javascript - 如何计算贝塞尔曲线的面积？

Question

给定以下路径(例如)，它描述了 SVG 三次贝塞尔曲线:

中号 300,140 中号 300,40, 500,40, 500,140

假设有一条直线连接端点300,140和500,140(封闭曲线下的面积)，是否可以计算出如此封闭的面积？

有人可以建议一个公式(或 JavaScript)来完成这个任务吗？

Answer 1

Convert the path to a polygon任意精度，然后 calculate the area of the polygon .

交互式演示:Area of Path via Subdivision _(已损坏)

                     

上述演示的核心使用 adaptively subdividing path into a polygon 的函数和 computing the area of a polygon :

// path:      an SVG <path> element
// threshold: a 'close-enough' limit (ignore subdivisions with area less than this)
// segments:  (optional) how many segments to subdivisions to create at each level
// returns:   a new SVG <polygon> element
function pathToPolygonViaSubdivision(path,threshold,segments){
  if (!threshold) threshold = 0.0001; // Get really, really close
  if (!segments)  segments = 3;       // 2 segments creates 0-area triangles

  var points = subdivide( ptWithLength(0), ptWithLength( path.getTotalLength() ) );
  for (var i=points.length;i--;) points[i] = [points[i].x,points[i].y];

  var doc  = path.ownerDocument;
  var poly = doc.createElementNS('http://www.w3.org/2000/svg','polygon');
  poly.setAttribute('points',points.join(' '));
  return poly;

  // Record the distance along the path with the point for later reference
  function ptWithLength(d) {
    var pt = path.getPointAtLength(d); pt.d = d; return pt;
  }

  // Create segments evenly spaced between two points on the path.
  // If the area of the result is less than the threshold return the endpoints.
  // Otherwise, keep the intermediary points and subdivide each consecutive pair.
  function subdivide(p1,p2){
    var pts=[p1];
    for (var i=1,step=(p2.d-p1.d)/segments;i<segments;i++){
      pts[i] = ptWithLength(p1.d + step*i);
    }
    pts.push(p2);
    if (polyArea(pts)<=threshold) return [p1,p2];
    else {
      var result = [];
      for (var i=1;i<pts.length;++i){
        var mids = subdivide(pts[i-1], pts[i]);
        mids.pop(); // We'll get the last point as the start of the next pair
        result = result.concat(mids)
      }
      result.push(p2);
      return result;
    }
  }

  // Calculate the area of an polygon represented by an array of points
  function polyArea(points){
    var p1,p2;
    for(var area=0,len=points.length,i=0;i<len;++i){
      p1 = points[i];
      p2 = points[(i-1+len)%len]; // Previous point, with wraparound
      area += (p2.x+p1.x) * (p2.y-p1.y);
    }
    return Math.abs(area/2);
  }
}

// Return the area for an SVG <polygon> or <polyline>
// Self-crossing polys reduce the effective 'area'
function polyArea(poly){
  var area=0,pts=poly.points,len=pts.numberOfItems;
  for(var i=0;i<len;++i){
    var p1 = pts.getItem(i), p2=pts.getItem((i+-1+len)%len);
    area += (p2.x+p1.x) * (p2.y-p1.y);
  }
  return Math.abs(area/2);
}

<小时/>

以下是原始答案，它使用不同的(非自适应)技术来转换 <path>到 <polygon> .

交互式演示:http://phrogz.net/svg/area_of_path.xhtml _(已损坏)

                 

上述演示的核心使用 approximating a path with a polygon 的函数和 computing the area of a polygon .

// Calculate the area of an SVG polygon/polyline
function polyArea(poly){
  var area=0,pts=poly.points,len=pts.numberOfItems;
  for(var i=0;i<len;++i){
    var p1 = pts.getItem(i), p2=pts.getItem((i+len-1)%len);
    area += (p2.x+p1.x) * (p2.y-p1.y);
  }
  return Math.abs(area/2);
}

// Create a <polygon> approximation for an SVG <path>
function pathToPolygon(path,samples){
  if (!samples) samples = 0;
  var doc = path.ownerDocument;
  var poly = doc.createElementNS('http://www.w3.org/2000/svg','polygon');

  // Put all path segments in a queue
  for (var segs=[],s=path.pathSegList,i=s.numberOfItems-1;i>=0;--i)
    segs[i] = s.getItem(i);
  var segments = segs.concat();

  var seg,lastSeg,points=[],x,y;
  var addSegmentPoint = function(s){
    if (s.pathSegType == SVGPathSeg.PATHSEG_CLOSEPATH){
      
    }else{
      if (s.pathSegType%2==1 && s.pathSegType>1){
        x+=s.x; y+=s.y;
      }else{
        x=s.x; y=s.y;
      }          
      var last = points[points.length-1];
      if (!last || x!=last[0] || y!=last[1]) points.push([x,y]);
    }
  };
  for (var d=0,len=path.getTotalLength(),step=len/samples;d<=len;d+=step){
    var seg = segments[path.getPathSegAtLength(d)];
    var pt  = path.getPointAtLength(d);
    if (seg != lastSeg){
      lastSeg = seg;
      while (segs.length && segs[0]!=seg) addSegmentPoint( segs.shift() );
    }
    var last = points[points.length-1];
    if (!last || pt.x!=last[0] || pt.y!=last[1]) points.push([pt.x,pt.y]);
  }
  for (var i=0,len=segs.length;i<len;++i) addSegmentPoint(segs[i]);
  for (var i=0,len=points.length;i<len;++i) points[i] = points[i].join(',');
  poly.setAttribute('points',points.join(' '));
  return poly;
}