Animation help needed with
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Animation help needed
I'm trying to display 10 graphs in a sequence as an animation, running from t=1 to t=10. Here's my code,
close all;
clear all;
l=0.33;R=100;T=68;m=0.000125;w0=0.1;t=0;mu=m/l;c=sqrt(T/mu);r=1/2;x0=l*r;
for t = 1:10
for x=1:34;
for n=1:5000;
wxt=((4*w0)/pi)*exp(-R*(t*0.001))*((l/(pi*n^2*x0))...
*sin((n*pi*x0)/l))*sin((n*pi*((x-1)*0.01))/l)*cos((sqrt((c*n*pi/l)^2-R^2)*(t*0.001)));
wxtrec(n)=wxt;
end
w=sum(wxtrec);
wrec(x)=w;
end
end
plot(1:34,wrec(t+10))
axis equal
M(t) = getframe
%plot(1:34,wrec)
The bottom plot works by its self, displaying the graph of t=10, but I can't seem to make the animation work.
Any help would be really appreciated.
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Weitere Antworten (4)
Tom
am 18 Jan. 2012
1 Stimme
1 Kommentar
Chandra Kurniawan
am 18 Jan. 2012
Hi,
You can write :
axis ([0 35 -0.05 0.05]);
before line
plot(1:34,wrec)
inside the loop.
Or you can also set the axis unvisible inside the loop by
axis off
Tom
am 18 Jan. 2012
1 Stimme
Chandra Kurniawan
am 18 Jan. 2012
l=0.33;R=100;T=68;m=0.000125;w0=0.1;t=0;mu=m/l;c=sqrt(T/mu);r=1/2;x0=l*r;
axis equal
for t = 1:10
for x=1:34;
for n=1:5000;
wxt=((4*w0)/pi)*exp(-R*(t*0.001))*((l/(pi*n^2*x0))...
*sin((n*pi*x0)/l))*sin((n*pi*((x-1)*0.01))/l)*cos((sqrt((c*n*pi/l)^2-R^2)*(t*0.001)));
wxtrec(n)=wxt;
end
w=sum(wxtrec);
wrec(x)=w;
end
plot(1:34,wrec(t+10));
F(t) = getframe;
end
movie(F)
owr
am 18 Jan. 2012
If you're not actually trying to make a movie, just visualize the animation within MATLAB, I'd highly recommend calling "plot" just once and capturing a handle to the plot object. You can then update this handle's "ydata" property within your loop without repeatedly calling plot. The result is that the animation will be much smoother and you'll be able to zoom, pan, rotate interactively during the animation without everything reseting. Here is a simple example - try rotating, etc. mid animation to see what Im talking about:
x = -1:0.01:1;
y = x.^2;
h = plot(x,y);
axis([-1,1,-1,1]);
for t = 1:500
set(h,'ydata',cos(t/10)*y);
pause(0.1);
end
Good luck!
1 Kommentar
David
am 25 Okt. 2013
Thanks for sharing this!
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