canvas中的三角运动(2) —— 旋转动画
<h2>一. 需求:</h2> <p>来一个挑战: 绘制一个物体,并让它随着鼠标旋转,使它总能指向鼠标。</p> <p>假设这个可供旋转的对象为箭头对象,箭头的构造函数如下:</p> <pre> <code class="language-java">// 箭头绘制的构造函数 function Arrow() { this.x = 0; this.y = 0; this.color = "#ffff00"; this.rotation = 0; } Arrow.prototype.draw = function(context) { context.save(); context.translate(this.x, this.y); context.rotate(this.rotation); context.lineWidth = 2; context.fillStyle = this.color; context.beginPath(); context.moveTo(-50, -25); context.lineTo(0, -25); context.lineTo(0, -50); context.lineTo(50, 0); context.lineTo(0, 50); context.lineTo(0, 25); context.lineTo(-50, 25); context.lineTo(-50, -25); context.closePath(); context.fill(); context.stroke(); context.restore(); }</code></pre> <h2>二. 解决思路:</h2> <p>鼠标的位置可以通过getMouse(e).x和getMouse(e).y属性获得它的坐标值。</p> <p>箭头的位置可以通过arrow.x和arrow.y得到。</p> <p>通过这两个坐标的差值,就可以计算到三角形两边的长度dx、dy。此时,只需要通过 <strong>Math.atan2(dy, dx)</strong> 方法即可计算出角度的大小,并将其赋值给箭头对象的rotation属性。</p> <p>箭头应旋转的角度如下图所示:</p> <p><img src="https://simg.open-open.com/show/ba03e0aec73843c475ecd7eceb5f2dad.png"></p> <p>该过程如下:</p> <pre> <code class="language-java">var dx = getMouse(e).x - arrow.x, dy = getMouse(e).y - arrow.y; arrow.rotation = Math.atan2(dy, dx); // 计算箭头旋转的弧度</code></pre> <p>完整代码如下:</p> <pre> <code class="language-java"><canvas id="canvas" width="200" height="200" style="background: #ccc;"></canvas> <script type="text/javascript"> // 箭头绘制的构造函数 function Arrow() { this.x = 0; this.y = 0; this.color = "#ffff00"; this.rotation = 0; } Arrow.prototype.draw = function(context) { context.save(); context.translate(this.x, this.y); context.rotate(this.rotation); context.lineWidth = 2; context.fillStyle = this.color; context.beginPath(); context.moveTo(-50, -25); context.lineTo(0, -25); context.lineTo(0, -50); context.lineTo(50, 0); context.lineTo(0, 50); context.lineTo(0, 25); context.lineTo(-50, 25); context.lineTo(-50, -25); context.closePath(); context.fill(); context.stroke(); context.restore(); } var canvas = document.getElementById("canvas"), context = canvas.getContext("2d"), arrow = new Arrow(); arrow.x = canvas.width / 2; arrow.y = canvas.height / 2; // 鼠标跟随事件 canvas.addEventListener("mousemove", function(e) { context.clearRect(0, 0, canvas.width, canvas.height); var dx = getMouse(e).x - arrow.x, dy = getMouse(e).y - arrow.y; arrow.rotation = Math.atan2(dy, dx); // 计算箭头旋转的弧度 arrow.draw(context); }, false); // 获取鼠标的当前位置 function getMouse(event) { var event = event || window.event; var mouse = {}; var x, y; if(event.pageX || event.pageY) { x = event.pageX; y = event.pageY; } else if(event.clientX || event.clientY) { var scrollLeft = document.documentElement.scrollLeft || document.body.scrollLeft; var scrollTop = document.documentElement.scrollTop || document.body.scrollTop; x = event.clientX + scrollLeft; y = event.clientY + scrollTop; } mouse.x = x; mouse.y = y; return mouse; } </script></code></pre> <p>演示如下: <a href="/misc/goto?guid=4959677072273599888" rel="nofollow,noindex">http://codepen.io/dengzhirong/pen/rLbkXj</a></p> <h2>三. 总结:</h2> <p>实际上,旋转功能不限于鼠标。可以将该功能演变成强制一个物体围绕特定的点旋转。</p> <p>旋转动画用到的三角函数是: Math.atan2(dy, dx) 。根据直角三角形的对边和邻边,获得角的弧度。从而计算出旋转的角度。</p> <p>朝鼠标(或任意一点)旋转的公式如下:</p> <pre> <code class="language-java">// 假设mouse为旋转跟随点,object为旋转物体 dx = mouse.x - object.x; dy = mouse.y - object.y; object.rotation = Math.atan2(dy, dx) * 180 / Math.PI;</code></pre> <p> </p> <p>来自:http://www.dengzhr.com/js/953</p> <p> </p> <p><span style="color:rgb(255, 255, 255)">Save</span></p>
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