关于在2D游戏中编写重力代码的问题

Question

我是一个初学者，正在学习android studio中编写flappy bird的教程。关于下面的代码，我有两个问题。第一帧，鸟下降了10个像素(=重力)。然后将帧的数量乘以每帧的10个像素，这样他每帧下降的速度就会更快。但是velocity.scl(1/dt)有什么用呢？也有可能我理解错了什么？为什么下落看起来很平滑？我希望它看起来更刺耳，因为鸟每一帧都移动了很多像素。

    if(position.y>0){
        velocity.add(0, GRAVITY, 0); // bird falls 15 pixels above ground
    }

    velocity.scl(dt); //multiply gravity with dt (frame) -> gravity gets larger
    position.add(0, velocity.y, 0);

    if(position.y<0){
        position.y=0; // bird doesn't fall through ground
    }

    velocity.scl(1/dt); // what does this do

Full Bird类：

private static final int GRAVITY = -10;
private Vector3 position;
private Vector3 velocity;
private Texture bird;

public Bird(int x, int y){
    position = new Vector3(x,y,0);
    velocity=new Vector3(0,0,0);
    bird = new Texture("Flappy-midflap.png");
}

public void update(float dt){
    if(position.y>0){
        velocity.add(0, GRAVITY, 0); // bird falls 15 pixels above ground
    }

    velocity.scl(dt); //multiply gravity with dt (frame) -> gravity gets larger
    position.add(0, velocity.y, 0);

    if(position.y<0){
        position.y=0; // bird doesn't fall through ground
    }

    velocity.scl(1/dt); // what does this do
}

public void jump(){
    velocity.y = 250;
}

public Vector3 getPosition() {
    return position;
}

public Texture getTexture() {
    return bird;
}

Answer 1

首先，dt是渲染第一帧和第二帧的渲染时间差。在1秒内，这个渲染方法大约执行60次(也就是每秒的帧数)。

因此，你应该用你的速度乘以增量来平滑移动。示例

First render loop:
Velocity-Y: 250px
dt: 0.018
Result: 4.5px

Second render loop:
Velocity-Y: 240px
dt: 0.025
Result: 4 px

In result this will become 
250 px in 1 second.

If you do not use scale this will become 
First Render Loop: 
Velocity-Y: 250px
dt: 0.018: 
Result: 250px

Second Render Loop:
Velocity-Y: 240px
dt: 0.025
Result: 240px

In result there will be 250 + 240 + .... 10 + 0 px for 1 second