Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
381 views
in Technique[技术] by (71.8m points)

javascript - Divide bezier curve into two equal halves

I have the bezier curves between 2 points. I'd like to cut all curves into two equal half. One of my idea is if I can control 't' value I'll draw 2 curves by t = [0,0.5] and t = [0.5,1] but I don't know how. Below is my code. I won't mind any other idea or suggestion

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>D3 test</title>
    <script src="http://d3js.org/d3.v3.min.js"></script>

    <script>
    var Over = function(){
        d3.select(this)
        .style("stroke-opacity", 0.25);
    }
    var Out = function(){
        d3.select(this)
        .transition().duration(200)
        .style("stroke-opacity", 0);
    }

    function curve(n,x1,y1,x2,y2){

        var xr = (x1+x2)/2,
            yr = (y1+y2)/2,
            euDist = Math.sqrt(Math.pow(x2-x1,2)+Math.pow(y2-y1,2)),
            x3 = -y1+xr+yr, x4 = -y2+xr+yr,
            y3 =  x1+yr-xr, y4 =  x2+yr-xr,
            ctrl , curveDescription;

        svg.append('path')
            .attr("stroke", 'blue')
            .attr('fill','none')
            .style("stroke-opacity",0.25)
            .attr('d', 'M'+x3+','+y3+'L'+x4+','+y4)
            .attr('stroke-width',strokeWidth);

        for(var j=0;j<=n;j++){
            ctrl = [(x4-x3)*j/n+x3 , (y4-y3)*j/n+y3] ,                  
            curveDescription=   
                    'M' +x1+','     +y1+ 
                    'Q' +ctrl[0]+','+ctrl[1]+','
                        +x2+','     +y2;

            svg.append('path')
                .attr("stroke", 'blue')
                .attr('fill','none')
                .style("stroke-opacity",0.25)
                .attr('d', curveDescription)
                .attr('stroke-width',strokeWidth);  

            svg.append('path')
                .attr("stroke", 'blue')
                .attr('fill','none')
                .style("stroke-opacity",0)
                .on("mouseover", Over)
                .on("mouseout", Out)
                .attr('d', curveDescription)
                .attr('stroke-width',strokeWidth*25);

        }

    }
    </script>

</head>

<body>
    <script>
    var w = 1268 , h = 680 , strokeWidth = 2;

    var svg = d3.select("body")
                .append("svg")
                .attr("width", w)
                .attr("height", h)

    curve(5, 100,100, 400,500);


    </script>
</body>
</html>
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Splitting a bezier into two curves is fairly simple. Look up De Casteljau's Algorithm. https://en.wikipedia.org/wiki/De_Casteljau%27s_algorithm

Update

De Casteljau is simpler than it looks. That WP article could be clearer for non-mathermeticians. So I'll explain more simply.

Imagine you have a bezier defined by the points A,B,C & D. Where A and D are the endpoints and B and C are the control points.

So, say you wanted to find the value of the curve at point 't' along the curve (where t is in the range 0..1. You can do it this way by geometry:

  1. Find the point E that is at 't' along the straight line AB.
  2. Find the point F that is at 't' along the straight line BC.
  3. Find the point G that is at 't' along the straight line CD.

  4. Find the point H that is at 't' along the straight line EF.

  5. Find the point J that is at 't' along the straight line FG.

  6. Finally, find the point K that is at 't' along the straight line HJ.

K is also equal to the point that is 't' along the bezier. This is De Casteljau's Algorithm.

But usefully, it also gives us the control points of the two beziers that would result if the curve was split at point K. The two bezier curves are: A,E,H,K and K,J,G,D.

In your case t=0.5, so finding the two curves is just a sequence of additions and divides-by-2.

  E = (A+B)/2
  F = (B+C)/2
  G = (C+D)/2
  H = (E+F)/2
  J = (F+G)/2
  K = (H+J)/2

Obviously each of these calculations has to be done for x and y.

Hope this helps.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...