this is my first time in this web, so I hope I'll do it properly.
I'm doing a heat diffusion simulation for a cylinder. Due to the heat source is outside the cylinder, this won't have cylindrical symmetry so I decided to create a X-Y grid and try to approximate it to a cyilinder. Some thing like this:
This is my first and simpliest try:
from numpy import empty,zeros,max
import matplotlib.pyplot as plt
# Constants
M = 10 # Grid squares on a side
V = 100.0 # Voltage at top wall
target = 1e-4 # Target accuracy
# Create arrays to hold potential values
phi = zeros([M+1,M+1],float)
phi[0,:] = V
phi[M,:]=V
phi[:,0]=V
phi[:,M]=V
phiprime = empty([M+1,M+1],float)
# Main loop
delta = 1.0
while delta>target:
# Calculate new values of the potential
for i in range(M+1):
for j in range(M+1):
if i==0 or i==M or j==0 or j==M:
phiprime[i,j] = phi[i,j]
elif i==1 or i==M-1:
if j==1 or j==2 or j==M-2 or j==M-1:
phiprime[i,j]=V-V/2
else:
phiprime[i,j] = (phi[i+1,j] + phi[i-1,j] \
+ phi[i,j+1] + phi[i,j-1])/4
elif i==2 or i==M-2:
if j==1 or j==M-1:
phiprime[i,j]=V-V/2
else:
phiprime[i,j] = (phi[i+1,j] + phi[i-1,j] \
+ phi[i,j+1] + phi[i,j-1])/4
else:
phiprime[i,j] = (phi[i+1,j] + phi[i-1,j] \
+ phi[i,j+1] + phi[i,j-1])/4
delta = max(abs(phi - phiprime))
phi, phiprime = phiprime, phi
plt.imshow(phiprime)
plt.gray()
plt.show()
This was quite easy, but this isn't a good approximation for a cylinder, I have to make it quite bigger. As you can see, in that attemp, I can code the "positions" by hand, but as I make it 100x100 or 1000x1000, it's impossible.
So, my question is: Is there any good parametritzation for that "indented square"(the black ones)?
Thanks
