我有这段代码可以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));
}
}
}
这段代码的问题是:我无法控制速度,而且速度非常慢...... 如何控制速度并保持选择到达位置的想法?我找到了相关主题,但我没有找到任何适合我的情况,当然是因为 1 像素的步长太小,但我不知道我该如何制作。
[编辑] 这就是我想要做的(当红圈缩小时,我必须在 2 秒内添加一条记录)。显然我是按照 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()
}
];
编辑:添加了另一个运动示例(查看青色方块)
要回答您关于如何在固定时间内到达“某处”的评论,您可以对大多数函数进行线性化,然后通过固定时间来求解方程。这对于线性运动很容易,但对于复杂的情况却很困难,例如沿着非线性函数(例如对数螺旋)移动。
对于在
(x0, y0)
时间内从 (x1, y1)
到 dt
的恒定速度(无加速/减速)线性运动,您可以使用线性插值:
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
};
};
}
此函数现在可用于“组装”路径。首先通过生成线段来定义路径:
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
)并将其转换为分段本地时间:
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
我刚刚为您准备了这个工作示例。 看看我如何不重做画布或上下文并一遍又一遍地在相同的画布或上下文上绘制。以及运动如何不依赖于帧速率,它是在利萨如函数中定义的。
"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>
您不应该依赖
requestAnimtionFrame()
进行此类运动。
你应该做的就是这个。
t
),在本例中是利萨如轨道:function orbit(t)
{
return { "x": 34 * Math.sin(t * 0.84), "y": 45 * Math.cos(t * 0.23) };
}
这些数字只是为了展示。您可以参数化它们并使用柯里化来固定它们并获得“orbit()”函数,如下所示:
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,你就可以生成任意的利萨如轨道:
let movement = generateLissajousOrbit(34, 0.84, 45, 0.23);
显然,任何移动功能都是有效的。唯一的限制是:
t
调用;x
接收具有 y
和 t
坐标的物体。现在如何实施更简单的动作应该是显而易见的。另请注意,通过这种方式可以非常轻松地插入任何运动。
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)
};
要在任何给定时间获取
x
和 y
,您现在可以执行以下操作:
let now = performance.now();
let pos = mymovingobject.movement(now - mymovingobject.started);
// pos.x and pos.y contain the current coordinates
您将获得完全依赖于实时的刷新(动画帧)独立运动,这是您的主观感知空间。
如果机器因任何原因出现故障或刷新率发生变化(用户刚刚重新校准了显示器,将桌面上的窗口从 120 Hz 移动到 60 Hz 显示器,或其他什么)....移动仍然会发生实时绑定且完全独立于帧速率。
在处理
requestAnimationFrame()
的函数中,您只需如上所示轮询位置,然后在 pos.x
和 pos.y
处绘制对象,而无需考虑实际的刷新率是多少。
您还可以跳帧来降低帧速率,并让用户通过计算帧数来决定频率,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
}
由于监视器频率较高,降低帧速率在当今尤为重要。只需更换显示器,您的应用程序就会从 60 像素/秒跃升至 120 像素/秒。这不是你想要的。
requestAnimationFrame()
设施看起来像是平滑滚动的灵丹妙药,但事实是你将自己绑定到完全未知的硬件限制(想想 2035 年的现代显示器......谁知道它们会是什么样子)。
该技术将物理帧频率与逻辑(游戏)速度要求分开。
希望这是有道理的。
常量运行(){
请求动画帧(运行);
让埃伦斯[8];
const ax (Math.cos(3 frm) rad 宽度) / 高度; (Math.sin(4 frm) rad 高度) / 宽度;
常量
e.x + (ax 指针.xe.x) / 10;
e.y+ (ay 指针.ye.y) / 10;
对于(设 11;i < N; i++) {
让 e=elems[1];
让 ep elems[i-1];
(ep.xe.x e.x (Math.cos(a) (1001))/5)/4;
const a Math.atan2(e.yep.y, e.x ep.х);
e.y + (ep.ye.y (Math.sin(a) (1001))/5)/4;
consts e.use.setAttributeNS(
(1624 (11))/50;
空,
“转换”,
翻译($((ep.x + e.x) / 2),$((ep.ye.y) / 2)) 旋转($(
(188/数学.PI)a
1)翻译(${0},$(0))缩放($(s),$(s))