我想在这些步骤中求解这些方程
q1*q3*k/(r^2)= F1_3(这是1和3之间的电场力)
q2*q3*k/r^2)= f12(质点1和质点2之间的力)
从这里我可以得到两个电荷之间的合力f1_3+f1_2=fnet
与净力i,然后用a=fnet/m (m为质量)找到加速度。(现在上面的一切我都可以做,但现在我感到困惑)
取刚求出的加速度并求出速度。v=-(at)为时间间隔(我从第6步的方程中推导出这个方程,初始方程为x(0.05)=at+v
取该速度和先前的加速度并找到新的位置:x=1/2at^2+v*t+x
x的值成为粒子3的新位置,现在我回到顶部以计算电力,然后加速度等冲洗并重复。
这些步骤就是我想要一遍又一遍地重复的,现在我的间隔,我想要它是1微秒或10e-6,所以从0.0-1.0微秒0.0-2.0微秒,我想要它在每个间隔存储位置。问题是,它必须重新计算力的值,使用我在前几步中列出的公式,然后再回去找到位置。我不知道是否应该用for来循环,但我不知道该怎么做。我在顶部有扫描方法,因为稍后当我完成代码时,我希望能够输入电荷的位置和它们的大小,然后它将使用方程,但是一次一步。
我想做的是不可能的吗?
import java.util.Scanner;
import javax.swing.JFrame;
public class Firstgui {
private static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
//declare variables
double postionq1= 0.0; // position of q1 is at origin i just put it here for reference
double distanceq3_q1=.01; //q3 is placed between q1 and q2
double distanceq3_q2= .014-distanceq3_q1; //distance from q3 to q2
double q1=5e-6;
double q2=-4e-6;
double q3=-2e-6;
double mq1=3e-5;
double k= 8.99e9;
double F1_3 = Force(q1, q3, k, distanceq3_q1);
double F2_3= -Force(q2, q3, k, distanceq3_q2);
double Fnet=F1_3+F2_3;
System.out.println(F1_3);
System.out.println(F2_3);
System.out.println(Fnet);
System.out.println("particle 3 position from 0.0-1micro-seconds is " + position(acceleration(Fnet,mq1), 0.000001 , velocity( 0.000001 , acceleration(Fnet,mq1)),.01));
// this print line above is the final the position of q3 a 1 microsecond
//now with that value that it prints how would i use that for the next
//position of q3 and recalculate the fnet then acceleration etc.
}
public static double Force(double q1, double q2, double k, double r) {
double electricforce=(q1*q2*k)/(Math.pow(r, 2));
return electricforce;
}
public static double acceleration(double f, double m) {
double acell=f/m;
return acell;
}
public static double position(double a, double t, double v, double x ) {
double postion=.5*a*(t*t)+(v*t)+x; // a- acceleration through out out this time interval is constant
return postion;
}
public static double velocity(double t, double a) {
double v=-(t*a); // a- acceleration through out out this time interval is constant
return v;
}
}
是的,这在物理模拟中很常见。通常你定义一个时间步长是如何产生的,然后你在整个时间内循环。对于涉及粒子的模拟,通常需要存储位置和速度来计算运动。
public static void main(String[] args) {
int nbrSteps = 1000;
// Store the data
PointArray points = // The data would look like this. Particle at index 0 corresponds with the data {x0, y0, z0}
0: {x0, y0, z0},
1: {x1, y1, z1},
...
VelocityArray velocities = // And here, very similarly
0: {vx0, vy0, vz0},
1: {vx1, vy1, vz1},
...
for (int i = 0; i < nbrSteps; i++) {
takeStep(points, velocities);
// You probably want to record the simulation somehow. Either you save your
// data to disk, or you render it on the screen. Both calls could be put here.
}
}
public static void takeStep(PointArray points, VelocityArray velocities) {
for (int i = 0; i < points.length(); i++) {
const double timestep = 1e-6;
// Here you implement the steps you described, and update position and velocity accordingly
velocities[i] = ...
points[i] = ...
}
}
编辑:作为一个例子,我们可以模拟重力。我们可以用这个
来改变takeStep
public static void takeStep(PointArray points, VelocityArray velocities) {
for (int i = 0; i < points.length(); i++) {
const double timestep = 1e-6;
// Doing something simple, lets simulate gravity
const double gravity = -9.82;
velocities[i].y += f_gravity*timestep;
points[i] = velocity[i]*timestep;
}
}
对于你的情况,你需要计算每个粒子的力(我们不需要,因为重力的值已经知道了),然后应用它类似于我用线性积分的方法。这意味着
velocity = force*change_in_time
position = velocity*change_in_time
注意这里force
, velocity
和position
是矢量