I am trying to solve this set of differential equations with solve_ivp in Python. Winds are horizontal two dimensional vectors, so I need differential equation solved in y dimension.
This is the error I get:
ValueError: not enough values to unpack (expected 6, got 4)
Is there any other way of solving it? Here is my code:(just projectile motion in x-z plane, distance as function of angle). For the example of two such initial angles (with which we hit approximately the same point) I am trying to calculate the dispersion of hits in the wind according to the point hit in no wind.
Thanks for any help or pointers
from random import randrange
from statistics import stdev
from turtle import title
import numpy as np
import scipy as sp
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt
m = 30
v0 = 301
g = 9.81
ro = 1.28
A = 0.0026
c = 1
k = 0.5 * c * ro * A
B = k / m
V = v0
c_max = 10
c_windx = 0
c_windy = 0
def dSdt(t, S, B, c_windx, c_windy):
x, vx, y, vy, z, vz = S
speed = np.hypot(np.hypot((vx + c_windx),(vy + c_windy)), vz)
return [vx, -B * speed * (vx + c_windx), vy, -B * speed * (vy + c_windy), vz, -B * speed * vz - g]
def distance(angle, B, c_windx, V=v0, t=600):
v0x = V * np.cos(np.radians(angle))
v0y = 0
v0z = V * np.sin(np.radians(angle))
sol = solve_ivp(dSdt, [0, t], y0 =[0,v0x,0,v0y,0,v0z], t_eval=np.linspace(0,t,10000), args=(B, c_windx,c_windy), atol=1e-7, rtol=1e-4)
justabove = np.where(np.diff(np.sign(sol.y[2])) < 0)[0][0]
justunder = justabove + 1
xloc = (sol.y[0][justabove] + sol.y[0][justunder])/2
return xloc
angles = np.linspace(0, 90, 200)
xlocs = np.vectorize(distance)(angles, B=B, c_windx = c_windx)
plt.plot(angles, xlocs)
plt.xlabel('angles []')
plt.ylabel('distance [m]')
plt.axvline(angles[np.argmax(xlocs)], ls='--', color='r')
plt.text(angles[np.argmax(xlocs)] + 0.2 ,0, round(angles[np.argmax(xlocs)], 2), rotation=90, color='r')
plt.title('Distance(angle)')
plt.show()
alfa = 15 #its optional
def same_distance_alfa(B,c_windx, V=v0, t=600):
for a in np.arange(alfa + 5, 90, 0.1):
if abs(distance(alfa, B, c_windx, V=v0, t=600) - distance(a, B, c_windx, V=v0, t=600) < 10):
print((distance(15, B, c_windx, V=v0, t=600) - distance(a, B, c_windx, V=v0, t=600)))
return a
beta = same_distance_alfa(B)
Related
I am trying to fit experimental data
with a function of the form:
A * np.sin(w*t + p) * np.exp(-g*t) + c
However, the fitted curve (the line in the following image) is not accurate:
If I leave out the exponential decay part, it works and I get a sinus function that is not decaying:
The function that I use is from this thread:
def fit_sin(tt, yy):
'''Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc"'''
tt = np.array(tt)
yy = np.array(yy)
ff = np.fft.fftfreq(len(tt), (tt[1]-tt[0])) # assume uniform spacing
Fyy = abs(np.fft.fft(yy))
guess_freq = abs(ff[np.argmax(Fyy[1:])+1]) # excluding the zero frequency "peak", which is related to offset
guess_amp = np.std(yy) * 2.**0.5
guess_offset = np.mean(yy)
guess_phase = 0.
guess_damping = 0.5
guess = np.array([guess_amp, 2.*np.pi*guess_freq, guess_phase, guess_offset, guess_damping])
def sinfunc(t, A, w, p, g, c): return A * np.sin(w*t + p) * np.exp(-g*t) + c
popt, pcov = scipy.optimize.curve_fit(sinfunc, tt, yy, p0=guess)
A, w, p, g, c = popt
f = w/(2.*np.pi)
fitfunc = lambda t: A * np.sin(w*t + p) * np.exp(-g*t) + c
return {"amp": A, "omega": w, "phase": p, "offset": c, "damping": g, "freq": f, "period": 1./f, "fitfunc": fitfunc, "maxcov": np.max(pcov), "rawres": (guess,popt,pcov)}
res = fit_sin(x, y)
x_fit = np.linspace(np.min(x), np.max(x), len(x))
plt.plot(x, y, label='Data', linewidth=line_width)
plt.plot(x_fit, res["fitfunc"](x_fit), label='Fit Curve', linewidth=line_width)
plt.show()
I am not sure if I implemented the code incorrectly or if the function is not able to describe my data correctly. I appreciate your help!
You can load the txt file from here:
GitHub
and manipulate the data like this to compare it with the post:
file = 'A2320_Data.txt'
column = 17
data = np.loadtxt(file, float)
start = 270
end = 36000
time_scale = 3600
x = []
y = []
for i in range(len(data)):
if start < data[i][0] < end:
x.append(data[i][0]/time_scale)
y.append(data[i][column])
x = np.array(x)
y = np.array(y)
plt.plot(x, y, label='Pyro Oscillations', linewidth=line_width)
Your fitted curve will look like this
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize
def sinfunc(t, A, w, p, g, c): return A * np.sin(w*t + p) * np.exp(-g*t) + c
tt = np.linspace(0, 10, 1000)
yy = sinfunc(tt, -1, 10, 2, 0.3, 2)
plt.plot(tt, yy)
g stretches the envelope horizontally, c moves the center vertically, w determines the oscillation frequency, A stretches the envelope vertically.
So it can't accurately model the data you have.
Also, you will not be able to reliably fit w, to determine the oscilation frequency it is better to try an FFT.
Of course you could adjust the function to look like your data by adding a few more parameters, e.g.
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize
def sinfunc(t, A, w, p, g, c1, c2, c3): return A * np.sin(w*t + p) * (np.exp(-g*t) - c1) + c2 * np.exp(-g*t) + c3
tt = np.linspace(0, 10, 1000)
yy = sinfunc(tt, -1, 20, 2, 0.5, 1, 1.5, 1)
plt.plot(tt, yy)
But you will still have to give a good guess for the frequency.
I'm trying to make a diffraction plot using Simpson integration but this just plots a straight line. I know it's just evaluating a single number but I'd like it to make a diffraction pattern with the strongest maxima in the middle.
import math
import cmath
import numpy as np
import sympy as sym
import matplotlib.pyplot as plt
from scipy.integrate import simps
from scipy import integrate
from math import sin, cos, pi
n = 100 #Number of areas to be calculated for integration approximation
λ = 1*10**-6 #Wavelength
z = 0.02 #Screen distance
k = (2*math.pi)/λ #wave number
j = complex(cmath.sqrt(-1))
n = 100
b = 2E-5
a = 0
def f(x):
return np.exp((1j*k)/(2*z)*(x)**2)
x = np.linspace(-0.005, 0.005, 100)
y = np.linspace(0, 0, 100, dtype= "complex")
def g(p, q):
h = (b - a) / n
k = 0
z = 0
for i in range(1, n // 2):
k += 2 * f(a + 2 * i * h)
for i in range(1, n // 2 + 1):
z += 4 * f(a + (2 * i - 1) * h)
return h * (z+k+f(a)+f(b)) / 3.0
for i in range(0,100,n):
if( x[i] == 0 ):
y[i] = 0
else:
q = np.linspace(0,x[i],100)
p = f(q)
y[i] = abs(np.conj(g(p, q))*(g(p, q)))*8.85E-12
plt.plot(x, np.imag(y))
This python code can solve one non- coupled differential equation:
import numpy as np
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
#numba.jit()
# A sample differential equation "dy / dx = (x - y**2)/2"
def dydx(x, y):
return ((x - y**2)/2)
# Finds value of y for a given x using step size h
# and initial value y0 at x0.
def rungeKutta(x0, y0, x, h):
# Count number of iterations using step size or
# step height h
n = (int)((x - x0)/h)
# Iterate for number of iterations
y = y0
for i in range(1, n + 1):
"Apply Runge Kutta Formulas to find next value of y"
k1 = h * dydx(x0, y)
k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1)
k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2)
k4 = h * dydx(x0 + h, y + k3)
# Update next value of y
y = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4)
# Update next value of x
x0 = x0 + h
return y
def dplot(start,end,steps):
Y=list()
for x in np.linspace(start,end,steps):
Y.append(rungeKutta(x0, y, x , h))
plt.plot(np.linspace(start,end,steps),Y)
print("Execution time:",time.clock() - start_time, "seconds")
plt.show()
start,end = 0, 10
steps = end* 100
x0 = 0
y = 1
h = 0.002
dplot(start,end,steps)
This code can solve this differential equation:
dydx= (x - y**2)/2
Now I have a system of coupled differential equations:
dydt= (x - y**2)/2
dxdt= x*3 + 3y
How can I implement these two as a system of coupled differential equations in the above code?
Is there any more generalized way for system of n-number of coupled differential equations?
With the help of others, I got to this:
import numpy as np
from math import sqrt
import matplotlib.pyplot as plt
import numba
import time
start_time = time.clock()
a=1
b=1
c=1
d=1
# Equations:
#numba.jit()
#du/dt=V(u,t)
def V(u,t):
x, y, vx, vy = u
return np.array([vy,vx,a*x+b*y,c*x+d*y])
def rk4(f, u0, t0, tf , n):
t = np.linspace(t0, tf, n+1)
u = np.array((n+1)*[u0])
h = t[1]-t[0]
for i in range(n):
k1 = h * f(u[i], t[i])
k2 = h * f(u[i] + 0.5 * k1, t[i] + 0.5*h)
k3 = h * f(u[i] + 0.5 * k2, t[i] + 0.5*h)
k4 = h * f(u[i] + k3, t[i] + h)
u[i+1] = u[i] + (k1 + 2*(k2 + k3 ) + k4) / 6
return u, t
u, t = rk4(V, np.array([1., 0., 1. , 0.]) , 0. , 10. , 100000)
x,y, vx,vy = u.T
# plt.plot(t, x, t,y)
plt.semilogy(t, x, t,y)
plt.grid('on')
print("Execution time:",time.clock() - start_time, "seconds")
plt.show()
I am trying to write a program using the Lotka-Volterra equations for predator-prey interactions. Solve Using ODE's:
dx/dt = a*x - B*x*y
dy/dt = g*x*y - s*y
Using 4th order Runge-Kutta method
I need to plot a graph showing both x and y as a function of time from t = 0 to t=30.
a = alpha = 1
b = beta = 0.5
g = gamma = 0.5
s = sigma = 2
initial conditions x = y = 2
Here is my code so far but not display anything on the graph. Some help would be nice.
#!/usr/bin/env python
from __future__ import division, print_function
import matplotlib.pyplot as plt
import numpy as np
def rk4(f, r, t, h):
""" Runge-Kutta 4 method """
k1 = h*f(r, t)
k2 = h*f(r+0.5*k1, t+0.5*h)
k3 = h*f(r+0.5*k2, t+0.5*h)
k4 = h*f(r+k3, t+h)
return (k1 + 2*k2 + 2*k3 + k4)/6
def f(r, t):
alpha = 1.0
beta = 0.5
gamma = 0.5
sigma = 2.0
x, y = r[2], r[2]
fxd = x*(alpha - beta*y)
fyd = -y*(gamma - sigma*x)
return np.array([fxd, fyd], float)
tpoints = np.linspace(0, 30, 0.1)
xpoints = []
ypoints = []
r = np.array([2, 2], float)
for t in tpoints:
xpoints += [r[2]]
ypoints += [r[2]]
r += rk4(f, r, t, h)
plt.plot(tpoints, xpoints)
plt.plot(tpoints, ypoints)
plt.xlabel("Time")
plt.ylabel("Population")
plt.title("Lotka-Volterra Model")
plt.savefig("Lotka_Volterra.png")
plt.show()
A simple check of your variable tpoints after running your script shows it's empty:
In [7]: run test.py
In [8]: tpoints
Out[8]: array([], dtype=float64)
This is because you're using np.linspace incorrectly. The third argument is the number of elements desired in the output. You've requested an array of length 0.1.
Take a look at np.linspace's docstring. You won't have a problem figuring out how to adjust your code.
1) define 'h' variable.
2) use
tpoints = np.arange(30) #array([0, 1, 2, ..., 30])
not
np.linspace()
and don't forget to set time step size equal to h:
h=0.1
tpoints = np.arange(0, 30, h)
3) be careful with indexes:
def f(r,t):
...
x, y=r[0], r[1]
...
for t in tpoints:
xpoints += [r[0]]
ypoints += [r[1]]
...
and better use .append(x):
for t in tpoints:
xpoints.append(r[0])
ypoints.append(r[1])
...
Here's tested code for python 3.7 (I've set h=0.001 for more presize)
import matplotlib.pyplot as plt
import numpy as np
def rk4(r, t, h): #edited; no need for input f
""" Runge-Kutta 4 method """
k1 = h*f(r, t)
k2 = h*f(r+0.5*k1, t+0.5*h)
k3 = h*f(r+0.5*k2, t+0.5*h)
k4 = h*f(r+k3, t+h)
return (k1 + 2*k2 + 2*k3 + k4)/6
def f(r, t):
alpha = 1.0
beta = 0.5
gamma = 0.5
sigma = 2.0
x, y = r[0], r[1]
fxd = x*(alpha - beta*y)
fyd = -y*(gamma - sigma*x)
return np.array([fxd, fyd], float)
h=0.001 #edited
tpoints = np.arange(0, 30, h) #edited
xpoints, ypoints = [], []
r = np.array([2, 2], float)
for t in tpoints:
xpoints.append(r[0]) #edited
ypoints.append(r[1]) #edited
r += rk4(r, t, h) #edited; no need for input f
plt.plot(tpoints, xpoints)
plt.plot(tpoints, ypoints)
plt.xlabel("Time")
plt.ylabel("Population")
plt.title("Lotka-Volterra Model")
plt.savefig("Lotka_Volterra.png")
plt.show()
You can also try to plot "cycles":
plt.xlabel("Prey")
plt.ylabel("Predator")
plt.plot(xpoints, ypoints)
plt.show()
https://i.stack.imgur.com/NB9lc.png
I am trying to find intersection points between an elipse and a line, but seem to be unable to do so. I have tried two approaches, one with shapely by trying to find intersection between LineString and LinearRing as in the code bellow, but did not get any usable values out of it. One of the problems, is the elipse will always be off center and at an small or high angle
# -*- coding: utf-8 -*-
"""
Created on Mon Aug 19 17:38:55 2013
#author: adudchenko
"""
from pylab import *
import numpy as np
from shapely.geometry.polygon import LinearRing
from shapely.geometry import LineString
def ellipse_polyline(ellipses, n=100):
t = np.linspace(0, 2*np.pi, n, endpoint=False)
st = np.sin(t)
ct = np.cos(t)
result = []
for x0, y0, a, b, angle in ellipses:
angle = np.deg2rad(angle)
sa = np.sin(angle)
ca = np.cos(angle)
p = np.empty((n, 2))
p[:, 0] = x0 + a * ca * ct - b * sa * st
p[:, 1] = y0 + a * sa * ct + b * ca * st
result.append(p)
return result
def intersections(a, line):
ea = LinearRing(a)
eb = LinearRing(b)
mp = ea.intersection(eb)
print mp
x = [p.x for p in mp]
y = [p.y for p in mp]
return x, y
ellipses = [(1, 1, 2, 1, 45), (2, 0.5, 5, 1.5, -30)]
a, b = ellipse_polyline(ellipses)
line=LineString([[0,0],[4,4]])
x, y = intersections(a, line)
figure()
plot(x, y, "o")
plot(a[:,0], a[:,1])
plot(b[:,0], b[:,1])
show()
I also tried using fsolve as in example bellow, but it finds the wrong intersect points ( or actually one wrong point.
from pylab import *
from scipy.optimize import fsolve
import numpy as np
def ellipse_polyline(ellipses, n=100):
t = np.linspace(0, 2*np.pi, n, endpoint=False)
st = np.sin(t)
ct = np.cos(t)
result = []
for x0, y0, a, b, angle in ellipses:
angle = np.deg2rad(angle)
sa = np.sin(angle)
ca = np.cos(angle)
p = np.empty((n, 2))
p[:, 0] = x0 + a * ca * np.cos(t) - b * sa * np.sin(t)
p[:, 1] = y0 + a * sa * np.cos(t) + b * ca * np.sin(t)
result.append(p)
return result
def ellipse_line(txy):
t,x,y=txy
x0, y0, a, b, angle,m,lb=1, 1, 2, 1, 45,0.5,0
sa = np.sin(angle)
ca = np.cos(angle)
return (x0 + a * ca * np.cos(t) - b * sa * np.sin(t)-x,y0 + a * sa * np.cos(t) + b * ca * np.sin(t)-y,m*x+lb-y)
a,b= ellipse_polyline([(1, 1, 2, 1, 45), (2, 0.5, 5, 1.5, -30)])
t,y,x=fsolve(ellipse_line,(0,0,0))
print t,y,x
#print a[:,0]
m=0.5
bl=0
xl,yl=[],[]
for i in range(10):
xl.append(i)
yl.append(m*i+bl)
figure()
plot(x, y, "o")
plot(a[:,0], a[:,1])
plot(xl,yl)
Any help would be appriceated?
With the Shapely part, def intersections(a, line) can be fixed. First, there is an error, where it references a global b instead of using the line parameter, which is ignored. Thus the comparison is between the two ellipses a and b, and not between each ellipse to line, as I think is intended.
Also, the intersection between two lines can be one of several outcomes: an empty set, a point (1 intersection) a multipoint (more than 1 intersection), a linestring (if any part(s) of the linestrings overlap and are parallel to each other), or a collection of points and lines. Assuming only the first three outcomes:
def intersections(a, line):
ea = LinearRing(a)
mp = ea.intersection(line)
if mp.is_empty:
print('Geometries do not intersect')
return [], []
elif mp.geom_type == 'Point':
return [mp.x], [mp.y]
elif mp.geom_type == 'MultiPoint':
return [p.x for p in mp], [p.y for p in mp]
else:
raise ValueError('something unexpected: ' + mp.geom_type)
So now these look correct:
>>> intersections(a, line)
([2.414213562373095], [2.414213562373095])
>>> intersections(b, line)
([0.0006681263405436677, 2.135895843256409], [0.0006681263405436642, 2.135895843256409])