我有这样的代码来将画布中的图像从一个位置移动到另一个位置:
class Target {
constructor(img, x_init, y_init, img_width = 100, img_height = 100) {
this.img = img;
this.x = x_init;
this.y = y_init;
this.img_width = img_width;
this.img_height = img_height;
}
get position() {
return this.x
}
move(canvas, x_dest, y_dest) {
ctx = canvas.getContext('2d');
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.drawImage(this.img, this.x, this.y, this.img_width, this.img_height);
if (this.x != x_dest) {
if (this.x > x_dest) {
this.x -=1;
} else {
this.x +=1;
}
}
if (this.y != y_dest) {
if (this.y > y_dest) {
this.y -=1;
} else {
this.y +=1;
}
}
if (this.x != x_dest || this.y != y_dest) {
//setTimeout(this.move.bind(target, canvas, x_dest, y_dest), 0);
window.requestAnimationFrame(this.move.bind(target, canvas, x_dest, y_dest));
}
}
}这段代码的特点是:我无法控制速度,而且速度很慢.我怎样才能控制速度,保持选择到达位置的想法?我找到了这方面的主题,但我没有找到任何在我的例子中有效的东西,当然,因为一个像素的步骤太小,但我看不出我是怎么做的。
编辑这里是我想要做的(当红色圆圈缩小时,我必须在2秒内添加一个记录)。我很明显是按照pid指令做的。再次感谢他。
(function() {
function sleep(ms) {
return new Promise(resolve => setTimeout(resolve, ms));
}
var canvas = document.getElementById("calibrator");
var ctx = canvas.getContext('2d');
canvas.width = window.innerWidth;
canvas.height = window.innerHeight;
const points = [{
"x": 0,
"y": 0
},
{
"x": canvas.width / 2 - 100,
"y": 0
},
{
"x": canvas.width - 100,
"y": 0
},
{
"x": 0,
"y": canvas.height / 2 - 100
},
{
"x": canvas.width / 2 - 100,
"y": canvas.height / 2 - 100
},
{
"x": canvas.width - 100,
"y": canvas.height / 2 - 100
},
{
"x": 0,
"y": canvas.height - 100,
},
{
"x": canvas.width / 2 - 100,
"y": canvas.height - 100
},
{
"x": canvas.width - 100,
"y": canvas.height - 100
}
]
function generateLinear(x0, y0, x1, y1, dt) {
return (t) => {
let f0, f1;
f0 = t >= dt ? 1 : t / dt; // linear interpolation (aka lerp)
f1 = 1 - f0;
return {
"x": f1 * x0 + f0 * x1, // actually is a matrix multiplication
"y": f1 * y0 + f0 * y1
};
};
}
function generateShrink(x0, y0, x1, y1, r0, dt) {
return (t) => {
var f0 = t >= dt ? 0 : dt - t;
var f1 = t >= dt ? 1 : dt / t;
var f2 = 1 - f1;
return {
"x": f2 * x0 + f1 * x1,
"y": f2 * y0 + f1 * y1,
"r": f0 * r0
};
};
}
function create_path_circle() {
var nbPts = points.length;
var path = [];
for (var i = 0; i < nbPts - 1; i++) {
path.push({
"duration": 2,
"segment": generateShrink(points[i].x, points[i].y, points[i].x, points[i].y, 40, 2)
});
path.push({
"duration": 0.5,
"segment": generateShrink(points[i].x, points[i].y, points[i + 1].x, points[i + 1].y, 0, 0.5)
});
}
path.push({
"duration": 2,
"segment": generateShrink(points[nbPts - 1].x, points[nbPts - 1].y, points[nbPts - 1].x, points[nbPts - 1].y, 40, 2)
})
return path;
}
function create_path_target() {
var nbPts = points.length;
var path = [];
for (var i = 0; i < nbPts - 1; i++) {
path.push({
"duration": 2,
"segment": generateLinear(points[i].x, points[i].y, points[i].x, points[i].y, 2)
});
path.push({
"duration": 0.5,
"segment": generateLinear(points[i].x, points[i].y, points[i + 1].x, points[i + 1].y, 0.5)
});
}
path.push({
"duration": 2,
"segment": generateLinear(points[nbPts - 1].x, points[nbPts - 1].y, points[nbPts - 1].x, points[nbPts - 1].y, 2)
})
return path;
}
const path_target = create_path_target();
const path_circle = create_path_circle();
function renderImage(img, img_width, img_height) {
return (pos) => {
ctx = canvas.getContext('2d');
ctx.drawImage(img, pos.x, pos.y, img_width, img_height);
}
}
function renderCircle() {
return (pos) => {
ctx = canvas.getContext('2d');
ctx.beginPath();
ctx.arc(pos.x + 50, pos.y + 50, pos.r, 0, 2 * Math.PI);
ctx.fillStyle = "#FF0000";
ctx.fill();
ctx.stroke();
}
}
function generatePath(path) {
let i, t;
// fixup timing
t = 0;
for (i = 0; i < path.length; i++) {
path[i].start = t;
t += path[i].duration;
path[i].end = t;
}
return (t) => {
while (path.length > 1 && t >= path[0].end) {
path.shift(); // remove old segments, but leave last one
}
return path[0].segment(t - path[0].start); // time corrected
};
}
var base_image = new Image();
base_image.src = 'https://www.pngkit.com/png/full/17-175027_transparent-crosshair-sniper-scope-reticle.png';
const sprites = [
{
"move": generatePath(path_circle),
"created": performance.now(),
"render": renderCircle()
},
{
"move": generatePath(path_target),
"created": performance.now(),
"render": renderImage(base_image, 100, 100)
}
];
const update = () => {
let now;
ctx.fillStyle = "#FFFFFF";
ctx.fillRect(0, 0, canvas.width, canvas.height);
// put aside so all sprites are drawn for the same ms
now = performance.now();
for (var sprite of sprites) {
sprite.render(sprite.move((now - sprite.created) / 1000));
}
window.requestAnimationFrame(update);
};
window.requestAnimationFrame(update);
})();<!DOCTYPE html>
<html>
<head>
<title>Calibration</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css">
</head>
<body>
<canvas id="calibrator"></canvas>
<video id="stream"></video>
<canvas id="picture"></canvas>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.16.0/umd/popper.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.2/js/bootstrap.min.js"></script>
<script src="calibration.js"></script>
</body>
</html>
对于拍摄,如果我们假设我有一个返回图片的takeSnapshot()函数,我会这样做:
function film(dt) {
return (t) => {
if (t >= dt) {
return false;
} else {
return true;
}
}
}
function create_video_timeline() {
var nbPts = points.length;
var path = [];
for (var i = 0 ; i < nbPts - 1; i++) {
path.push(
{
"duration": 2,
"segment": film(2)
}
);
path.push(
{
"duration":0.5,
"segment": film(0)
}
);
}
path.push(
{
"duration": 2,
"segment": film(2)
}
)
return path;
}
const video_timeline = create_video_timeline();
function getSnapshot() {
return (bool) => {
if (bool) {
data.push(takepicture());
}
}
}
const sprites = [
{
"move": generatePath(path_circle),
"created": performance.now(),
"render": renderCircle()
},
{
"move": generatePath(path_target),
"created": performance.now(),
"render": renderImage(base_image, 100, 100)
},
{
"render": getSnapshot(),
"move": generatePath(video_timeline),
"created": performance.now()
}
];发布于 2021-03-25 16:01:52
编辑:添加了另一个移动示例(查看青色方块)
要回答你关于如何在固定的时间内“到达某个地方”的评论,你可以线性化大部分函数,然后通过固定时间来求解方程。这对于线性运动来说很容易,但是对于复杂的情况来说却相当困难,比如沿着非线性函数(例如对数螺旋)移动。
对于从(x0, y0)到(x1, y1) in time dt的等速直线运动(不加/减速),可以使用线性插值:
function generateLinear(x0, y0, x1, y1, dt)
{
return (t) => {
let f0, f1;
f0 = t >= dt ? 1 : t / dt; // linear interpolation (aka lerp)
f1 = 1 - f;
return {
"x": f0 * x0 + f1 * x1, // actually is a matrix multiplication
"y": f0 * y0 + f1 * y1
};
};
}这个函数现在可以用来“组装”一个路径。首先,通过生成分段来定义路径:
const path = [
{
"duration": dt1,
"segment": generateLinear(x0, y0, x1, y1, dt1)
},
{
"duration": dt2,
"segment": generateLinear(x1, y1, x2, y2, dt2)
},
{
"duration": dt3,
"segment": generateLinear(x2, y2, x3, y3, dt3)
}
];注意,现在将如何处理总路径时间(使用duration),并将其转换为段本地时间:
function generatePath(path)
{
let t;
// fixup timing
t = 0;
for (i = 0; i < path.length; i++)
{
path[i].start = t;
t += path[i].duration;
path[i].end = t;
}
return (t) => {
while (path.length > 1 && t >= path[0].end)
{
path.shift(); // remove old segments, but leave last one
}
return path[0].segment(t - path[0].start); // time corrected
};
}编辑:工作示例
我刚给你举了个例子。看看我如何不重做画布或上下文,并在相同的基础上反复绘制。运动并不取决于框架,它是在lissajous函数中定义的。
"use strict";
const cvs = document.querySelector("#cvs");
const ctx = cvs.getContext("2d");
function generateLissajous(dx, dy, tx, ty)
{
return (t) => {
return {
"x": 150 + dx * Math.sin(tx * t),
"y": 75 + dy * Math.cos(ty * t)
};
};
}
function generateLinear(x0, y0, x1, y1, dt)
{
return (t) => {
let f0, f1;
f0 = t >= dt ? 1 : t / dt; // linear interpolation (aka lerp)
f1 = 1 - f0;
return {
"x": f1 * x0 + f0 * x1, // actually is a matrix multiplication
"y": f1 * y0 + f0 * y1
};
};
}
function generatePath(path)
{
let i, t;
// fixup timing
t = 0;
for (i = 0; i < path.length; i++)
{
path[i].start = t;
t += path[i].duration;
path[i].end = t;
}
return (t) => {
let audio;
while (path.length > 1 && t >= path[0].end)
{
path.shift(); // remove old segments, but leave last one
}
if (path[0].hasOwnProperty("sound"))
{
audio = new Audio(path[0].sound);
audio.play();
delete path[0].sound; // play only once
}
return path[0].segment(t - path[0].start); // time corrected
};
}
function generateRenderer(size, color)
{
return (pos) => {
ctx.fillStyle = color;
ctx.fillRect(pos.x, pos.y, size, size);
};
}
const path = [
{
"duration": 3,
"segment": generateLinear(20, 20, 120, 120, 3)
},
{
"sound": "boing.ogg",
"duration": 3,
"segment": generateLinear(120, 120, 120, 20, 3)
},
{
"sound": "boing.ogg",
"duration": 2,
"segment": generateLinear(120, 20, 20, 120, 2)
}
];
const sprites = [
{
"move": generateLissajous(140, 60, 1.9, 0.3),
"created": performance.now(),
"render": generateRenderer(10, "#ff0000")
},
{
"move": generateLissajous(40, 30, 3.23, -1.86),
"created": performance.now(),
"render": generateRenderer(15, "#00ff00")
},
{
"move": generateLissajous(80, 50, -2.3, 1.86),
"created": performance.now(),
"render": generateRenderer(5, "#0000ff")
},
{
"move": generateLinear(10, 150, 300, 20, 30), // 30 seconds
"created": performance.now(),
"render": generateRenderer(15, "#ff00ff")
},
{
"move": generatePath(path),
"created": performance.now(),
"render": generateRenderer(25, "#00ffff")
}
];
const update = () => {
let now, sprite;
ctx.fillStyle = "#000000";
ctx.fillRect(0, 0, 300, 150);
// put aside so all sprites are drawn for the same ms
now = performance.now();
for (sprite of sprites)
{
sprite.render(sprite.move((now - sprite.created) / 1000));
}
window.requestAnimationFrame(update);
};
window.requestAnimationFrame(update);canvas
{
border: 1px solid red;
}<canvas id="cvs"></canvas>
这种运动不应该依赖于requestAnimtionFrame()。
你应该做的就是这个。
t)的运动函数,在本例中是一个Lissajous轨道:function orbit(t)
{
return { "x": 34 * Math.sin(t * 0.84), "y": 45 * Math.cos(t * 0.23) };
}那些数字只是为了表演。您可以将它们参数化,并使用运行来固定它们,并获得如下所示的“轨道()”函数:
function generateLissajousOrbit(dx, tx, dy, ty)
{
return (t) => { // this is the orbit function
return { "x": dx * Math.sin(t * tx), "y": dy * Math.cos(t * ty) };
};
}这样,就可以生成任意的Lissajous轨道:
let movement = generateLissajousOrbit(34, 0.84, 45, 0.23);显然,任何运动功能都是有效的。唯一的制约因素是:
带有realtime;
t表示在time t.上具有x和y坐标的对象
到目前为止,在如何实施方面,更简单的动作应该是显而易见的。还请注意,这种方式,它是非常容易插入任何运动。
开始时,将当前的实时毫秒放在一边,如下所示:
let mymovingobject = {
"started": performance.now(),
"movement": generateLissajousOrbit(34, 0.84, 45, 0.23)
};要在任何给定的时间获得x和y,现在可以这样做:
let now = performance.now();
let pos = mymovingobject.movement(now - mymovingobject.started);
// pos.x and pos.y contain the current coordinates您将得到一个刷新(动画帧)独立的运动,完全依赖于实时,这是您的主观感知空间。
如果机器由于任何原因而有hickup或刷新速率的变化(用户刚刚重新调整了监视器,将窗口从120 Hz移动到60 Hz监视器,或者其他什么).运动仍然是实时的,完全独立于帧速率。
在处理requestAnimationFrame()的函数中,您只需轮询上面所示的位置,然后在pos.x和pos.y中绘制对象,而不考虑实际的刷新速率。
您还可以跳过帧以降低帧速率,并让用户通过计数帧来决定频率,如下所示:
let frame = 0;
function requestAnimationFrameHandler()
{
if (frame % 2 === 0)
{
window.requestAnimationFrame();
return; // quick bail-out for this frame, see you next time!
}
// update canvas at half framerate
}能够减少帧数是特别重要的今天,因为高频监视器。您的应用程序将从60像素/秒跳到120像素/秒,只需改变监视器。这不是你想要的。
requestAnimationFrame()工具看起来是平滑滚动的灵丹妙药,但事实是,您将自己绑定到完全未知的硬件约束中(想想2035年的现代监视器)。谁知道他们会怎样)。
这种技术将物理帧频率与逻辑(游戏)速度要求分开。
希望这是有意义的。
https://stackoverflow.com/questions/66802877
复制相似问题