info/BezierInfo-2
1
0
Fork 0
mirror of https://github.com/Pomax/BezierInfo-2.git synced 2025-08-30 11:40:27 +02:00
Code Issues Releases Wiki Activity

sections 21-30 converted for localization

Browse Source
This commit is contained in:
Pomax
2017-02-25 18:16:03 -08:00
parent 3894fddb17
commit a889d16f36
21 changed files with 819 additions and 1028 deletions
Download Patch File Download Diff File Expand all files Collapse all files

366
components/sections/catmullconv/content.en-GB.md Normal file
View File

@@ -0,0 +1,366 @@
# Bézier curves and Catmull-Rom curves
Taking an excursion to different splines, the other common design curve is the [Catmull-Rom spline](https://en.wikipedia.org/wiki/Cubic_Hermite_spline#Catmull.E2.80.93Rom_spline). Now, a Catmull-Rom spline is a form of cubic Hermite spline, and as it so happens the cubic Bézier curve is also a cubic Hermite spline, so maybe... maybe we can convert one into the other, and back, with some simple substitutions?
Unlike Bézier curves, Catmull-Rom splines pass through each point used to define the curve, except the first and last, which makes sense if you read the "natural language" descriptionfor how a Catmull-Rom spline works: a Catmull-Rom spline is a curve that, at each point P<sub>x</sub>, has a tangent along the line P<sub>x-1</sub> to P<sub>x+1</sub>. The curve runs from points P<sub>2</sub> to P<sub>n-1</sub>, and has a "tension" that determines how fast the curve passes through each point. The lower the tension, the faster the curve goes through each point, and the bigger its local tangent is.
I'll be showing the conversion to and from Catmull-Rom curves for the tension that the Processing language uses for its Catmull-Rom algorithm.
We start with showing the Catmull-Rom matrix form:
\[
CatmullRom(t) =
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-3 & 3 & -2 & -1 \\
2 & -2 & 1 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
V_1 \\ V_2 \\ V'_1 \\ V'_2
\end{bmatrix}
\]
However, there's something funny going on here: the coordinate column matrix looks weird. The reason is that Catmull-Rom curves are actually curve segments that are described by two points, and two tangents; the curve leaves a point V1 (if we have four coordinates instead, this is coordinate 2), arriving at a point V2 (coordinate 3), with the curve departing V1 with a tangent vector V'1 (equal to the tangent from coordinate 1 to coordinate 3) and arriving at V2 with tangent vector V'2 (equal to the tangent from coordinate 2 to coordinate 4). So if we want to express this as a matrix form based on four coordinates, we get this representation instead:
\[
\begin{bmatrix}
V_1 \\ V_2 \\ V'_1 \\ V'_2
\end{bmatrix}
=
T
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
=
\begin{bmatrix}
P_2 \\ P_3 \\ \frac{P_3 - P_1}{2} \\ \frac{P_4 - P_2}{2}
\end{bmatrix}
\ \Rightarrow \
T
=
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
-1 & 0 & 1 & 0 \\
0 & -1 & 0 & 1
\end{bmatrix}
\]
<div className="note">
## Where did that 2 come from?
Catmull-Rom splines are based on the concept of tension: the higher the tensions, the shorter the tangents at the departure and arrival points. The basic Catmull-Rom curve arrives and departs with tangents equal to half the distance between the two adjacent points, so that's where that 2 came from.
However, the "real" matrix is this:
\[
T
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
=
\begin{bmatrix}
P_2 \\ P_3 \\ \frac{P_3 - P_1}{2 \cdot τ} \\ \frac{P_4 - P_2}{2 \cdot τ}
\end{bmatrix}
\Rightarrow \
T
=
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
-τ & 0 & τ & 0 \\
0 & -τ & 0 & τ
\end{bmatrix}
\]
This bakes in the tension factor τ explicitly.
</div>
Plugging this into the "two coordinates and two tangent vectors" matrix form, we get:
\[
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-3 & 3 & -2 & -1 \\
2 & -2 & 1 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
V_1 \\ V_2 \\ V'_1 \\ V'_2
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-3 & 3 & -2 & -1 \\
2 & -2 & 1 & 1
\end{bmatrix}
\cdot
\left (
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
-τ & 0 & τ & 0 \\
0 & -τ & 0 & τ
\end{bmatrix}
\right )
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]
So let's find out which transformation matrix we need in order to convert from Catmull-Rom to Bézier:
\[
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
A
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]
The difference is somewhere in the actual hermite matrix, since the <em>t</em> and coordinate values are identical, so let's solve that matrix equasion:
\[
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
A
\]
We left-multiply both sides by the inverse of the Bézier matrix, to get rid of the Bézier matrix on the right side of the equals sign:
\[
{
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
}^{-1}
\cdot
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
=
{
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
}^{-1}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
A
\ =\
I \cdot A
\ =\
A
\]
Which gives us:
\[
\frac{1}{6}
\cdot
\begin{bmatrix}
0 & 6 & 0 & 0 \\
-τ & 6 & τ & 0 \\
0 & τ & 0 & -τ \\
0 & 0 & 6 & 0
\end{bmatrix}
=
A
\]
Multiplying this ***A*** with our coordinates will give us a proper Bézier matrix expression again:
\[
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
\frac{1}{6}
\cdot
\begin{bmatrix}
0 & 6 & 0 & 0 \\
-τ & 6 & τ & 0 \\
0 & τ & 0 & -τ \\
0 & 0 & 6 & 0
\end{bmatrix}
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
P_2 \\
P_2 + \frac{P_3-P_1}{6 \cdot τ} \\
P_3 - \frac{P_4-P_2}{6 \cdot τ} \\
P_3
\end{bmatrix}
\]
So a Catmull-Rom to Bézier conversion, based on coordinates, requires turning the Catmull-Rom coordinates on the left into the Bézier coordinates on the right (with τ being our tension factor):
\[
\begin{bmatrix}
P_1 \\
P_2 \\
P_3 \\
P_4
\end{bmatrix}_{CatmullRom}
\Rightarrow
\begin{bmatrix}
P_2 \\
P_2 + \frac{P_3-P_1}{6 \cdot τ} \\
P_3 - \frac{P_4-P_2}{6 \cdot τ} \\
P_3
\end{bmatrix}_{Bézier}
\]
And the other way around, a Bézier to Catmull-Rom conversion requires turning the Bézier coordinates on the left this time into the Catmull-Rom coordinates on the right. Note that there is no tension this time, because Bézier curves don't have any. Converting from Bézier to Catmull-Rom is simply a default-tension Catmull-Rom curve:
\[
\begin{bmatrix}
P_1 \\
P_2 \\
P_3 \\
P_4
\end{bmatrix}_{Bézier}
\Rightarrow
\begin{bmatrix}
P_4 + 6 \cdot (P_1 - P_2) \\
P_1 \\
P_4 \\
P_1 + 6 \cdot (P_4 - P_3)
\end{bmatrix}_{CatmullRom}
\]
Done. We can now draw the curves we want using either Bézier curves or Catmull-Rom splines, the choice mostly being which drawing algorithms we have natively available.

409
components/sections/catmullconv/index.js
View File

@@ -1,415 +1,18 @@
var React = require("react");
var SectionHeader = require("../../SectionHeader.jsx");
var Locale = require("../../../lib/locale");
var locale = new Locale();
var page = "catmullconv";
var CatmullRomConversion = React.createClass({
getDefaultProps: function() {
return {
title: "Bézier curves and Catmull-Rom curves"
title: locale.getTitle(page)
};
},
render: function() {
return (
<section>
<SectionHeader {...this.props} />
<p>Taking an excursion to different splines, the other common design curve is
the <a href="https://en.wikipedia.org/wiki/Cubic_Hermite_spline#Catmull.E2.80.93Rom_spline">Catmull-Rom
spline</a>. Now, a Catmull-Rom spline is a form of cubic Hermite spline, and as it so happens
the cubic Bézier curve is also a cubic Hermite spline, so maybe... maybe we can convert one
into the other, and back, with some simple substitutions?</p>
<p>Unlike Bézier curves, Catmull-Rom splines pass through each point used to define the curve,
except the first and last, which makes sense if you read the "natural language" description
for how a Catmull-Rom spline works: a Catmull-Rom spline is a curve that, at each point
P<sub>x</sub>, has a tangent along the line P<sub>x-1</sub> to P<sub>x+1</sub>. The curve
runs from points P<sub>2</sub> to P<sub>n-1</sub>, and has a "tension" that determines
how fast the curve passes through each point. The lower the tension, the faster the curve
goes through each point, and the bigger its local tangent is.</p>
<p>I'll be showing the conversion to and from Catmull-Rom curves for the tension that the
Processing language uses for its Catmull-Rom algorithm.</p>
<p>We start with showing the Catmull-Rom matrix form:</p>
<p>\[
CatmullRom(t) =
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-3 & 3 & -2 & -1 \\
2 & -2 & 1 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
V_1 \\ V_2 \\ V'_1 \\ V'_2
\end{bmatrix}
\]
</p>
<p>However, there's something funny going on here: the coordinate column matrix looks weird.
The reason is that Catmull-Rom curves are actually curve segments that are described by two
points, and two tangents; the curve leaves a point V1 (if we have four coordinates instead,
this is coordinate 2), arriving at a point V2 (coordinate 3), with the curve departing V1
with a tangent vector V'1 (equal to the tangent from coordinate 1 to coordinate 3) and
arriving at V2 with tangent vector V'2 (equal to the tangent from coordinate 2 to coordinate
4). So if we want to express this as a matrix form based on four coordinates, we get this
representation instead:</p>
<p>\[
\begin{bmatrix}
V_1 \\ V_2 \\ V'_1 \\ V'_2
\end{bmatrix}
=
T
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
=
\begin{bmatrix}
P_2 \\ P_3 \\ \frac{P_3 - P_1}{2} \\ \frac{P_4 - P_2}{2}
\end{bmatrix}
\ \Rightarrow \
T
=
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
-1 & 0 & 1 & 0 \\
0 & -1 & 0 & 1
\end{bmatrix}
\]</p>
<div className="note">
<h2>Where did that 2 come from?</h2>
<p>Catmull-Rom splines are based on the concept of tension: the higher the tensions,
the shorter the tangents at the departure and arrival points. The basic Catmull-Rom
curve arrives and departs with tangents equal to half the distance between the two
adjacent points, so that's where that 2 came from.</p>
<p>However, the "real" matrix is this:</p>
<p>\[
T
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
=
\begin{bmatrix}
P_2 \\ P_3 \\ \frac{P_3 - P_1}{2 \cdot τ} \\ \frac{P_4 - P_2}{2 \cdot τ}
\end{bmatrix}
\Rightarrow \
T
=
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
-τ & 0 & τ & 0 \\
0 & -τ & 0 & τ
\end{bmatrix}
\]</p>
<p>This bakes in the tension factor τ explicitly.</p>
</div>
<p>Plugging this into the "two coordinates and two tangent vectors" matrix form,
we get:</p>
<p>\[
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-3 & 3 & -2 & -1 \\
2 & -2 & 1 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
V_1 \\ V_2 \\ V'_1 \\ V'_2
\end{bmatrix}
\]</p>
<p>\[
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-3 & 3 & -2 & -1 \\
2 & -2 & 1 & 1
\end{bmatrix}
\cdot
\left (
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
-τ & 0 & τ & 0 \\
0 & -τ & 0 & τ
\end{bmatrix}
\right )
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]</p>
<p>\[
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]</p>
<p>So let's find out which transformation matrix we need in order to convert from Catmull-Rom to Bézier:</p>
<p>\[
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
A
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]</p>
<p>The difference is somewhere in the actual hermite matrix, since the <em>t</em> and coordinate
values are identical, so let's solve that matrix equasion:</p>
<p>\[
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
A
\]</p>
<p>We left-multiply both sides by the inverse of the Bézier matrix, to get rid of the
Bézier matrix on the right side of the equals sign:</p>
<p>\[
{
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
}^{-1}
\cdot
\frac{1}{2}
\cdot
\begin{bmatrix}
0 & 2 & 0 & 0 \\
-τ & 0 & τ & 0 \\
2τ & τ-6 & -2(τ-3) & -τ \\
-τ & 4-τ & τ-4 & τ
\end{bmatrix}
=
{
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
}^{-1}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
A
\ =\
I \cdot A
\ =\
A
\]
</p>
<p>Which gives us:</p>
<p>\[
\frac{1}{6}
\cdot
\begin{bmatrix}
0 & 6 & 0 & 0 \\
-τ & 6 & τ & 0 \\
0 & τ & 0 & -τ \\
0 & 0 & 6 & 0
\end{bmatrix}
=
A
\]</p>
<p>Multiplying this <strong><em>A</em></strong> with our coordinates
will give us a proper Bézier matrix expression again:</p>
<p>\[
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
\frac{1}{6}
\cdot
\begin{bmatrix}
0 & 6 & 0 & 0 \\
-τ & 6 & τ & 0 \\
0 & τ & 0 & -τ \\
0 & 0 & 6 & 0
\end{bmatrix}
\cdot
\begin{bmatrix}
P_1 \\ P_2 \\ P_3 \\ P_4
\end{bmatrix}
\]</p>
<p>\[
=
\begin{bmatrix}
1 & t & t^2 & t^3
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
-3 & 3 & 0 & 0 \\
3 & -6 & 3 & 0 \\
-1 & 3 & -3 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
P_2 \\
P_2 + \frac{P_3-P_1}{6 \cdot τ} \\
P_3 - \frac{P_4-P_2}{6 \cdot τ} \\
P_3
\end{bmatrix}
\]</p>
<p>So a Catmull-Rom to Bézier conversion, based on coordinates, requires turning
the Catmull-Rom coordinates on the left into the Bézier coordinates on the right
(with τ being our tension factor):</p>
<p>\[
\begin{bmatrix}
P_1 \\
P_2 \\
P_3 \\
P_4
\end{bmatrix}_{CatmullRom}
\Rightarrow
\begin{bmatrix}
P_2 \\
P_2 + \frac{P_3-P_1}{6 \cdot τ} \\
P_3 - \frac{P_4-P_2}{6 \cdot τ} \\
P_3
\end{bmatrix}_{Bézier}
\]</p>
<p>And the other way around, a Bézier to Catmull-Rom conversion requires turning
the Bézier coordinates on the left this time into the Catmull-Rom coordinates
on the right. Note that there is no tension this time, because Bézier curves
don't have any. Converting from Bézier to Catmull-Rom is simply a default-tension
Catmull-Rom curve:</p>
<p>\[
\begin{bmatrix}
P_1 \\
P_2 \\
P_3 \\
P_4
\end{bmatrix}_{Bézier}
\Rightarrow
\begin{bmatrix}
P_4 + 6 \cdot (P_1 - P_2) \\
P_1 \\
P_4 \\
P_1 + 6 \cdot (P_4 - P_3)
\end{bmatrix}_{CatmullRom}
\]</p>
<p>Done. We can now draw the curves we want using either Bézier curves or Catmull-Rom
splines, the choice mostly being which drawing algorithms we have natively available.</p>
</section>
);
return <section>{ locale.getContent(page, this) }</section>;
}
});