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Correction for precession

The plane of an east-west array coincides with the apparent equatorial plane at the date of the observation and this is tilted slightly with respect to the J2000.0 equatorial plane. If ( $\alpha_{p}^{}$ , $\delta_{p}^{}$ ) are the J2000.0 right ascension and declination of the apparent pole then

n_u	=	-cos $\displaystyle \delta_{p}^{}$ sin( $\displaystyle \alpha_{p}^{}$ - $\displaystyle \alpha_{0}^{}$ )	(15)
n_v	=	sin $\displaystyle \delta_{p}^{}$ cos $\displaystyle \delta_{0}^{}$ - cos $\displaystyle \delta_{p}^{}$ sin $\displaystyle \delta_{0}^{}$ cos( $\displaystyle \alpha_{p}^{}$ - $\displaystyle \alpha_{0}^{}$ )
n_w	=	sin $\displaystyle \delta_{p}^{}$ sin $\displaystyle \delta_{0}^{}$ + cos $\displaystyle \delta_{p}^{}$ cos $\displaystyle \delta_{0}^{}$ cos( $\displaystyle \alpha_{p}^{}$ - $\displaystyle \alpha_{0}^{}$ )

These may be substituted directly into equations (7) and (8). Precession from 1990 to 2000 amounts to about 3'. For $\alpha_{p}^{}$ = $\alpha_{0}^{}$ and $\delta_{0}^{}$ = 30^o we get

p₁	=	0
p₂	=	1.72857

Equations 12 with no precession correction give

p₁	=	0
p₂	=	1.73205

For a position 1^o from the field centre at this declination the difference between these amounts to about 0".1.

Next: Field shifts Up: No Title Previous: Special cases of the SYN projection: SIN and