Question

我在我制作的一个小型2D游戏中使用了RK4物理，但是当涉及到引力时，所有对象都被拉到0,0我怎么能改变这个？（对不起，没有评论也没有时间评论它）

    #include "Gravity.h"

gravity::accelerationReturn gravity::acceleration(state & state , float time) //Returns a struct instead easier to put into a vector.
{
    accelerationReturn return1;
    const float k =9.8;
    const float b = 1;
    return1.Xaccel =  -k * state.pos.getX() - b* state.vel.getX();
    return1.Yaccel = -k * state.pos.getY() - b* state.vel.getY();
    return return1;
}


gravity::derivative gravity::evaluate(state  & initial, float time, float deltaTime,derivative & Derivative)
{
    state state;
    state.pos.setXY(initial.pos.getX() + Derivative.dpos.getX()*deltaTime,
        initial.pos.getY() - Derivative.dpos.getY()*deltaTime);

    state.vel.setXY(initial.vel.getX() + Derivative.dvel.getX() * deltaTime,
        initial.vel.getY() + Derivative.dvel.getY() * deltaTime);

        derivative output;
        output.dpos.setXY(state.vel.getX(),state.vel.getY());
        output.dvel.setXY(acceleration(state,time + deltaTime).Xaccel,
            acceleration(state,time + deltaTime).Yaccel);
        return output;

}

void gravity::integrate(state & State , float time, float deltaTime)
{

    derivative a = evaluate(State ,time,0.0f,derivative());
    derivative b = evaluate(State,time,deltaTime * 0.5f,a);
    derivative c = evaluate(State,time,deltaTime * 0.5f,b);
    derivative d = evaluate(State,time,deltaTime,c);

    Vector dpdt;
    Vector dvdt;
    dpdt.setXY((1.0f/10.0f * (a.dpos.getX() + 2.0f * (b.dpos.getX() + c.dpos.getX()) + d.dpos.getX())),
              (1.0f/10.0f * (a.dpos.getY() + 2.0f * (b.dpos.getY() + c.dpos.getY()) + d.dpos.getY())));
    dvdt.setXY((1.0f/10.0f * (a.dvel.getX() + 2.0f * (b.dvel.getX() + c.dvel.getX()) + d.dvel.getX())),
               ( 1.0f/10.0f * (a.dvel.getY() + 2.0f * (b.dvel.getY() + c.dvel.getY()) + d.dvel.getY())));


    State.pos.setXY ((State.pos.getX() + dpdt.getX() * deltaTime),(State.pos.getY() + dpdt.getY() * deltaTime));
    State.vel.setXY ((State.vel.getX() + dvdt.getX() * deltaTime),(State.vel.getY() + dvdt.getY() * deltaTime));

}

cpp

#ifndef GRAVITY_H
#define GRAVITY_H

#include "2DVector.h"


class gravity :Vector 
{
private:



    float time; //current time
    float deltaTime; // previous time
public:
    struct accelerationReturn //for returning acceleration
    {
        float Xaccel;
        float Yaccel;
    };
    struct state //current state of object.
    { 
        Vector pos; //position
        Vector vel; //velocity
    };
    struct derivative
    {
        Vector dpos; // derivative of posistion is velocity.
        Vector dvel; // derivative of velocity is acceleration.
    };

    accelerationReturn acceleration (state &,float); //calculates the new acceleration
    derivative evaluate ( state & , float , float ,  derivative &); //gets derived values
    void integrate ( state &, float , float);




};
#endif

主

#include "Gravity.h"
#include <iostream>
#include "SFML\Graphics.hpp"
#include "SFML\System.hpp"

using namespace std;

int main()
{


    gravity grav;
    gravity::state marrio;
    marrio.pos.setXY(500,500); 


    sf::RenderWindow mwindow(sf::VideoMode(800, 800), "my window");
    sf::CircleShape shape(50);
    shape.setFillColor(sf::Color(100,250,250));

    sf::Time timeC;
    float time;
    float dTime = 0.01f;

    sf::Clock timePhysics;
    while(mwindow.isOpen())
    {

               timeC = timePhysics.getElapsedTime();
               time = timeC.asSeconds();
               grav.integrate(marrio,time,dTime);
               time += dTime;
               cout<< "pos     vel"<<endl<<marrio.pos.getX()<<" "<<marrio.pos.getY()<<" "<<
                marrio.vel.getX()<<" "<<marrio.vel.getY()<<endl;
                shape.setPosition(marrio.pos.getX(),marrio.pos.getY());



                mwindow.clear();
                mwindow.draw(shape);
                mwindow.display();
    }

    system("PAUSE");
    return 0;
}

Answer 1

您在第一个代码块中执行的物理操作存在一个小问题。如果你试图模拟引力，你将需要使用牛顿引力定律计算加速度。你现在所拥有的是胡克定律;弹簧力，线性阻力。要立即回答有关如何将吸引力中心从点（0,0）移开的问题，您可以简单地更改包含偏移的内容。即，

return1.Xaccel = -k * (state.pos.getX() - xOffset) - b* state.vel.getX();
return1.Yaccel = -k * (state.pos.getY() - yOffset) - b* state.vel.getY();

其中xOffset和yOffset定义了所需吸引中心的（x，y）位置。这将允许你移动你的力源，但这仍然没有解决你的力是弹性的问题，而不是重力（逆平方律）。要做到这一点，您将需要类似于以下内容的内容

float dx = state.pos.getX() - xOffset; // again xOffset is the source location
float dy = state.pos.getY() - yOffset;
float distSqrd = dx*dx + dy*dy;
float dist = sqrt( distSqrd );
return1.Xaccel = -G * mass * dx / (distSqrd * dist);
return1.Yaccel = -G * mass * dy / (distSqrd * dist);

请注意，为了考虑方向，我将F = G*m1*m2/dist^2乘以dx/dist和dy/dist。这相当于cos(angle)和sin(angle)，其中angle是未计算的最佳数量（指定力的方向）。 mass将是吸引粒子的物体的质量（？），因为我们只计算加速度，所以粒子质量被忽略了。如果你想将线性阻力加回到加速度中，你也可以轻松地做到这一点，这将使轨道更加稳定。

请注意，使用向量可以大大清除此代码。如果需要澄清，请告诉我。

RK4引力方向

1 个答案: