Here I collect a bunch of August 2015 notes. I now have a small OTW code (parang.sh) that does the job.
Most of this from the parang source code directories.
A fairly clear definition:
==========================
The parallactic angle is the angle between the great circle through
a celestial object and the zenith, and the hour circle of the object.
It is usually denoted q. In the triangle zenith—object—celestial
pole, the parallactic angle will be the angle at the celestial object.
I good web calculator is at:
http://irtfweb.ifa.hawaii.edu/cgi-bin/spex/parangle.cgi
I can set for APO (Apache Point, close to McD's latitude)
I specify HA (- for East, + for West)
I specify DEC
-----> These last two let me move to different AZIMUTH values
Q = parallactic angle
Here is what it gives:
HA DEC Q ~AZ(sco)
0.0 0.0 0.0 180
0.0 -5.0 0.0 180
-4.0 0.0 126.6 90
-3.0 0.0 132.0 90
-0.1 0.0 3.9 180
-4.0 +60.0 82.0 60
-1.0 -30.0 163.0 170
-0.2 -30.0 2.4 180
It seems that the HET parallactic angle is NOT a parallactic angle
at all. It is the direction of the parallactic vector, and is defined
on an angular scale that runs from 0 to 360 degrees. I belive it is
the vector that "tells you", from the prespective of the target being
observed, the direction to the Zenith point.
Why does my Q value (a direction, not an angle!) not agree with the
HET equations (or with Figure 3 in Wray #42)?
Azimuth,Expected,Computed:
65.4525 90.0 71.96599
I should get 90 but my code predicts 72.
I will go through each step in code (already printed to log file)
and check the calculations by hand.
I will illustrte the steps using JFs equations:
c Here is how the Tcs calculates paralactic angle. Telecentric
c azimuth is the structure azimuth (AZ) and telecentric elevation
c (telangle) is basically 55 degrees. Our latitude is 30:40:53.17
c Intermediate variables.
c P = cos( Telecentric Elevation ) * cos( Latitude )
c Q = sin( Telecentric Elevation ) * sin( Latitude )
c These parameters are dependent only on the Az/El of the
c telecentric axis.
c Tde: Telecentric Declination
c H0: Telecentric Hour Angle
c Tde = asin( P * cos( Telecentric Azimuth ) + Q )
c H0 = asin( -cos( Telecentric Elevation ) * sin( Telecentric Azimuth ) / cos( Tde ) )
c Q = Pa_Paralactic_Angle = acos( cos( H0 ) * cos( Telecentric Azimuth )
c + sin( H0 ) * sin( Telecentric Azimuth ) * sin( Latitude ) )
Numbers my code uses:
Q = 0.417863
P = 0.493336
telangle = 55.0
xlatitude = 30.6716667
For my, I start at AZ=65.45
My first cal: tde Tde = asin( P * cos( Telecentric Azimuth ) + Q )
Tde = asin ( 0.493336 * cos(65.45) + 0.417863)
Tde = asin ( 0.493336 * 0.41549 + 0.417863)
Tde = asin ( 0.20497 + 0.417863 )
Tde = asin ( 0.62284 )
Tde = 38.5238 degrees (My code tde = 38.52223)
Next, I calc H0 = angle between tele-axis and meridian
c H0 = asin(-cos( 55 ) * sin( 65.45 ) / cos( tde ) )
------------------------- ----------
arg2 arg3
---------------------------------------
arg4
arg2 = -cos( 55 ) * sin( 65.45 )
= -0.57358 * 0.90960
= -0.52173 My code: arg2 = -0.52174
arg3 = cos( tde )
= cos( 38.5238 )
= 0.78245 My code: arg3 = 0.78237
agr4 = arg2 / arg3
= -0.52174 / 0.78237
= -0.66687 My code: arg4 = -0.66687
H0 = asin( arg4 )
= asin( -0.66687 )
= -41.82595 deg My code: H0 = -41.82578
Finally we get tp Pallactic Angle = PA
PA = Pa_Paralactic_Angle
Lat = 30.6716667
PA = acos [ cos(H0)*cos(AZ) + sin(H0)*sin(AZ)*sin(Lat) ]
-------------- ------------------------
arg10 arg11
arg10 = cos(H0)*cos(AZ)
= cos(-41.82578) * cos(65.45)
= 0.74518 * 0.41549
= 0.30961 My code: arg10 = 0.30958
arg11 = Here was the typo!!!!!! But I am so close, I'll finish the
full sample calculation.
arg11 = sin(H0)*sin(AZ)*sin(Lat)
= sin(-41.82578) * sin(65.45) * sin(30.67166)
= -0.66687 * 0.90960 * 0.51012
= -0.30943 My code: arg11 = -0.30943
PA = acos ( arg10 + arg11 )
= acos ( 0.30958 - 0.30943 )
= acos ( 0.00015 )
PA = 89.9914 My code: PA = 89.99158
Also, I did a bunch of other cases:
(See: parangle/T/S1_handcals/RUN_HAND)
PA = Parallactic Angle
Azimuth Expected_PA Computed_PA
31.7 49.48 55.82590
311.07117 284.8037 284.71555
68.58792 92.5072 92.52967
65.4525 90.0 89.99158
0.0000 0.0 0.00000
180.0000 180.0 180.00000
65.4525 270.0 270.00842
parangle/Notes_parangle/README.parangle
From Matt (02/20/2015)
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So it should be a constant offset between the parallactic angle and the
position angle not a constant offset between the Az and position angle.
There is some web code that computes this:
/home/het/astronomer/HETweb_Programs/Instruments
rho.c:
#define R2D 57.29577958 /* degrees per radian */
#define Q 0.417863 /* sin alt * sin latitude */ Not alt!
#define P 0.493336 /* cos alt * cos latitude */ Not alt!
#define telangle 55.0
#define latitude 30.6716667
The explanations of Q,P above are INCORRECT
They should read:
#define Q 0.417863 /* sin telangle * sin latitude */
0.81915 * 0.51012 = 0.417864
#define P 0.493336 /* cos telangle * cos latitude */
0.57358 * 0.860105 = 0.493339
DEC2 = DEC2/R2D;
az = 1/P*(sin(DEC2)-Q);
tde = P*cos(az/R2D) + Q ;
tde = atan2(tde,sqrt(1-tde*tde)) * R2D;
h = (cos(telangle/R2D)*sin(az/R2D))/-cos(tde/R2D);
h = atan2(h,sqrt(1-h*h))*R2D;
parang=cos(h/R2D)*cos(az/R2D)+sin(h/R2D)*sin(az/R2D)*sin(latitude/R2D);
parang = atan2(sqrt(1-parang*parang),parang)*R2D;
Then if you are working in the West you have to do a 360-parang
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From Jim (02/04/2015)
===========================================================================
Steve,
Here is how the Tcs calculates paralactic angle. Telecentric azimuth
is the structure azimuth and telecentric elevation is basically
55 degrees. Our latitude is 30:40:53.17
Intermediate variables.
P = cos( Telecentric Elevation ) * cos( Latitude )
Q = sin( Telecentric Elevation ) * sin( Latitude )
These parameters are dependent only on the Az/El of the telecentric axis.
Tde: Telecentric Declination
H0: Telecentric Hour Angle
Tde = asin( P * cos( Telecentric Azimuth ) + Q )
H0 = asin(-cos( Telecentric Elevation ) * sin( Telecentric Azimuth ) / cos( Tde ) )
Pa Paralactic Angle = acos( cos( H0 ) * cos( Telecentric Azimuth )
+ sin( H0 ) * sin( Telecentric Azimuth ) * sin( Latitude ) )
Best wishes,
Jim
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