我正在尝试使用P5.js库在javascript中制作简单的动画.我想让一个球出现在某个高度,然后让它掉下来使其弹跳直至停止.
我不是在寻找外部库,只是P5.
我的代码是这样的:
function Ball() {
this.diameter = 50;
this.v_speed = 5;
this.ypos = height/2 - 100;
this.xpos = width/2;
this.update = function(){
this.ypos = this.ypos + this.v_speed;
this.ypos = constrain(this.ypos,this.diameter/2,height-this.diameter/2);
}
this.show = function(){
ellipse(this.xpos,this.ypos,this.diameter);
fill(255);
}
}
var ball;
function setup() {
createCanvas(600,600);
ball = new Ball();
}
function draw() {
background(0);
ball.update();
ball.show();
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.6.1/p5.js"></script>
有人能帮我吗?非常感谢
最佳答案
首先,您必须将起始速度设置为0并定义重力:
this.v_speed = 0;
this.gravity = 0.2;
可以直接应用于您的示例的有效更新方法如下所示:
this.starty = height/2 - 100;
this.endy = height-this.diameter/2;
this.update = function(){
this.v_speed = this.v_speed + this.gravity;
this.ypos = this.ypos + this.v_speed;
if (this.ypos >= this.endy){
this.ypos = this.endy;
this.v_speed *= -1.0; // change direction
this.v_speed = this.v_speed*0.9;
if ( Math.abs(this.v_speed) < 0.5 ) {
this.ypos = this.starty;
}
}
}
关键是当球在地面上反弹时降低速度并改变方向:
this.v_speed *= -1.0;
this.v_speed = this.v_speed*0.9;
另请参见Bouncing Balls,struggling with getting more than 1 (processing).
请参见示例,其中我将建议应用于原始代码:
function Ball() {
this.diameter = 50;
this.v_speed = 0;
this.gravity = 0.2;
this.starty = height/2 - 100;
this.endy = height-this.diameter/2;
this.ypos = this.starty;
this.xpos = width/2;
this.update = function(){
this.v_speed = this.v_speed + this.gravity;
this.ypos = this.ypos + this.v_speed;
if (this.ypos >= this.endy){
this.ypos = this.endy;
this.v_speed *= -1.0; // change direction
this.v_speed = this.v_speed*0.9;
if ( Math.abs(this.v_speed) < 0.5 ) {
this.ypos = this.starty;
}
}
}
this.show = function(){
ellipse(this.xpos,this.diameter);
fill(255);
}
}
var ball;
function setup() {
createCanvas(600,600);
ball = new Ball();
}
function draw() {
background(0);
ball.update();
ball.show();
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.7.3/p5.js"></script>