关于在2D游戏中编写重力代码的问题

Question

我是初学者，正在按照教程在 android studio 中编写 flappy bird。我对下面的代码有 2 个问题。第一帧鸟落下 10 像素（=重力）。然后帧数乘以每帧 10 个像素，因此他每帧下降得更快。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顺利？我希望它看起来更刺耳，因为鸟每帧移动了很多像素。

    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

满鸟class:

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delta time渲染第一帧和第二帧的渲染时间差。在 1 秒内，此渲染方法执行了 60 次（即每秒帧数）。

因此，您应该将 delta 乘以您的速度以平滑移动。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

关于在2D游戏中编写重力代码的问题

Question about writing gravity code in a 2D game

java

android-studio

libgdx

game-physics

2d-games