python triangle angles returns null - python

A function that returns the value of triangle angles in degrees from length of the sides, the results are integers sorted from lowest to highest
from math import acos, degrees
def angles(a, b, c):
alpha = beta = gamma = 0
if a !=0 and b != 0 and c != 0:
if a + b < c or a + c < b or b + c < a:
return [alpha, beta, gamma]
else:
alpha = int(degrees(acos(( c*c + b*b - a*a ) / (2 * c * b))))
beta = int(degrees(acos(( a*a + c*c - b*b ) / (2 * a * c))))
gamma = int(degrees(acos(( a*a + b*b - c*c ) / (2 * a * b))))
return list.sort([alpha, beta, gamma])
why do I get None as a result?

list.sort sorts the list in-place and returns None. You could use sorted instead:
return sorted([alpha, beta, gamma])

Try this code:
from math import acos, degrees
def angles(a, b, c):
alpha = beta = gamma = 0
if a !=0 and b != 0 and c != 0:
if a + b < c or a + c < b or b + c < a:
return [alpha, beta, gamma]
else:
alpha = int(degrees(acos(( c*c + b*b - a*a ) / (2 * c * b))))
beta = int(degrees(acos(( a*a + c*c - b*b ) / (2 * a * c))))
gamma = int(degrees(acos(( a*a + b*b - c*c ) / (2 * a * b))))
lst = [alpha, beta, gamma]
lst.sort()
return lst
print angles(4,4,4)

Related

How can I get the same performance from Cython as I am getting from Numba?

I am trying to speed up my function in python. My original function is:
def SMO(X, y, C, epsilon, tol, max_iter):
#Initialization
n = X.shape[0]
alpha = np.zeros((n))
b = 0
count = 0
while True:
count += 1
alpha_prev = np.copy(alpha)
for i in range(n):
E_i = np.dot(X[i, :], alpha*y) + b - y[i]
#if alpha_i doesn't satisfy conditions then we try to optimise it
if (E_i*y[i]<-tol and alpha[i]<C) or (E_i*y[i]>tol and alpha[i]>0):
#get a random integer not equal to what we already have (i)
j = i
while j == i:
j = np.random.randint(0, n-1)
eta = 2 * X[i,j] - 2
if eta == 0:
continue
# store current alpha values
alpha_j, alpha_i = alpha[j], alpha[i]
# calculate values for L and H
if y[i] != y[j]:
L, H = max(0, alpha_j - alpha_i), min(C, C - alpha_i + alpha_j)
else:
L, H = max(0, alpha_i + alpha_j - C), min(C, alpha_i + alpha_j)
E_j = np.dot(X[j,:], alpha*y) + b - y[j]
# Set new alpha values and b value
alpha[j] = alpha_j - y[j] * (E_i - E_j)/eta
alpha[j] = max(alpha[j], L)
alpha[j] = min(alpha[j], H)
alpha[i] = alpha_i + y[i]*y[j] * (alpha_j - alpha[j])
b1 = b - E_i - y[i] * (alpha[i] - alpha_i) - y[j] * (alpha[j] - alpha_j) * X[i,j]
b2 = b - E_j - y[j] * (alpha[j] - alpha_j) - y[i] * (alpha[i] - alpha_i) * X[i,j]
if alpha[i]>0 and alpha[i]<C:
b = b1
elif alpha[j]>0 and alpha[j]<C:
b = b2
else:
b = (b1 + b2)/2
# Check convergence
diff = np.linalg.norm(alpha - alpha_prev)
if diff < epsilon:
break
if count >= max_iter:
#print("Iteration number exceeded the max of %d iterations" % (max_iter))
return alpha, b
return alpha, b
and it takes around 26ms to run. I thought this function would be good to implement in Cython because of the for loop and there's a lot of indexing going on. Here's my Cython attempt:
%%cython --annotate
import numpy as np
cimport numpy as np
cimport cython
#cython.boundscheck(False) # Deactivate bounds checking
#cython.wraparound(False) # Deactivate negative indexing.
def C_SMO(double[:,:] X, double[:] y, float C, float epsilon, float tol, int max_iter):
cdef Py_ssize_t i, j
cdef int count
cdef double L, H, b, b1, b2, alpha_j, alpha_i, E_i, E_j, eta, diff
#Initialization
n = X.shape[1]
cdef double[:] alpha = np.zeros((n))
cdef double[:] alpha_y = np.zeros((n))
b = 0
count = 0
while True:
count += 1
alpha_prev = np.copy(alpha)
for i in range(n):
alpha_y[i] = alpha[i] * y[i]
for i in range(n):
E_i = np.dot(X[i, :], alpha_y) + b - y[i]
#if alpha_i doesn't satisfy conditions then we try to optimise it
if (E_i*y[i]<-tol and alpha[i]<C) or (E_i*y[i]>tol and alpha[i]>0):
#get a random integer not equal to what we already have (i)
j = i
while j == i:
j = np.random.randint(0, n-1)
eta = 2 * X[i,j] - 2
if eta == 0:
continue
# store current alpha values
alpha_j, alpha_i = alpha[j], alpha[i]
if y[i] != y[j]:
L, H = max(0, alpha_j - alpha_i), min(C, C - alpha_i + alpha_j)
else:
L, H = max(0, alpha_i + alpha_j - C), min(C, alpha_i + alpha_j)
E_j = np.dot(X[j,:], alpha_y) + b - y[j]
# Set new alpha values and b value
alpha[j] = alpha_j - y[j] * (E_i - E_j)/eta
alpha[j] = max(alpha[j], L)
alpha[j] = min(alpha[j], H)
alpha[i] = alpha_i + y[i]*y[j] * (alpha_j - alpha[j])
b1 = b - E_i - y[i] * (alpha[i] - alpha_i) - y[j] * (alpha[j] - alpha_j) * X[i,j]
b2 = b - E_j - y[j] * (alpha[j] - alpha_j) - y[i] * (alpha[i] - alpha_i) * X[i,j]
if alpha[i]>0 and alpha[i]<C:
b = b1
elif alpha[j]>0 and alpha[j]<C:
b = b2
else:
b = (b1 + b2)/2
# Check convergence
diff = np.linalg.norm(alpha - alpha_prev)
if diff < epsilon:
break
if count >= max_iter:
#print("Iteration number exceeded the max of %d iterations" % (max_iter))
return alpha, b
return alpha, b
This code takes around 30ms to run so if anything I've made it worse. Also deactivating bounds checking doesn't seem to help like I thought it would due to all the indexing.
Is there anything I can improve on here? Also when I write #jit(nopython=True) above the original function it runs in 680μs. So if there's not much I can do to my Cython code, why does Numba provide such a speed-up?

Fitting a line with gradient descent

I am trying to fit a line to a couple of points using gradient descent. I am no expert on this and tried to write down the mathematical algorithm for it in python. It runs for a couple of iterations, but my predictions seem to explode at some point. Here is the code:
import numpy as np
import matplotlib.pyplot as plt
def mean_squared_error(n, A, b, m, c):
e = 0
for i in range(n):
e += (b[i] - (m*A[i] + c)) ** 2
return e/n
def der_wrt_m(n,A,b,m,c):
d = 0
for i in range(n):
d += (2 * (b[i] - (m*A[i] + c)) * (-A[i]))
return d/n
def der_wrt_c(n,A,b,m,c):
d = 0
for i in range(n):
d += (2 * (b[i] - (m*A[i] + c)))
return d/n
def update(n,A,b,m,c,descent_rate):
return descent_rate * der_wrt_m(n,A,b,m,c)), descent_rate * der_wrt_c(n,A,b,m,c))
A = np.array(((0,1),
(1,1),
(2,1),
(3,1)))
x = A.T[0]
b = np.array((1,2,0,3), ndmin=2 ).T
y = b.reshape(4)
def descent(x,y):
m = 0
c = 0
descent_rate = 0.00001
iterations = 100
n = len(x)
plt.scatter(x, y)
u = np.linspace(0,3,100)
prediction = 0
for itr in range(iterations):
print(m,c)
prediction = prediction + m * x + c
m,c = update(n,x,y,m,c,descent_rate)
plt.plot(u, u * m + c, '-')
descent(x,y)
And that's my output:
0 0
19.25 -10.5
-71335.1953125 24625.9453125
5593771382944640.0 -2166081169939480.2
-2.542705027685638e+48 9.692684648057364e+47
2.40856742196228e+146 -9.202614421953049e+145
-inf inf
nan nan
nan nan
nan nan
nan nan
nan nan
nan nan
etc...
Update: The values aren't exploding anymore, but it's still not converging in a nice manner:
# We could also solve it using gradient descent
import numpy as np
import matplotlib.pyplot as plt
def mean_squared_error(n, A, b, m, c):
e = 0
for i in range(n):
e += ((b[i] - (m * A[i] + c)) ** 2)
#print("mse:",e/n)
return e/n
def der_wrt_m(n,A,b,m,c):
d = 0
for i in range(n):
# d += (2 * (b[i] - (m*A[i] + c)) * (-A[i]))
d += (A[i] * (b[i] - (m*A[i] + c)))
#print("Dm",-2 * d/n)
return (-2 * d/n)
def der_wrt_c(n,A,b,m,c):
d = 0
for i in range(n):
d += (2 * (b[i] - (m*A[i] + c)))
#print("Dc",d/n)
return d/n
def update(n,A,b,m,c, descent_rate):
return (m - descent_rate * der_wrt_m(n,A,b,m,c)),(c - descent_rate * der_wrt_c(n,A,b,m,c))
A = np.array(((0,1),
(1,1),
(2,1),
(3,1)))
x = A.T[0]
b = np.array((1,2,0,3), ndmin=2 ).T
y = b.reshape(4)
def descent(x,y):
m = 0
c = 0
descent_rate = 0.0001
iterations = 10000
n = len(x)
plt.scatter(x, y)
u = np.linspace(0,3,100)
prediction = 0
for itr in range(iterations):
prediction = prediction + m * x + c
m,c = update(n,x,y,m,c,descent_rate)
loss = mean_squared_error(n, A, b, m, c)
print(loss)
print(m,c)
plt.plot(u, u * m + c, '-')
descent(x,y)
And now the graph looks like this after about 10000 iterations with a learning rate of 0.0001:
[4.10833186 5.21468937]
1.503547594304175 -1.9947003678083184
Whereas the least square fit shows something like this:
In your update function, you should subtract calculated gradients from current m and c
def update(n,A,b,m,c,descent_rate):
return m - (descent_rate * der_wrt_m(n,A,b,m,c)), c - (descent_rate * der_wrt_c(n,A,b,m,c))
Update: Here is the working version. I got rid of A matrix after obtaining x,y since it confuses me =). For example in your gradient calculations you have an expression d += (A[i] * (b[i] - (m*A[i] + c))) but it should be d += (x[i] * (b[i] - (m*x[i] + c))) since x[i] gives you a single element whereas A[i] gives you a list.
Also you forgot a minus sign while calculating derivative with respect to c. If your expression is (y - (m*x + c))^2) than derivative with respect to c should be 2 * (-1) * (y - (m*x + c)) since there is a minus in front of c.
# We could also solve it using gradient descent
import numpy as np
import matplotlib.pyplot as plt
def mean_squared_error(n, x, y, m, c):
e = 0
for i in range(n):
e += (m*x[i]+c - y[i])**2
e = e/n
return e/n
def der_wrt_m(n, x, y, m, c):
d = 0
for i in range(n):
d += x[i] * (y[i] - (m*x[i] + c))
d = -2 * d/n
return d
def der_wrt_c(n, x, y, m, c):
d = 0
for i in range(n):
d += (y[i] - (m*x[i] + c))
d = -2 * d/n
return d
def update(n,x,y,m,c, descent_rate):
return (m - descent_rate * der_wrt_m(n,x,y,m,c)),(c - descent_rate * der_wrt_c(n,x,y,m,c))
A = np.array(((0,1),
(1,1),
(2,1),
(3,1)))
x = A.T[0]
b = np.array((1,2,0,3), ndmin=2 ).T
y = b.reshape(4)
print(x)
print(y)
def descent(x,y):
m = 0.0
c = 0.0
descent_rate = 0.01
iterations = 10000
n = len(x)
plt.scatter(x, y)
u = np.linspace(0,3,100)
prediction = 0
for itr in range(iterations):
prediction = prediction + m * x + c
m,c = update(n,x,y,m,c,descent_rate)
loss = mean_squared_error(n, x, y, m, c)
print(loss)
print(loss)
print(m,c)
plt.plot(u, u * m + c, '-')
plt.show()
descent(x,y)

Python scipy overload _stats function for beta distribution

I need to overload the _stats function for my beta distribution. This is my current code:
from scipy.stats import beta
import scipy.stats as st
class CustomBeta(st.rv_continuous):
def _stats(self, a, b):
# will add own code here
mn = a * 1.0 / (a + b)
var = (a * b * 1.0) / (a + b + 1.0) / (a + b) ** 2.0
g1 = 2.0 * (b - a) * sqrt((1.0 + a + b) / (a * b)) / (2 + a + b)
g2 = 6.0 * (a ** 3 + a ** 2 * (1 - 2 * b) + b ** 2 * (1 + b) - 2 * a * b * (2 + b))
g2 /= a * b * (a + b + 2) * (a + b + 3)
return mn, var, g1, g2
dist = beta(4, 6)
print dist.rvs() # works fine
dist = CustomBeta(4, 6)
print dist.rvs() # crashes
Getting _rvs() from my custom object gives me a long stacktrace and an error
RuntimeError: maximum recursion depth exceeded
This has nothing to do with overloading _stats. The same behavior is caused simply by
class CustomBeta(st.rv_continuous):
pass
dist = CustomBeta(4, 6)
print(dist.rvs()) # crashes
The documentation of rv_continuous states that
New random variables can be defined by subclassing the rv_continuous class and re-defining at least the _pdf or the _cdf method.
You will need to provide at least one of these methods to compute the probability density function (pdf) or the cumulative probability density function(cdf).
Furthermore,
[rv_continuous] cannot be used directly as a distribution.
It is used as follows:
class CustomBetaGen(st.rv_continuous):
...
CustomBeta = CustomBetaGen(name='CustomBeta')
dist = CustomBeta(4, 6)
Finally, rvs.() does not seem to work properly for the beta distribution if you do not provide a _rvs method.
Putting everything together and stealing the appropriate methods from the beta distribution:
from scipy.stats import beta
import scipy.stats as st
import numpy as np
class CustomBetaGen(st.rv_continuous):
def _cdf(self, x, a, b):
return beta.cdf(x, a, b)
def _pdf(self, x, a, b):
return beta.pdf(x, a, b)
def _rvs(self, a, b):
return beta.rvs(a, b)
def _stats(self, a, b):
# will add own code here
mn = a * 1.0 / (a + b)
var = (a * b * 1.0) / (a + b + 1.0) / (a + b) ** 2.0
g1 = 2.0 * (b - a) * np.sqrt((1.0 + a + b) / (a * b)) / (2 + a + b)
g2 = 6.0 * (a ** 3 + a ** 2 * (1 - 2 * b) + b ** 2 * (1 + b) - 2 * a * b * (2 + b))
g2 /= a * b * (a + b + 2) * (a + b + 3)
return mn, var, g1, g2
CustomBeta = CustomBetaGen(name='CustomBeta')
dist = beta(4, 6)
print(dist.rvs()) # works fine
print(dist.stats()) # (array(0.4), array(0.021818181818181816))
dist = CustomBeta(4, 6)
print(dist.rvs()) # works fine
print(dist.stats()) # (array(0.4), array(0.021818181818181816))

Plotting a NACA 4-series airfoil

I'm trying to plot an airfoil from the formula as described on this wikipedia page.
This Jupyter notebook can be viewed on this github page.
%matplotlib inline
import math
import matplotlib.pyplot as pyplot
def frange( start, stop, step ):
yield start
while start <= stop:
start += step
yield start
#https://en.wikipedia.org/wiki/NACA_airfoil#Equation_for_a_cambered_4-digit_NACA_airfoil
def camber_line( x, m, p, c ):
if 0 <= x <= c * p:
yc = m * (x / math.pow(p,2)) * (2 * p - (x / c))
#elif p * c <= x <= c:
else:
yc = m * ((c - x) / math.pow(1-p,2)) * (1 + (x / c) - 2 * p )
return yc
def dyc_over_dx( x, m, p, c ):
if 0 <= x <= c * p:
dyc_dx = ((2 * m) / math.pow(p,2)) * (p - x / c)
#elif p * c <= x <= c:
else:
dyc_dx = ((2 * m ) / math.pow(1-p,2)) * (p - x / c )
return dyc_dx
def thickness( x, t, c ):
term1 = 0.2969 * (math.sqrt(x/c))
term2 = -0.1260 * (x/c)
term3 = -0.3516 * math.pow(x/c,2)
term4 = 0.2843 * math.pow(x/c,3)
term5 = -0.1015 * math.pow(x/c,4)
return 5 * t * c * (term1 + term2 + term3 + term4 + term5)
def naca4( m, p, t, c=1 ):
for x in frange( 0, 1.0, 0.01 ):
dyc_dx = dyc_over_dx( x, m, p, c )
th = math.atan( dyc_dx )
yt = thickness( x, t, c )
yc = camber_line( x, m, p, c )
xu = x - yt * math.sin(th)
xl = x + yt * math.sin(th)
yu = yc + yt * math.cos(th)
yl = yc - yt * math.cos(th)
yield (xu, yu), (xl, yl)
#naca2412
m = 0.02
p = 0.4
t = 12
naca4points = naca4( m, p, t )
for (xu,yu),(xl,yl) in naca4points:
pyplot.plot( xu, yu, 'r,')
pyplot.plot( xl, yl, 'r,')
pyplot.ylabel('y')
pyplot.xlabel('x')
pyplot.axis('equal')
figure = pyplot.gcf()
figure.set_size_inches(16,16,forward=True)
The result looks like .
I expected it to look more like .
Questions: Why is the line not completely smooth? There seems to be a discontinuity where the beginning and end meet. Why does it not look like the diagram on wikipedia? How do I remove the extra loop at the trailing edge? How do I fix the chord so that it runs from 0.0 to 1.0?
First, t should be 0.12 not 12. Second, to make a smoother plot, increase the sample points.
It is also a good idea to use vectorize method in numpy:
%matplotlib inline
import math
import matplotlib.pyplot as plt
import numpy as np
#https://en.wikipedia.org/wiki/NACA_airfoil#Equation_for_a_cambered_4-digit_NACA_airfoil
def camber_line( x, m, p, c ):
return np.where((x>=0)&(x<=(c*p)),
m * (x / np.power(p,2)) * (2.0 * p - (x / c)),
m * ((c - x) / np.power(1-p,2)) * (1.0 + (x / c) - 2.0 * p ))
def dyc_over_dx( x, m, p, c ):
return np.where((x>=0)&(x<=(c*p)),
((2.0 * m) / np.power(p,2)) * (p - x / c),
((2.0 * m ) / np.power(1-p,2)) * (p - x / c ))
def thickness( x, t, c ):
term1 = 0.2969 * (np.sqrt(x/c))
term2 = -0.1260 * (x/c)
term3 = -0.3516 * np.power(x/c,2)
term4 = 0.2843 * np.power(x/c,3)
term5 = -0.1015 * np.power(x/c,4)
return 5 * t * c * (term1 + term2 + term3 + term4 + term5)
def naca4(x, m, p, t, c=1):
dyc_dx = dyc_over_dx(x, m, p, c)
th = np.arctan(dyc_dx)
yt = thickness(x, t, c)
yc = camber_line(x, m, p, c)
return ((x - yt*np.sin(th), yc + yt*np.cos(th)),
(x + yt*np.sin(th), yc - yt*np.cos(th)))
#naca2412
m = 0.02
p = 0.4
t = 0.12
c = 1.0
x = np.linspace(0,1,200)
for item in naca4(x, m, p, t, c):
plt.plot(item[0], item[1], 'b')
plt.plot(x, camber_line(x, m, p, c), 'r')
plt.axis('equal')
plt.xlim((-0.05, 1.05))
# figure.set_size_inches(16,16,forward=True)
Thanks for the code.
I have modified the code for symmetrical airfoils:
def naca4s(x, t, c=1):
yt = thickness(x, t, c)
return ((x, yt),
(x, -yt))

Intersections between Geodesics (shortest distance paths) on the surface of a sphere

I've searched far and wide but have yet to find a suitable answer to this problem. Given two lines on a sphere, each defined by their start and end points, determine whether or not and where they intersect. I've found this site (http://mathforum.org/library/drmath/view/62205.html) which runs through a good algorithm for the intersections of two great circles, although I'm stuck on determining whether the given point lies along the finite section of the great circles.
I've found several sites which claim they've implemented this, Including some questions here and on stackexchange, but they always seem to reduce back to the intersections of two great circles.
The python class I'm writing is as follows and seems to almost work:
class Geodesic(Boundary):
def _SecondaryInitialization(self):
self.theta_1 = self.point1.theta
self.theta_2 = self.point2.theta
self.phi_1 = self.point1.phi
self.phi_2 = self.point2.phi
sines = math.sin(self.phi_1) * math.sin(self.phi_2)
cosines = math.cos(self.phi_1) * math.cos(self.phi_2)
self.d = math.acos(sines - cosines * math.cos(self.theta_2 - self.theta_1))
self.x_1 = math.cos(self.theta_1) * math.cos(self.phi_1)
self.x_2 = math.cos(self.theta_2) * math.cos(self.phi_2)
self.y_1 = math.sin(self.theta_1) * math.cos(self.phi_1)
self.y_2 = math.sin(self.theta_2) * math.cos(self.phi_2)
self.z_1 = math.sin(self.phi_1)
self.z_2 = math.sin(self.phi_2)
self.theta_wraps = (self.theta_2 - self.theta_1 > PI)
self.phi_wraps = ((self.phi_1 < self.GetParametrizedCoords(0.01).phi and
self.phi_2 < self.GetParametrizedCoords(0.99).phi) or (
self.phi_1 > self.GetParametrizedCoords(0.01).phi) and
self.phi_2 > self.GetParametrizedCoords(0.99))
def Intersects(self, boundary):
A = self.y_1 * self.z_2 - self.z_1 * self.y_2
B = self.z_1 * self.x_2 - self.x_1 * self.z_2
C = self.x_1 * self.y_2 - self.y_1 * self.x_2
D = boundary.y_1 * boundary.z_2 - boundary.z_1 * boundary.y_2
E = boundary.z_1 * boundary.x_2 - boundary.x_1 * boundary.z_2
F = boundary.x_1 * boundary.y_2 - boundary.y_1 * boundary.x_2
try:
z = 1 / math.sqrt(((B * F - C * E) ** 2 / (A * E - B * D) ** 2)
+ ((A * F - C * D) ** 2 / (B * D - A * E) ** 2) + 1)
except ZeroDivisionError:
return self._DealWithZeroZ(A, B, C, D, E, F, boundary)
x = ((B * F - C * E) / (A * E - B * D)) * z
y = ((A * F - C * D) / (B * D - A * E)) * z
theta = math.atan2(y, x)
phi = math.atan2(z, math.sqrt(x ** 2 + y ** 2))
if self._Contains(theta, phi):
return point.SPoint(theta, phi)
theta = (theta + 2* PI) % (2 * PI) - PI
phi = -phi
if self._Contains(theta, phi):
return spoint.SPoint(theta, phi)
return None
def _Contains(self, theta, phi):
contains_theta = False
contains_phi = False
if self.theta_wraps:
contains_theta = theta > self.theta_2 or theta < self.theta_1
else:
contains_theta = theta > self.theta_1 and theta < self.theta_2
phi_wrap_param = self._PhiWrapParam()
if phi_wrap_param <= 1.0 and phi_wrap_param >= 0.0:
extreme_phi = self.GetParametrizedCoords(phi_wrap_param).phi
if extreme_phi < self.phi_1:
contains_phi = (phi < max(self.phi_1, self.phi_2) and
phi > extreme_phi)
else:
contains_phi = (phi > min(self.phi_1, self.phi_2) and
phi < extreme_phi)
else:
contains_phi = (phi > min(self.phi_1, self.phi_2) and
phi < max(self.phi_1, self.phi_2))
return contains_phi and contains_theta
def _PhiWrapParam(self):
a = math.sin(self.d)
b = math.cos(self.d)
c = math.sin(self.phi_2) / math.sin(self.phi_1)
param = math.atan2(c - b, a) / self.d
return param
def _DealWithZeroZ(self, A, B, C, D, E, F, boundary):
if (A - D) is 0:
y = 0
x = 1
elif (E - B) is 0:
y = 1
x = 0
else:
y = 1 / math.sqrt(((E - B) / (A - D)) ** 2 + 1)
x = ((E - B) / (A - D)) * y
theta = (math.atan2(y, x) + PI) % (2 * PI) - PI
return point.SPoint(theta, 0)
def GetParametrizedCoords(self, param_value):
A = math.sin((1 - param_value) * self.d) / math.sin(self.d)
B = math.sin(param_value * self.d) / math.sin(self.d)
x = A * math.cos(self.phi_1) * math.cos(self.theta_1) + (
B * math.cos(self.phi_2) * math.cos(self.theta_2))
y = A * math.cos(self.phi_1) * math.sin(self.theta_1) + (
B * math.cos(self.phi_2) * math.sin(self.theta_2))
z = A * math.sin(self.phi_1) + B * math.sin(self.phi_2)
new_phi = math.atan2(z, math.sqrt(x**2 + y**2))
new_theta = math.atan2(y, x)
return point.SPoint(new_theta, new_phi)
EDIT: I forgot to specify that if two curves are determined to intersect, I then need to have the point of intersection.
A simpler approach is to express the problem in terms of geometric primitive operations like the dot product, the cross product, and the triple product. The sign of the determinant of u, v, and w tells you which side of the plane spanned by v and w contains u. This enables us to detect when two points are on opposite sites of a plane. That's equivalent to testing whether a great circle segment crosses another great circle. Performing this test twice tells us whether two great circle segments cross each other.
The implementation requires no trigonometric functions, no division, no comparisons with pi, and no special behavior around the poles!
class Vector:
def __init__(self, x, y, z):
self.x = x
self.y = y
self.z = z
def dot(v1, v2):
return v1.x * v2.x + v1.y * v2.y + v1.z * v2.z
def cross(v1, v2):
return Vector(v1.y * v2.z - v1.z * v2.y,
v1.z * v2.x - v1.x * v2.z,
v1.x * v2.y - v1.y * v2.x)
def det(v1, v2, v3):
return dot(v1, cross(v2, v3))
class Pair:
def __init__(self, v1, v2):
self.v1 = v1
self.v2 = v2
# Returns True if the great circle segment determined by s
# straddles the great circle determined by l
def straddles(s, l):
return det(s.v1, l.v1, l.v2) * det(s.v2, l.v1, l.v2) < 0
# Returns True if the great circle segments determined by a and b
# cross each other
def intersects(a, b):
return straddles(a, b) and straddles(b, a)
# Test. Note that we don't need to normalize the vectors.
print(intersects(Pair(Vector(1, 0, 1), Vector(-1, 0, 1)),
Pair(Vector(0, 1, 1), Vector(0, -1, 1))))
If you want to initialize unit vectors in terms of angles theta and phi, you can do that, but I recommend immediately converting to Cartesian (x, y, z) coordinates to perform all subsequent calculations.
Intersection using plane trig can be calculated using the below code in UBasic.
5 'interx.ub adapted from code at
6 'https://rosettacode.org
7 '/wiki/Find_the_intersection_of_two_linesSinclair_ZX81_BASIC
8 'In U Basic by yuji kida https://en.wikipedia.org/wiki/UBASIC
10 XA=48.7815144526:'669595.708
20 YA=-117.2847245001:'2495736.332
30 XB=48.7815093807:'669533.412
40 YB=-117.2901673467:'2494425.458
50 XC=48.7824947147:'669595.708
60 YC=-117.28751374:'2495736.332
70 XD=48.77996737:'669331.214
80 YD=-117.2922957:'2494260.804
90 print "THE TWO LINES ARE:"
100 print "YAB=";YA-XA*((YB-YA)/(XB-XA));"+X*";((YB-YA)/(XB-XA))
110 print "YCD=";YC-XC*((YD-YC)/(XD-XC));"+X*";((YD-YC)/(XD-XC))
120 X=((YC-XC*((YD-YC)/(XD-XC)))-(YA-XA*((YB-YA)/(XB-XA))))/(((YB-YA)/(XB-XA))-((YD-YC)/(XD-XC)))
130 print "Lat = ";X
140 Y=YA-XA*((YB-YA)/(XB-XA))+X*((YB-YA)/(XB-XA))
150 print "Lon = ";Y
160 'print "YCD=";YC-XC*((YD-YC)/(XD-XC))+X*((YD-YC)/(XD-XC))

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