two sections left

components/Graphic.jsx

@@ -239,13 +239,17 @@ var Graphic = React.createClass({
     this.refs.canvas.height = h;
   },
 
+  setCurves: function(c) {
+    this.setCurve(c);
+  },
+
   setCurve: function(c) {
     var pts = [];
-    c = Array.prototype.slice.call(arguments);
+    c = (typeof c === "array") ? c : Array.prototype.slice.call(arguments);
     c.forEach(nc => {
       pts = pts.concat(nc.points);
     });
-    this.curve = c.length === 1 ? c[0] : c;
+    this.curve = (c.length === 1) ? c[0] : c;
     this.lpts = pts;
   },
 
@@ -333,8 +337,11 @@ var Graphic = React.createClass({
     if(pts.length>2) {
       this.ctx.strokeStyle = "lightgrey";
       this.drawLine(pts[0], pts[1], offset);
-      if(pts.length === 3) { this.drawLine(pts[1], pts[2], offset); }
-      else {this.drawLine(pts[2], pts[3], offset); }
+      var last = pts.length-2;
+      for (var i=1; i<last; i++) {
+        this.drawLine(pts[i], pts[i+1], offset);
+      }
+      this.drawLine(pts[last], pts[last+1], offset);
     }
     this.ctx.strokeStyle = "black";
     this.drawPoints(pts, offset);

components/sections/arcapproximation/index.js

@@ -1,3 +1,144 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var atan2 = Math.atan2, PI = Math.PI, TAU = 2*PI, cos = Math.cos, sin = Math.sin;
+
+var Introduction = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Approximating Bézier curves with circular arcs"
+    };
+  },
+
+  setupCircle: function(api) {
+    var curve = new api.Bezier(70,70, 140,40, 240,130);
+    api.setCurve(curve);
+  },
+
+  setupQuadratic: function(api) {
+    var curve = api.getDefaultQuadratic();
+    api.setCurve(curve);
+  },
+
+  setupCubic: function(api) {
+    var curve = api.getDefaultCubic();
+    api.setCurve(curve);
+  },
+
+  getCCenter: function(api, p1, p2, p3) {
+    // deltas
+    var dx1 = (p2.x - p1.x),
+        dy1 = (p2.y - p1.y),
+        dx2 = (p3.x - p2.x),
+        dy2 = (p3.y - p2.y);
+
+    // perpendiculars (quarter circle turned)
+    var dx1p = dx1 * cos(PI/2) - dy1 * sin(PI/2),
+        dy1p = dx1 * sin(PI/2) + dy1 * cos(PI/2),
+        dx2p = dx2 * cos(PI/2) - dy2 * sin(PI/2),
+        dy2p = dx2 * sin(PI/2) + dy2 * cos(PI/2);
+
+    // chord midpoints
+    var mx1 = (p1.x + p2.x)/2,
+        my1 = (p1.y + p2.y)/2,
+        mx2 = (p2.x + p3.x)/2,
+        my2 = (p2.y + p3.y)/2;
+
+    // midpoint offsets
+    var mx1n = mx1 + dx1p,
+        my1n = my1 + dy1p,
+        mx2n = mx2 + dx2p,
+        my2n = my2 + dy2p;
+
+    // intersection of these lines:
+    var i = api.utils.lli8(mx1,my1,mx1n,my1n, mx2,my2,mx2n,my2n);
+    var r = api.utils.dist(i,p1);
+
+    // arc start/end values, over mid point
+    var s = atan2(p1.y - i.y, p1.x - i.x),
+        m = atan2(p2.y - i.y, p2.x - i.x),
+        e = atan2(p3.y - i.y, p3.x - i.x);
+
+    // determine arc direction (cw/ccw correction)
+    var __;
+    if (s<e) {
+      // if s<m<e, arc(s, e)
+      // if m<s<e, arc(e, s + TAU)
+      // if s<e<m, arc(e, s + TAU)
+      if (s>m || m>e) { s += TAU; }
+      if (s>e) { __=e; e=s; s=__; }
+    } else {
+      // if e<m<s, arc(e, s)
+      // if m<e<s, arc(s, e + TAU)
+      // if e<s<m, arc(s, e + TAU)
+      if (e<m && m<s) { __=e; e=s; s=__; } else { e += TAU; }
+    }
+
+    // assign and done.
+    i.s = s;
+    i.e = e;
+    i.r = r;
+    return i;
+  },
+
+  drawCircle: function(api, curve) {
+    api.reset();
+    var pts = curve.points;
+
+    // get center
+    var C = this.getCCenter(api, pts[0], pts[1], pts[2]);
+    api.setColor("black");
+    pts.forEach(p => api.drawCircle(p,3));
+    api.drawCircle(C, 3);
+
+    // chords and perpendicular lines
+    api.setColor("blue");
+    api.drawLine(pts[0], pts[1]);
+    api.drawLine({x: (pts[0].x + pts[1].x)/2, y: (pts[0].y + pts[1].y)/2}, C);
+
+    api.setColor("red");
+    api.drawLine(pts[1], pts[2]);
+    api.drawLine({x: (pts[1].x + pts[2].x)/2, y: (pts[1].y + pts[2].y)/2}, C);
+
+    api.setColor("green");
+    api.drawLine(pts[2], pts[0]);
+    api.drawLine({x: (pts[2].x + pts[0].x)/2, y: (pts[2].y + pts[0].y)/2}, C);
+
+    api.setColor("grey");
+    api.drawCircle(C, api.utils.dist(C,pts[0]));
+  },
+
+  drawSingleArc: function(api, curve) {
+    api.reset();
+    var arcs = curve.arcs(0.5);
+    api.drawSkeleton(curve);
+    api.drawCurve(curve);
+
+    var a = arcs[0];
+    api.setColor("red");
+    api.setFill("rgba(200,0,0,0.4)");
+    api.debug = true;
+    api.drawArc(a);
+  },
+
+  drawArcs: function(api, curve) {
+    api.reset();
+    var arcs = curve.arcs(0.5);
+    api.drawSkeleton(curve);
+    api.drawCurve(curve);
+    arcs.forEach(a => {
+      api.setRandomColor(0.3);
+      api.setFill(api.getColor());
+      api.drawArc(a);
+    });
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
         <p>Let's look at converting Bézier curves into sequences of circular arcs. We already saw in the
           section on circle approximation that this will never yield a perfect equivalent, but sometimes
           you need circular arcs, such as when you're working with fabrication machinery, or simple vector
@@ -14,40 +155,25 @@
         <p>So: step 1, how do we find a circle through three points? That part is actually really simple.
           You may remember (if you ever learned it!) that a line between two points on a circle is called
           a <a href="https://en.wikipedia.org/wiki/Chord_%28geometry%29">chord</a>, and one property of
-          chords is that the line from the center of the chord, perpendicular to the chord, passes through
-          the center of the circle. So: if we have have three points, we have two (different) chords, and
-          consequently, two (different) lines that go from those chords through the center of the circle:
-          find the centers of the chords, find the perpendicular lines, find the intersection of those lines,
-          find the center of the circle that goes through all three points.</p>
+          chords is that the line from the center of any chord, perpendicular to that chord, passes through
+          the center of the circle.</p>
 
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Finding a circle through three points">
-        void setupCurve() {
-          setupDefaultQuadratic();
-        }
+        <p>So: if we have have three points, we have three (different) chords, and consequently, three
+          (different) lines that go from those chords through the center of the circle. So we find the
+          centers of the chords, find the perpendicular lines, find the intersection of those lines,
+          and thus find the center of the circle.</p>
 
-        void drawCurve(BezierCurve curve) {
-          curve.drawPoints();
-          CircleAbstractor ca = new CircleAbstractor(curve);
-          Point[] p = curve.points;
-          CircleAbstractor.Point cp = ca.getCCenter(p[0], p[1], p[2]);
-          stroke(0,100);
-          noFill();
-          ellipse(cp.x, cp.y, cp.r*2, cp.r*2);
-          ellipse(cp.x, cp.y, 5, 5);
-          fill(0);
-          text((int)cp.x+","+(int)cp.y, cp.x + 5, cp.y+5);
+        <p>The following graphic shows this procedure with a different colour for each chord and its
+          associated perpendicular through the center. You can move the points around as much as you
+          like, those lines will always meet!</p>
 
-          stroke(200,0,0);
-          line(cp.x, cp.y, (p[0].x+p[1].x)/2, (p[0].y+p[1].y)/2);
-          line(p[0].x,p[0].y,p[1].x,p[1].y);
-
-          stroke(0,0,255);
-          line(cp.x, cp.y, (p[1].x+p[2].x)/2, (p[1].y+p[2].y)/2);
-          line(p[2].x,p[2].y,p[1].x,p[1].y);
-        }</textarea>
+        <Graphic preset="simple" title="Finding a circle through three points" setup={this.setupCircle} draw={this.drawCircle} />
 
         <p>So, with the procedure on how to find a circle through three points, finding the arc through those points
-          is straight-forward. Let's apply this to a Bezier curve:</p>
+          is straight-forward: pick one of the three points as start point, pick another as an end point, and
+          the arc has to necessarily go from the start point, over the remaining point, to the end point.</p>
+
+        <p>So how can we convert a Bezier curve into a (sequence of) circular arc(s)?</p>
 
         <ul>
           <li>Start at <em>t=0</em></li>
@@ -67,8 +193,8 @@
         </ul>
 
         <p>The result of this is shown in the next graphic: we start at a guaranteed failure: s=0, e=1. That's
-          the entire curve. The midpoint is simply at <em>t=0.5</em>, and then we start performing a
-          <a href="https://en.wikipedia.org/wiki/Binary_search_algorithm">Binary Search</a>.</p>
+          the entire curve. The midpoint is simply at <em>t=0.5</em>, and then we start performing
+          a <a href="https://en.wikipedia.org/wiki/Binary_search_algorithm">Binary Search</a>.</p>
 
         <ol>
           <li>We start with {0, 0.5, 1}</li>
@@ -88,31 +214,7 @@
           and you can use your '+' and '-' keys to increase to decrease the error threshold, to see what the effect
           of a smaller or larger error threshold is.</p>
 
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Arc approximation of a Bézier curve">
-        void setupCurve() {
-          setupDefaultCubic();
-          offsetting();
-          offset = 5;
-        }
-
-        void drawCurve(BezierCurve curve) {
-          double threshold = 0.5;
-          if (offset < 1) offset = 1;
-          if (0 < offset && offset < 10) threshold = offset/10;
-          else if (10 < offset && offset < 110) threshold = offset-10;
-          else if (110 < offset) threshold = 100 + (offset-110)*10;
-
-          curve.draw();
-          CircleAbstractor ca = new CircleAbstractor(curve, threshold);
-          stroke(255,0,0);
-          fill(0,50);
-          for (CircleAbstractor.Point c : ca.getCircles()) {
-            arc(c.x, c.y, 2*c.r, 2*c.r, c.s, c.e);
-            break;
-          }
-          fill(0);
-          text("error threshold: "+ca.errorThreshold, 5, 15);
-        }</textarea>
+        <Graphic preset="simple" title="Arc approximation of a Bézier curve" setup={this.setupCubic} draw={this.drawSingleArc} />
 
         <p>With that in place, all that's left now is to "restart" the procedure by treating the found arc's
           end point as the new to-be-determined arc's starting point, and using points further down the curve. We
@@ -121,33 +223,7 @@
           so you can see how picking a different threshold changes the number of arcs that are necessary to
           reasonably approximate a curve:</p>
 
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Arc approximation of a Bézier curve">
-        void setupCurve() {
-          setupDefaultCubic();
-          offsetting();
-          offset = 5;
-        }
-
-        void drawCurve(BezierCurve curve) {
-          double threshold = 0.5;
-          if (offset < 1) offset = 1;
-          if (0 < offset && offset < 10) threshold = offset/10;
-          else if (10 < offset && offset < 110) threshold = offset-10;
-          else if (110 < offset) threshold = 100 + (offset-110)*10;
-
-          CircleAbstractor ca = new CircleAbstractor(curve, threshold);
-          stroke(255,0,0);
-          fill(0,50);
-          ArrayList<CircleAbstractor.Point> circles = ca.getCircles();
-          for (CircleAbstractor.Point c : circles) {
-            arc(c.x, c.y, 2*c.r, 2*c.r, c.s, c.e);
-          }
-          curve.drawControlLines();
-          curve.drawPoints();
-          fill(0);
-          text("error threshold: "+ca.errorThreshold, 5, 15);
-          text("Approximated the curve using " + circles.size() + " arcs.", 5, dim-5);
-        }</textarea>
+        <Graphic preset="simple" title="Arc approximation of a Bézier curve" setup={this.setupCubic} draw={this.drawArcs} />
 
         <p>So... what is this good for? Obviously, If you're working with technologies that can't do curves,
           but can do lines and circles, then the answer is pretty straight-forward, but what else? There are
@@ -161,3 +237,9 @@
           approximation are guaranteed "off" by some small value, and depending on how much precision you
           need, arc approximation is either going to be super useful, or completely useless. It's up to you
           to decide which, based on your application!</p>
+      </section>
+    );
+  }
+});
+
+module.exports = Introduction;

components/sections/circles/index.js

@@ -1,3 +1,74 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var sin = Math.sin, cos = Math.cos;
+
+var Circles = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Circles and quadratic Bézier curves"
+    };
+  },
+
+  setup: function(api) {
+    api.w = api.getPanelWidth();
+    api.h = api.getPanelHeight();
+    api.pad = 20;
+    api.r = api.w/2 - api.pad;
+    api.mousePt = false;
+    api.angle = 0;
+    var spt = { x: api.w-api.pad, y: api.h/2 };
+    api.setCurve(new api.Bezier(spt, spt, spt));
+  },
+
+  draw: function(api, curve) {
+    api.reset();
+    api.setColor("lightgrey");
+    api.drawGrid(1,1);
+    api.setColor("red");
+    api.drawCircle({x:api.w/2,y:api.h/2},api.r);
+    api.setColor("transparent");
+    api.setFill("rgba(100,255,100,0.4)");
+    var p = {
+      x: api.w/2,
+      y: api.h/2,
+      r: api.r,
+      s: api.angle < 0 ? api.angle : 0,
+      e: api.angle < 0 ? 0 : api.angle,
+    };
+    api.drawArc(p);
+    api.setColor("black");
+    api.drawSkeleton(curve);
+    api.drawCurve(curve);
+  },
+
+  onMouseMove: function(evt, api) {
+    var x = evt.offsetX - api.w/2,
+        y = evt.offsetY - api.h/2;
+    var angle = Math.atan2(y,x);
+    var pts = api.curve.points;
+    // new control
+    var r = api.r,
+        b = (cos(angle) - 1) / sin(angle);
+    pts[1] = {
+      x: api.w/2 + r * (cos(angle) - b * sin(angle)),
+      y: api.w/2 + r * (sin(angle) + b * cos(angle))
+    };
+    // new endpoint
+    pts[2] = {
+      x: api.w/2 + api.r * cos(angle),
+      y: api.w/2 + api.r * sin(angle)
+    };
+    api.setCurve(new api.Bezier(pts));
+    api.angle = angle;
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
         <p>Circles and Bézier curves are very different beasts, and circles are infinitely easier
         to work with than Bézier curves. Their formula is much simpler, and they can be drawn more
         efficiently. But, sometimes you don't have the luxury of using circles, or ellipses, or
@@ -26,32 +97,7 @@
         you can move the mouse around over a unit circle, to see how well, or poorly, a quadratic
         curve can approximate the arc from (1,0) to where your mouse cursor is:</p>
 
-        <textarea class="sketch-code" data-sketch-preset="arcfitting" data-sketch-title="Quadratic Bézier arc approximation">
-        void setupCurve() { order = 2; }
-
-        void checkConnect() {
-          if((s==0 && e>PI/2) || s<-PI/2) { connect(); }
-          else { noConnect(); }
-        }
-
-        void findArcFitting() {
-          Point[] points = {
-            new Point(dim/2 + dim/f2, dim/2),
-            new Point(dim/2, dim/2),
-            new Point(dim/2 + dim/f2*cos(ax), dim/2 + dim/f2*sin(ay))
-          };
-          Point c = comp.lli(new Point[]{
-            points[0],
-            new Point(points[0].x, points[0].y + 10),
-            points[2],
-            new Point(points[2].x + dx, points[2].y + dy)
-          });
-          if(c==null) return;
-          points[1] = c;
-          BezierCurve bc = new BezierCurve(points);
-          bc.draw();
-          bc.getPoint(0.5).draw();
-        }</textarea>
+        <Graphic preset="arcfitting" title="Quadratic Bézier arc approximation" setup={this.setup} draw={this.draw} onMouseMove={this.onMouseMove}/>
 
         <p>As you can see, things go horribly wrong quite quickly; even trying to approximate a quarter circle
         using a quadratic curve is a bad idea. An eighth of a turns might look okay, but how okay is okay?
@@ -82,10 +128,15 @@
         <p>\[ \begin{array}{l}
           1 = cos(φ) + b \cdot -sin(φ) \ → \
           1 - cos(φ) = -b \cdot sin(φ) \ → \
-          -1 + cos(φ) = b \cdot sin(φ) \\ \\
-          b = \frac{-1 + cos(φ)}{sin(φ)}
+          -1 + cos(φ) = b \cdot sin(φ)
         \end{array} \]</p>
 
+        <p>which yields:</p>
+
+        <p>\[
+          b = \frac{cos(φ)-1}{sin(φ)}
+        \]</p>
+
         <p>which we can then substitute in the expression for <i>a</i>:</p>
 
         <p>\[ \begin{align*}
@@ -109,11 +160,13 @@
         <p>We compute T, observing that if <i>t=0.5</i>, the polynomial values (1-t)², 2(1-t)t, and t²
         are 0.25, 0.5, and 0.25 respectively:</p>
 
+        <p>\[
+          T = \frac{1}{4}S + \frac{2}{4}C + \frac{1}{4}E = \frac{1}{4}(S + 2C + E)
+        \]</p>
+
+        <p>Which, worked out for the x and y components, gives:</p>
+
         <p>\[\begin{array}{l}
-          T = \frac{1}{4}S + \frac{2}{4}C + \frac{1}{4}E = \frac{1}{4}(S + 2C + E) \\
-
-          = \
-
           \left\{\begin{align*}
           T_x &= \frac{1}{4}(3 + cos(φ))\\
           T_y &= \frac{1}{4}\left(\frac{2-2cos(φ)}{sin(φ)} + sin(φ)\right)
@@ -123,29 +176,38 @@
 
         <p>And the distance between these two is the standard Euclidean distance:</p>
 
-        <p>\[\begin{array}{l}
-          d_x(φ) = T_x - P_x = \frac{1}{4}(3 + cos(φ)) - cos(\frac{φ}{2}) = 2sin^4\left(\frac{φ}{4}\right) \ , \\
-          d_y(φ) = T_y - P_y = \frac{1}{4}\left(2tan\left(\frac{φ}{2}\right) + sin(φ)\right) - sin(\frac{φ}{2}) \ , \\
-          d(φ) = \sqrt{d^2_x + d^2_y} = \ ... \  = 2sin^4(\frac{φ}{2})\sqrt{\frac{1}{cos^2(\frac{φ}{2})}}
-        \end{array}\]</p>
+        <p>\[\begin{align}
+          d_x(φ) &= T_x - P_x = \frac{1}{4}(3 + cos(φ)) - cos(\frac{φ}{2}) = 2sin^4\left(\frac{φ}{4}\right) \ , \\
+          d_y(φ) &= T_y - P_y = \frac{1}{4}\left(2tan\left(\frac{φ}{2}\right) + sin(φ)\right) - sin(\frac{φ}{2}) \ , \\
+          &⇓\\
+          d(φ) &= \sqrt{d^2_x + d^2_y} = \ ... \  = 2sin^4(\frac{φ}{2})\sqrt{\frac{1}{cos^2(\frac{φ}{2})}}
+        \end{align}\]</p>
 
         <p>So, what does this distance function look like when we plot it for a
         number of ranges for the angle φ, such as a half circle, quarter circle and eighth circle?</p>
 
-        <table><tr><td>
-          <p><a href="http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi"><img
-          src="images/arc-q-pi.gif"
-          ></a></p>
+        <table><tbody><tr><td>
+          <p>
+            <a href="http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi">
+              <img src="images/arc-q-pi.gif"/>
+            </a>
+          </p>
           <p>plotted for 0 ≤ φ ≤ π:</p>
         </td><td>
-          <p><a href="http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi%2F2"><img
-          src="images/arc-q-pi2.gif"></a></p>
+          <p>
+            <a href="http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi%2F2">
+              <img src="images/arc-q-pi2.gif"/>
+            </a>
+          </p>
           <p>plotted for 0 ≤ φ ≤ ½π:</p>
         </td><td>
-          <p><a href="http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi%2F4"><img
-          src="images/arc-q-pi4.gif"></a></p>
+          <p>
+            <a href="http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi%2F4">
+              <img src="images/arc-q-pi4.gif"/>
+            </a>
+          </p>
           <p>plotted for 0 ≤ φ ≤ ¼π:</p>
-        </td></tr></table>
+        </td></tr></tbody></table>
 
         <p>We now see why the eighth circle arc looks decent, but the quarter circle arc doesn't:
         an error of roughly 0.06 at <i>t=0.5</i> means we're 6% off the mark... we will already be
@@ -165,13 +227,20 @@
           φ = 4 \cdot arccos \left(\frac{\sqrt{2+ε-\sqrt{ε(2+ε)}}}{\sqrt{2}}\right)
         \]</p>
 
-        <p>Things are starting to look, frankly, a bit ridiculous at this point, but this is as far
-        as we need the math to take us. If we plug in the precisions 0.1, 0.01, 0.001 and 0.0001 we
-        get the values 1.748, 1.038, 0.594 and 0.3356; in degrees, roughly 100 (requiring four curves),
-        59.5 (requiring six curves), 34 (requiring 11 curves), and 19.2 (requiring a whopping nineteen
-        curves). </p>
+        <p>And frankly, things are starting to look a bit ridiculous at this point, we're doing way more
+        maths than we've ever done, but thankfully this is as far as we need the maths to take us:
+        If we plug in the precisions 0.1, 0.01, 0.001 and 0.0001 we get the radians values 1.748, 1.038, 0.594
+        and 0.3356; in degrees, that means we can cover roughly 100 degrees (requiring four curves),
+        59.5 degrees (requiring six curves), 34 degrees (requiring 11 curves), and 19.2 degrees (requiring
+        a whopping nineteen curves). </p>
 
         <p>The bottom line? <strong>Quadratic curves are kind of lousy</strong> if you want circular
         (or elliptical, which are circles that have been squashed in one dimension) curves. We
         can do better, even if it's just by raising the order of our curve once. So let's try the
-        same thing for cubic curves.</p>
+        same thing for cubic curves.</p>
+      </section>
+    );
+  }
+});
+
+module.exports = Circles;

components/sections/circles_cubic/index.js

@@ -0,0 +1,444 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var sin = Math.sin, cos = Math.cos, tan = Math.tan, abs = Math.abs;
+
+var CirclesCubic = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Circles and cubic Bézier curves"
+    };
+  },
+
+  setup: function(api) {
+    api.setSize(400,400);
+    api.w = api.getPanelWidth();
+    api.h = api.getPanelHeight();
+    api.pad = 80;
+    api.r = api.w/2 - api.pad;
+    api.mousePt = false;
+    api.angle = 0;
+    var spt = { x: api.w-api.pad, y: api.h/2 };
+    api.setCurve(new api.Bezier(spt, spt, spt, spt));
+  },
+
+  guessCurve: function(S, B, E)  {
+    var C = {
+          x: (S.x + E.x)/2,
+          y: (S.y + E.y)/2
+        },
+        A = {
+          x: B.x + (B.x-C.x)/3, // cubic ratio at t=0.5 is 1/3
+          y: B.y + (B.y-C.y)/3
+        },
+        bx = (E.x-S.x)/4,
+        by = (E.y-S.y)/4,
+        e1 = {
+          x: B.x - bx,
+          y: B.y - by
+        },
+        e2 = {
+          x: B.x + bx,
+          y: B.y + by
+        },
+
+        v1 = {
+          x: A.x + (e1.x-A.x)*2,
+          y: A.y + (e1.y-A.y)*2
+        },
+        v2 = {
+          x: A.x + (e2.x-A.x)*2,
+          y: A.y + (e2.y-A.y)*2
+        },
+
+        nc1 = {
+          x: S.x + (v1.x-S.x)*2,
+          y: S.y + (v1.y-S.y)*2
+        },
+        nc2 = {
+          x: E.x + (v2.x-E.x)*2,
+          y: E.y + (v2.y-E.y)*2
+        };
+    return [nc1, nc2];
+  },
+
+  draw: function(api, curve) {
+    api.reset();
+
+    api.setColor("lightgrey");
+    api.drawGrid(1,1);
+    api.setColor("rgba(255,0,0,0.4)");
+    api.drawCircle({x:api.w/2,y:api.h/2},api.r);
+    api.setColor("transparent");
+    api.setFill("rgba(100,255,100,0.4)");
+    var p = {
+      x: api.w/2,
+      y: api.h/2,
+      r: api.r,
+      s: api.angle < 0 ? api.angle : 0,
+      e: api.angle < 0 ? 0 : api.angle,
+    };
+    api.drawArc(p);
+
+    // guessed curve
+    var B = {
+      x: api.w/2 + api.r * cos(api.angle/2),
+      y: api.w/2 + api.r * sin(api.angle/2)
+    };
+    var S = curve.points[0],
+        E = curve.points[3],
+        nc = this.guessCurve(S,B,E);
+    var guess = new api.Bezier([S, nc[0], nc[1], E]);
+    api.setColor("rgb(140,140,255)");
+    api.drawLine(guess.points[0], guess.points[1]);
+    api.drawLine(guess.points[1], guess.points[2]);
+    api.drawLine(guess.points[2], guess.points[3]);
+    api.setColor("blue");
+    api.drawCurve(guess);
+    api.drawCircle(guess.points[1], 3);
+    api.drawCircle(guess.points[2], 3);
+
+    // real curve
+    var offset = {x:api.w, y:0};
+    api.drawSkeleton(curve);
+    api.setColor("black");
+    api.drawLine(curve.points[1], curve.points[2]);
+    api.drawCurve(curve);
+  },
+
+  onMouseMove: function(evt, api) {
+    var x = evt.offsetX - api.w/2,
+        y = evt.offsetY - api.h/2;
+    if (x>api.w/2) return;
+
+    var angle = Math.atan2(y,x);
+    if (angle < 0) {
+      angle = 2*Math.PI + angle;
+    }
+    var pts = api.curve.points;
+    // new control 1
+    var r = api.r,
+        f = (4 * tan(angle/4)) /3;
+    pts[1] = {
+      x: api.w/2 + r,
+      y: api.w/2 + r * f
+    };
+    // new control 2
+    pts[2] = {
+      x: api.w/2 + api.r * (cos(angle) + f*sin(angle)),
+      y: api.w/2 + api.r * (sin(angle) - f*cos(angle))
+    };
+    // new endpoint
+    pts[3] = {
+      x: api.w/2 + api.r * cos(angle),
+      y: api.w/2 + api.r * sin(angle)
+    };
+    api.setCurve(new api.Bezier(pts));
+    api.angle = angle;
+  },
+
+  drawCircle: function(api) {
+    api.setSize(325,325);
+    api.reset();
+    var w = api.getPanelWidth(),
+        h = api.getPanelHeight(),
+        pad = 60,
+        r = w/2 - pad,
+        k = 0.55228,
+        offset = {x: -pad/2, y:0};
+    var curve = new api.Bezier([
+      {x:w/2 + r,   y:h/2},
+      {x:w/2 + r,   y:h/2 + k*r},
+      {x:w/2 + k*r, y:h/2 + r},
+      {x:w/2,       y:h/2 + r}
+    ]);
+    api.setColor("lightgrey");
+    api.drawLine({x:0,y:h/2}, {x:w+pad,y:h/2}, offset);
+    api.drawLine({x:w/2,y:0}, {x:w/2,y:h}, offset);
+
+    var pts = curve.points;
+
+    api.setColor("red");
+    api.drawCircle(pts[0], 3, offset);
+    api.drawCircle(pts[1], 3, offset);
+    api.drawCircle(pts[2], 3, offset);
+    api.drawCircle(pts[3], 3, offset);
+    api.drawCurve(curve, offset);
+    api.setColor("rgb(255,160,160)");
+    api.drawLine(pts[0],pts[1],offset);
+    api.drawLine(pts[1],pts[2],offset);
+    api.drawLine(pts[2],pts[3],offset);
+
+    api.setFill("red");
+    api.text((pts[0].x - w/2) + "," + (pts[0].y - h/2), {x: pts[0].x + 7, y: pts[0].y + 3}, offset);
+    api.text((pts[1].x - w/2) + "," + (pts[1].y - h/2), {x: pts[1].x + 7, y: pts[1].y + 3}, offset);
+    api.text((pts[2].x - w/2) + "," + (pts[2].y - h/2), {x: pts[2].x + 7, y: pts[2].y + 7}, offset);
+    api.text((pts[3].x - w/2) + "," + (pts[3].y - h/2), {x: pts[3].x, y: pts[3].y + 13}, offset);
+
+    pts.forEach(p => {
+      p.x = -(p.x - w);
+    });
+    api.setColor("blue");
+    api.drawCurve(curve, offset);
+
+    pts.forEach(p => {
+      p.y = -(p.y - h);
+    });
+    api.setColor("green");
+    api.drawCurve(curve, offset);
+
+    pts.forEach(p => {
+      p.x = -(p.x - w);
+    });
+    api.setColor("purple");
+    api.drawCurve(curve, offset);
+
+    api.setColor("black");
+    api.setFill("black");
+    api.drawLine({x:w/2, y:h/2}, {x:w/2 + r -2, y:h/2}, offset);
+    api.drawLine({x:w/2, y:h/2}, {x:w/2, y:h/2 + r -2}, offset);
+    api.text("r = " + r, {x:w/2 + r/3, y:h/2 + 10}, offset);
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
+        <p>In the previous section we tried to approximate a circular arc with a quadratic curve,
+        and it mostly made us unhappy. Cubic curves are much better suited to this task, so what
+        do we need to do?</p>
+
+        <p>For cubic curves, we basically want the curve to pass through three points on the circle:
+        the start point, the mid point at "angle/2", and the end point at "angle". We then also need
+        to make sure the control points are such that the start and end tangent lines line up with the
+        circle's tangent lines at the start and end point.</p>
+
+        <p>The first thing we can do is "guess" what the curve should look like, based on the previously
+        outlined curve-through-three-points procedure. This will give use a curve with correct start, mid
+        and end points, but possibly incorrect derivatives at the start and end, because the control points
+        might not be in the right spot. We can then slide the control points along the lines that connect
+        them to their respective end point, until they effect the corrected derivative at the start and
+        end points.  However, if you look back at the section on fitting curves through three points, the
+        rules used were such that they optimized for a near perfect hemisphere, so using the same guess
+        won't be all that useful: guessing the solution based on knowing the solution is not really guessing.</p>
+
+        <p>So have a graphical look at a "bad" guess versus the true fit, where we'll be using the
+        bad guess and the description in the second paragraph to derive the maths for the true fit:</p>
+
+        <Graphic preset="arcfitting" title="Cubic Bézier arc approximation" setup={this.setup} draw={this.draw} onMouseMove={this.onMouseMove}/>
+
+        <p>We see two curves here; in blue, our "guessed" curve and its control points, and in grey/black,
+        the true curve fit, with proper control points that were shifted in, along line between our guessed
+        control points, such that the derivatives at the start and end points are correct.</p>
+
+        <p>We can already seethat cubic curves are a lot better than quadratic curves, and don't look all
+        that wrong until we go well past a quarter circle; ⅜th starts to hint at problems, and half a circle
+        has an obvious "gap" between the real circle and the cubic approximation. Anything past that just looks
+        plain ridiculous... but quarter curves actually look pretty okay!</p>
+
+        <p>So, maths time again: how okay is "okay"? Let's apply some more maths to find out.</p>
+
+        <p>Unlike for the quadratic curve, we can't use <i>t=0.5</i> as our reference point because by its
+        very nature it's one of the three points that are actually guaranteed to lie on the circular curve.
+        Instead, we need a different <i>t</i> value. If we run some analysis on the curve we find that the
+        actual <i>t</i> value at which the curve is furthest from what it should be is 0.211325 (rounded),
+        but we don't know "why", since finding this value involves root-finding, and is nearly impossible
+        to do symbolically without pages and pages of math just to express one of the possible solutions.</p>
+
+        <p>So instead of walking you through the derivation for that value, let's simply take that <i>t</i> value
+        and see what the error is for circular arcs with an angle ranging from 0 to 2π:</p>
+
+        <table><tbody><tr><td>
+          <p><img src="images/arc-c-2pi.gif"/></p>
+          <p>plotted for 0 ≤ φ ≤ 2π:</p>
+        </td><td>
+          <p><img src="images/arc-c-pi.gif"/></p>
+          <p>plotted for 0 ≤ φ ≤ π:</p>
+        </td><td>
+          <p><img src="images/arc-c-pi2.gif"/></p>
+          <p>plotted for 0 ≤ φ ≤ ½π:</p>
+        </td></tr></tbody></table>
+
+        <p>We see that cubic Bézier curves are much better when it comes to approximating circular arcs,
+        with an error of less than 0.027 at the two "bulge" points for a quarter circle (which had an
+        error of 0.06 for quadratic curves at the mid point), and an error near 0.001 for an eighth
+        of a circle, so we're getting less than half the error for a quarter circle, or: at a slightly
+        lower error, we're getting twice the arc. This makes cubic curves quite useful!</p>
+
+        <p>In fact, the precision of a cubic curve at a quarter circle is considered "good enough" by
+        so many people that it's generally considered "just fine" to use four cubic Bézier curves to
+        fake a full circle when no circle primitives are available; generally, people won't notice
+        that it's not a real circle unless you also happen to overlay an actual circle, so that
+        the difference becomes obvious.</p>
+
+        <p>So with the error analysis out of the way, how do we actually compute the coordinates
+        needed to get that "true fit" cubic curve? The first observation is that we already know
+        the start and end points, because they're the same as for the quadratic attempt:</p>
+
+        <p>\[ S = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \ , \ \  E = \begin{pmatrix} cos(φ) \\ sin(φ) \end{pmatrix} \]</p>
+
+        <p>But we now need to find two control points, rather than one. If we want the derivatives
+        at the start and end point to match the circle, then the first control point can only lie
+        somewhere on the vertical line through S, and the second control point can only lie somewhere
+        on the line tangent to point E, which means:</p>
+
+        <p>\[
+          C_1 = S + a \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix}
+        \]</p>
+
+        <p>where "a" is some scaling factor, and:</p>
+
+        <p>\[
+          C_2 = E + b \cdot \begin{pmatrix} -sin(φ) \\ cos(φ) \end{pmatrix}
+        \]</p>
+
+        <p>where "b" is also some scaling factor.</p>
+
+        <p>Starting with this information, we slowly maths our way to success, but I won't lie: the maths for
+        this is pretty trig-heavy, and it's easy to get lost if you remember (or know!) some of the core
+        trigonoetric identities, so if you just want to see the final result just skip past the next section!</p>
+
+        <div className="note">
+          <h2>Let's do this thing.</h2>
+
+          <p>Unlike for the quadratic case, we need some more information in order to compute <i>a</i> and <i>b</i>,
+          since they're no longer dependent variables. First, we observe that the curve is symmetrical, so whatever
+          values we end up finding for C<sub>1</sub> will apply to C<sub>2</sub> as well (rotated along its tangent),
+          so we'll focus on finding the location of C<sub>1</sub> only. So here's where we do something that you might
+          not expect: we're going to ignore for a moment, because we're going to have a much easier time if we just
+          solve this problem with geometry first, then move to calculus to solve a much simpler problem.</p>
+
+          <p>If we look at the triangle that is formed between our starting point, or initial guess C<sub>1</sub>
+          and our real C<sub>1</sub>, there's something funny going on: if we treat the line {start,guess} as
+          our opposite side, the line {guess,real} as our adjacent side, with {start,real} our hypothenuse, then
+          the angle for the corner hypothenuse/adjacent is half that of the arc we're covering. Try it: if you
+          place the end point at a quarter circle (pi/2, or 90 degrees), the angle in our triangle is half a
+          quarter (pi/4, or 45 degrees). With that knowledge, and a knowledge of what the length of any of
+          our lines segments are (as a function), we can determine where our control points are, and thus have
+          everything we need to find the error distance function. Of the three lines, the one we can easiest
+          determine is {start,guess}, so let's find out what the guessed control point is. Again geometrically,
+          because we have the benefit of an on-curve <i>t=0.5</i> value.</p>
+
+          <p>The distance from our guessed point to the start point is exactly the same as the projection distance
+          we looked at earlier. Using <i>t=0.5</i> as our point "B" in the "A,B,C" projection, then we know the
+          length of the line segment {C,A}, since it's d<sub>1</sub> = {A,B} + d<sub>2</sub> = {B,C}:</p>
+
+          <p>\[
+            ||{A,C}|| = d_2 + d_1 = d_2 + d_2 \cdot ratio_3 \left(\frac{1}{2}\right) = d_2 + \frac{1}{3}d_2 = \frac{4}{3}d_2
+          \]</p>
+
+          <p>So that just leaves us to find the distance from <i>t=0.5</i> to the baseline for an arbitrary
+          angle φ, which is the distance from the centre of the circle to our <i>t=0.5</i> point, minus the
+          distance from the centre to the line that runs from start point to end point. The first is the
+          same as the point P we found for the quadratic curve:</p>
+
+          <p>\[
+            P_x = cos(\frac{φ}{2}) \ , \ \  P_y = sin(\frac{φ}{2})
+          \]</p>
+
+          <p>And the distance from the origin to the line start/end is another application of angles,
+          since the triangle {origin,start,C} has known angles, and two known sides. We can find
+          the length of the line {origin,C}, which lets us trivially compute the coordinate for C:</p>
+
+          <p>\[\begin{array}{l}
+            l = cos(\frac{φ}{2}) \ , \\
+            \left\{\begin{array}{l}
+              C_x = l \cdot cos\left(\frac{φ}{2}\right) = cos^2\left(\frac{φ}{2}\right)\ , \\
+              C_y = l \cdot sin\left(\frac{φ}{2}\right) = cos(\frac{φ}{2}) \cdot sin\left(\frac{φ}{2}\right)\ , \\
+            \end{array}\right.
+          \end{array}\]</p>
+
+          <p>With the coordinate C, and knowledge of coordinate B, we can determine coordinate A, and get a vector
+          that is identical to the vector {start,guess}:</p>
+
+          <p>\[\left\{\begin{array}{l}
+            B_x - C_x = cos\left(\frac{φ}{2}\right) - cos^2\left(\frac{φ}{2}\right) \\
+            B_y - C_y = sin\left(\frac{φ}{2}\right) - cos(\frac{φ}{2}) \cdot sin\left(\frac{φ}{2}\right)
+                      = sin\left(\frac{φ}{2}\right) - \frac{sin(φ)}{2}
+          \end{array}\right.\]</p>
+
+          <p>\[\left\{\begin{array}{l}
+            \vec{v}_x = \{C,A\}_x = \frac{4}{3} \cdot (B_x - C_x) \\
+            \vec{v}_y = \{C,A\}_y = \frac{4}{3} \cdot (B_y - C_y)
+          \end{array}\right.\]</p>
+
+          <p>Which means we can now determine the distance {start,guessed}, which is the same as the distance
+            {C,A}, and use that to determine the vertical distance from our start point to our C<sub>1</sub>:</p>
+
+          <p>\[\left\{\begin{array}{l}
+            C_{1x} = 1 \\
+            C_{1y} = \frac{d}{sin\left(\frac{φ}{2}\right)}
+                   = \frac{\sqrt{\vec{v}^2_x + \vec{v}^2_y}}{sin\left(\frac{φ}{2}\right)}
+                   = \frac{4}{3} tan \left( \frac{φ}{4} \right)
+          \end{array}\right.\]</p>
+
+          <p>And after this tedious detour to find the coordinate for C<sub>1</sub>, we can
+          find C<sub>2</sub> fairly simply, since it's lies at distance -C<sub>1y</sub> along the end point's tangent:</p>
+
+          <p>\[\begin{array}{l}
+            E'_x = -sin(φ) \ , \ E'_y = cos(φ) \ , \ ||E'|| = \sqrt{ (-sin(φ))^2 + cos^2(φ)} = 1 \ , \\
+            \left\{\begin{array}{l}
+              C_2x = E_x - C_{1y} \cdot \frac{E_x'}{||E'||}
+                   = cos(φ) + C_{1y} \cdot sin(φ)
+                   = cos(φ) + \frac{4}{3} tan \left( \frac{φ}{4} \right) \cdot sin(φ) \\
+              C_2y = E_y - C_{1y} \cdot \frac{E_y'}{||E'||}
+                   = sin(φ) - C_{1y} \cdot cos(φ)
+                   = sin(φ) - \frac{4}{3} tan \left( \frac{φ}{4} \right) \cdot cos(φ)
+            \end{array}\right.
+          \end{array}\]</p>
+
+          <p>And that's it, we have all four points now for an approximation of an arbitrary
+            circular arc with angle φ.</p>
+        </div>
+
+        <p>So, to recap, given an angle φ, the new control coordinates are:</p>
+
+        <p>\[
+          C_1 = \left [ \begin{matrix}
+            1 \\
+            f
+          \end{matrix} \right ],\ with\ f = \frac{4}{3} tan \left( \frac{φ}{4} \right)
+        \]</p>
+
+        <p>and</p>
+
+        <p>\[
+          C_2 = \left [ \begin{matrix}
+            cos(φ) + f \cdot sin(φ) \\
+            sin(φ) - f \cdot cos(φ)
+          \end{matrix} \right ],\ with\ f = \frac{4}{3} tan \left( \frac{φ}{4} \right)
+        \]</p>
+
+        <p>And, because the "quarter curve" special case comes up so incredibly often, let's look at what
+        these new control points mean for the curve coordinates of a quarter curve, by simply filling in
+        φ = π/2:</p>
+
+        <p>\[\begin{array}{l}
+          S = (1, 0) \ , \
+          C_1 = \left ( 1, 4 \frac{\sqrt{2}-1}{3} \right ) \ , \
+          C_2 = \left ( 4 \frac{\sqrt{2}-1}{3} , 1 \right ) \ , \
+          E = (0, 1)
+        \end{array}\]</p>
+
+        <p>Which, in decimal values, rounded to six significant digits, is:</p>
+
+        <p>\[\begin{array}{l}
+          S = (1, 0) \ , \
+          C_1 = (1, 0.55228) \ , \
+          C_2 = (0.55228 , 1) \ , \
+          E = (0, 1)
+        \end{array}\]</p>
+
+        <p>Of course, this is for a circle with radius 1, so if you have a different radius circle,
+        simply multiply the coordinate by the radius you need. And then finally, forming a full curve
+        is now a simple a matter of mirroring these coordinates about the origin:</p>
+
+        <Graphic preset="simple" title="Cubic Bézier circle approximation" draw={this.drawCircle} static={true}/>
+      </section>
+    );
+  }
+});
+
+module.exports = CirclesCubic;

components/sections/graduatedoffset/index.js

@@ -1,3 +1,58 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var GraduatedOffsetting = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Graduated curve offsetting"
+    };
+  },
+
+  setup: function(api, curve) {
+    api.setCurve(curve);
+    api.distance = 20;
+  },
+
+  setupQuadratic: function(api) {
+    var curve = api.getDefaultQuadratic();
+    this.setup(api, curve);
+  },
+
+  setupCubic: function(api) {
+    var curve = api.getDefaultCubic();
+    this.setup(api, curve);
+  },
+
+  draw: function(api, curve) {
+    api.reset();
+    api.drawSkeleton(curve);
+    api.drawCurve(curve);
+
+
+    api.setColor("blue");
+    var outline = curve.outline(0,0,api.distance,api.distance);
+    outline.curves.forEach(c => api.drawCurve(c));
+  },
+
+  values: {
+    "38": 1,   // up arrow
+    "40": -1,  // down arrow
+  },
+
+  onKeyDown: function(e, api) {
+    var v = this.values[e.keyCode];
+    if(v) {
+      e.preventDefault();
+      api.distance += v;
+    }
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
         <p>What if we want to do graduated offsetting, starting at some distance <i>s</i> but ending
         at some other distance <i>e</i>? well, if we can compute the length of a curve (which we can
         if we use the Legendre-Gauss quadrature approach) then we can also determine how far "along the
@@ -21,46 +76,18 @@
           <li>end: <i>map(<strong>S+length(subcurve)</strong>, 0,L, s,e)</i></li>
         </ul>
 
-        At each of the relevant points (start, end, and the projections of the control points onto
+        <p>At each of the relevant points (start, end, and the projections of the control points onto
         the curve) we know the curve's normal, so offsetting is simply a matter of taking our original
         point, and moving it along the normal vector by the offset distance for each point. Doing so
         will give us the following result (these have with a starting width of 0, and an end width
-        of 40 pixels, but can be controlled with your + and - keys):</p>
+        of 40 pixels, but can be controlled with your up and down cursor keys):</p>
 
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Graduated offsetting a quadratic Bézier curve">
-        void setupCurve() {
-          setupDefaultQuadratic();
-          offsetting();
-          offset = 20;
-        }
+        <Graphic preset="simple" title="Offsetting a quadratic Bézier curve" setup={this.setupQuadratic} draw={this.draw} onKeyDown={this.onKeyDown}/>
+        <Graphic preset="simple" title="Offsetting a cubic Bézier curve" setup={this.setupCubic} draw={this.draw} onKeyDown={this.onKeyDown}/>
 
-        void drawCurve(BezierCurve curve) {
-          additionals();
-          curve.draw();
-          if(offset>0) {
-            noAdditionals();
-            BezierCurve[] offsetCurve = curve.offset(offset, 0, 1);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-            offsetCurve = curve.offset(-offset, 0, 1);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-          }
-        }</textarea>
+      </section>
+    );
+  }
+});
 
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Graduated offsetting a cubic Bézier curve">
-        void setupCurve() {
-          setupDefaultCubic();
-          offsetting();
-          offset = 20;
-        }
-
-        void drawCurve(BezierCurve curve) {
-          additionals();
-          curve.draw();
-          if(offset>0) {
-            noAdditionals();
-            BezierCurve[] offsetCurve = curve.offset(offset, 0, 1);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-            offsetCurve = curve.offset(-offset, 0, 1);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-          }
-        }</textarea>
+module.exports = GraduatedOffsetting;

components/sections/index.js

@@ -37,31 +37,23 @@ module.exports = {
   pointcurves: require("./pointcurves"),
 
   catmullconv: require("./catmullconv"),
-  catmullmoulding: require("./catmullmoulding")
-};
+  catmullmoulding: require("./catmullmoulding"),
 
+  /*
+    // This requires bezier.js to have a proper poly implementation
+    polybezier: require("./polybezier"),
+  */
 
-/*
-  polybezier: require("./polybezier"),
-  shapes: require("./shapes"),
+  /*
+    // This section is way too much work to port, and not worth implementing given paper.js etc.
+    shapes: require("./shapes"), //   Boolean shape operations
+  */
 
   projections: require("./projections"),
-
   offsetting: require("./offsetting"),
   graduatedoffset: require("./graduatedoffset"),
 
   circles: require("./circles"),
   circles_cubic: require("./circles_cubic"),
   arcapproximation: require("./arcapproximation")
-*/
-
-/*
-  Forming poly-Bézier curves
-  Boolean shape operations
-  Projecting a point onto a Bézier curve
-  Curve offsetting
-  Graduated curve offsetting
-  Circles and quadratic Bézier curves
-  Circles and cubic Bézier curves
-  Approximating Bézier curves with circular arcs
-*/
+};

components/sections/offsetting/index.js

@@ -1,3 +1,61 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var Offsetting = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Curve offsetting"
+    };
+  },
+
+  setup: function(api, curve) {
+    api.setCurve(curve);
+    api.distance = 20;
+  },
+
+  setupQuadratic: function(api) {
+    var curve = api.getDefaultQuadratic();
+    this.setup(api, curve);
+  },
+
+  setupCubic: function(api) {
+    var curve = api.getDefaultCubic();
+    this.setup(api, curve);
+  },
+
+  draw: function(api, curve) {
+    api.reset();
+    api.drawSkeleton(curve);
+    api.drawCurve(curve);
+
+    api.setColor("red");
+    var offset = curve.offset(api.distance);
+    offset.forEach(c => api.drawCurve(c));
+
+    api.setColor("blue");
+    offset = curve.offset(-api.distance);
+    offset.forEach(c => api.drawCurve(c));
+  },
+
+  values: {
+    "38": 1,   // up arrow
+    "40": -1,  // down arrow
+  },
+
+  onKeyDown: function(e, api) {
+    var v = this.values[e.keyCode];
+    if(v) {
+      e.preventDefault();
+      api.distance += v;
+    }
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
         <p>Perhaps you are like me, and you've been writing various small programs that use Bézier curves in some way or another,
         and at some point you make the step to implementing path extrusion. But you don't want to do it pixel based, you want to
         stay in the vector world. You find that extruding lines is relatively easy, and tracing outlines is coming along nicely
@@ -11,7 +69,7 @@
         <p>Bottom line: <strong>you can't</strong>. So you cheat. We're not going to do true curve scaling, or rather curve
         offsetting, because that's impossible. Instead we're going to try to generate 'looks good enough' offset curves.</p>
 
-        <div class="note">
+        <div className="note">
           <h2>"What do you mean, you can't. Prove it."</h2>
 
           <p>First off, when I say "you can't" what I really mean is "you can't offset a Bézier curve with another
@@ -95,47 +153,20 @@
         with respect to the curve's scaling origin (which is the intersection of the point normals at the start
         and end points).</p>
 
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Offsetting a quadratic Bézier curve">
-        void setupCurve() {
-          setupDefaultQuadratic();
-          offsetting();
-          offset = 20;
-        }
+        <p>The following graphics show off curve offsetting, and you can use your up and down cursor keys to control
+        the distance at which the curve gets offset:</p>
 
-        void drawCurve(BezierCurve curve) {
-          additionals();
-          curve.draw();
-
-          if(offset>0) {
-            noAdditionals();
-            BezierCurve[] offsetCurve = curve.offset(offset);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-            offsetCurve = curve.offset(-offset);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-          }
-        }</textarea>
-
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Offsetting a cubic Bézier curve">
-        void setupCurve() {
-          setupDefaultCubic();
-          offsetting();
-          offset = 20;
-        }
-
-        void drawCurve(BezierCurve curve) {
-          additionals();
-          curve.draw();
-
-          if(offset>0) {
-            noAdditionals();
-            BezierCurve[] offsetCurve = curve.offset(offset);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-            offsetCurve = curve.offset(-offset);
-            for(BezierCurve b: offsetCurve) { b.draw(); b.getPoint(0).draw(); b.getPoint(1).draw();}
-          }
-        }</textarea>
+        <Graphic preset="simple" title="Offsetting a quadratic Bézier curve" setup={this.setupQuadratic} draw={this.draw} onKeyDown={this.onKeyDown} />
+        <Graphic preset="simple" title="Offsetting a cubic Bézier curve" setup={this.setupCubic} draw={this.draw} onKeyDown={this.onKeyDown} />
 
         <p>You may notice that this may still lead to small 'jumps' in the sub-curves when moving the
         curve around. This is caused by the fact that we're still performing a naive form of offsetting,
         moving the control points the same distance as the start and end points. If the curve is large
         enough, this may still lead to incorrect offsets.</p>
+
+      </section>
+    );
+  }
+});
+
+module.exports = Offsetting;

components/sections/polybezier/index.js

@@ -0,0 +1,172 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var PolyBezier = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Forming poly-Bézier curves"
+    };
+  },
+
+  setupQuadratic: function(api) {
+    var w = api.getPanelWidth(),
+        h = api.getPanelHeight(),
+        cx = w/2, cy = h/2, pad = 40,
+        pts = [
+          // first curve:
+          {x:cx,y:pad}, {x:w-pad,y:pad}, {x:w-pad,y:cy},
+          // subsequent curve
+          {x:w-pad,y:h-pad}, {x:cx,y:h-pad},
+          // subsequent curve
+          {x:pad,y:h-pad}, {x:pad,y:cy},
+          // final curve control point
+          {x:pad,y:pad},
+        ];
+    api.lpts = pts;
+  },
+
+  setupCubic: function(api) {
+  },
+
+  draw: function(api, curves) {
+    api.reset();
+    var pts = api.lpts;
+
+    var c1 = new api.Bezier(pts[0],pts[1],pts[2]);
+    api.drawSkeleton(c1);
+    api.drawCurve(c1);
+
+    var c2 = new api.Bezier(pts[2],pts[3],pts[4]);
+    api.drawSkeleton(c2);
+    api.drawCurve(c2);
+
+    var c3 = new api.Bezier(pts[4],pts[5],pts[6]);
+    api.drawSkeleton(c3);
+    api.drawCurve(c3);
+
+    var c4 = new api.Bezier(pts[6],pts[7],pts[0]);
+    api.drawSkeleton(c4);
+    api.drawCurve(c4);
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
+        <p>Much like lines can be chained together to form polygons, Bézier curves can be chained together
+        to form poly-Béziers, and the only trick required is to make sure that: A) the end point of each
+        section is the starting point of the following section, and B) the derivatives across that
+        dual point line up. Unless, of course, you want discontinuities; then you don't even need (B).</p>
+
+        <p>We'll cover three forms of poly-Bézier curves in this section. First, we'll look at the kind
+        that enforces "the outgoing derivative is the same as the incoming derivative" across sections:</p>
+
+        <p>\[
+          B'(1)_n = B'(0)_{n+1}
+        \]</p>
+
+        <p>We can actually guarantee this really easily, because we know that the vector from a curve's
+        last control point to its last on-curve point is equal to the derivative vector. If we want to
+        ensure that the first control point of the next curve matches that, all we have to do is mirror
+        that last control point through the last on-curve point. And mirroring any point A through any
+        point B is really simple:</p>
+
+        <p>\[
+          Mirrored = \left [
+            \begin{matrix} B_x + (B_x - A_x) \\  B_y + (B_y - A_y) \end{matrix}
+          \right ] = \left [
+            \begin{matrix} 2B_x - A_x \\  2B_y - A_y \end{matrix}
+          \right ]
+        \]</p>
+
+        <p>So let's implement that and see what it gets us. The following two graphics show a quadratic
+        and a cubic poly-Bézier curve; both consist of multiple sub-curves, but because of our constraint,
+        not all points on the curves can be moved around freely. Some points, when moved, will move other
+        points by virtue of changing the curve across sections.</p>
+
+        <Graphic preset="poly" title="Forming a quadratic poly-Bézier" setup={this.setupQuadratic} draw={this.draw}/>
+
+        <textarea className="sketch-code" data-sketch-preset="poly" data-sketch-title="Forming a cubic poly-Bézier">
+        void setupCurve() {
+          setupDefaultCubicPoly();
+        }
+
+        void movePoint(PolyBezierCurve p, int pt, int mx, int my) {
+            p.movePointConstrained(pt, mx, my);
+        }</textarea>
+
+        <p>As you can see, quadratic curves are particularly ill-suited for poly-Bézier curves, as all
+        the control points are effectively linked. Move one of them, and you move all of them. This means
+        that we cannot use quadratic poly-Béziers for anything other than really, really simple shapes.
+        And even then, they're probably the wrong choice. Cubic curves are pretty decent, but the fact
+        that the derivatives are linked means we can't manipulate curves as well as we might if we
+        relaxed the constraints a little.</p>
+
+        <p>So: let's relax them!</p>
+
+        <p>We can change the constraint so that we still preserve the angle of the derivatives across
+        sections (so transitions from one section to the next will still look natural), but give up
+        the requirement that they should also have the same vector length. Doing so will give us
+        a much more a useful kind of poly-Bézier curve:</p>
+
+        <textarea className="sketch-code" data-sketch-preset="poly" data-sketch-title="A half-constrained quadratic poly-Bézier">
+        void setupCurve() {
+          setupDefaultQuadraticPoly();
+        }
+
+        void movePoint(PolyBezierCurve p, int pt, int mx, int my) {
+            p.movePointHalfConstrained(pt, mx, my);
+        }</textarea>
+
+        <textarea className="sketch-code" data-sketch-preset="poly" data-sketch-title="A half-constrained cubic poly-Bézier">
+        void setupCurve() {
+          setupDefaultCubicPoly();
+        }
+
+        void movePoint(PolyBezierCurve p, int pt, int mx, int my) {
+            p.movePointHalfConstrained(pt, mx, my);
+        }</textarea>
+
+        <p>Quadratic curves are still silly, but cubic curves are now much more controllable.</p>
+
+        <p>If we want even more control, we could just abandon the derivative constraints entirely,
+        and simply assure that the end point of one section is the same as the start point of the next section,
+        and then keep it at that. This gives us the greatest degree of freedom when it comes to modelling
+        shapes, but also means that our poly-Bézier constructs are no longer continuous curves. Sometimes
+        this is exactly what you want (because it lets you add corners to a shape, while still only using
+        Bézier curves).</p>
+
+        <textarea className="sketch-code" data-sketch-preset="poly" data-sketch-title="An unconstrained quadratic poly-Bézier">
+        void setupCurve() {
+          setupDefaultQuadraticPoly();
+        }
+
+        void movePoint(PolyBezierCurve p, int pt, int mx, int my) {
+          p.movePoint(pvt, mx, my);
+        }</textarea>
+
+        <textarea className="sketch-code" data-sketch-preset="poly" data-sketch-title="An unconstrained cubic poly-Bézier">
+        void setupCurve() {
+          setupDefaultCubicPoly();
+        }
+
+        void movePoint(PolyBezierCurve p, int pt, int mx, int my) {
+          p.movePoint(pvt, mx, my);
+        }</textarea>
+
+        <p>When doing any kind of modelling, you generally don't want a poly-Bézier that will only let you
+        pick one of the three forms for all your points; most graphics applications that deal with Bézier
+        curves will actually let you pick, per on-curve point, how to deal with the control points around it:
+        fully constrained, loosely constrained, or completely unconstrained. The best shape modelling comes
+        from having a curve that will let you pick what you need, when you need it, without having to start
+        a new poly-Bézier curve.</p>
+
+
+      </section>
+    );
+  }
+});
+
+module.exports = PolyBezier;

components/sections/projections/index.js

@@ -0,0 +1,102 @@
+var React = require("react");
+var Graphic = require("../../Graphic.jsx");
+var SectionHeader = require("../../SectionHeader.jsx");
+
+var Projections = React.createClass({
+  getDefaultProps: function() {
+    return {
+      title: "Projecting a point onto a Bézier curve"
+    };
+  },
+
+  setup: function(api) {
+    api.setSize(320,320);
+    var curve = new api.Bezier([
+      {x:248,y:188},
+      {x:218,y:294},
+      {x:45,y:290},
+      {x:12,y:236},
+      {x:14,y:82},
+      {x:186,y:177},
+      {x:221,y:90},
+      {x:18,y:156},
+      {x:34,y:57},
+      {x:198,y:18}
+    ]);
+    api.setCurve(curve);
+    api._lut = curve.getLUT();
+  },
+
+  findClosest: function(LUT, p, dist) {
+    var i,
+        end = LUT.length,
+        d,
+        dd = dist(LUT[0],p),
+        f = 0;
+    for(i=1; i<end; i++) {
+      d = dist(LUT[i],p);
+      if(d<dd) {f = i;dd = d;}
+    }
+    return f/(end-1);
+  },
+
+  draw: function(api, curve) {
+    api.reset();
+    api.drawSkeleton(curve);
+    api.drawCurve(curve);
+    if (api.mousePt) {
+      api.setColor("red");
+      api.drawCircle(api.mousePt, 3);
+      // naive t value
+      var t = this.findClosest(api._lut, api.mousePt, api.utils.dist);
+      // no real point in refining for illustration purposes
+      var p = curve.get(t);
+      api.drawLine(p, api.mousePt);
+      api.drawCircle(p, 3);
+    }
+  },
+
+  onMouseMove: function(evt, api) {
+    api.mousePt = {x: evt.offsetX, y: evt.offsetY };
+  },
+
+  render: function() {
+    return (
+      <section>
+        <SectionHeader {...this.props} />
+
+        <p>Say we have a Bézier curve and some point, not on the curve, of which we want to know
+        which <i>t</i> value on the curve gives us an on-curve point closest to our off-curve point.
+        Or: say we want to find the projection of a random point onto a curve. How do we do that?</p>
+
+        <p>If the Bézier curve is of low enough order, we might be able
+        to <a href="http://jazzros.blogspot.ca/2011/03/projecting-point-on-bezier-curve.html">work out
+        the maths for how to do this</a>, and get a perfect <i>t</i> value back, but in general this is
+        an incredibly hard problem and the easiest solution is, really, a numerical approach again. We'll
+        be finding our ideal <i>t</i> value using a <a href="https://en.wikipedia.org/wiki/Binary_search_algorithm">binary
+        search</a>. First, we do a coarse distance-check based on <i>t</i> values associated with the
+        curve's "to draw" coordinates (using a lookup table, or LUT). This is pretty fast. Then we run
+        this algorithm:</p>
+
+        <ol>
+          <li>with the <i>t</i> value we found, start with some small interval around <i>t</i> (1/length_of_LUT on either side is a reasonable start),</li>
+          <li>if the distance to <i>t ± interval/2</i> is larger than the distance to <i>t</i>, try again with the interval reduced to half its original length.</li>
+          <li>if the distance to <i>t ± interval/2</i> is smaller than the distance to <i>t</i>, replace <i>t</i> with the smaller-distance value.</li>
+          <li>after reducing the interval, or changing <i>t</i>, go back to step 1.</li>
+        </ol>
+
+        <p>We keep repeating this process until the interval is small enough to claim the difference
+        in precision found is irrelevant for the purpose we're trying to find <i>t</i> for. In this
+        case, I'm arbitrarily fixing it at 0.0001.</p>
+
+        <p>The following graphic demonstrates the result of this procedure.Simply move the cursor
+        around, and if it does not lie on top of the curve, you will see a line that projects the
+        cursor onto the curve based on an iteratively found "ideal" <i>t</i> value.</p>
+
+        <Graphic preset="simple" title="Projecting a point onto a Bézier curve" setup={this.setup} draw={this.draw} onMouseMove={this.onMouseMove}/>
+      </section>
+    );
+  }
+});
+
+module.exports = Projections;

components/sections/reordering/index.js

@@ -158,27 +158,3 @@ var Reordering = React.createClass({
 });
 
 module.exports = Reordering;
-
-/*
-void setupCurve() {
-          int d = dim - 2*pad;
-          int order = 10;
-          ArrayList<Point> pts = new ArrayList<Point>();
-
-          float dst = d/2.5, nx, ny, a=0, step = 2*PI/order, r;
-          for(a=0; a<2*PI; a+=step) {
-            r = random(-dst/4,dst/4);
-            pts.add(new Point(d/2 + cos(a) * (r+dst), d/2 + sin(a) * (r+dst)));
-            dst -= 1.2;
-          }
-
-          Point[] points = new Point[pts.size()];
-          for(int p=0,last=points.length; p<last; p++) { points[p] = pts.get(p); }
-          curves.add(new BezierCurve(points));
-          reorder();
-        }
-
-        void drawCurve(BezierCurve curve) {
-          curve.draw();
-        }</textarea>
- */

data/circles_cubic.jsx

@@ -1,270 +0,0 @@
-        <p>For cubic curves the control points must be each other's mirror around the line running
-        through the baseline midpoint, at a right angle, and again the derivatives at the start and
-        end points must agree. Again we don't have altogether that much choice: there is only one
-        pair of control points that guarantees correct derivatives for the start and end points,
-        while also making the midpoint of the curve lie on top of the curve.</p>
-
-        <p>In order to find a cubic curve, we first "guess" the curve, based on the previously outlined
-        curve-through-three-points procedure. This will give use a curve with correct start, mid and
-        end points, but incorrect derivatives for start and end, given the control points. We then
-        slide the control points along the line that connects them until they effect the corrected
-        derivative at the start and end points (you may remember that the derivative at the start
-        is aligned with the line from start point to control point 1, and that the derivative at the
-        end is aligned with the line from control point 2 to the end point).</p>
-
-        <textarea class="sketch-code" data-sketch-preset="arcfitting" data-sketch-title="Cubic Bézier arc approximation">
-        void setupCurve() { order = 3; }
-
-        void checkConnect() {
-          if(e < PI) { noConnect(); } else { connect(); }
-        }
-
-        void findArcFitting() {
-          // guess the curve based on the start/mid/end points:
-          Point p1 = new Point(dim/2 + dim/f2,dim/2),
-                p2 = new Point(dim/2 + dim/f2*cos((s+e)/2), dim/2 + dim/f2*sin((s+e)/2)),
-                p3 = new Point(dim/2 + dim/f2*cos(ax), dim/2 + dim/f2*sin(ay));
-          BezierCurve guess = comp.generateCurve(3,p1,p2,p3);
-
-          Point oc1 = guess.points[1];
-
-          drawGuess(guess);
-
-          // then, move the control points so that B'(0) and B'(1) are correct:
-          Point c1 = comp.lli(new Point[]{
-            guess.points[0],
-            new Point(p1.x, p1.y+10),
-            guess.points[1],
-            guess.points[2],
-          });
-
-          // taking advantage of symmetry, we trivially know c2 now, too:
-          dx = guess.points[1].x - c1.x;
-          dy = guess.points[1].y - c1.y;
-          Point c2 = new Point(guess.points[2].x + dx,
-                               guess.points[2].y + dy);
-
-          // replace, update, and draw.
-          guess.points[1] = c1;
-          guess.points[2] = c2;
-          guess.update();
-          guess.draw();
-
-          float a = 0.211325;
-          Point pa = guess.getPoint(a);
-          pa.draw();
-          guess.getPoint(1-a).draw();
-        }</textarea>
-
-        <p>We see two curves here; very faintly the "guessed" curve, and drawn normally, the proper curve
-        with the control points shifted along the control line so that the derivatives at the start and end
-        points are correct. With this, we can see that cubic curves are actually a lot better than quadratic
-        curves, and don't look all that wrong until we go past a quarter circle; ⅜th starts to hint at
-        problems, and half a circle has an obvious "gap" between the real circle and the cubic approximation.
-        Anything past that just looks plain ridiculous... but quarter curves actually look pretty okay!
-        Again, how okay is okay? Let's apply some more maths to find out.</p>
-
-        <p>Unlike for the quadratic curve, we can't use <i>t=0.5</i> as our reference point because by its
-        very nature it's one of the three points that are actually guaranteed to lie on the circular curve.
-        Instead, we need a different <i>t</i> value. If we run some analysis on the curve we find that the
-        actual <i>t</i> value at which the curve is furthest from what it should be is 0.211325 (rounded),
-        but we don't know "why", since finding this value involves root-finding, and is nearly impossible
-        to do symbolically without pages and pages of math just to express one of the possible solutions.</p>
-
-        <p>So instead, let's simply take that <i>t</i> value and see what the error is for circular arcs
-        with an angle ranging from 0 to 2π:</p>
-
-        <table><tr><td>
-          <p><img src="images/arc-c-2pi.gif"></p>
-          <p>plotted for 0 ≤ φ ≤ 2π:</p>
-        </td><td>
-          <p><img src="images/arc-c-pi.gif"></p>
-          <p>plotted for 0 ≤ φ ≤ π:</p>
-        </td><td>
-          <p><img src="images/arc-c-pi2.gif"></p>
-          <p>plotted for 0 ≤ φ ≤ ½π:</p>
-        </td></tr></table>
-
-        <p>We see that cubic Bézier curves are much better when it comes to approximating circular arcs,
-        with an error of less than 0.027 at the two "bulge" points for a quarter circle (which had an
-        error of 0.06 for quadratic curves at the mid point), and an error near 0.001 for an eighth
-        of a circle, so we're getting less than half the error for a quarter circle, or: at a slightly
-        lower error, we're getting twice the arc. This makes cubic curves quite useful.
-        In fact, the precision of a cubic curve at a quarter circle is considered "good enough" by many
-        to justify using four cubic Bézier curves to fake a full circle when no circle primitives are
-        available; generally, people will not notice it's not a real circle unless you overlay the
-        actual circle so they can see the difference.</p>
-
-        <p>So if we want to use a cubic Bézier curve, where do the curve's points go?
-        The start and end point are the same as before:</p>
-
-        <p>\[ S = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \ , \ \  E = \begin{pmatrix} cos(φ) \\ sin(φ) \end{pmatrix} \]</p>
-
-        <p>But we now need to find two control points, rather than one:</p>
-
-        <p>\[
-          C_1 = S + a \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix} \ \ , \ \
-          C_2 = E + b \cdot \begin{pmatrix} -sin(φ) \\ cos(φ) \end{pmatrix}
-        \]</p>
-
-        <div class="note">
-                <h2>Let's do this thing.</h2>
-        
-                <p>Unlike for the quadratic case, we need some more information in order to compute <i>a</i> and <i>b</i>,
-                since they're no longer dependent variables. First, we observe that the curve is symmetrical, so whatever
-                values we end up finding for C<sub>1</sub> will apply to C<sub>2</sub> as well (rotated along its tangent),
-                so we'll focus on finding the location of C<sub>1</sub> only. So here's where we do something that you might
-                not expect: we're going to ignore for a moment, because we're going to have a much easier time if we just
-                solve this problem with geometry first, then move to calculus to solve a much simpler problem.</p>
-        
-                <p>If we look at the triangle that is formed between our starting point, or initial guess C<sub>1</sub>
-                and our real C<sub>1</sub>, there's something funny going on: if we treat the line {start,guess} as
-                our opposite side, the line {guess,real} as our adjacent side, with {start,real} our hypothenuse, then
-                the angle for the corner hypothenuse/adjacent is half that of the arc we're covering. Try it: if you
-                place the end point at a quarter circle (pi/2, or 90 degrees), the angle in our triangle is half a
-                quarter (pi/4, or 45 degrees). With that knowledge, and a knowledge of what the length of any of
-                our lines segments are (as a function), we can determine where our control points are, and thus have
-                everything we need to find the error distance function. Of the three lines, the one we can easiest
-                determine is {start,guess}, so let's find out what the guessed control point is. Again geometrically,
-                because we have the benefit of an on-curve <i>t=0.5</i> value.</p>
-        
-                <p>The distance from our guessed point to the start point is exactly the same as the projection distance
-                we looked at earlier. Using <i>t=0.5</i> as our point "B" in the "A,B,C" projection, then we know the
-                length of the line segment {C,A}, since it's d<sub>1</sub> = {A,B} + d<sub>2</sub> = {B,C}:</p>
-        
-                <p>\[
-                  ||{A,C}|| = d_2 + d_1 = d_2 + d_2 \cdot ratio_3 \left(\frac{1}{2}\right) = d_2 + \frac{1}{3}d_2 = \frac{4}{3}d_2
-                \]</p>
-        
-                <p>So that just leaves us to find the distance from <i>t=0.5</i> to the baseline for an arbitrary
-                angle φ, which is the distance from the centre of the circle to our <i>t=0.5</i> point, minus the
-                distance from the centre to the line that runs from start point to end point. The first is the
-                same as the point P we found for the quadratic curve:</p>
-        
-                <p>\[
-                  P_x = cos(\frac{φ}{2}) \ , \ \  P_y = sin(\frac{φ}{2})
-                \]</p>
-        
-                <p>And the distance from the origin to the line start/end is another application of angles,
-                since the triangle {origin,start,C} has known angles, and two known sides. We can find
-                the length of the line {origin,C}, which lets us trivially compute the coordinate for C:</p>
-        
-                <p>\[\begin{array}{l}
-                  l = cos(\frac{φ}{2}) \ , \\
-                  \left\{\begin{array}{l}
-                    C_x = l \cdot cos\left(\frac{φ}{2}\right) = cos^2\left(\frac{φ}{2}\right)\ , \\
-                    C_y = l \cdot sin\left(\frac{φ}{2}\right) = cos(\frac{φ}{2}) \cdot sin\left(\frac{φ}{2}\right)\ , \\
-                  \end{array}\right.
-                \end{array}\]</p>
-        
-                <p>With the coordinate C, and knowledge of coordinate B, we can determine coordinate A, and get a vector
-                that is identical to the vector {start,guess}:</p>
-        
-                <p>\[\left\{\begin{array}{l}
-                  B_x - C_x = cos\left(\frac{φ}{2}\right) - cos^2\left(\frac{φ}{2}\right) \\
-                  B_y - C_y = sin\left(\frac{φ}{2}\right) - cos(\frac{φ}{2}) \cdot sin\left(\frac{φ}{2}\right)
-                            = sin\left(\frac{φ}{2}\right) - \frac{sin(φ)}{2}
-                \end{array}\right.\]</p>
-        
-                <p>\[\left\{\begin{array}{l}
-                  \vec{v}_x = \{C,A\}_x = \frac{4}{3} \cdot (B_x - C_x) \\
-                  \vec{v}_y = \{C,A\}_y = \frac{4}{3} \cdot (B_y - C_y)
-                \end{array}\right.\]</p>
-        
-                <p>Which means we can now determine the distance {start,guessed}, which is the same as the distance
-                  {C,A}, and use that to determine the vertical distance from our start point to our C<sub>1</sub>:</p>
-        
-                <p>\[\left\{\begin{array}{l}
-                  C_{1x} = 1 \\
-                  C_{1y} = \frac{d}{sin\left(\frac{φ}{2}\right)}
-                         = \frac{\sqrt{\vec{v}^2_x + \vec{v}^2_y}}{sin\left(\frac{φ}{2}\right)}
-                         = \frac{4}{3} tan \left( \frac{φ}{4} \right)
-                \end{array}\right.\]</p>
-        
-                <p>And after this tedious detour to find the coordinate for C<sub>1</sub>, we can find C<sub>2</sub>
-                fairly simply, since it's lies at distance -C<sub>1y</sub> along the end point's tangent:</p>
-        
-                <p>\[\begin{array}{l}
-                  E'_x = -sin(φ) \ , \ E'_y = cos(φ) \ , \ ||E'|| = \sqrt{ (-sin(φ))^2 + cos^2(φ)} = 1 \ , \\
-                  \left\{\begin{array}{l}
-                    C_2x = E_x - C_{1y} \cdot \frac{E_x'}{||E'||}
-                         = cos(φ) + C_{1y} \cdot sin(φ)
-                         = cos(φ) + \frac{4}{3} tan \left( \frac{φ}{4} \right) \cdot sin(φ) \\
-                    C_2y = E_y - C_{1y} \cdot \frac{E_y'}{||E'||}
-                         = sin(φ) - C_{1y} \cdot cos(φ)
-                         = sin(φ) - \frac{4}{3} tan \left( \frac{φ}{4} \right) \cdot cos(φ)
-                  \end{array}\right.
-                \end{array}\]</p>
-        
-                <p>And that's it, we have all four points now for an approximation of an arbitrary
-                  circular arc with angle φ.</p>
-        </div>
-
-        <p>If you skipped the derivation for the "true" formula in the hopes of finding some useful
-        information, then you're in luck: there are many possible angles with which to approximate
-        sections of a circle, but by far the most common one is using one curve for each quarter
-        circle, and it turns that while deriving the function to get the right value is a bit of a
-        pain, the actual final values are pretty simple, even as precise functions. So, which
-        values do we plug in?:</p>
-
-        <p>\[\begin{array}{l}
-          S = (1, 0) \ , \
-          C_1 = \left ( 1, 4 \frac{\sqrt{2}-1}{3} \right ) \ , \
-          C_2 = \left ( 4 \frac{\sqrt{2}-1}{3} , 1 \right ) \ , \
-          E = (0, 1)
-        \end{array}\]</p>
-
-        <p>Which, in decimal values, rounded to six significant digits, is:</p>
-
-        <p>\[\begin{array}{l}
-          S = (1, 0) \ , \
-          C_1 = (1, 0.55228) \ , \
-          C_2 = (0.55228 , 1) \ , \
-          E = (0, 1)
-        \end{array}\]</p>
-        
-        <p>Note that this is for a circle with radius 1, so if you have a different radius circle,
-        simply multiply the coordinate by the radius you need, and done! Finally, forming the
-        full curve is  now a simple a matter of mirroring these coordinates about the origin:</p>
-
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Cubic Bézier circle approximation">
-        void setupCurve() {
-          setupDefaultCubic();
-          noLabels();
-        }
-
-        void drawCurve(BezierCurve curve) {
-          float ox = dim/2, oy = dim/2, r = dim/2.5;
-
-          stroke(150);
-          line(0,dim/2,dim,dim/2);
-          line(dim/2,0,dim/2,dim);
-
-          (new BezierCurve(new Point[]{
-            new Point(ox + r*1,       oy + r*0),
-            new Point(ox + r*1,       oy + r*0.55228),
-            new Point(ox + r*0.55228, oy + r*1),
-            new Point(ox + r*0,       oy + r*1)
-          })).draw(color(255,0,0));
-
-          (new BezierCurve(new Point[]{
-            new Point(ox + r*0,       oy + r*1),
-            new Point(ox - r*0.55228, oy + r*1),
-            new Point(ox - r*1,       oy + r*0.55228),
-            new Point(ox - r*1,       oy + r*0)
-          })).draw();
-
-          (new BezierCurve(new Point[]{
-            new Point(ox - r*1,       oy + r*0),
-            new Point(ox - r*1,       oy - r*0.55228),
-            new Point(ox - r*0.55228, oy - r*1),
-            new Point(ox + r*0,       oy - r*1)
-          })).draw();
-
-          (new BezierCurve(new Point[]{
-            new Point(ox + r*0,       oy - r*1),
-            new Point(ox + r*0.55228, oy - r*1),
-            new Point(ox + r*1,       oy - r*0.55228),
-            new Point(ox + r*1,       oy + r*0)
-          })).draw();
-        }</textarea>

data/intersections.jsx

@@ -1,137 +0,0 @@
-        <p>Let's look at some more things we will want to do with Bézier curves. Almost immediately after figuring out how to
-        get bounding boxes to work, people tend to run into the problem that even though the minimal bounding box (based on
-        rotation) is tight, it's not sufficient to perform collision detection ("<i>does curve C touch, or pass through, curve
-        or line L?</i>"). In order to do this, we need to know whether or not there's an intersection on the actual curve.</p>
-
-        <p>We'll do this in steps, because it's a bit of a journey to get to curve/curve intersection checking. First, let's
-        start simple, by implementing a line-line intersection checker. While we can solve this the traditional calculus way
-        (determine the functions for both lines, then compute the intersection by equating them and solving for two unknowns),
-        linear algebra actually offers a nicer solution:</p>
-
-        <p id="intersection_ll">if we have two line segments with two coordinates each, segments A-B and C-D, we can find the
-        intersection of the lines these segments are an intervals on by linear algebra, using the procedure outlined in this
-        <a href="http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=geometry2#line_line_intersection">top coder</a> article.
-        Of course, we need to make sure that the intersection isn't just on the lines our line segments lie on, but also on our
-        line segments themselves, so after we find the intersection we need to verify it lies without the bounds of our original
-        line segments.</p>
-
-        <p>The following graphic implements this intersection detection, showing a red point for an intersection on the lines
-        our segments lie on (thus being a virtual intersection point), and a green point for an intersection that lies on
-        both segments (being a real intersection point).</p>
-
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Line/line intersections">
-        Point p1, p2, p3, p4;
-
-        void setupCurve() {
-          p1 = new Point(50,50);
-          p2 = new Point(150,110);
-          curves.add(new BezierCurve(new Point[]{p1,p2}, false));
-          p3 = new Point(50,250);
-          p4 = new Point(170,170);
-          curves.add(new BezierCurve(new Point[]{p3,p4}, false));
-        }
-
-        void drawCurve(BezierCurve curve) {
-          // draw the lines through p1/p2 and p3/p4
-          stroke(0,50);
-          float dx = 10*(p2.x-p1.x), dy = 10*(p2.y-p1.y);
-          line(p1.x-dx,p1.y-dy,p2.x+dx,p2.y+dy);
-          dx = 10*(p4.x-p3.x); dy = 10*(p4.y-p3.y);
-          line(p3.x-dx,p3.y-dy,p4.x+dx,p4.y+dy);
-
-          // show the line segments
-          curves.get(0).draw();
-          curves.get(1).draw();
-
-          // show the intersection point
-          Point ntr = comp.getProjection(p1,p2,p3,p4);
-
-          // red if virtual intersection, green if real
-          boolean oncurves = true;
-          if(min(p1.x,p2.x) > ntr.x || ntr.x > max(p1.x,p2.x) ||
-             min(p1.y,p2.y) > ntr.y || ntr.y > max(p1.y,p2.y)) oncurves = false;
-          if(oncurves) {
-            if(min(p3.x,p4.x) > ntr.x || ntr.x > max(p3.x,p4.x) ||
-              min(p3.y,p4.y) > ntr.y || ntr.y > max(p3.y,p4.y)) oncurves = false; }
-
-          stroke(oncurves?0:255, oncurves?255:0, 0);
-          ellipse(ntr.x,ntr.y,5,5);
-        }</textarea>
-
-        <p>Curve/line intersection is more work, but we've already seen the techniques we need to use in order
-        to perform it: first we translate/rotate both the line and curve together, in such a way that the line
-        coincides with the x-axis. This will position the curve in a way that makes it cross the line at
-        points where its y-function is zero. By doing this, the problem of finding intersections between a
-        curve and a line has now become the problem of performing root finding on our translated/rotated curve.
-        One Newton-Raphson root finding round later and the intersections have been found:</p>
-
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Quadratic curve/line intersections">
-        Point p1, p2;
-
-        void setupCurve() {
-          p1 = new Point(40,60);
-          p2 = new Point(260,200);
-          curves.add(new BezierCurve(new Point[]{
-            p1, p2
-          }, false));
-          curves.add(new BezierCurve(new Point[]{
-            new Point(25,150),
-            new Point(180,30),
-            new Point(230,250)
-          }));
-        }
-
-        void drawCurve(BezierCurve curve) {
-          curves.get(0).draw();
-          curves.get(1).draw();
-
-          BezierCurve aligned = curves.get(1).align(p1,p2);
-          float[] roots = comp.findAllRoots(0, aligned.y_values);
-          fill(150,0,150);
-          float x, y;
-          for(float t: roots) {
-            if(t<0 || t>1) continue;
-            x = curves.get(1).getXValue(t);
-            y = curves.get(1).getYValue(t);
-            ellipse(x,y,5,5);
-            text(""+round(1000*t)/1000,x+10,y);
-          }
-        }</textarea>
-
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Cubic curve/line intersections">
-        Point p1, p2;
-
-        void setupCurve() {
-          p1 = new Point(100,20);
-          p2 = new Point(195,255);
-          curves.add(new BezierCurve(new Point[]{
-            p1, p2
-          }, false));
-          curves.add(new BezierCurve(new Point[]{
-            new Point(150,125),
-            new Point(40,30),
-            new Point(270,115),
-            new Point(145,200)
-          }));
-        }
-
-        void drawCurve(BezierCurve curve) {
-          curves.get(0).draw();
-          curves.get(1).draw();
-
-          BezierCurve aligned = curves.get(1).align(p1,p2);
-          float[] roots = comp.findAllRoots(0, aligned.y_values);
-          fill(150,0,150);
-          float x, y;
-          for(float t: roots) {
-            if(t<0 || t>1) continue;
-            x = curves.get(1).getXValue(t);
-            y = curves.get(1).getYValue(t);
-            ellipse(x,y,5,5);
-            text(""+round(1000*t)/1000,x+10,y);
-          }
-        }</textarea>
-
-        <p>Curve/curve intersection, however, is more complicated. Since we have no straight line to align to, we
-        can't simply align one of the curves and be left with a simple procedure. Instead, we'll need to apply two
-        techniques we've not covered yet: de Casteljau's algorithm, and curve splitting.</p>

data/projections.jsx

@@ -1,57 +0,0 @@
-        <p>Say we have a Bézier curve and some point, not on the curve, of which we want to know which
-        <i>t</i> value on the curve gives us an on-curve point closest to our off-curve point. Or: say
-        we want to find the projection of a random point onto a curve. How do we do that?</p>
-
-        <p>If the Bézier curve is of low enough order, we might be able to <a href="http://jazzros.blogspot.ca/2011/03/projecting-point-on-bezier-curve.html">work out the maths for how
-        to do this</a>, and get a perfect <i>t</i> value back, but in general this is an incredibly hard
-        problem and the easiest solution is, really, a numerical approach again. We'll be finding our
-        ideal <i>t</i> value using a <a href="https://en.wikipedia.org/wiki/Binary_search_algorithm">binary
-        search</a>. First, we do a coarse distance-check based on <i>t</i> values associated with the
-        curve's "to draw" coordinates (using a lookup table, or LUT). This is pretty fast. Then we run
-        this algorithm:</p>
-
-        <ol>
-          <li>with the <i>t</i> value we found, start with some small interval around <i>t</i>
-          (1/length_of_LUT on either side is a reasonable start),</li>
-          <li>if the distance to <i>t ± interval/2</i> is larger than the distance to <i>t</i>,
-          try again with the interval reduced to half its original length.</li>
-          <li>if the distance to <i>t ± interval/2</i> is smaller than the distance to <i>t</i>,
-          replace <i>t</i> with the smaller-distance value.</li>
-          <li>after reducing the interval, or changing <i>t</i>, go back to step 1.</li>
-        </ol>
-
-        <p>We keep repeating this process until the interval is small enough to claim the difference
-        in precision found is irrelevant for the purpose we're trying to find <i>t</i> for. In this
-        case, I'm arbitrarily fixing it at 0.0001.</p>
-
-        <p>The following graphic demonstrates the result of this procedure.Simply move the cursor
-        around, and if it does not lie on top of the curve, you will see a line that projects the
-        cursor onto the curve based on an iteratively found "ideal" <i>t</i> value.</p>
-
-        <textarea class="sketch-code" data-sketch-preset="simple" data-sketch-title="Projecting a point onto a Bézier curve">
-        void setupCurve() {
-          int d = dim - 2*pad;
-          int order = 10;
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+ <use xlink:href="#E1-LATINMODERNNORMAL-1D45B" x="0" y="0"></use>
+ <use transform="scale(0.707)" xlink:href="#E1-LATINMODERNMAIN-34" x="855" y="583"></use>
+</g>
+ <use xlink:href="#E1-LATINMODERNMAIN-28" x="12667" y="0"></use>
+<g transform="translate(13061,0)">
+<g transform="translate(120,0)">
+<rect stroke="none" width="766" height="60" x="0" y="220"></rect>
+ <use xlink:href="#E1-LATINMODERNMAIN-3C6" x="60" y="747"></use>
+ <use xlink:href="#E1-LATINMODERNMAIN-32" x="130" y="-696"></use>
+</g>
+</g>
+ <use xlink:href="#E1-LATINMODERNMAIN-29" x="14067" y="0"></use>
+<g transform="translate(14461,0)">
+ <use xlink:href="#E1-LATINMODERNSIZE4-221A" x="0" y="-285"></use>
+<rect stroke="none" width="3823" height="60" x="1005" y="1415"></rect>
+<g transform="translate(1005,0)">
+<g transform="translate(120,0)">
+<rect stroke="none" width="3583" height="60" x="0" y="220"></rect>
+ <use xlink:href="#E1-LATINMODERNMAIN-31" x="1539" y="676"></use>
+<g transform="translate(60,-945)">
+ <use xlink:href="#E1-LATINMODERNNORMAL-1D450" x="0" y="0"></use>
+ <use xlink:href="#E1-LATINMODERNNORMAL-1D45C" x="438" y="0"></use>
+<g transform="translate(928,0)">
+ <use xlink:href="#E1-LATINMODERNNORMAL-1D460" x="0" y="0"></use>
+ <use transform="scale(0.707)" xlink:href="#E1-LATINMODERNMAIN-32" x="670" y="408"></use>
+</g>
+ <use xlink:href="#E1-LATINMODERNMAIN-28" x="1859" y="0"></use>
+<g transform="translate(2253,0)">
+<g transform="translate(120,0)">
+<rect stroke="none" width="576" height="60" x="0" y="220"></rect>
+ <use transform="scale(0.707)" xlink:href="#E1-LATINMODERNMAIN-3C6" x="84" y="856"></use>
+ <use transform="scale(0.707)" xlink:href="#E1-LATINMODERNMAIN-32" x="155" y="-598"></use>
+</g>
+</g>
+ <use xlink:href="#E1-LATINMODERNMAIN-29" x="3069" y="0"></use>
+</g>
+</g>
+</g>
+</g>
+</g>
+</g>
+</g>
+</g>
+</svg>

lib/p-loader.js

@@ -0,0 +1,17 @@
+var blockLoader = require("block-loader");
+
+var options = {
+  start: "<p>",
+  end: "</p>",
+
+  /**
+   * JSX curly brace replacement.
+   */
+  process: function fixPreBlocks(p) {
+    var replaced = p.replace(options.start,'').replace(options.end,'');
+    if(replaced.indexOf("\\[")>-1) return p;
+    return options.start + replaced.replace(/([{}])/g,"{'$1'}") + options.end;
+  }
+};
+
+module.exports = blockLoader(options);

lib/textarea-loader.js

@@ -0,0 +1,19 @@
+var blockLoader = require("block-loader");
+
+var options = {
+  start: "<textarea",
+  end: "</textarea>",
+
+  /**
+   * We want to be able to stick arbitrary text in a <textarea>
+   */
+  process: function fixPreBlocks(textarea) {
+    var fpos = textarea.indexOf('>');
+    var start = textarea.substring(0,fpos+1);
+    var replaced = textarea.replace(start,'').replace(options.end,'').replace(/"/g,'\\"').replace(/\n/g,'\\n');
+    var rewritten = start.substring(0,fpos) + ' defaultValue={"' + replaced +'"}/>';
+    return rewritten;
+  }
+};
+
+module.exports = blockLoader(options);

webpack.config.js

@@ -1,12 +1,26 @@
 var webpack = require('webpack');
 
-// Hot Reload server when we're in dev mode,
-// otherwise build it the normal way.
+// Bundle entry point
 var entry = ['./components/App.jsx'];
-if(!process.argv.indexOf("--prod")) {
+
+// Necessary webpack loaders for converting our content:
+var webpackLoaders = [
+  'babel-loader',
+  __dirname + '/lib/latex-loader',
+  __dirname + '/lib/pre-loader',
+  __dirname + '/lib/p-loader'
+];
+
+// Dev mode: make certain concessions to speed up dev work:
+if(process.argv.indexOf("--prod") === -1) {
+  // use the webpack hot Reload server:
   entry.push('webpack/hot/dev-server');
+
+  // allow textareas in  dev mode:
+  webpackLoaders.push(__dirname + '/lib/textarea-loader');
 }
 
+// And the final config that webpack will read in.
 module.exports = {
   entry:  entry,
   output: {
@@ -20,11 +34,7 @@ module.exports = {
       {
         test: /.jsx?$/,
         include: /components/,
-        loaders: [
-          'babel-loader',
-          __dirname + '/lib/latex-loader',
-          __dirname + '/lib/pre-loader'
-        ]
+        loaders: webpackLoaders
       }
     ]
   },