Leapfrog integration with multiple particles

3 次查看（过去 30 天）
AFHG
AFHG 2018-9-8
评论： Star Strider 2018-9-8
Hi all i'm new to programming and I wanted to ask a quick question: Consider the following leapfrog integration for one particle:
time=100;
steps=1000000;
dt=time/(steps-1);
v0=5
theta=2*pi*rand(1);
vx(1)=v0*sin(theta);
vy(1)=v0*cos(theta);
t(1)=0;
x(1)=50;
y(1)=50;
for i=1:steps-1
t(i+1)=t(i)+dt;
x(i+1)=x(i)+vx(1)*dt;
y(i+1)=y(i)+vy(1)*dt;
end
How do I increase the number of particles in the leapfrog integration? eg 1000? Thanks in advance!
Some info (edited after request): Ok so, basically the particle starts at time t=0, at position x,y and at velocity vx,vy.
The simulation lasts for 100 seconds and consists in a million step. Leapfrog integration, in this case, consists in the for loop that updates the particle position every 100/1000000.
The problem is that i dont know how to scale this for loop for multiple particles. Thanks!
  2 个评论
Geoff Hayes
Geoff Hayes 2018-9-8
Frederico - how is the single particle used in the above code? I guess some background on leapfrog integration would help. :)
AFHG
AFHG 2018-9-8
编辑：AFHG 2018-9-8
Ok so, basically the particle starts at time t=0, at position x,y and at velocity vx,vy.
The simulation lasts for 100 seconds and consists in a million step. Leapfrog integration, in this case, consists in the for loop that updates the particle position every 100/1000000.
The problem is that i dont know how to scale this for loop for multiple particles. Thanks!

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2018-9-8
编辑：Star Strider 2018-9-8
This works (with a much smaller value for ‘steps’):
time=100;
steps=1000000;
% steps = 100;
dt=time/(steps-1);
v0=5;
NrParticles = 1000;
theta=2*pi*rand(1,NrParticles);
vx=v0*sin(theta);
vy=v0*cos(theta);
t(1)=0;
x(1,1:NrParticles)=50;
y(1,1:NrParticles)=50;
for k = 1:NrParticles
for i=1:steps-1
t(i+1)=t(i)+dt;
x(i+1,k+1)=x(i,k)+vx(i)*dt;
y(i+1,k+1)=y(i,k)+vy(i)*dt;
end
end
figure
hold all
for k1 = 1:NrParticles
plot(x(:,k1), y(:,k1))
end
hold off
You will likely have to experiment with your code. I changed it slightly to use different values for ‘vx’ and ‘vy’ in the loop.
EDIT — Corrected error in ‘k’ loop.
  12 个评论
显示 10更早的评论
AFHG
AFHG 2018-9-8
This was the correct loop, btw:
e
for k = 1:NrParticles
for i=1:steps-1
t(i+1)=t(i)+dt;
x(i+1,k)=x(i,k)+vx(1,k)*dt;
y(i+1,k)=y(i,k)+vy(1,k)*dt;
end
end
Star Strider
Star Strider 2018-9-8
As always, my pleasure.
I was not certain how to scale up ‘vx’ and ‘vy’.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Programming 的更多信息

标签

产品


版本

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by