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Special cases of the SYN projection: SIN and NCP

Note in equations (7) that since theta is approximately 90^o the terms involving p₁ and p₂ are small; neglecting them, as is usually done, gives us the equations for the ``SIN'' projection:

x	=	-cos $\displaystyle \theta$ sin $\displaystyle \phi$	(10)
y	=	-cos $\displaystyle \theta$ cos $\displaystyle \phi$

From equations (7), for an array which lies in the J2000.0 equatorial plane, we have

n_u	=	0	(11)
n_v	=	cos $\displaystyle \delta_{0}^{}$
n_w	=	sin $\displaystyle \delta_{0}^{}$

where $\delta_{0}^{}$ is the declination of the field centre, whence

p₁	=	0	(12)
p₂	=	cot $\displaystyle \delta_{0}^{}$

and

x	=	- [cos $\displaystyle \theta$ sin $\displaystyle \phi$ ]	(13)
y	=	- [cos $\displaystyle \theta$ cos $\displaystyle \phi$ + cot $\displaystyle \delta_{0}^{}$ (sin $\displaystyle \theta$ - 1)]

These are the equations for the ``NCP'' projection. To first order the difference between equations (10b) and (13b) is

$\displaystyle {\frac{r^2}{2}}$ cot $\displaystyle \delta_{0}^{}$

(14)

where r is the distance from the field centre in radians. This amounts to nearly 1' for a position 1^o from the field centre at $\delta_{0}^{}$ = 30^o.

Next: Correction for precession Up: No Title Previous: SYN projection equations in equatorial coordinates