Python: Intersection of plots that intersect at points interpolated by matplotlib.pyplot - python

I am looking for the points of intersection of a vertical line with a plot that I have made that has pyplot's interpolated values.
I think the code and plot below will make my question more clear. Below is some example code, and the resulting plot. What I am looking for is all intersection points between the red vertical line and the blue lines (so there should be 3 such points in this case).
I am at a loss for how to do this - does anyone know how?
The code:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
t = np.linspace(-np.pi, np.pi, 512, endpoint=False) + 0.0001 # 0.0001 to get rid of 0 values.
# normalized square wave
u = np.sign(np.sin(2 * np.pi * t))
u = u - np.min(u)
u = u / np.max(u)
# rotate the square wave
phi = - np.pi / 3.0
t_rot = t * np.cos(phi) - u * np.sin(phi)
u_rot = u * np.cos(phi) + t * np.sin(phi)
# level the rotated square wave
u_rot_leveled = u_rot + np.tan(-phi) * t_rot
plt.plot(t_rot, u_rot_leveled, '.-')
plt.axvline(x=-1.1, linestyle=':', color='red')
The plot:
Thanks for any help!

Instead of interpolating the values of y where x==x0, you may actually find the roots(zeros) of x-x0 with respect to y.
import numpy as np
import matplotlib.pyplot as plt
t = np.linspace(-np.pi, np.pi, 512, endpoint=False) + 0.0001 # 0.0001 to get rid of 0 values.
# normalized square wave
u = np.sign(np.sin(2 * np.pi * t))
u = u - np.min(u)
u = u / np.max(u)
# rotate the square wave
phi = - np.pi / 3.0
t_rot = t * np.cos(phi) - u * np.sin(phi)
u_rot = u * np.cos(phi) + t * np.sin(phi)
# level the rotated square wave
u_rot_leveled = u_rot + np.tan(-phi) * t_rot
def find_roots(x,y):
s = np.abs(np.diff(np.sign(y))).astype(bool)
return x[:-1][s] + np.diff(x)[s]/(np.abs(y[1:][s]/y[:-1][s])+1)
x0 = -1.1
z = find_roots(u_rot_leveled, t_rot-x0)
plt.plot(t_rot, u_rot_leveled, '.-')
plt.axvline(x=x0, linestyle=':', color='red')
plt.plot(np.ones_like(z)*x0, z, marker="o", ls="", ms=4, color="limegreen")
plt.show()
Part of the solution here is taken from my answer to How to get values from a graph?

Related

How can I solve a contourf array problem?

I have trouble with plt.contourf.
The program is supposed to calculate the distance between s1 and m called s1m and S2 and m called s2m than using s1m and s2m we calculate the wave functions psi and psiP than we multiply them to get the intensity of light, we use what we get in contourf to see the results in a screen.
When I run the program I
import numpy as np
import matplotlib.pyplot as plt
S1 = np.array([100,0,-1])
S2 = np.array([-100,0,-1])
M = np.array([1,1,0])
Lambda = 633
s1m= np.substract(m,S1)#vector S1M
s2m= np.substract(m,S2)#vector S2M
SM1= np.multiply(s1m,s1m)
SM2= np.multiply(s2m,s2m)
S1M= np.sqrt(SM1)#distance s1m
S2M= np.sqrt(SM2)#distance s2m
def intensity (S1M,S2M):
Phi1=(2 * np pi * S1M)/lambda
Phi2=(2 * np.pi * S2M)/lambda
Tet1=(-2 * np.pi * S1M)/lambda
Tet2=(-2 * np.pi * S2M)/lambda
Psi1 = np.exp(Phi1)
Psi2 = np.exp(Phi2)
Psi1P = np.exp(Tet1)
Psi2P = np.exp(Tet2)
Psi = Psi1 + Psi2
PsiP = Psi1P + Psi2P
I = Psi * PsiP
x = np.linspace(1,5,5)
y = np.linspace(1,5,5)
XX,YY = np.meshgrid(x,y)
ZZ = intensity (S1M, S2M)
plt.contourf (XX, YY, ZZ)
plt.show()

only size-1 arrays can be converted to Python scalars: Trying to plot circular orbit with equation of. circle

I'm relatively new to python so forgive me for any nonsense in my code. I am trying to program a circular orbit of a planet (I just used the mass of Uranus and the Sun) in python using equations from my Classical Mechanics textbook (John R. Taylor's Classical Mechanics). I figured I could just use the equations and graph a function, y, that equals the equations of a circle with c_squared being the radius and x being an array of values being used to plot the circle. Let me know how I can improve the code or I am even going in the right direction.
...
import matplotlib.pyplot as plt
import numpy as np
import matplotlib
import math
fig = plt.figure()
ax = fig.add_subplot()
m_uranus = 8.681 * 10**(25)
m_sun = 1.989 * 10 **(30)
G = 6.67430 * 10**(-11)
mu = (m_uranus * m_sun)/(m_uranus + m_sun)
l = (1.7 * 10**(42)) * 1000 * 24 * 60 * 60
ang_squared = l ** 2
c = (ang_squared)/(G * m_uranus * m_sun * mu)
c_squared = c**2
print(m_sun, mu, m_uranus, ang_squared, c)
x = np.arange(-100, 100, 1)
y = math.sqrt(c_squared - x)
plt.plot(x, y)
plt.show()
...
As mentioned by #JohanC, use numpy np.sqrt() instead of math.sqrt() will fix your error, here the fix with (unnecessary libraries removed):
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.add_subplot()
m_uranus = 8.681 * 10 ** 25
m_sun = 1.989 * 10 ** 30
G = 6.67430 * 10 ** (-11)
mu = (m_uranus * m_sun) / (m_uranus + m_sun)
l = (1.7 * 10 ** 42) * 1000 * 24 * 60 * 60
ang_squared = l ** 2
c = ang_squared / (G * m_uranus * m_sun * mu)
c_squared = c ** 2
print(m_sun, mu, m_uranus, ang_squared, c)
x = np.arange(-100, 100, 1)
y = np.sqrt(c_squared - x)
plt.plot(x, y)
plt.show()
Hope, this will help you by!

Python mayavi : How to draw spheres at random positions in 3D space

This is a code for generating random sized spheres with mayavi,
I want to make the spheres to be connected with each other by the surface or with a bond line:
Spheres must be at random positions in 3D space
Spheres must be with the same radius
from mayavi import mlab
import numpy as np
[phi,theta] = np.mgrid[0:2*np.pi:12j,0:np.pi:12j]
x = np.cos(phi)*np.sin(theta)
y = np.sin(phi)*np.sin(theta)
z = np.cos(theta)
def plot_sphere(p):
r,a,b,c = p
r=1
return mlab.mesh(r*x+a, r*y+b, r*z )
for k in range(8):
c = np.random.rand(4)
c[0] /= 10.
plot_sphere(c)
mlab.show()
From sphere equation:
So when passing arguments to mlab.mesh we would like to set [x_0, y_0, z_0] for each sphere such as they are at different positions from the axis.
The problem was that the numbers generated by np.random.rand(4) are random, but not distinct.
Let's modify so that the arguments [x_0, y_0, z_0] are random and distinct:
We use sample to get distinct index numbers in a cube
We convert using index_to_3d the index to an (x, y, z) coordinates
The radius, r, can be adjusted to have more or less spacing between the spheres.
Spheres at 3D space
Code:
import random
from itertools import product
from mayavi import mlab
import numpy as np
[phi, theta] = np.mgrid[0:2 * np.pi:12j, 0:np.pi:12j]
x = np.cos(phi) * np.sin(theta)
y = np.sin(phi) * np.sin(theta)
z = np.cos(theta)
def plot_sphere(x_0, y_0, z_0):
r = 0.5
return mlab.mesh(r * x + x_0, r * y + y_0, r * z + z_0)
SPHERES_NUMBER = 200
CUBE_SIZE = 10
def index_to_3d(i, SIZE):
z = i // (SIZE * SIZE)
i -= (z * SIZE * SIZE)
y = i // SIZE
x = i % SIZE
return x, y, z
random_tuples = [index_to_3d(i, CUBE_SIZE) for i in random.sample(range(CUBE_SIZE ** 3), SPHERES_NUMBER)]
for k in range(SPHERES_NUMBER):
x_0, y_0, z_0 = random_tuples[k]
plot_sphere(x_0, y_0, z_0)
mlab.show()
Output:
Spheres cluster
Let's utilize gauss to create coordinates for the cluster points.
Code:
import random
from itertools import product
from mayavi import mlab
import numpy as np
[phi, theta] = np.mgrid[0:2 * np.pi:12j, 0:np.pi:12j]
x = np.cos(phi) * np.sin(theta)
y = np.sin(phi) * np.sin(theta)
z = np.cos(theta)
def plot_sphere(x_0, y_0, z_0):
r = 0.5
return mlab.mesh(r * x + x_0, r * y + y_0, r * z + z_0)
SPHERES_NUMBER = 200
def create_cluster(CLUSTER_SIZE):
means_and_deviations = [(1, 1.5), (1, 1.5), (1, 1.5)]
def generate_point(means_and_deviations):
return tuple(random.gauss(mean, deviation) for mean, deviation in means_and_deviations)
cluster_points = set()
while len(cluster_points) < CLUSTER_SIZE:
cluster_points.add(generate_point(means_and_deviations))
return list(cluster_points)
cluster_points = create_cluster(SPHERES_NUMBER)
for k in range(SPHERES_NUMBER):
x_0, y_0, z_0 = cluster_points[k]
plot_sphere(x_0, y_0, z_0)
mlab.show()
Output:
What about just using the mayavi points3d function? By default the mode parameter is set to sphere and you can set the diameter by using the scale_factor parameter. You can also increase the resolution of the sphere by varying the resolution parameter.
Here is the code:
def draw_sphere(
center_coordinates,
radius,
figure_title,
color,
background,
foreground
):
sphere = mlab.figure(figure_title)
sphere.scene.background = background
sphere.scene.foreground = foreground
mlab.points3d(
center_coordinates[0],
center_coordinates[1],
center_coordinates[2],
color=color,
resolution=256,
scale_factor=2*radius,
figure=sphere
)
Regarding the connected with each other by the surface issue, your explanation is poor. Maybe you mean just tangent spheres, but I would need more details.

How to draw cycloid on curve of other function (other cycloid)?

I would like to draw cycloid that is going on other cycloid but I don't know exactly how to do this. Here is my code.
import numpy as np
import matplotlib.pyplot as plt
import math
from matplotlib import animation
#r = float(input('write r\n'))
#R = float(input('write R\n'))
r = 1
R = 1
x = []
y = []
x2 = []
y2 = []
x3 = []
y3 = []
length=[0]
fig, ax = plt.subplots()
ln, = plt.plot([], [], 'r', animated=True)
f = np.linspace(0, 10*r*math.pi, 1000)
def init():
ax.set_xlim(-r, 12*r*math.pi)
ax.set_ylim(-4*r, 4*r)
return ln,
def update2(frame):
#parametric equations of cycloid
x0 = r * (frame - math.sin(frame))
y0 = r * (1 - math.cos(frame))
x.append(x0)
y.append(y0)
#derivative of cycloid
dx = r * (1 - math.cos(frame))
dy = r * math.sin(frame)
#center of circle
a = dy * dy + dx * dx
b = (-2 * x0 * dy) - (2 * frame * dy * dy) + (2 * y0 * dx) - (2 * frame * dx * dx)
c = (x0 * x0) + (2 * frame * x0 * dy) + (frame * frame * dy * dy) + (y0 * y0) - (2 * frame * y0 * dx) + (frame * frame * dx * dx) -1
t1 = (-b - math.sqrt(b * b - 4 * a * c)) / (2 * a)
#t2 = (-b + math.sqrt(b * b - 4 * a * c)) / (2 * a)
center1x=(x0-dy*(t1-x0))*R
center1y=(y0+dx*(t1-x0))*R
#center2x=(x0-dy*(t2-x0))*R
#center2y=(y0+dx*(t2-x0))*R
#length of cycloid
length.append(math.sqrt(x0*x0 + y0*y0))
dl=sum(length)
param = dl / R
W1x = center1x + R * math.cos(-param)
W1y = center1y + R * math.sin(-param)
#W2x = center2x + R * math.cos(-param)
#W2y = center2y + R * math.sin(-param)
x2.append(W1x)
y2.append(W1y)
#x3.append(W2x)
#y3.append(W2y)
ln.set_data([x, x2], [y, y2])
return ln,
ani = animation.FuncAnimation(fig, update2, frames=f,init_func=init, blit=True, interval = 0.1, repeat = False)
plt.show()
In my function update2 I created parametric equations of first cycloid and then tried to obtain co-ordinates of points of second cycloid that should go on the first one.
My idea is based on that that typical cycloid is moving on straight line, and cycloid that is moving on other curve must moving on tangent of that curve, so center of circle that's creating this cycloid is always placed on normal of curve. From parametric equations of normal I have tried to obtain center of circle that creating cycloid but I think that isn't good way.
My goal is to get something like this:
Here is one way. Calculus gives us the formulas to find the direction angles at any point on the cycloid and the arc lengths along the cycloid. Analytic Geometry tells us how to use that information to find your desired points.
By the way, a plot made by rolling a figure along another figure is called a roulette. My code is fairly simple and could be optimized, but it works now, can be used for other problems, and is broken up to make the math and algorithm easier to understand. To understand my code, use this diagram. The cycloid is the blue curve, the black circles are the rolling circle on the cycloid, point A is an "anchor point" (a point where the rim point touches the cycloid--I wanted to make this code general), and point F is the moving rim point. The two red arcs are the same length, which is what we mean by rolling the circle along the cycloid.
And here is my code. Ask if you need help with the source of the various formulas, but the direction angles and arc lengths use calculus.
"""Numpy-compatible routines for a standard cycloid (one caused by a
circle of radius r above the y-axis rolling along the positive x-axis
starting from the origin).
"""
import numpy as np
def x(t, r):
"""Return the x-coordinate of a point on the cycloid with parameter t."""
return r * (t - np.sin(t))
def y(t, r):
"""Return the y-coordinate of a point on the cycloid with parameter t."""
return r * (1.0 - np.cos(t))
def dir_angle_norm_in(t, r):
"""Return the direction angle of the vector normal to the cycloid at
the point with parameter t that points into the cycloid."""
return -t / 2.0
def dir_angle_norm_out(t, r):
"""Return the direction angle of the vector normal to the cycloid at
the point with parameter t that points out of the cycloid."""
return np.pi - t / 2.0
def arclen(t, r):
"""Return the arc length of the cycloid between the origin and the
point on the cycloid with parameter t."""
return 4.0 * r * (1.0 - np.cos(t / 2.0))
# Roulette problem
def xy_roulette(t, r, T, R):
"""Return the x-y coordinates of a rim point on a circle of radius
R rolling on a cycloid of radius r starting at the anchor point
with parameter T currently at the point with parameter t. (Such a
rolling curve on another curve is called a roulette.)
"""
# Find the coordinates of the contact point P between circle and cycloid
px, py = x(t, r), y(t, r)
# Find the direction angle of PC from the contact point to circle's center
a1 = dir_angle_norm_out(t, r)
# Find the coordinates of the center C of the circle
cx, cy = px + R * np.cos(a1), py + R * np.sin(a1)
# Find cycloid's arc distance AP between anchor and current contact points
d = arclen(t, r) - arclen(T, r) # equals arc PF
# Find the angle φ the circle turned while rolling from the anchor pt
phi = d / R
# Find the direction angle of CF from circle's center to rim point
a2 = dir_angle_norm_in(t, r) - phi # subtract: circle rolls clockwise
# Find the coordinates of the final point F
fx, fy = cx + R * np.cos(a2), cy + R * np.sin(a2)
# Return those coordinates
return fx, fy
import matplotlib.pyplot as plt
r = 1
R = 0.75
T = np.pi / 3
t_array = np.linspace(0, 2*np.pi, 201)
cycloid_x = x(t_array, r)
cycloid_y = y(t_array, r)
roulette_x, roulette_y = xy_roulette(t_array, r, T, R)
fig, ax = plt.subplots()
ax.set_aspect('equal')
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
ax.plot(cycloid_x, cycloid_y)
ax.plot(roulette_x, roulette_y)
plt.show()
And here is the resulting graphic. You can pretty this up as you choose. Note that this only has the circle rolling along one arch of the cycloid. If you clarify what should happen at the cusps, this could be extended.
Or, if you want a smaller circle and a curve that ends at the cusps (here r = 1, T = 0 n = 6 (the number of little arches), and R = 4 * r / np.pi / n),
You can generate coordinates of the center of rolling circle using parallel curve definition. Parametric equations for this center are rather simple (if I did not make mistakes):
for big cycloid:
X = R(t - sin(t))
Y = R(1 - cos(t))
X' = R(1 - cos(t))
Y' = R*sin(t)
parallel curve (center of small circle):
Sqrt(X'^2+Y'^2)=R*Sqrt(1-2*cos(t)+cos^2(t)+sin^2(t)) =
R*Sqrt(2-2*cos(t))=
R*Sqrt(4*sin^2(t/2))=
2*R*sin(t/2)
x(t) = X(t) + r*R*sin(t)/(2R*sin(t/2)) =
R(t - sin(t)) + r*2*sin(t/2)*cos(t/2) / (2*sin(t/2)) =
R(t - sin(t)) + r*cos(t/2)
y(t) = Y(t) - r*R*(1-cos(t))/(2*R*sin(t/2)) =
R(1 - cos(t)) - r*(2*sin^2(t/2)/(2*sin(t/2)) =
R(1 - cos(t)) - r*sin(t/2)
But trajectory of point on the circumference is superposition of center position and rotation around it with angular velocity that depends on length of main cycloid plus rotation of main tangent.
Added from dicussion in comments:
cycloid arc length
L(t) = 4R*(1-cos(t/2))
to use it for small circle rotation, divide by r
tangent rotation derivation
fi(t) = atan(Y'/X') = atan(sin(t)/(1-cos(t)) =
atan(2*sin(t/2)*cos(t/2)/(2(sin^2(t/2))) =
atan(ctg(t/2)) = Pi/2 - t/2
so tangent direction change is proportional to big cycloid parameter
and final result is (perhaps some signs are not correct)
theta(t) = L(t)/r + t/2 + Phase
ox(t) = x(t) + r * cos(theta(t))
oy(t) = y(t) + r * sin(theta(t))
Thanks everyone. Somehow I have managed to accomplish that. Solution is maybe ugly but sufficient for me.
import numpy as np
import matplotlib.pyplot as plt
import math
from matplotlib import animation
r = float(input('write r\n'))
R = float(input('write R\n'))
#r=1
#R=0.1
x = []
y = []
x2 = []
y2 = []
x_1=0
x_2=0
lengthX=[0]
lengthY=[0]
lengthabs=[0]
fig, ax = plt.subplots()
ln, = plt.plot([], [], 'r', animated=True)
f = np.linspace(0, 2*math.pi, 1000)
def init():
ax.set_xlim(-r, 4*r*math.pi)
ax.set_ylim(0, 4*r)
return ln,
def update2(frame):
#cycloid's equations
x0 = r * (frame - math.sin(frame))
y0 = r * (1 - math.cos(frame))
x.append(r * (frame - math.sin(frame)))
y.append(r * (1 - math.cos(frame)))
#arc's length
lengthabs.append(math.sqrt((x0-lengthX[-1])*(x0-lengthX[-1])+(y0-lengthY[-1])*(y0-lengthY[-1])))
lengthX.append(x0)
lengthY.append(y0)
dl=sum(lengthabs)
param = dl / R
#center of circle
center1x = r * (frame - math.sin(frame)) + R * math.cos((frame+2*math.pi) / 2)
center1y = r * (1 - math.cos(frame)) - R * math.sin((frame+2*math.pi) / 2)
if(frame<2*math.pi):
W1x = center1x + R * math.cos(-param)
W1y = center1y + R * math.sin(-param)
else:
W1x = center1x + R * math.cos(param)
W1y = center1y + R * math.sin(param)
x2.append(W1x)
y2.append(W1y)
ln.set_data([x,x2], [y,y2])
return ln,
ani = animation.FuncAnimation(fig, update2, frames=f,init_func=init, blit=True, interval = 0.1, repeat = False)
plt.show()

pylab plot showing asymptotes

How can I remove the asymptote?
import numpy as np
e = 1.26
beta = .705 * np.pi
rph = 7000
re = 6378
def r(nuh):
return rph * (1 + e) / (1 + e * np.cos(nuh + beta))
theta = np.linspace(-np.pi, np.pi, 50000)
fig2 = pylab.figure()
ax2 = fig2.add_subplot(111)
ax2.plot(r(theta) * np.cos(theta), r(theta) * np.sin(theta))
ax2.plot(rph * np.cos(theta), rph * np.sin(theta), 'r')
# adding the Earth
earth2 = pylab.Circle((0, 0), radius = re, color = 'b')
ax2.add_patch(earth2)
pylab.xlim((-50000, 100000))
pylab.ylim((-50000, 100000))
pylab.show()
As you can see here, setting the divergent points to np.nan will cause them not to be plotted.
In your problem, it is r(theta) which diverges. Define r and theta in the usual way, but then, you want to set the extrema of r(theta) to np.nan.
To do this, make an array first, then change its extrema to np.nan:
rt = r(theta)
ext = [np.argmin(rt), np.argmax(rt)]
rt[ext] = np.nan
Now, be sure to plot with the modified rt array not the original function:
ax2.plot(rt * np.cos(theta), rt * np.sin(theta))

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