Rotating a precession vector

Assume \hat x and \hat y are orthogonal ( at right angles ) vectors in the equatorial plane, and \hat z is an orthogonal vector perpendicular to the plane. These can be expressed as:

\hat x = [ x, 0, 0 ]

\hat y = [ 0, y, 0 ]

\hat z = [ 0, 0, z ]


Matrix multiplication works like this:

[ ax+by+cz,~ dx+ey+fz,~ gx+hy+yz ] = \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \times [x, y, z]

The new vector elements are made of linear sums of the old vector elements.


Rotations can be expressed as multiplication matrices, which are multiplied by a vector to yield a new vector. If we rotate a point, the coordinates (expressed as vectors) are multiplied by the rotation matrix to give a rotationally transformed coordinate. Here is the rotation matrix for a rotation around the y axis by angle \phi :

\left( \begin{array}{ccc} \cos( \phi ) & 0 & -\sin( \phi ) \\ 0 & 1 & 0 \\ \sin( \phi ) & 0 & \cos( \phi ) \end{array} \right)

And here is a rotation by angle \beta around the z axis :

\left( \begin{array}{ccc} \cos( \beta ) & -\sin( \beta ) & 0 \\ \sin( \beta ) & \cos( \beta ) & 0 \\ 0 & 0 & 1 \end{array} \right)

These rotational tranformations always preserve distance from the origin - the Pythagorean sum of the components sqrt( x^2 + y^2 + z^2 ) remains constant.


The equatorial plane can be represented as a unit vector in z (pointed towards the north pole):

Plane = [ 0, 0, 1 ]

The plane has an axial tilt \phi = 23.439281° . We can represent this as a matrix multiplication around the y axis:

Tilted = [ \sin( \phi ), 0, \cos( \phi ) ] = \left( \begin{array}{ccc} \cos( \phi ) & 0 & -\sin( \phi ) \\ 0 & 1 & 0 \\ \sin( \phi ) & 0 & \cos( \phi ) \end{array} \right) \times [ 0, 0, 1 ]

This is the way the north pole vector appears from the sun at the vernal (spring) equinox. If \beta is the time of year, that vector is rotated around the z axis:

Tilted and rotated = [ \cos( \beta ) \sin( \phi ), \sin( \beta ) \sin( \phi ), cos( \phi ) ] = \left( \begin{array}{ccc} \cos( \beta ) & -\sin( \beta ) & 0 \\ \sin( \beta ) & \cos( \beta ) & 0 \\ 0 & 0 & 1 \end{array} \right) \times [ \sin( \phi ), 0, \cos( \phi ) ]

Astronomers looking at the moon and other planets and stars are concerned about absolute orientation in space. For server sky, we are only concerned about the equator and sun angle relative to the equatorial plane. The sun angle can be computed by the " dot product " of our north pole vector with the sun, which is the cosine of the angle between the y axis and the north pole. The full dot product is more complicated, but in this simple case it is just the y term of the vector: \cos( Pole Angle ) = \sin( Equatorial Angle ) = \sin( \beta ) \sin( \phi )

\Large \sin( \theta_{eq} ) = \sin( \beta ) \sin( \phi )

In other words, the effect of precession is to tilt the equatorial plane by \theta_{eq} = \arcsin( \sin( \beta ) \sin( \phi ) ), where \beta is an angle representing the time of year, and \phi is the axial tilt. This can be approximated within 1.1 percent as \theta_{eq} \approx \phi \sin( \beta )

Precession (last edited 2011-03-30 23:29:29 by KeithLofstrom)