Double pendulum animation in python - numerical-methods

I am using the following code to solve the system of differential equations numerically and then animate it:
for i in range(steps-1):
Theta1 = t1[i]
Theta2 = t2[i]
dTheta1 = w1[i]
dTheta2 = w2[i]
a1 = (g*(np.sin(Theta2)*np.cos(Theta1-Theta2)-mu*np.sin(Theta1))-(l2*dTheta2*dTheta2+l1*dTheta1*dTheta1*np.cos(Theta1-Theta2))*np.sin(Theta1-Theta2))/(l1*(mu-np.cos(Theta1-Theta2)*np.cos(Theta1-Theta2)))
a2 = (mu*g*(np.sin(Theta1)*np.cos(Theta1-Theta2)-np.sin(Theta2))+(mu*l1*dTheta1*dTheta1+l2*dTheta2*dTheta2*np.cos(Theta1-Theta2))*np.sin(Theta1-Theta2))/(l2*(mu-np.cos(Theta1-Theta2)*np.cos(Theta1-Theta2)))
w1[i+1] = w1[i] + dt*a1
w2[i+1] = w2[i] + dt*a2
t1[i+1] = t1[i] + dt*w1[i]
t2[i+1] = t2[i] + dt*w2[i]
This gives me errors such as
RuntimeWarning: invalid value encountered in longdouble_scalars
and
RuntimeWarning: overflow encountered in longdouble_scalars.
At first, I thought this might be happening due to precisions errors, so I tried using longdouble. But that didn't help at all. When I run the animation, I get a few frames of unphysical systems and then it shuts downs. I have checked my equation from multiple sources and I ended up using their equations too and the problem still persists. What am I doing wrong?
Edit: Okay so simply decreasing the step size seemed to not give me the error any more, but now animation is unphysical. The pendulum seems to pick up speed over time, and also performs full rotations, even when it started from rest at an extreme. I am using a 4th order RK method and still getting this problem. Any ideas? I have provided my new code below
def diff_eq_a1(theta1,theta2,w1,w2):
a11 = mu
a12 = np.cos(theta1-theta2)
a21 = np.cos(theta1-theta2)
a22 = 1
b1 = (mu*g*l1*np.sin(theta1)) + (l2*np.sin(theta1-theta2)*w2*w2)
b2 = (g*l2*np.sin(theta2))-(l1*np.sin(theta1-theta2)*w1*w1)
return -(1/l1)*(b1*a22-b2*a12)/(a11*a22-a21*a12)
def diff_eq_a2(theta1,theta2,w1,w2):
a11 = mu
a12 = np.cos(theta1-theta2)
a21 = np.cos(theta1-theta2)
a22 = 1
b1 = (mu*g*l1*np.sin(theta1)) + (l2*np.sin(theta1-theta2)*w2*w2)
b2 = (g*l2*np.sin(theta2))-(l1*np.sin(theta1-theta2)*w1*w1)
return -(1/l2)*(b2*a11-b1*a21)/(a11*a22-a21*a12)
for i in range(steps-1):
a1_k1 = dt*diff_eq_a1(t1[i],t2[i],w1[i],w2[i])
a1_k2 = dt*diff_eq_a1((t1[i]+0.5*a1_k1),t2[i],w1[i],w2[i])
a1_k3 = dt*diff_eq_a1((t1[i]+0.5*a1_k2),t2[i],w1[i],w2[i])
a1_k4 = dt*diff_eq_a1((t1[i]+a1_k3),t2[i],w1[i],w2[i])
w1[i+1] = w1[i] + (a1_k1/6)+(a1_k2/3)+(a1_k3/3)+(a1_k4/6)
a2_k1 = dt*diff_eq_a1(t1[i],t2[i],w1[i],w2[i])
a2_k2 = dt*diff_eq_a1(t1[i],(t2[i]+0.5*a2_k1),w1[i],w2[i])
a2_k3 = dt*diff_eq_a1(t1[i],(t2[i]+0.5*a2_k2),w1[i],w2[i])
a2_k4 = dt*diff_eq_a1(t1[i],(t2[i]+a2_k3),w1[i],w2[i])
w2[i+1] = w2[i] + (a2_k1/6)+(a2_k2/3)+(a2_k3/3)+(a2_k4/6)
t1[i+1] = t1[i] + dt*w1[i]
t2[i+1] = t2[i] + dt*w2[i]
x1 = l1*np.sin(t1)
x2 = x1 + l2*np.sin(t2)
y1 = - l1*np.cos(t1)
y2 = y1 - l2*np.cos(t2)
I know its not very neat right now, but I really just want to get it working first

Related

Interp1 function in ODE45

I am trying to solve a 3 Degree Of Freedom vessel-crane system in MATLAB using ode45. In my ode function, the vessel motion is determined by simple spring and damper constants, but the device below the main hoist system, a cranemaster, has a displacement-dependent stiffness and thus depends on one of the Degrees Of Freedom. To determine this value, I use the interp1 function to determine the stiffness from a dataset:
vq = interp1(x,v,xq,method,extrapolation)
This all works fine, but only when I add the extrapolation option in the interp1 function, even though the xq values all fall inside the dataset x-values. If I do not specify extrapolation, the function returns NAN.
I have checked if the stiffness values in the dataset make sense by taking the median value as a constant value and then the ode45 solver works just fine and the displacement never falls outside of the dataset. I have also noticed that if I switch to ode23 as a solver, it also works. This all makes me believe it has something to do with the solving method and the stepsize the ode solver takes as its initial value.
I have provided (a simplified version of) my code and a picture of the system to get a feel for the problem. I know that switching to ode23 solves the problem but I would like to understand why and, if possible, stay with ode45 as it is a more general solver.
Any help would be appreciated!
Regards,
Diederik
Overview of the system I am solving
close all
tspan = [0 100];
x0 = [0.0; 0; 0.0; 0; 0.0; 0]; % initial disp. and vel. is 0. for all 3 DOF.
% x = [z; zdot; theta; thetadot; u; udot]
[t,x] = ode45(#Three_DOF,tspan,x0);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Plots
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
figure()
subplot(2,1,1)
plot(t,x(:,5),'color','b','LineWidth',1)
grid on
xlabel('Time (s)')
ylabel('Displacement (m)')
title('Displacement Vs Time')
subplot(2,1,2)
plot(t,x(:,6),'color','#D95319','LineWidth',1)
grid on
xlabel('Time (s)')
ylabel('Velocity (m/s)')
title('Velocity Vs Time')
function [dxdt] = Three_DOF(t,x)
% vessel parameters
m = 570.5e6;
m55 = 4.591e12;
mh = 200e3;
c33 = 3.83952e8;
c55 = 3.66116e12;
wnh = 0.360;
wnp = 0.596;
a33 = (c33-wnh^2*m)/wnh^2;
a55 = (c55-wnp^2*m55)/wnp^2;
zeta33 = 0.386;
zeta55 = 0.121;
b33 = zeta33*2*sqrt(c33*(m+a33));
b55 = zeta55*2*sqrt(c55*(m55+a55));
Fa = 153e6;
Ma = 3.013e9;
w1 = 0.21/0.36;
w2 = 0.18/0.28;
%%%% Cable parameters
k_c = 496e6;
b_c = 5.01e6;
%%% CRANEMASTER parameters
x_data = [-2.0; -1.87; -1.73; -1.60; -1.47; -1.33; -1.20; -1.07; -0.93; -0.8; -0.67; -0.53; -0.4; -0.27; -0.13; 0.0; 0.13; 0.27; 0.40; 0.53; 0.67; 0.80; 0.93; 1.07; 1.20; 1.33; 1.47; 1.60; 1.73; 1.87; 2.0];
k_data = [760; 792.1; 784.8; 813.9; 828.4; 850.2; 879.3; 901.1; 922.9; 959.2; 981.0; 1010.1; 1039.1; 1068.2; 1104.5; 1133.6; 1177.2; 1206.3; 1242.6; 1286.2; 1329.8; 1373.4; 1409.7; 1460.6; 1511.5; 1562.3; 1620.5; 1671.3; 1729.5; 1794.9; 1853];
k_cm = 30000*interp1(x_data,k_data,x(5));%,'linear','extrap');
%k_cm = 30000*1133.61; % middle value of dataset k_data
b_cm = 6e4;
l = 217.5;
% Matrices & solve
Mass = [m+a33 0 0; 0 m55+a55 0; 0 0 mh];
k = [c33+k_c -k_c*l -k_c; -k_c*l c55+k_c*l^2 k_c*l; -k_c k_c*l k_c+k_cm];
b = [b33+b_c -b_c*l -b_c; -b_c*l b55+b_c*l^2 b_c*l; -b_c b_c*l b_c+b_cm];
F = [Fa*sin(w1*t); Ma*sin(w2*t); 0];
EOM = Mass\(F-b*([x(2); x(4); x(6)])-k*([x(1); x(3); x(5)]));
dxdt = [x(2); EOM(1); x(4); EOM(2); x(6); EOM(3)];
% x = [z; zdot; theta; thetadot; u; udot]
end

Trouble Calling Terms from Function

I have a code that defines a function, then I try to use the variables I defined within that function in another expression. When I do this, I get an error that is:
Undefined function or variable 'phi'.
I'm not sure why phi is undefined, since I have it in the if/else statement.
It might be better to explain with my (shortened) code:
global I11 I22 I33 Mx My Mz w10 w20 w30 eps10 eps20 eps30 eps40...
C110 C120 C130 C210 C220 C230 C310 C320 C330 IC K0
IC = [w10 w20 w30...
eps10 eps20 eps30 eps40...
C110 C120 C130 C210 C220 C230 C310 C320 C330];
opts = odeset('RelTol', 1*10^(-10),'AbsTol', 1*10^(-10));
[t, y] = ode45(#(t,y) DynEqn2(t,y,I11,I22,I33,Mx,My,Mz), [ti tf], IC, opts);
N = sqrt(sum(y(:,4:7).^2,2));
kap = acosd(1-2*y(:,5).^2-2*y(:,7).^2);
phi1 = acosd((2.*(y(:,4).*y(:,5)+y(:,6).*y(:,7)))/sind(kap));
phi2 = asind((2.*(y(:,6).*y(:,4)-y(:,5).*y(:,7)))/sind(kap));
if phi1==phi2
phi = phi1;
elseif phi1==180-phi2
phi = phi1;
elseif -phi1==phi2
phi = -phi1;
elseif -phi1==180-phi2
phi = -phi1;
else
disp('Something is wrong with phi')
end
figure (1)
plot(t,phi)
figure (2)
plot(t,kap)
function soln = DynEqn2(t,y,I11,I22,I33,Mx,My,Mz)
w1 = y(1);
w2 = y(2);
w3 = y(3);
eps1 = y(4);
eps2 = y(5);
eps3 = y(6);
eps4 = y(7);
C11 = y(8);
C12 = y(9);
C13 = y(10);
C21 = y(11);
C22 = y(12);
C23 = y(13);
C31 = y(14);
C32 = y(15);
C33 = y(16);
w1_dot = (Mx - w2*w3*(I33-I22))/I11;
w2_dot = (My - w1*w3*(I11-I33))/I22;
w3_dot = (Mz - w1*w2*(I22-I11))/I33;
eps1_dot = .5*(w1*eps4-w2*eps3+w3*eps2);
eps2_dot = .5*(w1*eps3+w2*eps4-w3*eps1);
eps3_dot = .5*(-w1*eps2+w2*eps1+w3*eps4);
eps4_dot = -.5*(w1*eps1+w2*eps2+w3*eps3);
C11_dot = C12*w3-C13*w2;
C12_dot = C13*w1-C11*w3;
C13_dot = C11*w2-C12*w1;
C21_dot = C22*w3-C23*w2;
C22_dot = C23*w1-C21*w3;
C23_dot = C21*w2-C22*w1;
C31_dot = C32*w3-C33*w2;
C32_dot = C33*w1-C31*w3;
C33_dot = C31*w2-C32*w1;
soln = [w1_dot; w2_dot; w3_dot; ...
eps1_dot; eps2_dot; eps3_dot; eps4_dot; ...
C11_dot; C12_dot; C13_dot; C21_dot; C22_dot; C23_dot; C31_dot; C32_dot; C33_dot];
end
My lines where I calculate phi1, phi2, and then the if/else statement to find phi, are what I am struggling with.
I made sure that the variables defined in the function work, so for example, in the command window I typed in 'y(:,4)' and got the correct output. But whenever I try to use this within the functions i.e. 'phi1', it repeatedly outputs an incorrect value of '90.0000' until I stop it.
Where I define the 'N' variable, it is something similar, yet that one works without errors.
Does anyone have any ideas how to amend this issue?
Any help is appreciated, thanks.
Edit: The complete error message is as follows:
Undefined function or variable 'phi'.
Error in HW6_Q1 (line 85)
plot(t,phi)
I figured out my solution with the help from a colleague not on Stack Overflow.
I forgot the ./, which turned my phi into a matrix, rather than a vector which is what I wanted from it.

Recall from nltk.metrics.score returning None

I'm trying to calculate the precision and recall using the nltk.metrics.score (http://www.nltk.org/_modules/nltk/metrics/scores.html) with my NLTK.NaiveBayesClassifier.
However, I stumble upon the error:
"unsupported operand type(s) for +: 'int' and 'NoneType".
which I suspect is from my 10-fold cross-validation where in some reference sets, there are zero negative (the data set is a bit imbalanced where 87% of it is positive).
According to nltk.metrics.score,
def precision(reference, test):
"Given a set of reference values and a set of test values, return
the fraction of test values that appear in the reference set.
In particular, return card(``reference`` intersection
``test``)/card(``test``).
If ``test`` is empty, then return None."
It seems that some of my 10-fold set is returning recall as None since there are no Negative in the reference set. Any idea on how to approach this problem?
My full code is as follow:
trainfeats = negfeats + posfeats
n = 10 # 5-fold cross-validation
subset_size = len(trainfeats) // n
accuracy = []
pos_precision = []
pos_recall = []
neg_precision = []
neg_recall = []
pos_fmeasure = []
neg_fmeasure = []
cv_count = 1
for i in range(n):
testing_this_round = trainfeats[i*subset_size:][:subset_size]
training_this_round = trainfeats[:i*subset_size] + trainfeats[(i+1)*subset_size:]
classifier = NaiveBayesClassifier.train(training_this_round)
refsets = collections.defaultdict(set)
testsets = collections.defaultdict(set)
for i, (feats, label) in enumerate(testing_this_round):
refsets[label].add(i)
observed = classifier.classify(feats)
testsets[observed].add(i)
cv_accuracy = nltk.classify.util.accuracy(classifier, testing_this_round)
cv_pos_precision = precision(refsets['Positive'], testsets['Positive'])
cv_pos_recall = recall(refsets['Positive'], testsets['Positive'])
cv_pos_fmeasure = f_measure(refsets['Positive'], testsets['Positive'])
cv_neg_precision = precision(refsets['Negative'], testsets['Negative'])
cv_neg_recall = recall(refsets['Negative'], testsets['Negative'])
cv_neg_fmeasure = f_measure(refsets['Negative'], testsets['Negative'])
accuracy.append(cv_accuracy)
pos_precision.append(cv_pos_precision)
pos_recall.append(cv_pos_recall)
neg_precision.append(cv_neg_precision)
neg_recall.append(cv_neg_recall)
pos_fmeasure.append(cv_pos_fmeasure)
neg_fmeasure.append(cv_neg_fmeasure)
cv_count += 1
print('---------------------------------------')
print('N-FOLD CROSS VALIDATION RESULT ' + '(' + 'Naive Bayes' + ')')
print('---------------------------------------')
print('accuracy:', sum(accuracy) / n)
print('precision', (sum(pos_precision)/n + sum(neg_precision)/n) / 2)
print('recall', (sum(pos_recall)/n + sum(neg_recall)/n) / 2)
print('f-measure', (sum(pos_fmeasure)/n + sum(neg_fmeasure)/n) / 2)
print('')
Perhaps not the most elegant, but guess the most simple fix would be setting it to 0 and the actual value if not None, e.g.:
cv_pos_precision = 0
if precision(refsets['Positive'], testsets['Positive']):
cv_pos_precision = precision(refsets['Positive'], testsets['Positive'])
And for the others as well, of course.

Octave index out of bounds

The program is an optimized gradient descent.
Here is the code :
clear all
close all
[x,y] = meshgrid(-2:0.1:2);
z = x.^2 + 100*y.^2;
n = 1;
k(n)=0.01;
arret = 0.0001;
mesh(x,y,z);
[x1(n),y1(n)] = ginput(1);
diff_x(n) = 2*x1(n);
diff_y(n) = 200*y1(n);
while sqrt(diff_x(n)^2 + diff_y(n)^2) > arret && n < 30
k(n) = sqrt(diff_x(n)^2 + diff_y(n)^2)/(8*x1(n).^2+2*10.^6*y1(n).^2);
x1(n+1) = x1(n) - k(n)*diff_x(n);
y1(n+1) = y1(n) - k(n)*diff_y(n);
n = n+1;
diff_x(n) = 2*x1(n);
diff_y(n) = 200*y1(n);
z1(n) = x1(n).^2 + 100*y1(n).^2;
plot3(x1(n),y1(n),z1(n));
end
x1(n)
y1(n)
n
So I got this and I don't understand why.
error: GradientPasOptFinal2: A(I): index out of bounds; value 2 out of bound 1
error: called from:
error: error: C:\Octave\octave-3.8.2\GradientPasOptFinal2.m at line 13, column 8
SOLVED : the k(n) between y1(n) and n was the reason, i don't know why, but now the program works, thank you !
You have only defined k(1). k(2) is undefined.
Move this line:
k(n) = sqrt(diff_x(n)^2 + diff_y(n)^2)/(8*x1(n).^2+2*10.^6*y1(n).^2);
Ahead of these lines:
x1(n+1) = x1(n) - k(n)*diff_x(n);
y1(n+1) = y1(n) - k(n)*diff_y(n);
And it should work.

Plotting a function in matlab involving an integral

I'm trying to plot a function that contains a definite integral. My code uses all anonymous functions. When I run the file, it gives me an error. My code is below:
%%% List of Parameters %%%
gamma_sp = 1;
cap_gamma = 15;
gamma_ph = 0;
omega_0 = -750;
d_omega_0 = 400;
omega_inh = 100;
d_omega_inh = 1000;
%%% Formulae %%%
gamma_t = gamma_sp/2 + cap_gamma/2 + gamma_ph;
G = #(x) exp(-(x-omega_inh).^2./(2*d_omega_inh.^2))./(sqrt(2*pi)*d_omega_inh);
F = #(x) exp(-(x-omega_0).^2./(2*d_omega_0.^2))./(sqrt(2*pi)*d_omega_0);
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
Q_integral = #(x,y) F(x)./(y - x + 1i*gamma_t);
A = #(y) integral(#(x)A_integral(x,y),-1000,1000);
Q = #(y) integral(#(x)Q_integral(x,y),-3000,0);
P1 = #(y) -1./(1i.*(gamma_sp + cap_gamma)).*(1./(y + 2.*1i.*gamma_t)*(A(y)-conj(A(0)))-1./y.*(A(y)-A(0))+cap_gamma./gamma_sp.*Q(y).*(A(0)-conj(A(0))));
P2 = #(y) conj(P1(y));
P = #(y) P1(y) - P2(y);
sig = #(y) abs(P(y)).^2;
rng = -2000:0.05:1000;
plot(rng,sig(rng))
It seems to me that when the program runs the plot command, it should put each value of rng into sig(y), and that value will be used as the y value in A_integral and Q_integral. However, matlab throws an error when I try to run the program.
Error using -
Matrix dimensions must agree.
Error in #(x,y)G(x)./(y-x+1i*gamma_t)
Error in #(x)A_integral(x,y)
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);
Error in #(y)integral(#(x)A_integral(x,y),-1000,1000)
Error in
#(y)-1./(1i.*(gamma_sp+cap_gamma)).*(1./(y+2.*1i.*gamma_t)*(A(y)-conj(A(0)))-1. /y.*(A(y)-A(0))+cap_gamma./gamma_sp.*Q(y).*(A(0)-conj(A(0))))
Error in #(y)P1(y)-P2(y)
Error in #(y)abs(P(y)).^2
Error in fwm_spec_diff_paper_eqn (line 26)
plot(rng,sig(rng))
Any ideas about what I'm doing wrong?
You have
>> rng = -2000:0.05:1000;
>> numel(rng)
ans =
60001
all 60001 elements get passed down to
A = #(y) integral(#(x)A_integral(x,y),-1000,1000);
which calls
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
(similar for Q). The thing is, integral is an adaptive quadrature method, meaning (roughly) that the amount of x's it will insert into A_integral varies with how A_integral behaves at certain x.
Therefore, the amount of elements in y will generally be different from the elements in x in the call to A_integral. This is why y-x +1i*gamma_t fails.
Considering the complexity of what you're trying to do, I think it is best to re-define all anonymous functions as proper functions, and integrate a few of them into single functions. Look into the documentation of bsxfun to see if that can help (e.g., bsxfun(#minus, y.', x) instead of y-x could perhaps fix a few of these issues), otherwise, vectorize only in x and loop over y.
Thanks Rody, that made sense to me. I keep trying to use matlab like mathematica and I forget how matlab does things. I modified the code a bit, and it produces the right result. The integrals are evaluated very roughly, but it should be easy to fix that. I've posted my modified code below.
%%% List of Parameters %%%
gamma_sp = 1;
cap_gamma = 15;
gamma_ph = 0;
omega_0 = -750;
d_omega_0 = 400;
omega_inh = 100;
d_omega_inh = 1000;
%%% Formulae %%%
gamma_t = gamma_sp/2 + cap_gamma/2 + gamma_ph;
G = #(x) exp(-(x-omega_inh).^2./(2*d_omega_inh.^2))./(sqrt(2*pi)*d_omega_inh);
F = #(x) exp(-(x-omega_0).^2./(2*d_omega_0.^2))./(sqrt(2*pi)*d_omega_0);
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
Q_integral = #(x,y) F(x)./(y - x + 1i*gamma_t);
w = -2000:0.05:1000;
sigplot = zeros(size(w));
P1plot = zeros(size(w));
P2plot = zeros(size(w));
Pplot = zeros(size(w));
aInt_range = -1000:0.1:1200;
qInt_range = -2000:0.1:100;
A_0 = sum(A_integral(aInt_range,0).*0.1);
for k=1:size(w,2)
P1plot(k) = -1./(1i*(gamma_sp + cap_gamma)).*(1./(w(k)+2.*1i.*gamma_t).*(sum(A_integral(aInt_range,w(k)).*0.1)-conj(A_0))-1./w(k).*(sum(A_integral(aInt_range,w(k)).*0.1)-A_0)+cap_gamma./gamma_sp.*sum(Q_integral(qInt_range,w(k)).*0.1).*(A_0-conj(A_0)));
P2plot(k) = conj(P1plot(k));
Pplot(k) = P1plot(k) - P2plot(k);
sigplot(k) = abs(Pplot(k)).^2;
end
plot(w,sigplot)