needed so that the distance between
the rows of two adjacent pixels is exactly
. For a receiver of
pixels, we end up with
groups of
lines, the distance between two group of lines being noted
. A
bit of geometry gives

we obtain

(10) |

For HERA, and the value is fixed to by the observing wavelenght mm. gives , and gives , . Current optical design implies a minimum distance between the pixels which is only compatible with the solution.

In summary, by setting an angle of between one of the main axes
of a multi-pixels and the scanning direction, we can sweep in a
fully sampled mode a given portion of the sky with two parallel scans
separated by
. The region of the sky fully sampled will
then be rectangular: the length of the rectangular side perpendicular to
the scanning direction is then
, while the length
of the rectangular size parallel to the scanning direction,
, will
depend on the observing strategy. However, there is an edge effect, due to
the rotation of the array from the scanning direction. Indeed, the edges of
the maps are not fully sampled: Thus must thus be considered as overheads.
The area of the scanned sky must thus be larger than the targeted area,
which must be fully sampled. Let's assume that the targeted area
(
) is swept as a succession of
rectangles of size
. We get

(12) |

(13) |

We now define the mapping efficiency
as

(15) |

A given area
will be mapped in chunks whose area (
) is
defined by the linear scanning speed and the time of stability of the
system (
). This gives

(17) |

This equation of the 2nd order has only one physical solution

(19) |

(20) |

(21) |

Previous equations give the impression that the aspect ratio is a free
parameter. This is not fully true because,
must be an integer.
The following algorithm ensures that we get an integer value for
with the value of
and closest to 1. Starting with
, Eq. gives a value of
. We enforce the integer
nature of
with

(22) |

(23) |

In summary, the time spent in edges is counted as overheads. It translates into a multiplicative efficiency ( ) because we enforce a mapping pattern through rectangular chunks. Although it is not intuitive (edge sizes are in general unrelated to area), this is not a big assumption because the use of a square multi-pixel anyway enforces mapping in rectangular chunks. We now summarize the algorithm to compute :

**Step #1: Computation of input quantities**-

(24)

(25)

(26)

(27) **Step #2: Computation of and**-
**Case with**-

(28) (29) (30)

**Case**-

(31) (32) (33)

**Step #3: Computation of**-

(34) **Step #4: Recomputation of and when**-

(35) (36) (37)

If minute, the targeted area is too small and the PI should use raster mapping instead of OTF mapping.