Develop Reference

how to create a square on top of a given Vector. This square should be orthogonal to the given vector

Programming in Python (Blender):
I want to create a square and print all vertices (A;B;C;D) into my console on top of a given Vector. The square should be orthogonal to this vector, like this:
def create_verts_around_point(radius, vert):
# given Vector
vec = np.array([vert[0], vert[1], vert[2]])
# side_length of square
side_length = radius
# Vctor x-direction (1,0,0)
x_vec = np.array([1,0,0])
# Vekctor y-direction (0,1,0)
y_vec = np.array([0,1,0])
# Vector z-direction (0,0,1)
z_vec = np.array([0,0,1])
p1 = vec + (side_length/2) * x_vec + (side_length/2) * y_vec + (side_length/2) * z_vec
p2 = vec - (side_length/2) * x_vec + (side_length/2) * y_vec + (side_length/2) * z_vec
p3 = vec - (side_length/2) * x_vec - (side_length/2) * y_vec + (side_length/2) * z_vec
p4 = vec + (side_length/2) * x_vec - (side_length/2) * y_vec + (side_length/2) * z_vec
But my output looks like this in the end (Square is always parallel to my x-axis and y-axis):

I don't think you're really thinking about this problem in 3D, but see if this is close.
I create a square, perpendicular to the X axis. I then rotate that square based on the angles in x, y, and z. I then position the square at the end of the vector and plot it. I add plot points for the origin and the end of the vector, and I duplicate the last point in the square do it draws all the lines.
import math
import numpy as np
from mpl_toolkits import mplot3d
import matplotlib.pyplot as plt
def create_verts_around_point(sides, vert):
x0, y0, z0 = vert
# Here is the unrotated square.
half = sides/2
square = [
[0, -half,-half],
[0, -half, half],
[0, half, half],
[0, half,-half],
]
# Now find the rotation in each direction.
thetax = math.atan2( z0, y0 )
thetay = math.atan2( z0, x0 )
thetaz = math.atan2( y0, x0 )
# Now rotate the cube, first in x.
cubes = []
txcos = math.cos(thetax)
txsin = math.sin(thetax)
tycos = math.cos(thetay)
tysin = math.sin(thetay)
tzcos = math.cos(thetaz)
tzsin = math.sin(thetaz)
for x,y,z in square:
x,y,z = (x, y * txcos - z * txsin, y * txsin + z * txcos)
x,y,z = (x * txcos - z * txsin, y, x * txsin + z * txcos)
x,y,z = (x * txcos - y * txsin, x * txsin + y * txcos, z)
cubes.append( (x0+x, y0+y, z0+z) )
return cubes
point = (10,10,10)
square = create_verts_around_point(5, point)
points = [(0,0,0),point] + square + [square[0]]
x = [p[0] for p in points]
y = [p[1] for p in points]
z = [p[2] for p in points]
ax = plt.figure().add_subplot(111, projection='3d')
ax.plot( x, y, z )
plt.show()
Output:

Holoviz Polygon orientation settings

I am using holoviews+bokeh, and I would like to encircle my scatter plot data with a measure of standard deviation. Unfortunately I can't seem to get the orientation setting right. I am confused by the available descriptions:
Orientation in the Cartesian coordinate system, the
counterclockwise angle in radians between the first axis and the
horizontal
and
you can set the orientation (in radians, rotating anticlockwise)
My script and data example:
def create_plot(x, y, nstd=5):
x, y = np.asarray(x), np.asarray(y)
cov_matrix = np.cov([x, y])
eigenvalues, eigenvectors = np.linalg.eig(cov_matrix)
order = eigenvalues.argsort()[0]
angle = np.arctan2(eigenvectors[1, order], eigenvectors[1, order])
x0 = np.mean(x)
y0 = np.mean(y)
x_dir = np.cos(angle) * x - np.sin(angle) * y
y_dir = np.sin(angle) * x + np.cos(angle) * y
w = nstd * np.std(x_dir)
h = nstd * np.std(y_dir)
return hv.Ellipse(x0, y0, (w, h), orientation=-angle) * hv.Scatter((x, y))
c2x = np.random.normal(loc=-2, scale=0.6, size=200)
c2y = np.random.normal(loc=-2, scale=0.1, size=200)
combined = create_plot(c2x, c2y)
combined.opts(shared_axes=False)

Here is a solution, which draws Ellipse around the data. You math is just simplified.
import numpy as np
import holoviews as hv
from holoviews import opts
hv.extension('bokeh')
x = np.random.normal(loc=-2, scale=0.6, size=200)
y = np.random.normal(loc=-2, scale=0.1, size=200)
def create_plot(x, y, nstd=5):
x, y = np.asarray(x), np.asarray(y)
x0 = np.mean(x)
y0 = np.mean(y)
w = np.std(x)*nstd
h = np.std(y)*nstd
return hv.Ellipse(x0, y0, (w, h)) * hv.Scatter((x, y))
combined = create_plot(c2x, c2y)
combined.opts()
This gives you a plot which looks like a circle. To make it more visiable that it is a Ellipse your could genereate the plot calling
def hook(plot, element):
plot.handles['x_range'].start = -4
plot.handles['x_range'].end = 0
plot.handles['y_range'].start = -2.5
plot.handles['y_range'].end = -1
combined.opts(hooks=[hook])
which set fixed ranges and deactivates the auto focus.
In your example w and h were nearly the same, that means, you drawed a cercle. The orientation didn't have any effect. With the code above you can turn the Ellipse like
hv.Ellipse(x0, y0, (w, h), orientation=np.pi/2)
to see that it is working, but there is no need to do it anymore.

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()

Python: heat density plot in a disk

My goal is to make a density heat map plot of sphere in 2D. The plotting code below the line works when I use rectangular domains. However, I am trying to use the code for a circular domain. The radius of sphere is 1. The code I have so far is:
from pylab import *
import numpy as np
from matplotlib.colors import LightSource
from numpy.polynomial.legendre import leggauss, legval
xi = 0.0
xf = 1.0
numx = 500
yi = 0.0
yf = 1.0
numy = 500
def f(x):
if 0 <= x <= 1:
return 100
if -1 <= x <= 0:
return 0
deg = 1000
xx, w = leggauss(deg)
L = np.polynomial.legendre.legval(xx, np.identity(deg))
integral = (L * (f(x) * w)[None,:]).sum(axis = 1)
c = (np.arange(1, 500) + 0.5) * integral[1:500]
def r(x, y):
return np.sqrt(x ** 2 + y ** 2)
theta = np.arctan2(y, x)
x, y = np.linspace(0, 1, 500000)
def T(x, y):
return (sum(r(x, y) ** l * c[:,None] *
np.polynomial.legendre.legval(xx, identity(deg)) for l in range(1, 500)))
T(x, y) should equal the sum of c the coefficients times the radius as a function of x and y to the l power times the legendre polynomial where the argument is of the legendre polynomial is cos(theta).
In python: integrating a piecewise function, I learned how to use the Legendre polynomials in a summation but that method is slightly different, and for the plotting, I need a function T(x, y).
This is the plotting code.
densityinterpolation = 'bilinear'
densitycolormap = cm.jet
densityshadedflag = False
densitybarflag = True
gridflag = True
plotfilename = 'laplacesphere.eps'
x = arange(xi, xf, (xf - xi) / (numx - 1))
y = arange(yi, yf, (yf - yi) / (numy - 1))
X, Y = meshgrid(x, y)
z = T(X, Y)
if densityshadedflag:
ls = LightSource(azdeg = 120, altdeg = 65)
rgb = ls.shade(z, densitycolormap)
im = imshow(rgb, extent = [xi, xf, yi, yf], cmap = densitycolormap)
else:
im = imshow(z, extent = [xi, xf, yi, yf], cmap = densitycolormap)
im.set_interpolation(densityinterpolation)
if densitybarflag:
colorbar(im)
grid(gridflag)
show()
I made the plot in Mathematica for reference of what my end goal is

If you set the values outside of the disk domain (or whichever domain you want) to float('nan'), those points will be ignored when plotting (leaving them in white color).