irfpy.util.vectortools

Vector (numpy array) related functions

Vector tools for numpy array. Most of the funcitons (are intended to) support the 2-D (or >2-D) array, namely, (N, 3) (or (N0, N1, …, 3)) array.

Related modules:

• irfpy.util.cone

Code author: Yoshifumi Futaana

distance_to_cylinder_center(vector, center_axis)	Calculate the distance to the center axis of the cylinder.
cone_angle(vector, cone_axis[, cone_center])	Return cone angle in radians.

irfpy.util.vectortools.distance_to_cylinder_center(vector, center_axis, cylinder_center=(0, 0, 0))

Calculate the distance to the center axis of the cylinder.

Calculate the distance between a given vector, v, and the center of the given cylinder. The cylinder is defined by a vector, cylinder_center p, together with the axis vector, center_axis n.

Parameters:

• vector – Vector(s) for which the distance(s) are calculated. (3,) or (…, 3) array.

• cylinder_center – Specifies the center of the cylinder. A single point on the axis. (3,) array

• cylinder_axis – Specifies the axis vector of the cylinder. A single vector. (3,) array.

Examples follow.

>>> p = (2, 1, 0)
>>> n = (0, 0.5, 0)
>>> v = (0, 0, 0)
>>> distance_to_cylinder_center(v, n, cylinder_center=p)
np.float64(2.0)

>>> v = (2, 0, 0)
>>> distance_to_cylinder_center(v, n, cylinder_center=p)
np.float64(0.0)

>>> v = [(0, 0, 0), (0, 1, 0), (1, 0, 0), (0, 0, 1)]
>>> distance_to_cylinder_center(v, n, cylinder_center=p)
array([2. ,  2.  ,  1. ,  2.23606798])

The implementaion is as follows.

First, the cylinder_center is translated to the origin (0, 0, 0). The transformation does not change the center axis, n, but the given vector v is converted to ${\overrightarrow{v}}^{\prime }=\overrightarrow{v}-\overrightarrow{p}$.

The distance d is now calculated by $d=|v^{\prime }|\sin \alpha$ where $\alpha$ is the angle between the v’ and n vectors. Thus, $\cos \alpha =\overrightarrow{v^{\prime }}\cdot n/|v^{\prime }||n|$, and therefore, the distance is

$$\begin{array}{r}d=|v^{\prime }|\sin \alpha =|v^{\prime }|\sqrt{1-{\cos }^2\alpha }\\ =\sqrt{|v^{\prime }{|}^2-\frac{\overrightarrow{n}\cdot \overrightarrow{v^{\prime }}}{|n{|}^2}}\end{array}$$

irfpy.util.vectortools.cone_angle(vector, cone_axis, cone_center=(0, 0, 0))

Return cone angle in radians.

Parameters:

• vector – Vector for input. (3, ) or (…, 3) array

• cone_axis – Vector of the axis, along the line.

• cone_center – The top of the cone.

Returns:

The cone angle in radians. Scalar or (…) shaped.

>>> p = (1, 3, 0)
>>> a = (0, 2, 0)

>>> v = (1, 0, 0)
>>> print(np.rad2deg(cone_angle(v, a, cone_center=p)))
180.0

>>> v = (1, 4, 0)
>>> print(np.rad2deg(cone_angle(v, a, cone_center=p)))
0.0

>>> v = [(1, 0, 0), (1, 1, 0), (1, 2, 0), (1, 4, 0), (2, 3, 0)]
>>> print(np.rad2deg(cone_angle(v, a, cone_center=p)))
[180.  180.  180.  0.  90.]

>>> v = [2, 3, 0]