# streamline_sample¶

Sample script of stream lines

'''Sample script of stream lines
'''
import matplotlib.pyplot as plt
import numpy as np

import irfpy.util.streamline

def draw_vortex(ax):

# In x, y plane, vector field at (x, y) is (-y, x).
def vector_field_vortex(pos):
return np.array([-pos, pos, 0])

# Initialize
strm = irfpy.util.streamline.SimpleTracing((1, 0, 0), vector_field_vortex, 0.01)

for t in range(400):
strm.step_forward()

# Backward
strm2 = irfpy.util.streamline.SimpleTracing((1, 0, 0), vector_field_vortex, -0.01)

for t in range(200):
strm2.step_forward()

ax.plot(strm.xlist, strm.ylist, 'b')
ax.plot(strm2.xlist, strm2.ylist, 'r')
ax.set_aspect('equal')

def draw_dipole(ax):

# In x, z plane, the vector field is r = 2cos t and t = sin t.
def vector_field_dipole(pos):
x, y, z = pos
r2 = x * x + z * z
th = np.arccos(z / np.sqrt(r2))
r = 2 * np.cos(th)
t = np.sin(th)
lx = r * np.sin(th) + t * np.cos(th)
lz = r * np.cos(th) - t * np.sin(th)
return np.array([lx, 0, lz])

strm = irfpy.util.streamline.SimpleTracing((0.5, 0, 0), vector_field_dipole, 0.001)
# The vectof_field length is of the order of 1.
# dt is 0.001, thus, the segment length per step is about 0.001.

for t in range(30000):
strm.step_forward()
if strm.x ** 2 + strm.y ** 2 < 0.001 ** 2:
break
ax.plot(strm.xlist, strm.zlist, 'b')

strm = irfpy.util.streamline.SimpleTracing((0.5, 0, 0), vector_field_dipole, -0.001)
for t in range(30000):
strm.step_forward()
if strm.x ** 2 + strm.y ** 2 < 0.001 ** 2:
break
ax.plot(strm.xlist, strm.zlist, 'r-x')

# Indeed, direction only matters for vector field.
# Thus, the following is more physical, but practically rather
# difficult to handle, as the segment per step changes by r^2.

# In x, z plane, the vector field is r = 2cos t / r^2 and t = sin t / r^2.
def vector_field_dipole2(pos):
x, y, z = pos
r2 = x * x + z * z
th = np.arccos(z / np.sqrt(r2))
r = 2 * np.cos(th) / r2
t = np.sin(th) / r2
lx = r * np.sin(th) + t * np.cos(th)
lz = r * np.cos(th) - t * np.sin(th)
return np.array([lx, 0, lz])

strm = pyana.util.streamline.SimpleTracing((0.5, 0, 0), vector_field_dipole2, 0.001)
# The vectof_field length is of the order of 1.
# dt is 0.001, thus, the segment length per step is about 0.001.

for t in range(30000):
strm.step_forward()
if strm.x ** 2 + strm.y ** 2 < 0.1 ** 2:
break
ax.plot(strm.xlist, strm.zlist, 'bo')

ax.set_aspect('equal')

def draw_multidipole(ax):
# In x, z plane, the vector field is r = 2cos t and t = sin t.
def vector_field_dipole(pos):
x, y, z = pos
r2 = x * x + z * z
th = np.arccos(z / np.sqrt(r2))
r = 2 * np.cos(th)
t = np.sin(th)
lx = r * np.sin(th) + t * np.cos(th)
lz = r * np.cos(th) - t * np.sin(th)
return np.array([lx, 0, lz])

for x0 in np.arange(0.1, 3, 0.3):
print(x0)
strm = irfpy.util.streamline.SimpleTracing((x0, 0, 0), vector_field_dipole, 0.001)
for t in range(30000):
strm.step_forward()
if strm.x ** 2 + strm.y ** 2 < 0.001 ** 2:
break
ax.plot(strm.xlist, strm.zlist, 'b')

def main():
fig = plt.figure()

draw_vortex(ax1)