Positional Angle

In astronomy, position angle (usually abbreviated PA) is the convention for measuring angles on the sky. The International Astronomical Union defines it as the angle measured relative to the north celestial pole (NCP), turning positive into the direction of the right ascension. In the standard (non-flipped) images, this is a counterclockwise measure relative to the axis into the direction of positive declination.

In the case of observed visual binary stars, it is defined as the angular offset of the secondary star from the primary relative to the north celestial pole.

As the example illustrates, if one were observing a hypothetical binary star with a PA of 135°, that means an imaginary line in the eyepiece drawn from the north celestial pole to the primary (P) would be offset from the secondary (S) such that the angle would be 135°.

When graphing visual binaries, the NCP is, as in the illustration, normally drawn from the center point (origin) that is the Primary downward–that is, with north at bottom–and PA is measured counterclockwise. Also, the direction of the proper motion can, for example, be given by its position angle.

The definition of position angle is also applied to extended objects like galaxies, where it refers to the angle made by the major axis of the object with the NCP line.

Nautics

The concept of the position angle is inherited from nautical navigation on the oceans, where the optimum compass course is the course from a known position to a target position with minimum effort. Setting aside the influence of winds and ocean currents, the optimum course is the course of smallest distance between the two positions on the ocean surface. Computing the compass course is known as the ''inverse problem'' of geodesics.

This article considers only the abstraction of minimizing the distance between and traveling on the surface of a sphere with some radius : In which direction angle relative to North should the ship steer to reach the target position?

Global geocentric coordinate system

Detailed evaluation of the optimum direction is possible if the sea surface is approximated by a sphere surface. The standard computation places the ship at a geodetic latitude and geodetic longitude , where is considered positive if north of the equator, and where is considered positive if east of Greenwich. In the global coordinate system centered at the center of the sphere, the Cartesian components are

::$=R\left(\begin \cos\varphi_s \cos\lambda_s \\ \cos\varphi_s \sin\lambda_s \\ \sin\varphi_s \end\right)$
and the target position is

::$=R\left(\begin \cos\varphi_t \cos\lambda_t \\ \cos\varphi_t \sin\lambda_t \\ \sin\varphi_t \end\right).$
The North Pole is at
::$=R\left(\begin 0 \\ 0 \\ 1 \end\right).$
The minimum distance is the distance along a great circle that runs through and . It is calculated in a plane that contains the sphere center and the great circle,
::$d_=R\theta_$
where is the angular distance of two points viewed from the center of the sphere, measured in radians. The cosine of the angle is calculated by the dot product of the two vectors
::$\mathbf\cdot \mathbf = R^2\cos \theta_ = R^2(\sin\varphi_s\sin\varphi_t+\cos\varphi_s\cos\varphi_t\cos(\lambda_t-\lambda_s))$
If the ship steers straight to the North Pole, the travel distance is
::$d_ = R\theta_ = R(\pi/2-\varphi_s)$
If a ship starts at and swims straight to the North Pole, the travel distance is
::$d_ = R\theta_ =R(\pi/2-\varphi_t)$

Brief Derivation

The ''cosine formula'' of spherical trigonometry yields for the
angle between the great circles through that point to the North on one hand and to on the other hand
::$\cos\theta_ = \cos\theta_\cos\theta_+\sin\theta_\sin\theta_\cos p.$
::$\sin\varphi_t = \cos\theta_\sin\varphi_s +\sin\theta_\cos\varphi_s\cos p.$
The ''sine formula'' yields
::$\frac = \frac.$
Solving this for and insertion in the previous formula gives an expression for the tangent of the position angle,
::$\sin\varphi_t = \cos\theta_\sin\varphi_s +\frac\cos\varphi_t\cos\varphi_s\cos p;$
::$\tan p = \frac.$

Long Derivation

Because the brief derivation gives an angle between 0 and which does not reveal the sign (west or east of north ?), a more explicit derivation is desirable which yields separately the sine and the cosine of such that use of the correct branch of the inverse tangent allows to produce an angle in the full range .

The computation starts from a construction of the great circle between and . It lies in the plane that contains the sphere center, and and is constructed rotating by the angle around an axis . The axis is perpendicular to the plane of the great circle and computed by the normalized vector cross product of the two positions:
::$\mathbf = \frac\mathbf\times \mathbf = \frac\left(\begin \cos\varphi_s\sin\lambda_s\sin\varphi_t -\sin\varphi_s\cos\varphi_t\sin\lambda_t \\ \sin\varphi_s\cos\lambda_t\cos\varphi_t -\cos\varphi_s\sin\varphi_t\cos\lambda_s \\ \cos\varphi_s\cos\varphi_t\sin(\lambda_t-\lambda_s) \end\right).$
A right-handed tilted coordinate system with the center at the center of the sphere is given by the
following three axes: the
axis , the axis
::\mathbf_\perp = \omega \times \frac\mathbf =
\frac
\left(\begin
\cos\varphi_t\cos\lambda_t(\sin^2\varphi_s+\cos^2\varphi_s\sin^2\lambda_s)-\cos\lambda_s(\sin\varphi_s\cos\varphi_s\sin\varphi_t+\cos^2\varphi_s\sin\lambda_s\cos\varphi_t\sin\lambda_t)\\
\cos\varphi_t\sin\lambda_t(\sin^2\varphi_s+\cos^2\varphi_s\cos^2\lambda_s)-\sin\lambda_s(\sin\varphi_s\cos\varphi_s\sin\varphi_t+\cos^2\varphi_s\cos\lambda_s\cos\varphi_t\cos\lambda_t)\\
\cos\varphi_s cos\varphi_s\sin\varphi_t-\sin\varphi_s\cos\varphi_t\cos(\lambda_t-\lambda_s)\end\right)

and the axis .
A position along the great circle is
::$\mathbf(\theta) = \cos\theta \mathbf+\sin\theta \mathbf_\perp,\quad 0\le\theta\le 2\pi.$
The compass direction is given by inserting the two vectors and and computing the gradient of the vector with respect to at .
::$\frac\mathbf_=\mathbf_\perp.$
The angle is given by splitting this direction along two orthogonal directions in the plane tangential to the sphere at the point . The two directions are given by the partial derivatives of with respect to and with respect to , normalized to unit length:
::$\mathbf_N = \left( \begin -\sin\varphi_s\cos\lambda_s\\ -\sin\varphi_s\sin\lambda_s\\ \cos\varphi_s \end\right);$
::$\mathbf_E = \left(\begin -\sin\lambda_s\\ \cos\lambda_s\\ 0 \end \right);$
::$\mathbf_N\cdot \mathbf = \mathbf_E\cdot \mathbf_N =0$
points north and points east at the position .
The position angle projects
into these two directions,
::$\mathbf_\perp = \cos p \,\mathbf_N+\sin p\, \mathbf_E$,
where the positive sign means the positive position angles are defined to be north over east. The values of the cosine and sine of are computed by multiplying this equation on both sides with the two unit vectors,
::\cos p = \mathbf_\perp \cdot \mathbf_N
=\frac cos\varphi_s\sin\varphi_t - \sin\varphi_s\cos\varphi_t\cos(\lambda_t-\lambda_s)

::\sin p = \mathbf_\perp \cdot \mathbf_E
=\frac cos\varphi_t\sin(\lambda_t-\lambda_s)

Instead of inserting the convoluted expression of , the evaluation may employ that the triple product is invariant under a circular shift
of the arguments:
::$\cos p = (\mathbf\times \frac\mathbf)\cdot \mathbf_N = \omega\cdot(\frac\mathbf\times \mathbf_N).$

If atan2 is used to compute the value, one can reduce both expressions by division through
and multiplication by ,
because these values are always positive and that operation does not change signs; then effectively
::$\tan p = \frac.$

See also

*Parallactic angle
*Angular distance

Further reading

*

References

{{Reflist

External links

The Orbits of 150 Visual Binary Stars, by Dibon Smith (Accessed 2/26/06)

Astronomical coordinate systems
Angle
Observational astronomy