Tha*_*var 10
由于缺少间隔速度,你的小提琴链接对我不起作用,也应该使用getElementById(因为它在Internet Explorer中工作不会使它跨浏览器).
在这里,我分叉了,用这个代替:
http://jsfiddle.net/spechackers/hJhCz/
我还清理了第一个链接中的代码:
<pre id="p">
<script type="text/javascript">
var charMap=['p','.'];
var n=0;
function myInterval()
{
n+=7;//this is the amount of screen to "scroll" per interval
var outString="";
//this loop will execute exactly 4096 times. Once for each character we will be working with.
//Our display screen will consist of 32 lines or rows and 128 characters on each line
for(var i=64; i>0; i-=1/64)
{
//Note mod operations can result in numbers like 1.984375 if working with non-integer numbers like we currently are
var mod2=i%2;
if(mod2==0)
{
outString+="\n";
}else{
var tmp=(mod2*(64/i))-(64/i);//a number between 0.9846153846153847 and -4032
tmp=tmp+(n/64);//still working with floating points.
tmp=tmp^(64/i);//this is a bitwise XOR operation. The result will always be an integer
tmp=tmp&1;//this is a bitwise AND operation. Basically we just want to know if the first bit is a 1 or 0.
outString+=charMap[tmp];
}
}//for
document.getElementById("p").innerHTML=outString;
}
myInterval();
setInterval(myInterval,64);
</script>
</pre>
Run Code Online (Sandbox Code Playgroud)
您提供的两个链接中的代码结果彼此非常不同.但是代码中的逻辑非常相似.两者都使用for循环遍历所有字符,对非整数进行mod操作,以及bitwisexor操作.
它是如何工作的,基本上都是如此I can tell you is to pay attention to the variables changing as the input and output change.
所有逻辑似乎都是某种bitwise神秘的方式来决定将2个字符或换行符中的哪一个添加到页面中.
从某种角度来说,我并不完全遵循它calculus or trigonometry.
小智 6
考虑到扫过矩形区域的每条线实际上是围绕固定原点的(4?)线的旋转.
根据视错觉,背景似乎"移动".实际发生的是扫描线之间的区域在线条旋转时在两个焦点之间切换.
这是2维的旋转方程:
首先,在旋转(运动)之前和之后,在其中一行中可视化(x,y)坐标对:

因此,您可以为每条线创建一个点集合,并根据您想要动画的"平滑"程度将它们旋转到任意大小的角度.