Документ взят из кэша поисковой машины. Адрес оригинального документа : http://www.naic.edu/~phil/mbeam/mbeamrotation.html
Дата изменения: Wed Dec 8 17:32:20 2004
Дата индексирования: Tue Oct 2 07:07:52 2012
Кодировка:

Поисковые слова: п п п п п п п п п п

lband multibeam system.

(last modified 18feb04)
The lband multi beam system is envisioned to  consist of 7 feeds.  The feed system would have to rotate if the outer 6 feeds were to track the same spot on the sky for extended periods of time.

Do we really need to rotate the feed system and if so, how fast..

The Figure shows the parallactic angle in relation to the source position at (RA,DEC), the zenith of the telescope at (LST,AoLatitude), the hour Angle HA, and the azimuth direction AZ (in the picture the source has already transitted). When the feed is pointing at the source, the azimuth angle points at the (RA,DEC) position and the zenith angle moves along this arc till it is at the (RA,DEC) location (on the sky).  As the LST approaches the RA of the source (ie the source gets closer to transit) the source azimuth angle goes to zero. Now suppose that a second feed was placed 1 beam width away from the central horn parallel to the azimuth arm. At time LST, it would point 1 beam beyond RA,DEC in the az direction. At delta T seconds  later the offset horn is still pointing on the the azimuth direction but this direction has now rotated about the point (RA,DEC) by the change in the parallactic angle.  The important thing to remember is that if the central horn points at (RA,DEC), then it is pointing at the location (RA,DEC) on the celestial sphere. Any rotation about the central horn will be a "parallactic angle" rotation.  The equation for the parallactic angle is:
 
sin(parallactic angle)=sin(azimuth) * cos(latitude)/ cos(declination)
(see Spherical astronomy, Small pg. 49).

Rotating the feed system by the parallactic angle rate will then cause a horn on the  outer ring to stay fixed on the sky. Suppose we don't rotate the feeds. How long will it take before the outer horn moves by  1/2 of a beam? A linear motion in the focal plane has a plate scale (magnification or demagnification) that is different in the 2 directions:
 

  • 14 asecs on the sky per cm  perpendicular to the azimuth arm
  • 17 asecs on the sky per cm  parallel to the azimuth arm.

  •  

     
     
     
     
     
     
     
     
     

    The asymmetry comes from the secondary and tertiary mirrors.  The computation is then:

    1. Decide on the largest angle error you will accept (probably 1/2 a beam of 3.2 arc minutes).
    2. Compute how many centimeters this is in the focal plane using the worst case plate scale (17"/cm).
    3. Given the radius to the center of the outer horn (30 cm) and the arc Length from 2., compute the angle MaxDeltaPa (10.78 degrees). This is the largest parallactic angle change allowed.
    4. Compute the parallactic angle versus hour angle PA(ha). For each PA(ha) search forward to PA(ha+dt)=PA(ha)+MaxDeltaPa. dt will be the maximum integration time starting at ha. This needs to be done for each declination.
    The FIGURES are:
    1. Figure 1. shows the parallactic angle rate versus azimuth angle for declinations 8 through 28 (spaced 1 degree steps from Arecibo's latitude). The maximum rotation rate is .225 degrees per second. The left set (northern sources) and right set (southern sources) each have declinations stepping 1 degrees from the AO latitude.
    2. Figure 2. shows the same rate versus hour angle for declinations 0 through 17 degrees (the scale is blown up a bit here).
    3. Figure 3 shows how long you can integrate before the motion equals 1/2 of a beam. I used a beam width of 3.2 arc minutes, an outer horn location of 30 cm, and a plate scale of 17asecs/cm. I did not take into account the differential motion caused by the different plate scales.  Using these values, the maximum allowable rotation angle is: maxAngle=bmWd/2*60/plateScale/radius*radToDeg

    4. or 10.87 degrees. The slanted lines on the right are because the source will set before a longer integration can complete.
    5. Figure 4 is a blowup of Figure 3. For sources 12 degrees from our declination (5. deg dec and below) a 10 minute integration can be done anywhere on the dish (as long as you don't set). For a 172114 dec source you can do a 10 minute integration until 20 minutes before transit and any time 10 minutes after transit.
    The integration time plots show that 10 minute integrations (with up to half a beam motion) can be done at any declination provided the hour angle keeps you above the horizontal line in figure 4. Long integrations are primarily needed for the pulsar search program (since they must be coherent). In this mode you would track all your sources within the hour angle range that gives you the needed integration time.
        If you wanted to do 2 ten minute integrations at the same 7 positions on the sky, you would integrate, move the telescope by 60 degrees in parallactic angle to rotate the feeds by 1, and then do the second integration.
    processing: x101/mbeam/parangrate.pro
     home_~phil