I'm trying to plot a diagram of Coulomb damping mass-spring model, see image bellow. mi*N is the damping friction force.
Becuase the damping force switches direction depending on the direction of speed vector, we have 2 different equations of motion x(t), I named them x_a(t) and x_b(t). If you look closely on the diagram above, the use of x(t) depends on the period, for the first period which is from 0 to pi/omega_n x_a(t) is used, for the second period which is from pi/omega_n to 2*pi/omega_n we use x_b(t) and so on.
I was thinking to use a piece of code which would go something like
t = i * (pi/omega_n)
for i = 1, 3, 5, 7,...
x_a(t)
for i = 2, 4, 6, 8,...
x_b(t)
But I have no clue how to implement this.
I managed to define the x_a(t) part of the code and plot it with the undampled model, see bellow.
import numpy as np
import matplotlib.pyplot as plt
#constants
k = 2 #(N/m), spring coef
m = 0.04 #(kg), mass
x0 = -0.1 #(m), preload
mi = 0.3 #(), dry dynamic friction coef. ABS-ABS
N = 0.3 #(N), normal contact force
f_tr = mi * N / k #friction force/pring coef - equivalent distance
omega_0 = np.sqrt(k/m)
#time
t = np.linspace(0,5,100)
#undamped model
x_undamp = x0*np.cos(omega_0*t)
dx_undamp = -omega_0*x0*np.sin(omega_0*t)
#damped model
x_damp = (x0+f_tr)*np.cos(omega_0*t)-f_tr
dx_damp = -omega_0*(x0+f_tr)*np.sin(omega_0*t)
#time to x=0
t0 = np.arccos(f_tr/(x0+f_tr))/omega_0
print t0
#plotting
fig, (ax1, ax2) = plt.subplots(1, 2)
fig.suptitle('position on left, velocity on right')
ax1.plot(t, x_undamp,'r', label='x_undamp')
ax1.plot(t, x_damp, 'b', label = 'x_damp')
ax2.plot(t, dx_undamp, 'r', label = 'dx_undamp')
ax2.plot(t, dx_damp, 'b', label = 'dx_damp')
#grids, titles, legends, axis labels
ax1.grid()
ax2.grid()
ax1.set_title('Position vs time')
ax2.set_title('Velocity vs time')
ax1.legend()
ax2.legend()
ax1.set_xlabel('t(s)')
ax1.set_ylabel('x(m)')
ax2.set_xlabel('t(s)')
ax2.set_ylabel('dx(m/s)')
plt.show()
I'm gonna put plot a diagram of Coulomb damping mass-spring model solution here till someone comes with a better one:
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt
m = 0.2
k = 2.0
c = 0.1
mus = 0.3
muk = 0.2
g = 9.8
vf = 0.01
v0 = 0.0
t1 = 10
sign = lambda x: np.tanh(100*x)
def Xi(t):
if t < 1 :
return 0
else:
return 1
vXi = np.vectorize(Xi)
def eq(X, t, Xi):
Ff = k * (Xi(t) - X[0])
if np.abs(X[1]) < vf and np.abs(Ff) < mus * m * g :
Ff = k * (Xi(t) - X[0])
else:
Ff = sign(X[1]) * muk * m * g
d2x = (k * (Xi(t) - X[0]) - Ff) / m
return [X[1], d2x]
t = np.linspace(0, t1, 1000)
X0 = [v0, 0]
sol = odeint(func = eq, y0 = X0, t = t, args = (Xi, ), mxstep = 50000, atol = 1e-5)
plt.plot(t, sol[:, 0], 'r-', label = 'Output (x(t))')
plt.plot(t, vXi(t), 'b-', label = 'Input (xi(t))')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
Related
I wanted to solve this system of differential equations using Python, and I was wondering how to do it. I've tried nothing yet since I'm new to these systems, though I have already solved some individual diff. eqs.. Right now, I'm familiar with solve_ivp and that's practically it.
Here's the code:
import numpy as np
from numpy import sin,cos,sign,sqrt,pi
from scipy.integrate import solve_ivp
from matplotlib import pyplot as plt
from matplotlib.pyplot import figure
g = 9.807
h = 1.8
l = 0.56 * h
R_sc = 14
mu = 0.1
k = 0.3
m = 80
v_0 = 12
phi_0 = np.pi/4
def velphi(t,y):
dy=np.zeros([3])
dy[1] = y[2]
dy[0] = -mu*np.sqrt(g**2 + ((y[0])**2/(R_sc*np.cos(y[1])))**2)-(k/m)*(y[0])**2
dy[2] = (g/l)*np.sin(y[1])-((y[0])**2/(l*R_sc))*np.sign(y[1])
return dy
y0 = np.array([v_0, phi_0, 0])
time = np.linspace(0, 10, 10000)
sol = solve_ivp(velphi, (0,10), y0, method='RK45', t_eval=time, dense_output=True, rtol=1e-8, atol=1e-10)
t, pos, vel, phi, omega = sol.t, sol.y[0], sol.y[1], sol.y[2], sol.y[3]
plt.plot(t.T, phi.T,"b",linewidth = 0.65, label = "${\Phi}$")
plt.plot(t.T, omega.T,"g",linewidth = 0.65, label = "${\Omega}$")
s = 'Condicions inicials: ${\Phi}_{o}$ ='+str(np.degrees(phi_0))+'º, ${\Omega}_{o}$ = 0 rad/s'
plt.title('Pèndol centrífug invertit | '+ s)
plt.xlabel('Temps (s)')
plt.ylabel(u'${\Phi}$ (rad) | ${\Omega}$ (rad / s)')
plt.grid(False)
plt.legend(loc = "upper right")
plt.show()
plt.plot(t.T, vel.T,"r",linewidth = 0.65, label = "v")
s = 'Condicions inicials: ${\Phi}_{o}$ ='+str(np.degrees(phi_0))+'º, ${\Omega}_{o}$ = 0 rad/s'
plt.title('Pèndol centrífug invertit | '+ s)
plt.xlabel('Temps (s)')
plt.ylabel('v (m/s)')
plt.grid(False)
plt.legend(loc = "upper right")
plt.show()
And here the solutions:
Velocity
Phi and the angular velocity
import numpy as np
import matplotlib.pyplot as plot
from IPython.display import HTML
from matplotlib import animation
#setup fig with axis
fig, ax = plot.subplots(figsize=(8,8))
#set axis limits
ax.set(xlim=(-2,2), ylim=(0,600), xlabel="position, metres", ylabel="height, metres", title="falling apple")
#initial params
T = 100.
m = 3
g = 9.81
v0x = 10
H = 553.
#setting calc interval
dt = 0.1
N = int(T/dt)
#arrays
v = np.zeros((N+1 , 2))
x = np.zeros((N+1 , 2))
f = np.zeros((N+1 , 2))
#array start [x ,y] format
v[0] = np.array([0. , H])
x[0] = np.array([v0x , 0.])
# the only force is gravity
f[:] = np.array([0., m * g])
#running the dynamics sim
for n in range(N):
v[n+1] = v[n] + ((f[n]/m) * dt)
x[n+1] = x[n] + (v[n+1] * dt)
#scatter plot
scat_plt = ax.scatter(x[0,0], x[0,1], marker='o', c='#1f77b4', s=200)
## animating
def animate(i):
scat_plt.set_offsets(x[i])
ani = animation.FuncAnimation(fig, func=animate, frames=N)
ani.save('ball.html', writer=animation.HTMLWriter(fps= 1//dt))
plot.close()
ani.save('ball.mp4', fps= 1//dt)
HTML('ball.html')
The out put is just a circle going straight up where as this is supposed to simulate a ball being thrown horizontally off a tower
It would be highly appreciated if someone could suggest any changes to be made to the logic/physics or the code.
Thank you!!
I think you mixed x with v at some point. Also the force should be negative in y. I tried this and it seems to work:
import numpy as np
import matplotlib.pyplot as plot
from IPython.display import HTML
from matplotlib import animation
#setup fig with axis
fig, ax = plot.subplots(figsize=(8,8))
#set axis limits
ax.set(xlim=(-200,200), ylim=(0,600), xlabel="position, metres", ylabel="height, metres", title="falling apple")
#initial params
T = 100.
m = 3
g = 9.81
v0x = 10
H = 553.
#setting calc interval
dt = 0.1
N = int(T/dt)
#arrays
v = np.zeros((N+1 , 2))
x = np.zeros((N+1 , 2))
f = np.zeros((N+1 , 2))
#array start [x ,y] format
x[0] = np.array([0. , H])
v[0] = np.array([v0x , 0.])
# the only force is gravity
f[:] = np.array([0., -m * g])
#running the dynamics sim
for n in range(N):
v[n+1] = v[n] + ((f[n]/m) * dt)
x[n+1] = x[n] + (v[n+1] * dt)
#scatter plot
scat_plt = ax.scatter(x[0,0], x[0,1], marker='o', c='#1f77b4', s=200)
## animating
def animate(i):
scat_plt.set_offsets(x[i])
ani = animation.FuncAnimation(fig, func=animate, frames=N)
ani.save('ball.html', writer=animation.HTMLWriter(fps= 1//dt))
plot.close()
ani.save('ball.gif', fps= 1//dt)
HTML('ball.html')
i want to plot a 1D Heat Diffusion with implicit PDE. here is the problem Heat Diffusion. here's the PDE equation.
my code is fine but it creates a new figure for each iteration.
import numpy as np
import matplotlib.pyplot as plt
L = 0.05 #length
n = 5 #number of segment
T0 = 20 #initial temperature
T1s = 100 #boundary condition in segment 0
T2s = 25 #boundary condition in segment 5
dx = L/n #delta x
alpha = 0.000014129 #heat coeff
t_final = 9
dt = 3
x = np.linspace(0, L, n+1)
T = np.ones(n+1)*T0 #array of Temperature initial condition
dTdt = np.empty(n+1) #dTdt delta T /delta t
t = np.arange(0, t_final, dt)
for j in range(1,len(t)+1):
for i in range(1, n):
dTdt[i] = alpha*(T[i+1]-2*T[i]+T[i-1])/dx**2 #PDE iterative function for segment i
dTdt[0] = alpha*(T[1]-2*T[0]+T1s)/dx**2
dTdt[n-1] = alpha*(T[n-2]-2*T[n-1]+T2s)/dx**2
T = T + dTdt*dt
T[0]=T1s
T[5]=T2s
plt.figure(1)
plt.plot(x,T)
plt.axis([-0.01, L+0.01, 0, 120])
plt.xlabel('Distance (m)')
plt.ylabel('Temperature (C)')
plt.show()
i've added plt.ion() and plt.pause() but it didnt work. i also tried similar PDE plotting. but didnt work for me either. also is it possible to give each plot a different color and label in the figure?
any suggestion would help !
You can create the figure instance before the loop and add the figures + labels in the loop as follows:
import numpy as np
import matplotlib.pyplot as plt
L = 0.05 #length
n = 5 #number of segment
T0 = 20 #initial temperature
T1s = 100 #boundary condition in segment 0
T2s = 25 #boundary condition in segment 5
dx = L/n #delta x
alpha = 0.000014129 #heat coeff
t_final = 9
dt = 3
x = np.linspace(0, L, n+1)
T = np.ones(n+1)*T0 #array of Temperature initial condition
dTdt = np.empty(n+1) #dTdt delta T /delta t
t = np.arange(0, t_final, dt)
fig, ax = plt.subplots(figsize = (8,6))
for j in range(1,len(t)+1):
for i in range(1, n):
dTdt[i] = alpha*(T[i+1]-2*T[i]+T[i-1])/dx**2 #PDE iterative function for segment i
dTdt[0] = alpha*(T[1]-2*T[0]+T1s)/dx**2
dTdt[n-1] = alpha*(T[n-2]-2*T[n-1]+T2s)/dx**2
T = T + dTdt*dt
T[0]=T1s
T[5]=T2s
ax.plot(x,T,label = 'Iteration' + ' ' + str(j))
ax.legend()
ax.axis([-0.01, L+0.01, 0, 120])
ax.set_xlabel('Distance (m)')
ax.set_ylabel('Temperature (C)');
I don't understand why this code (reference):
from numpy import zeros, linspace
import matplotlib.pyplot as plt
# Time unit: 1 h
beta = 10./(40*8*24)
gamma = 3./(15*24)
dt = 0.1 # 6 min
D = 30 # Simulate for D days
N_t = int(D*24/dt) # Corresponding no of hours
t = linspace(0, N_t*dt, N_t+1)
S = zeros(N_t+1)
I = zeros(N_t+1)
R = zeros(N_t+1)
# Initial condition
S[0] = 50
I[0] = 1
R[0] = 0
# Step equations forward in time
for n in range(N_t):
S[n+1] = S[n] - dt*beta*S[n]*I[n]
I[n+1] = I[n] + dt*beta*S[n]*I[n] - dt*gamma*I[n]
R[n+1] = R[n] + dt*gamma*I[n]
fig = plt.figure()
l1, l2, l3 = plt.plot(t, S, t, I, t, R)
fig.legend((l1, l2, l3), ('S', 'I', 'R'), 'upper left')
plt.xlabel('hours')
plt.show()
doesn't produce the same results as this code I made:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
beta = 10. / (40 * 8 * 24)
gamma = 3. / (15 * 24)
def flu(y, t):
S = y[0]
P = y[1]
R = y[2]
S = - beta * S * P
P = beta * S * P - gamma * P
R = gamma * P
return [S, P, R]
C_I = [50, 1, 0]
t = np.linspace(0, 1000, 1000)
y = odeint(flu, C_I, t)
S = y[:, 0]
P = y[:, 1]
R = y[:, 2]
fig, ax = plt.subplots()
ax.plot(t, S, 'b--', label='S')
ax.plot(t, P, 'r--', label='I')
ax.plot(t, R, 'g--', label='R')
legend = ax.legend(loc='upper right', shadow=True, fontsize='x-large')
legend.get_frame().set_facecolor('#FFFCCC')
plt.show()
I used P instead of I to avoid confusion.
The equations solved with odeint should be the same as the ones provided in the reference link above. And if the equations I use are correct, which I am convinced they are, I don't understand where the mistake(s) lie(s).
Thank you for your help
You set S=y[0] then you set S=- beta * S * P. This overwrites y[0]!!! Similar problems for P and R
Try this:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
beta = 10. / (40 * 8 * 24)
gamma = 3. / (15 * 24)
def flu(y, t):
S = y[0]
P = y[1]
R = y[2]
dS = - beta * S * P
dP = beta * S * P - gamma * P
dR = gamma * P
return [dS, dP, dR]
C_I = [50, 1, 0]
t = np.linspace(0, 1000, 1000)
y = odeint(flu, C_I, t)
S = y[:, 0]
P = y[:, 1]
R = y[:, 2]
fig, ax = plt.subplots()
ax.plot(t, S, 'b--', label='S')
ax.plot(t, P, 'r--', label='I')
ax.plot(t, R, 'g--', label='R')
legend = ax.legend(loc='upper right', shadow=True, fontsize='x-large')
legend.get_frame().set_facecolor('#FFFCCC')
plt.show()
My code has been modified according to many great suggestions from people in this forum. However, I still have some questions about the code.
My code is:
from pylab import *
from numpy import *
N = 100 #lattice points per axis
dt = 1 #time step
dx = 1 #lattice spacing
t = arange(0, 1000000*dt, dt) #time
a = 1 #cofficient
epsilon = 100 #cofficient
M = 1.0 #cofficient
every = 100 #dump an image every
phi_0 = 0.5 #initial mean value of the order parameter
noise = 0.1 #initial amplitude of thermal fluctuations in the order parameter
th = phi_0*ones((N, N)) + noise*(rand(N, N) - 0.5) #initial condition
x, y = meshgrid(fftfreq(int(th.shape[0]), dx), fftfreq(int(th.shape[1]), dx))
k2 = (x*x + y*y) #k is a victor in the Fourier space, k2=x^2+y^2
g = lambda th, a: 4*a*th*(1-th)*(1-2*th) #function g
def update(th, dt, a, k2):
return ifft2((fft2(th)-dt*M*k2*fft2(g(th,a)))/(1+2*epsilon*M*dt*k2**2))
for i in range(size(t)):
print t[i]
if mod(i, every)==0:
imshow(abs(th), vmin=0.0, vmax=1.0)
colorbar()
show()
#savefig('t'+str(i/every).zfill(3)+'.png', dpi=100)
clf()
th=update(th, dt, a, k2)
When I run it, I have to close the figures one by one to see the changes. But I want to demonstrate the changes of the images in one figure. Any good ideas?
Use the "animation" feature of matplotlib, like in
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation
def update_line(num, data, line):
line.set_data(data[...,:num])
return line,
fig1 = plt.figure()
data = np.random.rand(2, 25)
l, = plt.plot([], [], 'r-')
plt.xlim(0, 1)
plt.ylim(0, 1)
plt.xlabel('x')
plt.title('test')
line_ani = animation.FuncAnimation(fig1, update_line, 25, fargs=(data, l),
interval=50, blit=True)
plt.show()
Tutorial at :
http://matplotlib.org/1.3.1/examples/animation/index.html