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Topic: Help: Springs, instability, and explosions! (Read 2156 times) |
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mflux
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Re: Help: Springs, instability, and explosions!
« Reply #15 on: Mar 4th, 2005, 11:08am » |
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Hey guys!
Thanks for all your guidance! I think I got it.
It's not Runge-Kutta "as described", but I took the concept and tried it myself and it more or less works!
My spring sets now have a MUCH higher threshold for density than before (on the order of about 10 times..)
Check it out:
http://users.design.ucla.edu/~mflux/p5/phenome/applet/
Press C to cycle through random spring sets.
The old version would blow up as soon as a high density spring set is created. Now it's quite stable, with the exception of a few monstrous spring sets...
My semi-Runge-Kutta code is pretty simple. Here is the Euler version:
Code:
position displace(float angle,float magnitude) //returns the position displaced by an angle and distance
{
float ra=radians(angle);
position newP=new position(x,y);
newP.x+=(cos(ra)*magnitude);
newP.y-=(sin(ra)*magnitude);
return newP;
}
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Here is the semi-Runge-Kutta version:
Code:
position displaceRK(vector v)
{
position newP=new position(x,y);
position k[]=new position[int(rkIterations)];
k[0]=new position(newP);
for(int i=1;i<rkIterations;i++)
{
k[i]=new position(k[i-1].displace(v.a,v.m/rkIterations));
}
float kx=0;
float ky=0;
for(int x=0;x<rkIterations;x++)
{
kx+=2*k[x].x;
ky+=2*k[x].y;
}
kx=kx/rkIterations/2;
ky=ky/rkIterations/2;
vector vec=new vector(newP,new position(kx,ky));
vec.m*=rkIterations/rkIterations*2;
newP=newP.displace(vec.a,vec.m);
return newP;
}
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It allows for multiple orders of Runge-Kutta. I think...
The difference is that, from all my reading about Runge-Kutta, they have numerical weights to the middle numbers. So for example...
In Runge-Kutta you would have
p1 (initial point), 2*p2 (midpoint solved from p1), 2*p3 (midpoint solved from p2), and p4 (the endpoint).
Next, you add these together and divide...
(p1+2*p2+2*p3*p4)/6
At least that's what's called a "fourth order Runge-Kutta". Described better here for math geeks:
http://mathworld.wolfram.com/Runge-KuttaMethod.html
However... the results I get from that are erratic. The whole idea of averaging these numbers out is so that the last point (where p4 is) would have a minimal amount of numerical error. Wouldn't multiplying 2 to the midpoints, in addition to having TWO midpoints when averaging, get you something that's NOT P4?
That's something I don't really understand about the descriptions of Runge-Kutta. At any rate, the technique I'm using is more-or-less working. If anyone wants to enlighten me further on Runge-Kutta please do so because I'm very curious now...
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| « Last Edit: Mar 4th, 2005, 11:15am by mflux » |
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mflux
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Re: Help: Springs, instability, and explosions!
« Reply #16 on: Mar 4th, 2005, 11:54am » |
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Ah nevermind. It works, but I was being stupid.
My code was actually decreasing spring tension, thus giving the illusion that it's working. I thought the slow-down was due to calculation but it was not.
It now actually does Runge-Kutta, by the book!
Woot: http://users.design.ucla.edu/~mflux/p5/phenome/applet/
Code:
position displaceRK(vector v)
{
position newP=new position(x,y);
position k1=newP;
position k2=k1.displace(v.a,v.m/2);
position k3=k2.displace(v.a,v.m/2);
position k4=k3.displace(v.a,v.m/2);
float kx=(k1.x+2*k2.x+2*k3.x+k4.x)/6;
float ky=(k1.y+2*k2.y+2*k3.y+k4.y)/6;
vector vec=new vector(newP,new position(kx,ky));
newP=newP.displace(vec.a,vec.m);
return newP;
}
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I can't believe how easy that was. I'm smacking myself in the head right now. Ofcourse now it's possible to write an N-order Runge-Kutta...
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| « Last Edit: Mar 4th, 2005, 12:33pm by mflux » |
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Quasimondo
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Re: Help: Springs, instability, and explosions!
« Reply #17 on: Mar 4th, 2005, 6:36pm » |
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Bravo - great work! And isn't it amazing how the code looks immediately more attractive?
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Mario Klingemann | Quasimondo | Incubator | côdeazur
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