theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
ax.plot(r_ECI[:, 0] / 1000,
r_ECI[:, 1] / 1000,
r_ECI[:, 2] / 1000,
label='Orbital Position',
color='k')
# Sphere to represent Earth
# Make data
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x1 = rEarth / 1000 * np.outer(np.cos(u), np.sin(v))
y1 = rEarth / 1000 * np.outer(np.sin(u), np.sin(v))
z1 = rEarth / 1000 * np.outer(np.ones(np.size(u)), np.cos(v))
# Plot the surface
ax.plot_surface(x1, y1, z1, cmap='GnBu')
# Legend and labels
ax.legend()
ax.set_xlabel('X Pos (km)')
ax.set_ylabel('Y Pos (km)')
ax.set_zlabel('Z Pos (km)')
ax.set_aspect('equal')
plt.show()
t=25 #days
def f(x, y):
return np.sin(np.sqrt(x ** 2 + y ** 2))
x = np.linspace(-7, 7, 30)
y = np.linspace(-5, 5, 30)
X, Y = np.meshgrid(x, y)
Z = f(X, Y)
fig = plt.figure()
ax = plt.axes(projection='3d')
ax.contour3D(X, Y, Z, 50, cmap='binary')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z');
# ## Models of bumblebee population growth over many seasons
#
# ### Infinite resources (exponential growth)
# Assuming infinite resources in the environment (probably never the case in nature for bees) and nothing systematically killing populations, bumblebee populations will increase at $\frac{dx}{dt}= kx$, which solves to $x(t)= Ce^{kt}$. k is just some growth constant.
#
# ### Finite resources (logistic growth)
# With finite resources, bee populations will grow according to $\frac {dx}{dt}=kx(1-\frac{x}{C})$, where k is a growth constant and C is the carrying capacity of the environment. An analytic solution is $x=\frac{C}{1+Ae^{-kt}}$ where $A=\frac{C-X_{0}}{X_{0}}$.
#
# ### Finite resources + big bad pesticides.
# Now let's introduce neonicotinoid into the equation once the bees have reached carrying capacity. These pesticides both kill bees directly and decrease their reproductive capacity, so here I'll have a lower k as well as some constant d representing a proportion of bee deaths.
# This scenario is represented by the differential equation $\frac{dx}{dt}= (k-d)x$, which solves to $x(t)= Ce^{(k-d)t}$
# In[3]:
from matplotlib import axes as ax
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
x, y, z, vx, vy, vz = np.loadtxt('./coordAa.dat.txt', unpack=True)
print(len(x))
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x, y, z, c='r', marker='.', s=0.1)
#Axes3D.scatter(x,y,z)
ax.set_xlabel('X [kpc]')
ax.set_ylabel('Y [kpc]')
ax.set_zlabel('Z [kpc]')
#plt.axis([-1500., 1500., -1500., 1500, -1500, 1500])
ax.set_xlim3d(-1500, 1500)
ax.set_ylim3d(-1500, 1500)
ax.set_zlim3d(-1500, 1500)
plt.show()
radius = sorted([
np.sqrt(i**2 + j**2 + k**2) for i, j, k in zip(x, y, z)
if np.sqrt(i**2 + j**2 + k**2) <= 300
])
max_radius = max(radius)
min_radius = min(radius)
print(max_radius, min_radius)
total += x * y
new[i][j] = total
return new
elif self.m == other.n:
return other * self
else:
raise ValueError(
"Cannot multiply two matrices of incompatable dimensions")
if __name__ == "__main__":
fig = plt.figure()
ax = fig.gca(projection="3d")
v3 = AVector("sin(x)", "cos(y)", "tan(z)")
v3.quiver(ax)
a, b, c = v3.evaluate(1, 1, 1)
print(a, b, c)
ax.set_xlabel("sin(x)")
ax.set_ylabel("cos(y)")
ax.set_zlabel("tan(z)")
plt.show()