我正在尋找等式來將圓轉換爲橢圓，以便我可以找到從點到橢圓邊界的最短距離。我找到了一個圓和一個點之間的距離的等式，但不知道如何將它轉換爲橢圓。將圓轉換爲橢圓，以便從橢圓邊框計算點的距離
像素和PY是點和x和y是圓原點和射線是半徑
closestCirclePoint: function(px, py, x, y, ray) {
var tg = (x += ray, y += ray, 0);
return function(x, y, x0, y0) {
return Math.sqrt((x -= x0) * x + (y -= y0) * y);
}(px, py, x, y) > ray
? {x: Math.cos(tg = Math.atan2(py - y, px - x)) * ray + x,
y: Math.sin(tg) * ray + y}
: {x: px, y: py};
}
[加法回答：如何近似橢圓上的最近點]
如果你願意犧牲完美的實用性...
這裏是一種方法來計算一個橢圓點是「接近ish」到你的目標點。
的方法：
缺點：
優點：
性能注意：
下面是這個方法的代碼：
// calc a point on the ellipse that is "near-ish" the target point
// uses "brute force"
function getEllipsePt(targetPtX,targetPtY){
// calculate which ellipse quadrant the targetPt is in
var q;
if(targetPtX>cx){
q=(targetPtY>cy)?0:3;
}else{
q=(targetPtY>cy)?1:2;
}
// calc beginning and ending radian angles to check
var r1=q*halfPI;
var r2=(q+1)*halfPI;
var dr=halfPI/steps;
var minLengthSquared=200000000;
var minX,minY;
// walk the ellipse quadrant and find a near-point
for(var r=r1;r<r2;r+=dr){
// get a point on the ellipse at radian angle == r
var ellipseX=cx+radiusX*Math.cos(r);
var ellipseY=cy+radiusY*Math.sin(r);
// calc distance from ellipsePt to targetPt
var dx=targetPtX-ellipseX;
var dy=targetPtY-ellipseY;
var lengthSquared=dx*dx+dy*dy;
// if new length is shortest, save this ellipse point
if(lengthSquared<minLengthSquared){
minX=ellipseX;
minY=ellipseY;
minLengthSquared=lengthSquared;
}
}
return({x:minX,y:minY});
}
這裏是代碼和一個小提琴：http://jsfiddle.net/m1erickson/UDBkV/
<!doctype html>
<html>
<head>
<link rel="stylesheet" type="text/css" media="all" href="css/reset.css" /> <!-- reset css -->
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<style>
body{ background-color: ivory; padding:20px; }
#wrapper{
position:relative;
width:300px;
height:300px;
}
#canvas{
position:absolute; top:0px; left:0px;
border:1px solid green;
width:100%;
height:100%;
}
#canvas2{
position:absolute; top:0px; left:0px;
border:1px solid red;
width:100%;
height:100%;
}
</style>
<script>
$(function(){
// get canvas references
var canvas=document.getElementById("canvas");
var ctx=canvas.getContext("2d");
var canvas2=document.getElementById("canvas2");
var ctx2=canvas2.getContext("2d");
// calc canvas position on page
var canvasOffset=$("#canvas").offset();
var offsetX=canvasOffset.left;
var offsetY=canvasOffset.top;
// define the ellipse
var cx=150;
var cy=150;
var radiusX=50;
var radiusY=25;
var halfPI=Math.PI/2;
var steps=8; // larger == greater accuracy
// get mouse position
// calc a point on the ellipse that is "near-ish"
// display a line between the mouse and that ellipse point
function handleMouseMove(e){
mouseX=parseInt(e.clientX-offsetX);
mouseY=parseInt(e.clientY-offsetY);
// Put your mousemove stuff here
var pt=getEllipsePt(mouseX,mouseY);
// testing: draw results
drawResults(mouseX,mouseY,pt.x,pt.y);
}
// calc a point on the ellipse that is "near-ish" the target point
// uses "brute force"
function getEllipsePt(targetPtX,targetPtY){
// calculate which ellipse quadrant the targetPt is in
var q;
if(targetPtX>cx){
q=(targetPtY>cy)?0:3;
}else{
q=(targetPtY>cy)?1:2;
}
// calc beginning and ending radian angles to check
var r1=q*halfPI;
var r2=(q+1)*halfPI;
var dr=halfPI/steps;
var minLengthSquared=200000000;
var minX,minY;
// walk the ellipse quadrant and find a near-point
for(var r=r1;r<r2;r+=dr){
// get a point on the ellipse at radian angle == r
var ellipseX=cx+radiusX*Math.cos(r);
var ellipseY=cy+radiusY*Math.sin(r);
// calc distance from ellipsePt to targetPt
var dx=targetPtX-ellipseX;
var dy=targetPtY-ellipseY;
var lengthSquared=dx*dx+dy*dy;
// if new length is shortest, save this ellipse point
if(lengthSquared<minLengthSquared){
minX=ellipseX;
minY=ellipseY;
minLengthSquared=lengthSquared;
}
}
return({x:minX,y:minY});
}
// listen for mousemoves
$("#canvas").mousemove(function(e){handleMouseMove(e);});
// testing: draw the ellipse on the background canvas
function drawEllipse(){
ctx2.beginPath()
ctx2.moveTo(cx+radiusX,cy)
for(var r=0;r<2*Math.PI;r+=2*Math.PI/60){
var ellipseX=cx+radiusX*Math.cos(r);
var ellipseY=cy+radiusY*Math.sin(r);
ctx2.lineTo(ellipseX,ellipseY)
}
ctx2.closePath();
ctx2.lineWidth=5;
ctx2.stroke();
}
// testing: draw line from mouse to ellipse
function drawResults(mouseX,mouseY,ellipseX,ellipseY){
ctx.clearRect(0,0,canvas.width,canvas.height);
ctx.beginPath();
ctx.moveTo(mouseX,mouseY);
ctx.lineTo(ellipseX,ellipseY);
ctx.lineWidth=1;
ctx.strokeStyle="red";
ctx.stroke();
}
}); // end $(function(){});
</script>
</head>
<body>
<div id="wrapper">
<canvas id="canvas2" width=300 height=300></canvas>
<canvas id="canvas" width=300 height=300></canvas>
</div>
</body>
</html>
原來的答案
這裏的圓形和橢圓形如何相關的
對於水平對準橢圓：
（X X）/（一個一）+（Y Y）/（B B）== 1;
其中a是水平頂點的長度，其中b是垂直頂點的長度。
如何圓和橢圓涉及：
如果== B，橢圓是圓！
但是......！
計算從任意點到橢圓上某個點的最小距離涉及到多於的計算。
下面是計算一個鏈接（點擊DistancePointEllipseEllipsoid.cpp）：
謝謝，這正是我正在尋找的。 – Zero
一個橢圓有兩個半徑和旋轉角度旁邊的原點，所以你是什麼意思「*射線意思是半徑的一半*「？ – Bergi
抱歉，ray是圓的半徑。我不知道這轉化爲一個橢圓。 – Zero