I am having an issue with plotting my airfoil on matlab. It appears that there are some abberations with only one or two values. I have checked again and again with the formulas so they should be good. I have not converted to degrees so that's not the issue.
%% setup
clc
clear all
%type of airfoil
typeNACA = '2312';
%extract values of airfoil type
M = str2double(typeNACA(1)); % max camber %chord
P = str2double(typeNACA(2)); % chordwise position
Thmax = str2double(typeNACA(3:4));
M=M/100;
P=P/100
Thmax=Thmax/100;
% gridpoints
gridpoints = 1000;
% Airfoil grid
x = linspace(0,1,gridpoints);
%camber equations
%yc1(x) = (M/P^2)*(2*P*x-x^2);
%yc(x) = (M/(1-P)^2*(1-2*P+2*Px-x^2); % (for P =< x <1)
%dyc1 = (2*M/P^2)*(P-x)
%dyc2 = (2*M)/(1-P^2)*(P-x)
%Camber and gradient
yc = ones(gridpoints,1);
dyc= ones(gridpoints,1);
theta=ones(gridpoints,1);
for i=1:gridpoints
if (x(i) >= 0 && x(i) < P)
yc(i) = (M/P^2)*(2*P*x(i)-x(i)^2); % (for 0 =< x < P)
dyc(i)= ((2*M)/P^2)*(P - x(i)); %dyc/dx
elseif (x(i) >= P && x(i) <= 1) % (for P =< x <1)
yc(i)=(M/(1-P)^2)*(1-2*P+2*P*x(i)-x(i)^2);
dyc(i) = (2*M)/((1-P)^2*(P-x(i)));
end
theta(i) = atan(dyc(i)); %angle theta
end
% thickness coefficients
a0 = 0.2969;
a1 = -0.126;
a2 = -0.3516;
a3 = 0.2843;
a4 = -0.1036; % -0.1015 for open TE -0.1036 for closed TE
%thickness distribution
yt = ones(gridpoints,1);
for i=1:1:gridpoints
yt(i) = (Thmax/0.2)*(a0*x(i)^0.5 + a1*x(i) + a2*x(i)^2 + a3*x(i)^3 + a4*x(i)^4);
end
% Upper surface points
xu = ones(gridpoints,1);
yu = ones(gridpoints,1);
for i= 1:gridpoints
xu(i) = x(i) - yt(i)*sin(theta(i));
yu(i) = yc(i) + yt(i)*cos(theta(i));
end
% Lower surface points
xl = ones(gridpoints,1);
yl = ones(gridpoints,1);
for i=1:1:gridpoints
xl(i) = x(i) + yt(i)*sin(theta(i));
yl(i) = yc(i) - yt(i)*cos(theta(i));
end
%PLOT
f1 = figure(1);
hold on; grid on;
axis equal
plot(xu,yu,'r';
plot(xl,yl,'b');
well! The first problem is you've forgotten the closing parenthesis in the second line from the end. But if you want me to check your code, I need to have your typeNACA() function .m file to revise it. Or u can post its code here.
%PLOT
f1 = figure(1);
hold on; grid on;
axis equal
plot(xu,yu,'r';
plot(xl,yl,'b');
Related
I have a problem with my MATLAB code to display a figure of a Bifurcation diagram of discrete SIR model.
My model is:
S(n+1) = S(n) - h*(0.01+beta*S(n)*I(n)+d*S(n)-gamma*R(n))
I(n+1) = I(n) + h*beta*S(n)*I(n)-h*(d+r)*I(n)
R(n+1) = R(n) + h*(r*I(n)-gamma*R(n));
I tried out the code below but it keeps MATLAB busy for almost 30 mins and showed up no figure.
MATLAB code:
close all;
clear all;
clc;
%Model parameters
beta = 1/300;
gamma = 1/100;
D = 30; % Simulate for D days
N_t = floor(D*24/0.1); % Corresponding no of hours
d = 0.001;
r = 0.07;
%Time parameters
dt = 0.01;
N = 10000;
%Set-up figure and axes
figure;
ax(1) = subplot(2,1,1);
hold on
xlabel ('h');
ylabel ('S');
ax(2) = subplot(2,1,2);
hold on
xlabel ('h');
ylabel ('I');
%Main loop
for h = 2:0.01:3
S = zeros(N,1);
I = zeros(N,1);
R = zeros(N,1);
S(1) = 8;
I(1) = 5;
R(1) = 0;
for n = 1:N_t
S(n+1) = S(n) - h*(0.01+beta*S(n)*I(n)+d*S(n)-gamma*R(n));
I(n+1) = I(n) + h*beta*S(n)*I(n)-h*(d+r)*I(n);
R(n+1) = R(n) + h*(r*I(n)-gamma*R(n));
end
plot(ax(1),h,S,'color','blue','marker','.');
plot(ax(2),h,I,'color','blue','marker','.');
end
Any suggestions?
It was very slow because you were plotting a single value of h versus a vector S which had 7200 points. I assumed that you only want to plot the last value of S versus h. So replacing S with S(end) in the plot command changes everything. You really didn't need to use hold and it's better call plot once for each axis, so here is how I would do it:
beta = 1/300;
gamma = 1/100;
D = 30; % Simulate for D days
N_t = floor(D*24/0.1); % Corresponding no of hours
d = 0.001;
r = 0.07;
%%Time parameters
dt = 0.01;
N = 10000;
%%Main loop
h = 2:0.01:3;
S_end = zeros(size(h));
I_end = zeros(size(h));
for idx = 1:length(h)
S = zeros(N_t,1);
I = zeros(N_t,1);
R = zeros(N_t,1);
S(1) = 8;
I(1) = 5;
R(1) = 0;
for n=1:(N_t - 1)
S(n+1) = S(n) - h(idx)*(0.01+beta*S(n)*I(n)+d*S(n)-gamma*R(n));
I(n+1) = I(n) + h(idx)*beta*S(n)*I(n) - h(idx)*(d+r)*I(n);
R(n+1) = R(n) + h(idx)*(r*I(n)-gamma*R(n));
end
S_end(idx) = S(end);
I_end(idx) = I(end);
end
figure(1)
subplot(2,1,1);
plot(h,S_end,'color','blue','marker','.');
xlabel ('h');
ylabel ('S');
subplot(2,1,2);
plot(h,I_end,'color','blue','marker','.');xlabel ('h');
xlabel ('h');
ylabel ('I');
This now runs in 0.2 seconds on my computer.
I have a code that creates the correct xy plot for elastic pendulum with spring. I would like to show an animation of the elastic spring pendulum on an xy plot as the system marches forward in time. How can this be done?
Here is my simulation code:
clc
clear all;
%Define parameters
global M K L g;
M = 1;
K = 25.6;
L = 1;
g = 9.8;
% define initial values for theta, thetad, del, deld
theta_0 = 0;
thetad_0 = .5;
del_0 = 1;
deld_0 = 0;
initialValues = [theta_0, thetad_0, del_0, deld_0];
% Set a timespan
t_initial = 0;
t_final = 36;
dt = .1;
N = (t_final - t_initial)/dt;
timeSpan = linspace(t_final, t_initial, N);
% Run ode45 to get z (theta, thetad, del, deld)
[t, z] = ode45(#OdeFunHndlSpngPdlmSym, timeSpan, initialValues);
% initialize empty column vectors for theta, thetad, del, deld
M_loop = zeros(N, 1);
L_loop = zeros(N, 1);
theta = zeros(N, 1);
thetad = zeros(N, 1);
del = zeros(N, 1);
deld = zeros(N, 1);
T = zeros(N, 1);
x = zeros(N, 1);
y = zeros(N, 1);
% Assign values for variables (theta, thetad, del, deld)
for i = 1:N
M_loop(i) = M;
L_loop(i) = L;
theta(i) = z(i, 1);
thetad(i) = z(i, 2);
del(i) = z(i, 3);
deld(i) = z(i, 4);
T(i) = (M*(thetad(i)^2*(L + del(i))^2 + deld(i)^2))/2;
V(i) = (K*del(i)^2)/2 + M*g*(L - cos(theta(i))*(L + del(i)));
E(i) = T(i) + V(i);
x(i) = (L + del(i))*sin(theta(i));
y(i) = -(L + del(i))*cos(theta(i));
end
figure(1)
plot(x, y,'r');
title('XY Plot');
xlabel('x position');
ylabel('y position');
Here is my function code:
function dz = OdeFunHndlSpngPdlmSym(~, z)
% Define Global Parameters
global M K L g
% Take output from SymDevFElSpringPdlm.m file for fy1 and fy2 and
% substitute into z2 and z4 respectively
%fy1=thetadd=z(2)= -(M*g*sin(z1)*(L + z3) + M*z2*z4*(2*L + 2*z3))/(M*(L + z3)^2)
%fy2=deldd=z(4)=((M*(2*L + 2*z3)*z2^2)/2 - K*z3 + M*g*cos(z1))/M
% return column vector [thetad; thetadd; deld; deldd]
dz = [z(2);
-(M*g*sin(z(1))*(L + z(3)) + M*z(2)*z(4)*(2*L + 2*z(3)))/(M*(L + z(3))^2);
z(4);
((M*(2*L + 2*z(3))*z(2)^2)/2 - K*z(3) + M*g*cos(z(1)))/M];
You can "simulate" animation in a plot with a continous updating FOR loop and assignig graphic object to variables.
Something like (I assume only to use x,y as arrays function of time array t)
%In order to block the axis and preventing continuous zoom, choose proper axes limit
x_lim = 100; %these values depends on you, these are examples
y_lim = 100;
axis equal
axis([-x_lim x_lim -y_lim y_lim]); %now x and y axes are fixed from -100 to 100
ax = gca;
for i=1:length(t)
if i > 1
delete(P);
delete(L);
end
P = plot(x(i),y(i)); %draw the point
hold on
L = quiver(0,0,x(i),y(i)); %draw a vector from 0,0 to the point
hold on
%other drawings
drawnow
pause(0.1) %pause in seconds needed to simulate animaton.
end
"Hold on" instruction after every plot instruction.
This is only a basic animation, of course.
I am trying to use Matlab's ifftn in 3-dimensions to get solution in physical space. In particular I am trying to use ifftn on 1/k^2. The analytical solution to that in physical space is 1/(4*pi*r). However I am not recovering that. Please note: $r = sqrt(x^2 + y^2 + z^2)$ and $k = sqrt(kx^2 + ky^2 + kz^2)$.
clc; clear;
n = 128; % no. of points for ifft
L = 2*pi; % size of the periodic domain
x = linspace(-L/2,L/2,n); y = x; z = x; % creating vectors for x-y-z direction
[X,Y,Z] = meshgrid(x,y,z); % creating meshgrid for physical space
R = sqrt(X.^2 + Y.^2 + Z.^2); % use for 1/(4*pi*r)
k = (2*pi/L)*[0:n/2 -n/2+1:-1]; % wave vector;
[k1,k2,k3] = meshgrid(k,k,k);
denom = (k1.^2 + k2.^2 + k3.^2); % This is k^2
F = 1./denom; F(1,1,1) = 0; % The first value is set to zero as it is infinite
phi = 1./(4*pi*R); % physical domain solution
phys = fftshift(ifftn(F)); % Using ifftn
ph_abs = abs(phys);
mid = ph_abs(n/2,:,:); % looking at the midplane of the output
mid = permute(mid,[3 2 1]); % permuting for contourplot.
PHI = phi(n/2,:,:); %looking at the midplane of the physical space.
PHI = permute(PHI,[3 2 1]);
figure(1)
surf(x,z,log(mid))
shading flat
colorbar();
figure(2)
surf(x,z,log10(abs(PHI)))
shading flat
colorbar();
I am plotting the azimuth and elevation of some satellites, where the color of each trajectory represents the S4 index, from low (blue) to high (red). I would like however, to be able to format the colors and corresponding values, so that more of the lower effects of scintillation can actually be distinguished. This is because the high end of scintillation (red) only shows one or two points. Here is a picture of the trajectory and the code.
clc; close all; clear all
load combo_323_full
circular_plot
x = [];
y = [];
for s = 1:samples
% plot each satellite location for that sample
for sv = 1:sats
% check if positive or negative elevation
if (elevation((s - 1) * sats + sv) < 0)
elNeg = 1;
else
elNeg = 0;
end
% convert to plottable cartesian coordinates
el = elevation((s - 1) * sats + sv);
az = azimuth((s - 1) * sats + sv);
x = [x;(pi/2-abs(el))/(pi/2).*cos(az-pi/2)];
y = [y;-1*(pi/2-abs(el))/(pi/2).*sin(az-pi/2)];
% check for final sample
% if (s == samples)
% plot(x,y,'r*');
% text(x,y+.07,int2str(SVs(sv)), ...
% 'horizontalalignment', ...
% 'center','color','r');
% else
% check for +/- elevation
% if (elNeg == 0)
% plot(x,y,'.','color',rgb('DarkBlue'));
% else
% plot(x,y,'g.');
% % end
% end
end
end
z = combo(:,5);
for j = 1:10
% hold on
% circular_plot
lRef = length(x);
l1 = floor(lRef*(1/100));
l2 = floor(l1*j);
x_time = x(1:l2);
y_time = y(1:l2);
zr = z(1:l2);
navConstants;
% find out from 'plotMat' if plotting satellite locations or trajectories in
% addition determine how many satellites are being tracked and how many
% samples for each satellite (# samples / satellite must always be equal)
gpsTime = combo(1,2);
i = 1;
t = gpsTime;
while ((i ~= size(combo,1)) & (t == gpsTime))
i = i + 1;
t = combo(i,2);
end
if (t == gpsTime)
sats = i;
else
sats = i - 1;
end;
samples = size(combo,1) / sats;
SVs = combo(1:sats,1);
elevation = combo(:,20).*pi/180;
azimuth = combo(:,19).*pi/180;
% initialize polar - plotting area
figure(j);
axis([-1.4 1.4 -1.1 1.1]);
axis('off');
axis(axis);
hold on;
% plot circular axis and labels
th = 0:pi/50:2*pi;
x_c = [ cos(th) .67.*cos(th) .33.*cos(th) ];
y_c = [ sin(th) .67.*sin(th) .33.*sin(th) ];
plot(x_c,y_c,'color','w');
text(1.1,0,'90','horizontalalignment','center');
text(0,1.1,'0','horizontalalignment','center');
text(-1.1,0,'270','horizontalalignment','center');
text(0,-1.1,'180','horizontalalignment','center');
% plot spoke axis and labels
th = (1:6)*2*pi/12;
x_c = [ -cos(th); cos(th) ];
y_c = [ -sin(th); sin(th) ];
plot(x_c,y_c,'color','w');
text(-.46,.93,'0','horizontalalignment','center');
text(-.30,.66,'30','horizontalalignment','center');
text(-.13,.36,'60','horizontalalignment','center');
text(.04,.07,'90','horizontalalignment','center');
scatter(x_time,y_time,3,zr)
colorbar
axis equal
end
You can make your own colormap, it's just a N by 3 matrix where the columns are the red, green and blue components respectively.
The default colormap is jet. If you type for instance
>>> jet(16)
you will get a 16 by 3 matrix and you can see how it is made.
Then use colormap(your_own_colormap) to change it.
I'm trying to model projectile motion with air resistance.
This is my code:
function [ time , x , y ] = shellflightsimulator(m,D,Ve,Cd,ElAng)
% input parameters are:
% m mass of shell, kg
% D caliber (diameter)
% Ve escape velocity (initial velocity of trajectory)
% Cd drag coefficient
% ElAng angle in RADIANS
A = pi.*(D./2).^2; % m^2, shells cross-sectional area (area of circle)
rho = 1.2 ; % kg/m^3, density of air at ground level
h0 = 6800; % meters, height at which density drops by factor of 2
g = 9.8; % m/s^2, gravity
dt = .1; % time step
% define initial conditions
x0 = 0; % m
y0 = 0; % m
vx0 = Ve.*cos(ElAng); % m/s
vy0 = Ve.*sin(ElAng); % m/s
N = 100; % iterations
% define data array
x = zeros(1,N + 1); % x-position,
x(1) = x0;
y = zeros(1,N + 1); % y-position,
y(1) = y0;
vx = zeros(1,N + 1); % x-velocity,
vx(1) = vx0;
vy = zeros(1,N + 1); % y-velocity,
vy(1) = vy0;
i = 1;
j = 1;
while i < N
ax = -Cd*.5*rho*A*(vx(i)^2 + vy(i)^2)/m*cos(ElAng); % acceleration in x
vx(i+1) = vx(i) + ax*dt; % Find new x velocity
x(i+1) = x(i) + vx(i)*dt + .5*ax*dt^2; % Find new x position
ay = -g - Cd*.5*rho*A*(vx(i)^2 + vy(i)^2)/m*sin(ElAng); % acceleration in y
vy(i+1) = vy(i) + ay*dt; % Find new y velocity
y(i+1) = y(i) + vy(i)*dt + .5*ay*dt^2; % Find new y position
if y(i+1) < 0 % stops when projectile reaches the ground
i = N;
j = j+1;
else
i = i+1;
j = j+1;
end
end
plot(x,y,'r-')
end
This is what I am putting into Matlab:
shellflightsimulator(94,.238,1600,.8,10*pi/180)
This yields a strange plot, rather than a parabola. Also it appears the positions are negative values. NOTE: ElAng is in radians!
What am I doing wrong? Thanks!
You have your vx and vy incorrect... vx= ve*sin(angle in radians) and opposite for vy. U also u do not need a dot in between ur initial velocity and the *... That is only used for element by element multiplication and initial velocity is a constant variable. However the dot multiplier will not change the answer, it just isn't necessary..