所以去年我寫了一個C程序,用於演示使用Lorenz方程和Runge Kutta方法求解微分方程的混沌。Visual Studio,在用戶點擊按鈕時繪製Lorenz混沌
我最近決定,我想重新考慮這個,並製作出能夠繪製出粒子軌跡的程序。我得到這個成功的工作,但現在想擴大這個,使用戶可以輸入參數,如粒子的起始位置,和其他參數(在我的情況下,a,b和r)。目前我的程序運行的時候軌跡就被繪製出來了,但是我想延遲這個軌跡直到用戶輸入他們的參數到一些文本框然後按下一個按鈕。要做到這一點,我想我應該創建一個新的類,並將我的當前代碼放到那裏,然後在btn1_Click方法下的主.cs文件中調用它。不過,我在這方面遇到了相當大的麻煩,主要還是不知道如何去做。在我迄今爲止的最佳嘗試中,我對包含「createGraphics()」的一行有錯誤,即在類文件中沒有定義它。我在頂部使用部分的頂部具有所有相同的引用,就像它在主要文件中工作正常的那樣。
此外,如果任何人都可以給我任何有關我的代碼(即任何不良行爲或我過於複雜的事情的地方)或任何建議,使其更好,我會非常感激,如果你需要任何更多我的幫助信息,然後我會盡我所能來回答!
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace Lorenz_chaos
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void Form1_Paint(object sender, PaintEventArgs e)
{
double a = 10, b = (8/3), r = 28; //standard values for lorenz model
/*m defines the number of iterations of the for loop so the number of lines drawn
good idea to keep m inversely proportional to dt (the time interval). A smaller dt will
mean smaller lines so smoother overall drawing m=50000 and dt=0.0005 is a good starting point
that demonstrates chaos well*/
double m = 500000, dt = 0.00005;
//EVOLUTION VALUE FOR RUNGE_KUTTA METHOD
//values for first particle
double y11, y12, y13;
double y21, y22, y23;
double y31, y32, y33;
double y41, y42, y43;
double y51, y52, y53;
double xi, yi, xf, yf; //coordinates for drawing particle 1 trajectory
double f10, f11, f12, f13; //function values to be calculated,
double f20, f21, f22, f23; //for fxy (x>1) these are intermediate fn calculations at different
double f30, f31, f32, f33; //times in Runga Kutta
//values for second particle
double z11, z12, z13;
double z21, z22, z23;
double z31, z32, z33;
double z41, z42, z43;
double z51, z52, z53;
double ai, bi, af, bf; //coordinates for drawing particle 2 trajectory (these are badly named...)
double g10, g11, g12, g13; //equivalent of f values for particle 2
double g20, g21, g22, g23;
double g30, g31, g32, g33;
//OTHER NEEDED QUANTITIES
int i; //for loop iteration integer
int k1 = 20; //scaling factors to make drawing fill form
int k2 = 9;
int y0 = 450; //offset values to centre drawing on form
int x0 = 550;
int start = 10; //starting position for calculations
double diff = 0.01;//initial displacement between two particles
//starting positions for particles
y11 = start;//particle 1
y12 = start;
y13 = start;
z11 = start + diff;//particle 2
z12 = start + diff;
z13 = start + diff;
//initial coords for particles at t=0
xi = (y11) * k1 + x0;
yi = (y12) * k2 + y0;
ai = (z11) * k1 + x0;
bi = (z12) * k2 + y0;
for (i = 1; i <= m; i++)
{
f10 = a * (y12 - y11);
f20 = r * y11 - y12 - y11 * y13;
f30 = y11 * y12 - b * y13;
y21 = y11 + f10 * dt/2; //finding y1 y2 y3 at the first
y22 = y12 + f20 * dt/2; //fraction of dt
y23 = y13 + f30 * dt/2;
f11 = a * (y22 - y21);
f21 = r * y21 - y22 - y21 * y23;
f31 = y21 * y22 - b * y23;
y31 = y11 + f11 * dt/2; //finding y1 y2 y3 at the second
y32 = y12 + f21 * dt/2; //fraction of dt
y33 = y13 + f31 * dt/2;
f12 = a * (y32 - y31);
f22 = r * y31 - y32 - y31 * y33;
f32 = y31 * y32 - b * y33;
y41 = y11 + f12 * dt; //finding y1 y2 y3 at the third
y42 = y12 + f22 * dt; //fraction of dt
y43 = y13 + f32 * dt;
f13 = a * (y42 - y41);
f23 = r * y41 - y42 - y41 * y43;
f33 = y41 * y42 - b * y43;
y51 = y11 + (f10 + 2 * f11 + 2 * f12 + f13) * dt/6; //final y values at y(t+dt)
y52 = y12 + (f20 + 2 * f21 + 2 * f22 + f23) * dt/6; //then to be repesated in for loop for all steps
y53 = y13 + (f30 + 2 * f31 + 2 * f32 + f33) * dt/6;
xf = (y51) * k1 + x0;
yf = (y52) * k2 + y0;
//second particle calculation
g10 = a * (z12 - z11);
g20 = r * z11 - z12 - z11 * z13;
g30 = z11 * z12 - b * z13;
z21 = z11 + g10 * dt/2; //finding y1 y2 y3 at the first
z22 = z12 + g20 * dt/2; //fraction of dt
z23 = z13 + g30 * dt/2;
g11 = a * (z22 - z21);
g21 = r * z21 - z22 - z21 * z23;
g31 = z21 * z22 - b * z23;
z31 = z11 + g11 * dt/2; //finding y1 y2 y3 at the second
z32 = z12 + g21 * dt/2; //fraction of dt
z33 = z13 + g31 * dt/2;
g12 = a * (z32 - z31);
g22 = r * z31 - z32 - z31 * z33;
g32 = z31 * z32 - b * z33;
z41 = z11 + g12 * dt; //finding y1 y2 y3 at the third
z42 = z12 + g22 * dt; //fraction of dt
z43 = z13 + g32 * dt;
g13 = a * (z42 - z41);
g23 = r * z41 - z42 - z41 * z43;
g33 = z41 * z42 - b * z43;
z51 = z11 + (g10 + 2 * g11 + 2 * g12 + g13) * dt/6; //final y values at y(t+dt)
z52 = z12 + (g20 + 2 * g21 + 2 * g22 + g23) * dt/6; //then to be repesated in for loop for all steps
z53 = z13 + (g30 + 2 * g31 + 2 * g32 + g33) * dt/6;
af = (z51) * k1 + x0;
bf = (z52) * k2 + y0;
//DRAWING LINE JUST CALCULATED
System.Drawing.Graphics graphicsObj;
graphicsObj = this.CreateGraphics();
Pen myPen = new Pen(System.Drawing.Color.Red, 1);
//myPen.DashStyle = System.Drawing.Drawing2D.DashStyle.DashDotDot;
graphicsObj.DrawLine(myPen, (int)xi, (int)yi, (int)xf, (int)yf);
myPen.Color = System.Drawing.Color.RoyalBlue;
graphicsObj.DrawLine(myPen, (int)ai, (int)bi, (int)af, (int)bf);
//REDEFINING COORDS AND VALUES FOR NEXT LOOP
//first particle
xi = (y51) * k1 + x0;
yi = (y52) * k2 + y0;
y11 = y51;
y12 = y52;
y13 = y53;
//second particle
ai = (z51) * k1 + x0;
bi = (z52) * k2 + y0;
z11 = z51;
z12 = z52;
z13 = z53;
/*even at 1 the below makes the program far too slow, need an alternative
intention was for it to allow user to see the particle trajectories better*/
//System.Threading.Thread.Sleep(1);
}
}
}
}
好吧這看起來不錯,我沒有得到一個問題(我認爲)該程序正在執行的事情的順序。所以我可以設置一切從三個文本框中讀取值,但是當我運行時,我只是得到一個大的紅十字,但是如果我只是將代碼中的值設置爲10,8/3和28,就像之前運行良好。另外由於某種原因,彈道在第一次完成後再次被拉伸,我不知道爲什麼因爲只有一條線可以繪製軌跡。從文本框中讀取我起訴引用和 aVal = double.Parse(aBox.Text); – Dmist
大的紅叉意味着繪圖時出現一些錯誤。你的解析是可以的。你不應該在paint方法中解析,而是在事件處理器之外解析。 –
刪除Thread.Sleep。只要提一下,你不能解析8/3。但是,你可以解析2.6666666 ... –