Java 如何循环和存储物理方程中的值
所以我想在这些步骤中做这些方程Java 如何循环和存储物理方程中的值,java,loops,for-loop,physics,Java,Loops,For Loop,Physics,所以我想在这些步骤中做这些方程 q1*q3*k/(r^2)=F1_3(这是介于1和3之间的电力) q2*q3*k/r^2)=F1_2(粒子1和2之间的力) 从那里我可以找到两个电荷之间的净作用力,f1_3+f1_2=fnet 对于净力,我会用a=fnet/m(m是质量)来求加速度。(现在上面的一切我都可以做,但现在我弄糊涂了) 取刚得到的加速度,求出速度。v=-(at)是时间间隔(我从第6步的方程中得到了方程aver,初始方程是x(.05)=at+v 取那个速度和先前的加速度,找到新的位置:x=
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,这在物理模拟中非常常见。通常你定义一个时间步的制作方式,然后在整个时间内循环。对于涉及粒子的模拟,你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] = ...
}
}
在你的例子中,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),然后用类似于我用线性积分的方法来应用它
velocity = force*change_in_time
position = velocity*change_in_time
请注意,
force
、velocity
和position
在我看来是可能的。请详细说明您的具体位置好吗?我以前从未真正使用过数组,所以我不确定您的意思是什么?我将在哪里实施我曾经想过的步骤?据我所知,您说我必须是否每年更新位置和速度?我真的不明白你想用这些代码做什么,或者我错过了什么?可能是示例吗?@Cosmik11添加了一个示例。
velocity = force*change_in_time
position = velocity*change_in_time