Skew, size and rotate a rectangle to fit triangle

问题:

I'm trying to make half of a rectangle - devided diagonally - to fit inside a triangle. Rotation works well, so does sizing of the rectangle. But once I try to skew it, it all gets messed up. Basically I want to simulate a 3D surface.

That means I have to find the angle of abc, where b is the center point. And then apply this angle as a skew to the rectangle. But for some reason that doesn't work as intended.

Here is a simple illustration of what I want to accomplish:

You will probably understand more once you take a look at the fiddle: http://jsfiddle.net/p7g7Y/11/ EDIT: Got the width right at least: http://jsfiddle.net/p7g7Y/12/

The piece of code you need to look at is at line 63 - 95. Try comment out the transform, and you will see that rotation and size works well.

function triangle(a, b, c){

context.save();

//Draw the triangle
context.beginPath();
context.moveTo(a[0], a[1]);
context.lineTo(b[0], b[1]);
context.lineTo(c[0], c[1]);
context.lineTo(a[0], a[1]);
context.closePath();
context.stroke();

//Lets find the distance between a and b to set height of the image
var imgHeight = lineDistance(a, b);

//And the width b to c
var imgWidth = lineDistance(b, c);

//Now we gotta skew it acording to the rad between ba and bc
var skewAngle = find_angle(a,c,b); //Find angle and make it rad

//Find the angle of b to a line
var theta = Math.atan2(a[1] - b[1], a[0] - b[0]);
context.translate(a[0], a[1]); //Set origin of rotation
context.rotate(theta + 1.57079633); //Had to rotate it some more 1.57079633 = 90deg
context.transform(1, skewAngle, 0, 1, 0, 0);

context.rect( 0, 0, imgHeight, imgWidth);
context.stroke();

context.restore();
}

If anything is unclear, please ask! I would love some help on this!

回答1:

It's easier if you solve the problem more generally: find a, b, c, d, e and f so that

  // (x0, y0) maps to (x_0, y_0)
  a*x0 + b*y0 + c = x_0
  d*x0 + e*y0 + f = y_0

  // (x1, y1) maps to (x_1, y_1)
  a*x1 + b*y1 + c = x_1
  d*x1 + e*y1 + f = y_1

  // (x2, y2) maps to (x_2, y_2)
  a*x2 + b*y2 + c = x_2
  d*x2 + e*y2 + f = y_2

This 6x6 linear system is composed of two independent 3x3 linear systems:

  a*x0 + b*y0 + c = x_0
  a*x1 + b*y1 + c = x_1
  a*x2 + b*y2 + c = x_2

  d*x0 + e*y0 + f = y_0
  d*x1 + e*y1 + f = y_1
  d*x2 + e*y2 + f = y_2

Solving them gives you the 6 numbers to pass to setTransform to map any three points to other three points.

delta = x0*y1 + y0*x2 + x1*y2 - y1*x2 - y0*x1 - x0*y2

delta_a = x_0*y1 + y0*x_2 + x_1*y2 - y1*x_2 - y0*x_1 - x_0*y2
delta_b = x0*x_1 + x_0*x2 + x1*x_2 - x_1*x2 - x_0*x1 - x0*x_2
delta_c = x0*y1*x_2 + y0*x_1*x2 + x_0*x1*y2 - x_0*y1*x2 - y0*x1*x_2 - x0*x_1*y2

delta_d = y_0*y1 + y0*y_2 + y_1*y2 - y1*y_2 - y0*y_1 - y_0*y2
delta_e = x0*y_1 + y_0*x2 + x1*y_2 - y_1*x2 - y_0*x1 - x0*y_2
delta_f = x0*y1*y_2 + y0*y_1*x2 + y_0*x1*y2 - y_0*y1*x2 - y0*x1*y_2 - x0*y_1*y2

a = delta_a / delta
b = delta_b / delta
c = delta_c / delta
d = delta_d / delta
e = delta_e / delta
f = delta_f / delta

For a full description of 3d texture mapping using 2d canvas context see this more detailed answer.

回答2:

Here’s how to calculate transforms necessary to fit a rectangle to a triangle:

• Translate to the “pivot point” of your triangle – point B.
• Rotate by the angle of side BC.
• Skew in the X direction by the angle of corner B.

So, first translate:

        // transform translate = pt2

        var translate = pt2;

Then rotate:

        // transform rotation = angleBC (based on slope of BC)

        var rotation = Math.atan2((pt3.y-pt2.y),(pt3.x-pt2.x));

Finally skewX:

        // transform skewX, based on angleB 

        var skewX = Math.tan(angleB-Math.PI/2);

Here’s how to get angleB for use in skewX:

        // calculate segment lengths

        var AB = Math.sqrt(Math.pow(pt2.x-pt1.x,2)+ Math.pow(pt2.y-pt1.y,2));    
        var BC = Math.sqrt(Math.pow(pt2.x-pt3.x,2)+ Math.pow(pt2.y-pt3.y,2)); 
        var AC = Math.sqrt(Math.pow(pt3.x-pt1.x,2)+ Math.pow(pt3.y-pt1.y,2));

        // calculate angleB using law of cosines

        var angleB = Math.acos((BC*BC+AB*AB-AC*AC)/(2*BC*AB));

You’ll also need the width and height of the rectangle to draw:

        // rectangle height = triangle altitude

        var rectHeight = AB * Math.sin(angleB);

        // rectangle width = triangle BC

        var rectWidth = BC;

A small “gotcha”:

Your translate point is B, but rectangles are drawn starting at top-left.

This means you must offset your rectangle vertically by the rectHeight:

        ctx.rect(0,  -rectHeight,  rectWidth,  rectHeight);

Also, not really a “gotcha”, but more of a natual limitation:

The angle at corner B must be <180.

So, if your triangle “inverts”, I you’ll have to compensate by flipping points A and C.

Interesting project you have there!

Would you share a bit when you’re done?

Here is code and a Fiddle: http://jsfiddle.net/m1erickson/KKELu/

<!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; }
    #canvas{border:1px solid red;}
</style>

<script>
$(function(){

    var canvas=document.getElementById("canvas");
    var ctx=canvas.getContext("2d");

    var pt1={x:100,y:100};
    var pt2={x:150,y:225};
    var pt3={x:250,y:150};


    drawTriangle();

    drawRectangle();


    function drawRectangle(){

        // calc transform info
        var info=analyzeTriangle();

        ctx.save();
        ctx.translate(info.translate.x,info.translate.y);
        ctx.rotate(info.rotation);
        ctx.transform(1,0,info.skewX,1,0,0);
        ctx.beginPath();
        // since rects origin is top left, must offset y by -height
        ctx.rect(0,-info.rectHeight,info.rectWidth,info.rectHeight);
        ctx.strokeStyle="purple";
        ctx.stroke();
        ctx.restore();
    }


    function drawTriangle(){
        ctx.beginPath();
        ctx.strokeStyle="blue";
        ctx.moveTo(pt1.x,pt1.y);
        ctx.lineTo(pt2.x,pt2.y);
        ctx.lineTo(pt3.x,pt3.y);
        ctx.closePath();
        ctx.stroke();
        ctx.fillStyle="rgba(255,255,0,0.10)";
        ctx.fill();
    }


    function analyzeTriangle(){

        // segment lengths
        var AB = Math.sqrt(Math.pow(pt2.x-pt1.x,2)+ Math.pow(pt2.y-pt1.y,2));    
        var BC = Math.sqrt(Math.pow(pt2.x-pt3.x,2)+ Math.pow(pt2.y-pt3.y,2)); 
        var AC = Math.sqrt(Math.pow(pt3.x-pt1.x,2)+ Math.pow(pt3.y-pt1.y,2));

        // angleB = using law of cosines
        var angleB = Math.acos((BC*BC+AB*AB-AC*AC)/(2*BC*AB));

        // transform translate = pt2
        var translate = pt2;

        // transform rotation = angleBC (based on slope of BC)
        var rotation = Math.atan2((pt3.y-pt2.y),(pt3.x-pt2.x));

        // transform skewX, based on angleB 
        var skewX = Math.tan(angleB-Math.PI/2);

        // rectangle height = triangle altitude
        var rectHeight = AB * Math.sin(angleB);

        // rectangle width = triangle BC
        var rectWidth = BC;

        return({
            translate:translate,
            rotation:rotation,
            skewX:skewX,
            rectHeight:rectHeight,
            rectWidth:rectWidth
        });

    }

}); // end $(function(){});
</script>

</head>

<body>
    <canvas id="canvas" width=350 height=350></canvas>
</body>
</html>