commit 70c323565b57d8914bb197f8997226bfd773371d
parent 305acef0a717a0293a287379ceb95aacbb0bfffd
Author: Kenji Matsuda <ftvda283@gmail.com>
Date: Tue, 26 Oct 2021 14:46:45 +0900
modify mathjax
Diffstat:
3 files changed, 72 insertions(+), 48 deletions(-)
diff --git a/translation/Getting-started/Camera.html b/translation/Getting-started/Camera.html
@@ -101,8 +101,8 @@ glm::vec3 cameraUp = <function id='61'>glm::cross</function>(cameraDirection, ca
\[LookAt = \begin{bmatrix} \color{red}{R_x} & \color{red}{R_y} & \color{red}{R_z} & 0 \\ \color{green}{U_x} & \color{green}{U_y} & \color{green}{U_z} & 0 \\ \color{blue}{D_x} & \color{blue}{D_y} & \color{blue}{D_z} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} * \begin{bmatrix} 1 & 0 & 0 & -\color{purple}{P_x} \\ 0 & 1 & 0 & -\color{purple}{P_y} \\ 0 & 0 & 1 & -\color{purple}{P_z} \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
- Where \(\color{red}R\) is the right vector, \(\color{green}U\) is the up vector, \(\color{blue}D\) is the direction vector and \(\color{purple}P\) is the camera's position vector. Note that the rotation (left matrix) and translation (right matrix) parts are inverted (transposed and negated respectively) since we want to rotate and translate the world in the opposite direction of where we want the camera to move. Using this LookAt matrix as our view matrix effectively transforms all the world coordinates to the view space we just defined. The LookAt matrix then does exactly what it says: it creates a view matrix that <em>looks</em> at a given target.
- ここで、\(\color{red}R\)はカメラの右方向のベクトル、\(\color{green}U\)は上方向のベクトル、\(\color{blue}D\)はカメラの方向ベクトル、そして\)\color{purple}P\)はカメラの位置ベクトルです。左側の行列の回転と、右側の行列の平行移動がそれぞれ逆向きになっていることに注意して下さい。転置行列による回転と、符号が反転した平行移動にそれぞれなっています。カメラを移動させるのと反対方向に世界全体を回転、平行移動する為です。この視点行列を視野行列として用いることで、大域座標全体を今しがた定義した視野空間に変換できます。視点行列は与えられた位置を<em>見る</em>ような視野行列を作成します。視点行列は名前の通りの仕事をするのです。
+ Where \({\color{red}R}\) is the right vector, \({\color{green}U}\) is the up vector, \({\color{blue}D}\) is the direction vector and \({\color{purple}P}\) is the camera's position vector. Note that the rotation (left matrix) and translation (right matrix) parts are inverted (transposed and negated respectively) since we want to rotate and translate the world in the opposite direction of where we want the camera to move. Using this LookAt matrix as our view matrix effectively transforms all the world coordinates to the view space we just defined. The LookAt matrix then does exactly what it says: it creates a view matrix that <em>looks</em> at a given target.
+ ここで、\({\color{red}R}\)はカメラの右方向のベクトル、\({\color{green}U}\)は上方向のベクトル、\({\color{blue}D}\)はカメラの方向ベクトル、そして\({\color{purple}P}\)はカメラの位置ベクトルです。左側の行列の回転と、右側の行列の平行移動がそれぞれ逆向きになっていることに注意して下さい。転置行列による回転と、符号が反転した平行移動にそれぞれなっています。カメラを移動させるのと反対方向に世界全体を回転、平行移動する為です。この視点行列を視野行列として用いることで、大域座標全体を今しがた定義した視野空間に変換できます。視点行列は与えられた位置を<em>見る</em>ような視野行列を作成します。視点行列は名前の通りの仕事をするのです。
</p>
<p>
@@ -318,8 +318,8 @@ void processInput(GLFWwindow *window)
<img src="/img/getting-started/camera_triangle.png" class="clean"/>
<p>
- If we define the hypotenuse to be of length <code>1</code> we know from trigonometry (soh cah toa) that the adjacant side's length is \(\cos \ \color{red}x/\color{purple}h = \cos \ \color{red}x/\color{purple}1 = \cos\ \color{red}x\) and that the opposing side's length is \(\sin \ \color{green}y/\color{purple}h = \sin \ \color{green}y/\color{purple}1 = \sin\ \color{green}y\). This gives us some general formulas for retrieving the length in both the <code>x</code> and <code>y</code> sides on right triangles, depending on the given angle. Let's use this to calculate the components of the direction vector.
- 斜辺の長さを<code>1</code>とした場合、三角関数を使うと、底辺の長さは\(\cos \ \color{red}x/\color{purple}h = \cos \ \color{red}x/\color{purple}1 = \cos\ \color{red}x\)、高さは\(\sin \ \color{green}y/\color{purple}h = \sin \ \color{green}y/\color{purple}1 = \sin\ \color{green}y\)となります。
+ If we define the hypotenuse to be of length <code>1</code> we know from trigonometry (soh cah toa) that the adjacant side's length is \(\cos \ {\color{red}x}/{\color{purple}h} = \cos \ {\color{red}x}/{\color{purple}1} = \cos\ {\color{red}x}\) and that the opposing side's length is \(\sin \ {\color{green}y}/{\color{purple}h} = \sin \ {\color{green}y}/{\color{purple}1} = \sin\ {\color{green}y.}\) This gives us some general formulas for retrieving the length in both the <code>x</code> and <code>y</code> sides on right triangles, depending on the given angle. Let's use this to calculate the components of the direction vector.
+ 斜辺の長さを<code>1</code>とした場合、三角関数を使うと、底辺の長さは\(\cos \ {\color{red}x}/{\color{purple}h} = \cos \ {\color{red}x}/{\color{purple}1} = \cos \ {\color{red}x}\)、高さは\(\sin \ {\color{green}y}/{\color{purple}h} = \sin \ {\color{green}y}/{\color{purple}1} = \sin \ {\color{green}y}\)となります。
</p>
<p>
diff --git a/translation/Getting-started/Review.html b/translation/Getting-started/Review.html
@@ -15,7 +15,7 @@
<h1 id="content-url" style='display:none;'>Getting-started/Review</h1>
<p>
Congratulations on reaching the end of the <em>Getting started</em> chapters. By now you should be able to create a window, create and compile shaders, send vertex data to your shaders via buffer objects or uniforms, draw objects, use textures, understand vectors and matrices and combine all that knowledge to create a full 3D scene with a camera to play around with.
-おめでとうございます。これにて<em>第一節</em>は完了です。ここまで読んだ方はウィンドウを作成し、シェーダーを作成及びコンパイルし、バッファオブジェクトやユニフォームを通して頂点のデータをシェーダーに送信し、物体を描画し、テクスチャを利用し、ベクトルや行列を理解しそれらを組み合わせて3次元の世界をカメラと共に作成することができるようになりました。
+おめでとうございます。これにて<em>入門</em>の節は完了です。ここまで読んだ方はウィンドウを作成し、シェーダーを作成及びコンパイルし、バッファオブジェクトやユニフォームを通して頂点のデータをシェーダーに送信し、物体を描画し、テクスチャを利用し、ベクトルや行列を理解しそれらを組み合わせて3次元の世界をカメラと共に作成することができるようになりました。
</p>
<p>
@@ -34,29 +34,53 @@
<li><code>Viewport</code>: the 2D window region where we render to. </li>
<li><code>可視領域（Viewport）</code>: 映像を描画するウィンドウの2次元領域。</li>
<li><code>Graphics Pipeline</code>: the entire process vertices have to walk through before ending up as one or more pixels on the screen. </li>
+ <li><code>グラフィックパイプライン（Graphics Pipeline）</code>: 頂点データが画面上のピクセルに変換されるまでに行われる処理。</li>
<li><code>Shader</code>: a small program that runs on the graphics card. Several stages of the graphics pipeline can use user-made shaders to replace existing functionality.</li>
+ <li><code>シェーダ（Shader）</code>: グラフィックカード上で実行される小さなプログラム。グラフィックパイプライン上のいくつかの処理はユーザーが独自に作成したシェーダにより置き換えることが出来る。</li>
<li><code>Vertex</code>: a collection of data that represent a single point. </li>
+ <li><code>頂点（Vertex）</code>: 点を表わすデータの集合。</li>
<li><code>Normalized Device Coordinates</code>: the coordinate system your vertices end up in after perspective division is performed on clip coordinates. All vertex positions in NDC between <code>-1.0</code> and <code>1.0</code> will not be discarded or clipped and end up visible. </li>
+ <li><code>正規化デバイス座標（Normalized Device Coordinates）</code>: クリップ座標における透視除算の後に頂点が納まるべき座標系。各要素が<code>-1.0</code>と<code>1.0</code>の間にあれば切り落されずに残り、画面上に表示される。</li>
<li><code>Vertex Buffer Object</code>: a buffer object that allocates memory on the GPU and stores all the vertex data there for the graphics card to use. </li>
+ <li><code>頂点バッファオブジェクト（Vertex Buffer Object）</code>: GPU上のメモリを確保しそこに頂点データを保存するバッファオブジェクト。</li>
<li><code>Vertex Array Object</code>: stores buffer and vertex attribute state information.</li>
+ <li><code>頂点配列オブジェクト（Vertex Array Object）</code>: バッファと頂点属性の状態を保存する。</li>
<li><code>Element Buffer Object</code>: a buffer object that stores indices on the GPU for indexed drawing. </li>
+ <li><code>要素バッファオブジェクト（Element Buffer Object）</code>: インデックスによる描画に利用するインデックスをGPUに保存するバッファオブジェクト。</li>
<li><code>Uniform</code>: a special type of GLSL variable that is global (each shader in a shader program can access this uniform variable) and only has to be set once. </li>
+ <li><code>ユニフォーム（Uniform）</code>: GLSLの特別な変数。大域的（シェーダープログラム中の各シェーダーからこの変数にアクセスできる）で一度値が決まれば変更できない。</li>
<li><code>Texture</code>: a special type of image used in shaders and usually wrapped around objects, giving the illusion an object is extremely detailed. </li>
+ <li><code>テクスチャ（Texture）</code>: シェーダーで利用される画像。通常物体に貼り付けられその物体を微細に描くために用いられる。</li>
<li><code>Texture Wrapping</code>: defines the mode that specifies how OpenGL should sample textures when texture coordinates are outside the range: (<code>0</code>, <code>1</code>). </li>
+ <li><code>テクスチャの繰り返し（Texture Wrapping）</code>: <code>0</code>と<code>1</code>の外のテクスチャ座標においてOpenGLがどのようにテクスチャをサンプリングするかを定める。</li>
<li><code>Texture Filtering</code>: defines the mode that specifies how OpenGL should sample the texture when there are several texels (texture pixels) to choose from. This usually occurs when a texture is magnified. </li>
+ <li><code>テクスチャフィルタリング（Texture Filtering）</code>: テクスチャのピクセルであるテクセルが複数ある場合にどのようにサンプリングするかを定める。テクスチャが拡大される時に利用される。</li>
<li><code>Mipmaps</code>: stored smaller versions of a texture where the appropriate sized version is chosen based on the distance to the viewer. </li>
+ <li><code>ミップマップ（Mipmaps）</code>: テクスチャを縮小したものを保存。カメラからの距離に応じて適切なサイズのテクスチャを選ぶために利用。</li>
<li><code>stb_image</code>: image loading library. </li>
- <li><code>Texture Units</code>: allows for multiple textures on a single shader program by binding multiple textures, each to a different texture unit. </li>
- <li><code>Vector</code>: a mathematical entity that defines directions and/or positions in any dimension. </li>
+ <li><code>stb_image</code>: 画像を読み込むためのライブラリ。</li>
+ <li><code>Texture Units）</code>: allows for multiple textures on a single shader program by binding multiple textures, each to a different texture unit. </li>
+ <li><code>テクスチャユニット（Texture Units）</code>: 単一のシェーダープログラムに対して複数のテクスチャを紐付けるためのもの。それぞれのテクスチャをそれぞれのテクスチャユニットに割り当てる。</li>
+ <li><code>Vector</code>: a mathematical entity that defines directions and/or positions in any dimension. </li>
+ <li><code>ベクトル（Vector）</code>: 任意の次元の空間において方向や位置を定める数学の道具。</li>
<li><code>Matrix</code>: a rectangular array of mathematical expressions with useful transformation properties. </li>
+ <li><code>行列（Matrix）</code>: 数式を四角に並べたもの。座標変換において便利。</li>
<li><code>GLM</code>: a mathematics library tailored for OpenGL. </li>
+ <li><code>GLM</code>: OpenGLに特化した数学ライブラリ。</li>
<li><code>Local Space</code>: the space an object begins in. All coordinates relative to an object's origin. </li>
+ <li><code>局所空間（Local Space）</code>: 各物体に固有の座標空間。</li>
<li><code>World Space</code>: all coordinates relative to a global origin. </li>
+ <li><code>大域空間（World Space）</code>: 大域的な原点に相対的な座標。</li>
<li><code>View Space</code>: all coordinates as viewed from a camera's perspective. </li>
+ <li><code>視野空間（View Space）</code>: カメラから見た座標。</li>
<li><code>Clip Space</code>: all coordinates as viewed from the camera's perspective but with projection applied. This is the space the vertex coordinates should end up in, as output of the vertex shader. OpenGL does the rest (clipping/perspective division). </li>
+ <li><code>クリップ空間（Clip Space）</code>: カメラから見た座標を投影したもの。頂点座標は最終的にこの座標に変換されシェーダーからの出力となる。この後OpenGLがクリッピングと透視除算を行う。</li>
<li><code>Screen Space</code>: all coordinates as viewed from the screen. Coordinates range from <code>0</code> to screen width/height. </li>
+ <li><code>スクリーン空間（Screen Space）</code>: スクリーンから見た座標。座標は<code>0</code>とスクリーンの幅、高さの間。</li>
<li><code>LookAt</code>: a special type of view matrix that creates a coordinate system where all coordinates are rotated and translated in such a way that the user is looking at a given target from a given position. </li>
- <li><code>Euler Angles</code>: defined as <code>yaw</code>, <code>pitch</code> and <code>roll</code> that allow us to form any 3D direction vector from these 3 values. </li>
+ <li><code>視点行列（LookAt）</code>: 視野行列のひとつ。与えられた場所から与えられた位置を見つめるように座標変換するもの。</li>
+ <li><code>Euler Angles</code>: defined as <code>yaw</code>, <code>pitch</code> and <code>roll</code> that allow us to form any 3D direction vector from these 3 values. </li>
+ <li><code>オイラー角（Euler Angles）</code>: <code>yaw</code>（方位角）、<code>pitch</code>（仰角）、<code>roll</code>（傾斜角）によって3次元空間のあらゆる方向を定める。</li>
</ul>
</p>
diff --git a/translation/Getting-started/Transformations.html b/translation/Getting-started/Transformations.html
@@ -36,8 +36,8 @@
</p>
<p>
- Below you'll see 3 vectors where each vector is represented with <code>(x,y)</code> as arrows in a 2D graph. Because it is more intuitive to display vectors in 2D (rather than 3D) you can think of the 2D vectors as 3D vectors with a <code>z</code> coordinate of <code>0</code>. Since vectors represent directions, the origin of the vector does not change its value. In the graph below we can see that the vectors \(\color{red}{\bar{v}}\) and \(\color{blue}{\bar{w}}\) are equal even though their origin is different:
- 下の図は2次元の座標空間上に<code>(x, y)</code>と共に矢印で図示した3つのベクトルです。ベクトルは3次元より2次元で図示する方が直感的なので、2次元ベクトルは<code>z</code>座標が<code>0</code>である3次元ベクトルだと見做せます。ベクトルは方向を表わしたものなので、始点が変化してもベクトル自身は同じものです。下の図において2つのベクトル\(\color{red}{\bar{v}}\)と\(\color{blue}{\bar{w}}\)は始点は違いますがベクトルとしては同じものです:
+ Below you'll see 3 vectors where each vector is represented with <code>(x,y)</code> as arrows in a 2D graph. Because it is more intuitive to display vectors in 2D (rather than 3D) you can think of the 2D vectors as 3D vectors with a <code>z</code> coordinate of <code>0</code>. Since vectors represent directions, the origin of the vector does not change its value. In the graph below we can see that the vectors \({\color{red}\bar{v}}\) and \({\color{blue}\bar{w}}\) are equal even though their origin is different:
+ 下の図は2次元の座標空間上に<code>(x, y)</code>と共に矢印で図示した3つのベクトルです。ベクトルは3次元より2次元で図示する方が直感的なので、2次元ベクトルは<code>z</code>座標が<code>0</code>である3次元ベクトルだと見做せます。ベクトルは方向を表わしたものなので、始点が変化してもベクトル自身は同じものです。下の図において2つのベクトル\({\color{red}\bar{v}}\)と\({\color{blue}\bar{w}}\)は始点は違いますがベクトルとしては同じものです:
</p>
<img src="/img/getting-started/vectors.png" class="clean" />
@@ -46,7 +46,7 @@
When describing vectors mathematicians generally prefer to describe vectors as character symbols with a little bar over their head like \(\bar{v}\). Also, when displaying vectors in formulas they are generally displayed as follows:
数学者はよく\(\bar{v}\)のように文字の上に小さな棒を書いてベクトルを表わす書き方をします。そして数式の中にベクトルを書く際は一般に以下のような形になります:
- \[\bar{v} = \begin{pmatrix} \color{red}x \\ \color{green}y \\ \color{blue}z \end{pmatrix} \]
+ \[\bar{v} = \begin{pmatrix} {\color{red}x} \\ {\color{green}y} \\ {\color{blue}z} \end{pmatrix} \]
</p>
<p>
@@ -65,7 +65,7 @@
A <def>scalar</def> is a single digit. When adding/subtracting/multiplying or dividing a vector with a scalar we simply add/subtract/multiply or divide each element of the vector by the scalar. For addition it would look like this:
<def>スカラ</def>というのは数字です。ベクトルに対してスカラを足す／引く／掛ける演算は単にベクトルの各要素に対してスカラを足す／引く／掛けるだけです。足し算は以下のようになります:
- \[ \begin{pmatrix} \color{red}1 \\ \color{green}2 \\ \color{blue}3 \end{pmatrix} + x \rightarrow \begin{pmatrix} \color{red}1 \\ \color{green}2 \\ \color{blue}3 \end{pmatrix} + \begin{pmatrix} x \\ x \\ x \end{pmatrix} = \begin{pmatrix} \color{red}1 + x \\ \color{green}2 + x \\ \color{blue}3 + x \end{pmatrix} \]
+ \[ \begin{pmatrix} {\color{red}1} \\ {\color{green}2} \\ {\color{blue}3} \end{pmatrix} + x \rightarrow \begin{pmatrix} {\color{red}1} \\ {\color{green}2} \\ {\color{blue}3} \end{pmatrix} + \begin{pmatrix} x \\ x \\ x \end{pmatrix} = \begin{pmatrix} {\color{red}1} + x \\ {\color{green}2} + x \\ {\color{blue}3} + x \end{pmatrix} \]
Where \(+\) can be \(+\),\(-\),\(\cdot\) or \(\div\) where \(\cdot\) is the multiplication operator.
\(+\)になっているところは\(+\)、\(-\)、\(\cdot\)あるいは\(\div\)が使えます。ただし\(\cdot\)は掛け算です。
@@ -77,7 +77,7 @@
Negating a vector results in a vector in the reversed direction. A vector pointing north-east would point south-west after negation. To negate a vector we add a minus-sign to each component (you can also represent it as a scalar-vector multiplication with a scalar value of <code>-1</code>):
ベクトルを反転させるというのはベクトルの向きを逆にするということです。北東を向いていたベクトルは反転により南西を向くようになります。ベクトルを反転させるには各要素にマイナスの記号を付ければよいです（あるいはベクトルに対して<code>-1</code>を掛けることでも反転できます）:
- \[-\bar{v} = -\begin{pmatrix} \color{red}{v_x} \\ \color{blue}{v_y} \\ \color{green}{v_z} \end{pmatrix} = \begin{pmatrix} -\color{red}{v_x} \\ -\color{blue}{v_y} \\ -\color{green}{v_z} \end{pmatrix} \]
+ \[-\bar{v} = -\begin{pmatrix} {\color{red}v_x} \\ {\color{blue}v_y} \\ {\color{green}v_z} \end{pmatrix} = \begin{pmatrix} -{\color{red}v_x} \\ -{\color{blue}v_y} \\ -{\color{green}v_z} \end{pmatrix} \]
</p>
<h2>Addition and subtraction</h2>
@@ -86,7 +86,7 @@
Addition of two vectors is defined as <def>component-wise</def> addition, that is each component of one vector is added to the same component of the other vector like so:
2つのベクトルの足し算は<def>要素毎</def>の足し算として定義されます。つまり以下のように、一方のベクトルの各要素が他方のベクトルの同じ要素と足し合せられるということです:
- \[\bar{v} = \begin{pmatrix} \color{red}1 \\ \color{green}2 \\ \color{blue}3 \end{pmatrix}, \bar{k} = \begin{pmatrix} \color{red}4 \\ \color{green}5 \\ \color{blue}6 \end{pmatrix} \rightarrow \bar{v} + \bar{k} = \begin{pmatrix} \color{red}1 + \color{red}4 \\ \color{green}2 + \color{green}5 \\ \color{blue}3 + \color{blue}6 \end{pmatrix} = \begin{pmatrix} \color{red}5 \\ \color{green}7 \\ \color{blue}9 \end{pmatrix} \]
+ \[\bar{v} = \begin{pmatrix} {\color{red}1} \\ {\color{green}2} \\ {\color{blue}3} \end{pmatrix}, \bar{k} = \begin{pmatrix} {\color{red}4} \\ {\color{green}5} \\ {\color{blue}6} \end{pmatrix} \rightarrow \bar{v} + \bar{k} = \begin{pmatrix} {\color{red}1} + {\color{red}4} \\ {\color{green}2} + {\color{green}5} \\ {\color{blue}3} + {\color{blue}6} \end{pmatrix} = \begin{pmatrix} {\color{red}5} \\ {\color{green}7} \\ {\color{blue}9} \end{pmatrix} \]
Visually, it looks like this on vectors <code>v=(4,2)</code> and <code>k=(1,2)</code>, where the second vector is added on top of the first vector's end to find the end point of the resulting vector (head-to-tail method):
視覚的には以下の図のように、2つのベクトル<code>v=(4, 2)</code>と<code>k=(1, 2)</code>を足す場合、2つ目のベクトルが1つ目のベクトルの終点から始まり、結果として足し合せられたベクトルは1つ目のベクトルの始点から2つ目のベクトルの終点を差すようになります:
@@ -98,7 +98,7 @@
Just like normal addition and subtraction, vector subtraction is the same as addition with a negated second vector:
通常の加法、減法と同様に、ベクトルの引き算は2つ目のベクトルを反転させたものを足し合せるだけです:
- \[\bar{v} = \begin{pmatrix} \color{red}{1} \\ \color{green}{2} \\ \color{blue}{3} \end{pmatrix}, \bar{k} = \begin{pmatrix} \color{red}{4} \\ \color{green}{5} \\ \color{blue}{6} \end{pmatrix} \rightarrow \bar{v} + -\bar{k} = \begin{pmatrix} \color{red}{1} + (-\color{red}{4}) \\ \color{green}{2} + (-\color{green}{5}) \\ \color{blue}{3} + (-\color{blue}{6}) \end{pmatrix} = \begin{pmatrix} -\color{red}{3} \\ -\color{green}{3} \\ -\color{blue}{3} \end{pmatrix} \]
+ \[\bar{v} = \begin{pmatrix} {\color{red}1} \\ {\color{green}2} \\ {\color{blue}3} \end{pmatrix}, \bar{k} = \begin{pmatrix} {\color{red}4} \\ {\color{green}5} \\ {\color{blue}6} \end{pmatrix} \rightarrow \bar{v} + -\bar{k} = \begin{pmatrix} {\color{red}1} + (-{\color{red}4}) \\ {\color{green}2} + (-{\color{green}5}) \\ {\color{blue}3} + (-{\color{blue}6}) \end{pmatrix} = \begin{pmatrix} -{\color{red}3} \\ -{\color{green}3} \\ -{\color{blue}3} \end{pmatrix} \]
</p>
@@ -120,20 +120,20 @@
<img src="/img/getting-started/vectors_triangle.png" class="clean"/>
<p>
- Since the length of the two sides <code>(x, y)</code> are known and we want to know the length of the tilted side \(\color{red}{\bar{v}}\) we can calculate it using the Pythagoras theorem as:
+ Since the length of the two sides <code>(x, y)</code> are known and we want to know the length of the tilted side \({\color{red}\bar{v}}\) we can calculate it using the Pythagoras theorem as:
2辺の長さ<code>(x, y)</code>は既知なので、斜辺の長さはピタゴラスの定理より:
- \[||\color{red}{\bar{v}}|| = \sqrt{\color{green}x^2 + \color{blue}y^2} \]
+ \[||{\color{red}\bar{v}}|| = \sqrt{{\color{green}x}^2 + {\color{blue}y}^2} \]
- Where \(||\color{red}{\bar{v}}||\) is denoted as <em>the length of vector \(\color{red}{\bar{v}}\)</em>. This is easily extended to 3D by adding \(z^2\) to the equation.
- となります。ここで\(||\color{red}{\bar{v}}||\)は<em>ベクトル\(\color{red}{\bar{v}}\)の長さ</em>を表わします。ここに\(z^2\)を加えることで簡単に3次元に拡張できます。
+ Where \(||{\color{red}\bar{v}}||\) is denoted as <em>the length of vector \({\color{red}\bar{v}}\)</em>. This is easily extended to 3D by adding \(z^2\) to the equation.
+ となります。ここで\(||{\color{red}\bar{v}}||\)は<em>ベクトル\({\color{red}\bar{v}}\)の長さ</em>を表わします。ここに\(z^2\)を加えることで簡単に3次元に拡張できます。
</p>
<p>
In this case the length of vector <code>(4, 2)</code> equals:
これを用いると、ベクトル<code>(4, 2)</code>の長さは:
- \[||\color{red}{\bar{v}}|| = \sqrt{\color{green}4^2 + \color{blue}2^2} = \sqrt{\color{green}16 + \color{blue}4} = \sqrt{20} = 4.47 \]
+ \[||{\color{red}\bar{v}}|| = \sqrt{{\color{green}4}^2 + {\color{blue}2}^2} = \sqrt{{\color{green}16} + {\color{blue}4}} = \sqrt{20} = 4.47 \]
Which is <code>4.47</code>.
となり、これは<code>4.47</code>です。
@@ -184,7 +184,7 @@
So how do we calculate the dot product? The dot product is a component-wise multiplication where we add the results together. It looks like this with two unit vectors (you can verify that both their lengths are exactly <code>1</code>):
ではどのようにして内積を計算するのでしょう。内積は各要素毎に積を取り、それらを足し合わせて計算できます。2つの単位ベクトルでは以下のようになります（各ベクトルの長さが<code>1</code>であることを確認してください）:
- \[ \begin{pmatrix} \color{red}{0.6} \\ -\color{green}{0.8} \\ \color{blue}0 \end{pmatrix} \cdot \begin{pmatrix} \color{red}0 \\ \color{green}1 \\ \color{blue}0 \end{pmatrix} = (\color{red}{0.6} * \color{red}0) + (-\color{green}{0.8} * \color{green}1) + (\color{blue}0 * \color{blue}0) = -0.8 \]
+ \[ \begin{pmatrix} {\color{red}0.6} \\ -{\color{green}0.8} \\ {\color{blue}0} \end{pmatrix} \cdot \begin{pmatrix} {\color{red}0} \\ {\color{green}1} \\ {\color{blue}0} \end{pmatrix} = ({\color{red}0.6} * {\color{red}0}) + (-{\color{green}0.8} * {\color{green}1}) + ({\color{blue}0} * {\color{blue}0}) = -0.8 \]
To calculate the degree between both these unit vectors we use the inverse of the cosine function \(cos^{-1}\) and this results in <code>143.1</code> degrees. We now effectively calculated the angle between these two vectors. The dot product proves very useful when doing lighting calculations later on.
2つのベクトルのなす角を求めるにはコサインの逆関数\(cos^{-1}\)を利用します。今回の場合この値は<code>143.1</code>度です。2つのベクトルのなす角度を効率よく計算できるようになりました。内積は後程でてくる照明の計算において非常に役立ちます。
@@ -203,7 +203,7 @@
Unlike the other operations, the cross product isn't really intuitive without delving into linear algebra so it's best to just memorize the formula and you'll be fine (or don't, you'll probably be fine as well). Below you'll see the cross product between two orthogonal vectors A and B:
他の演算と違い、外積の計算は線形代数に踏み込まなければあまり直感的ではありません。とりあえず公式を暗記すれば十分です（覚えられなくてもおそらく大丈夫です）。以下の式は2つの直交するベクトルAとBの外積を表すものです:
- \[\begin{pmatrix} \color{red}{A_{x}} \\ \color{green}{A_{y}} \\ \color{blue}{A_{z}} \end{pmatrix} \times \begin{pmatrix} \color{red}{B_{x}} \\ \color{green}{B_{y}} \\ \color{blue}{B_{z}} \end{pmatrix} = \begin{pmatrix} \color{green}{A_{y}} \cdot \color{blue}{B_{z}} - \color{blue}{A_{z}} \cdot \color{green}{B_{y}} \\ \color{blue}{A_{z}} \cdot \color{red}{B_{x}} - \color{red}{A_{x}} \cdot \color{blue}{B_{z}} \\ \color{red}{A_{x}} \cdot \color{green}{B_{y}} - \color{green}{A_{y}} \cdot \color{red}{B_{x}} \end{pmatrix} \]
+ \[\begin{pmatrix} {\color{red}A_{x}} \\ {\color{green}A_{y}} \\ {\color{blue}A_{z}} \end{pmatrix} \times \begin{pmatrix} {\color{red}B_{x}} \\ {\color{green}B_{y}} \\ {\color{blue}B_{z}} \end{pmatrix} = \begin{pmatrix} {\color{green}A_{y}} \cdot {\color{blue}B_{z}} - {\color{blue}A_{z}} \cdot {\color{green}B_{y}} \\ {\color{blue}A_{z}} \cdot {\color{red}B_{x}} - {\color{red}A_{x}} \cdot {\color{blue}B_{z}} \\ {\color{red}A_{x}} \cdot {\color{green}B_{y}} - {\color{green}A_{y}} \cdot {\color{red}B_{x}} \end{pmatrix} \]
As you can see, it doesn't really seem to make sense. However, if you just follow these steps you'll get another vector that is orthogonal to your input vectors.
これを見てもよく意味が分からないでしょう。しかしこの手順に従えば2つのベクトルからそれらに直交するベクトルが得られます。
@@ -233,12 +233,12 @@
Matrix addition and subtraction between two matrices is done on a per-element basis. So the same general rules apply that we're familiar with for normal numbers, but done on the elements of both matrices with the same index. This does mean that addition and subtraction is only defined for matrices of the same dimensions. A 3x2 matrix and a 2x3 matrix (or a 3x3 matrix and a 4x4 matrix) cannot be added or subtracted together. Let's see how matrix addition works on two 2x2 matrices:
2つの行列の和と差は要素毎に行われます。なので通常の数字と同じような規則が各要素毎に成り立ちます。要素毎に行われるため、和と差は同じ型の行列に対してのみ定義できます。3x2の行列と2x3の行列、あるいは3x3の行列と4x4の行列は足したり引いたりすることができません。2x2の行列どうしの足し算を以下に示します:
- \[\begin{bmatrix} \color{red}1 & \color{red}2 \\ \color{green}3 & \color{green}4 \end{bmatrix} + \begin{bmatrix} \color{red}5 & \color{red}6 \\ \color{green}7 & \color{green}8 \end{bmatrix} = \begin{bmatrix} \color{red}1 + \color{red}5 & \color{red}2 + \color{red}6 \\ \color{green}3 + \color{green}7 & \color{green}4 + \color{green}8 \end{bmatrix} = \begin{bmatrix} \color{red}6 & \color{red}8 \\ \color{green}{10} & \color{green}{12} \end{bmatrix} \]
+ \[\begin{bmatrix} {\color{red}1} & {\color{red}2} \\ {\color{green}3} & {\color{green}4} \end{bmatrix} + \begin{bmatrix} {\color{red}5} & {\color{red}6} \\ {\color{green}7} & {\color{green}8} \end{bmatrix} = \begin{bmatrix} {\color{red}1} + {\color{red}5} & {\color{red}2} + {\color{red}6} \\ {\color{green}3} + {\color{green}7} & {\color{green}4} + {\color{green}8} \end{bmatrix} = \begin{bmatrix} {\color{red}6} & {\color{red}8} \\ {\color{green}10} & {\color{green}12} \end{bmatrix} \]
The same rules apply for matrix subtraction:
同様にして差は以下のようになります:
- \[\begin{bmatrix} \color{red}4 & \color{red}2 \\ \color{green}1 & \color{green}6 \end{bmatrix} - \begin{bmatrix} \color{red}2 & \color{red}4 \\ \color{green}0 & \color{green}1 \end{bmatrix} = \begin{bmatrix} \color{red}4 - \color{red}2 & \color{red}2 - \color{red}4 \\ \color{green}1 - \color{green}0 & \color{green}6 - \color{green}1 \end{bmatrix} = \begin{bmatrix} \color{red}2 & -\color{red}2 \\ \color{green}1 & \color{green}5 \end{bmatrix} \]
+ \[\begin{bmatrix} {\color{red}4} & {\color{red}2} \\ {\color{green}1} & {\color{green}6} \end{bmatrix} - \begin{bmatrix} {\color{red}2} & {\color{red}4} \\ {\color{green}0} & {\color{green}1} \end{bmatrix} = \begin{bmatrix} {\color{red}4} - {\color{red}2} & {\color{red}2} - {\color{red}4} \\ {\color{green}1} - {\color{green}0} & {\color{green}6} - {\color{green}1} \end{bmatrix} = \begin{bmatrix} {\color{red}2} & -{\color{red}2} \\ {\color{green}1} & {\color{green}5} \end{bmatrix} \]
</p>
@@ -248,7 +248,7 @@ The same rules apply for matrix subtraction:
A matrix-scalar product multiples each element of the matrix by a scalar. The following example illustrates the multiplication:
行列とスカラの積は行列の各要素にスカラを掛け合わせることとして定義されます。以下にこの積の例を示します:
- \[\color{green}2 \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} \color{green}2 \cdot 1 & \color{green}2 \cdot 2 \\ \color{green}2 \cdot 3 & \color{green}2 \cdot 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\]
+ \[{\color{green}2} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} {\color{green}2} \cdot 1 & {\color{green}2} \cdot 2 \\ {\color{green}2} \cdot 3 & {\color{green}2} \cdot 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}\]
Now it also makes sense as to why those single numbers are called scalars. A scalar basically <em>scales</em> all the elements of the matrix by its value. In the previous example, all elements were scaled by <code>2</code>.
これを見るとなぜ数字がスカラと呼ばれるのか分かります。スカラは行列の各要素をその値で<em>拡大、縮小</em>（英語でscaleと言います）するのです。前の例では各要素を<code>2</code>倍に拡大しています。
@@ -277,7 +277,7 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
Let's get started with an example of a matrix multiplication of 2 <code>2x2</code> matrices:
まずは2つの<code>2x2</code>の行列の積を例示しましょう:
- \[ \begin{bmatrix} \color{red}1 & \color{red}2 \\ \color{green}3 & \color{green}4 \end{bmatrix} \cdot \begin{bmatrix} \color{blue}5 & \color{purple}6 \\ \color{blue}7 & \color{purple}8 \end{bmatrix} = \begin{bmatrix} \color{red}1 \cdot \color{blue}5 + \color{red}2 \cdot \color{blue}7 & \color{red}1 \cdot \color{purple}6 + \color{red}2 \cdot \color{purple}8 \\ \color{green}3 \cdot \color{blue}5 + \color{green}4 \cdot \color{blue}7 & \color{green}3 \cdot \color{purple}6 + \color{green}4 \cdot \color{purple}8 \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \]
+ \[ \begin{bmatrix} {\color{red}1} & {\color{red}2} \\ {\color{green}3} & {\color{green}4} \end{bmatrix} \cdot \begin{bmatrix} {\color{blue}5} & {\color{purple}6} \\ {\color{blue}7} & {\color{purple}8} \end{bmatrix} = \begin{bmatrix} {\color{red}1} \cdot {\color{blue}5} + {\color{red}2} \cdot {\color{blue}7} & {\color{red}1} \cdot {\color{purple}6} + {\color{red}2} \cdot {\color{purple}8} \\ {\color{green}3} \cdot {\color{blue}5} + {\color{green}4} \cdot {\color{blue}7} & {\color{green}3} \cdot {\color{purple}6} + {\color{green}4} \cdot {\color{purple}8} \end{bmatrix} = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix} \]
Right now you're probably trying to figure out what the hell just happened? Matrix multiplication is a combination of normal multiplication and addition using the left-matrix's rows with the right-matrix's columns. Let's try discussing this with the following image:
いったい何ごとかと身構えることでしょう。行列の積は左の行列の行と右の行列の列を用いて通常の和と積を組み合わせることで計算されます。下図で説明します:
@@ -309,7 +309,7 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
Let's finish the discussion of matrix-matrix multiplication with a larger example. Try to visualize the pattern using the colors. As a useful exercise, see if you can come up with your own answer of the multiplication and then compare them with the resulting matrix (once you try to do a matrix multiplication by hand you'll quickly get the grasp of them).
もう少し大きな例で行列の積の議論を終えましょう。色分けにより計算法を可視化しています。いい演習として自分で行列の積を計算して答えを示し合わせて下さい。実際に手を動かすことで、計算の流れがよく掴めます。
- \[ \begin{bmatrix} \color{red}4 & \color{red}2 & \color{red}0 \\ \color{green}0 & \color{green}8 & \color{green}1 \\ \color{blue}0 & \color{blue}1 & \color{blue}0 \end{bmatrix} \cdot \begin{bmatrix} \color{red}4 & \color{green}2 & \color{blue}1 \\ \color{red}2 & \color{green}0 & \color{blue}4 \\ \color{red}9 & \color{green}4 & \color{blue}2 \end{bmatrix} = \begin{bmatrix} \color{red}4 \cdot \color{red}4 + \color{red}2 \cdot \color{red}2 + \color{red}0 \cdot \color{red}9 & \color{red}4 \cdot \color{green}2 + \color{red}2 \cdot \color{green}0 + \color{red}0 \cdot \color{green}4 & \color{red}4 \cdot \color{blue}1 + \color{red}2 \cdot \color{blue}4 + \color{red}0 \cdot \color{blue}2 \\ \color{green}0 \cdot \color{red}4 + \color{green}8 \cdot \color{red}2 + \color{green}1 \cdot \color{red}9 & \color{green}0 \cdot \color{green}2 + \color{green}8 \cdot \color{green}0 + \color{green}1 \cdot \color{green}4 & \color{green}0 \cdot \color{blue}1 + \color{green}8 \cdot \color{blue}4 + \color{green}1 \cdot \color{blue}2 \\ \color{blue}0 \cdot \color{red}4 + \color{blue}1 \cdot \color{red}2 + \color{blue}0 \cdot \color{red}9 & \color{blue}0 \cdot \color{green}2 + \color{blue}1 \cdot \color{green}0 + \color{blue}0 \cdot \color{green}4 & \color{blue}0 \cdot \color{blue}1 + \color{blue}1 \cdot \color{blue}4 + \color{blue}0 \cdot \color{blue}2 \end{bmatrix}
+ \[ \begin{bmatrix} {\color{red}4} & {\color{red}2} & {\color{red}0} \\ {\color{green}0} & {\color{green}8} & {\color{green}1} \\ {\color{blue}0} & {\color{blue}1} & {\color{blue}0} \end{bmatrix} \cdot \begin{bmatrix} {\color{red}4} & {\color{green}2} & {\color{blue}1} \\ {\color{red}2} & {\color{green}0} & {\color{blue}4} \\ {\color{red}9} & {\color{green}4} & {\color{blue}2} \end{bmatrix} = \begin{bmatrix} {\color{red}4} \cdot {\color{red}4} + {\color{red}2} \cdot {\color{red}2} + {\color{red}0} \cdot {\color{red}9} & {\color{red}4} \cdot {\color{green}2} + {\color{red}2} \cdot {\color{green}0} + {\color{red}0} \cdot {\color{green}4} & {\color{red}4} \cdot {\color{blue}1} + {\color{red}2} \cdot {\color{blue}4} + {\color{red}0} \cdot {\color{blue}2} \\ {\color{green}0} \cdot {\color{red}4} + {\color{green}8} \cdot {\color{red}2} + {\color{green}1} \cdot {\color{red}9} & {\color{green}0} \cdot {\color{green}2} + {\color{green}8} \cdot {\color{green}0} + {\color{green}1} \cdot {\color{green}4} & {\color{green}0} \cdot {\color{blue}1} + {\color{green}8} \cdot {\color{blue}4} + {\color{green}1} \cdot {\color{blue}2} \\ {\color{blue}0} \cdot {\color{red}4} + {\color{blue}1} \cdot {\color{red}2} + {\color{blue}0} \cdot {\color{red}9} & {\color{blue}0} \cdot {\color{green}2} + {\color{blue}1} \cdot {\color{green}0} + {\color{blue}0} \cdot {\color{green}4} & {\color{blue}0} \cdot {\color{blue}1} + {\color{blue}1} \cdot {\color{blue}4} + {\color{blue}0} \cdot {\color{blue}2} \end{bmatrix}
\\ = \begin{bmatrix} 20 & 8 & 12 \\ 25 & 4 & 34 \\ 2 & 0 & 4 \end{bmatrix}\]
</p>
@@ -341,10 +341,10 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
In OpenGL we usually work with <code>4x4</code> transformation matrices for several reasons and one of them is that most of the vectors are of size 4. The most simple transformation matrix that we can think of is the <def>identity matrix</def>. The identity matrix is an <code>NxN</code> matrix with only 0s except on its diagonal. As you'll see, this transformation matrix leaves a vector completely unharmed:
OpenGLにおいては通常<code>4x4</code>の変換行列を利用します。そのひとつの理由はほとんどの場合4次元のベクトルを扱うからです。想像し得る最も単純な変換行列は<def>単位行列</def>です。単位行列は対角成分以外の要素が0である<code>NxN</code>の行列です。この変換行列は以下のようにベクトルを全く変化させません:
- \[ \begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} = \begin{bmatrix} \color{red}1 \cdot 1 \\ \color{green}1 \cdot 2 \\ \color{blue}1 \cdot 3 \\ \color{purple}1 \cdot 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \]
+ \[ \begin{bmatrix} {\color{red}1} & {\color{red}0} & {\color{red}0} & {\color{red}0} \\ {\color{green}0} & {\color{green}1} & {\color{green}0} & {\color{green}0} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}1} & {\color{blue}0} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} = \begin{bmatrix} {\color{red}1} \cdot 1 \\ {\color{green}1} \cdot 2 \\ {\color{blue}1} \cdot 3 \\ {\color{purple}1} \cdot 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \]
- The vector is completely untouched. This becomes obvious from the rules of multiplication: the first result element is each individual element of the first row of the matrix multiplied with each element of the vector. Since each of the row's elements are 0 except the first one, we get: \(\color{red}1\cdot1 + \color{red}0\cdot2 + \color{red}0\cdot3 + \color{red}0\cdot4 = 1\) and the same applies for the other 3 elements of the vector.
- ベクトルは完全に元のままです。これは積の定義から明らかです。結果となるベクトルの1つ目の要素は行列の1行目にベクトルの各要素を掛けたものです。行列の一行目は1つめの要素を除いて0なので、得られる値は\(\color{red}1\cdot1 + \color{red}0\cdot2 + \color{red}0\cdot3 + \color{red}0\cdot4 = 1\)となり、残りの3つの要素についても同様です。
+ The vector is completely untouched. This becomes obvious from the rules of multiplication: the first result element is each individual element of the first row of the matrix multiplied with each element of the vector. Since each of the row's elements are 0 except the first one, we get: \({\color{red}1\cdot1} + {\color{red}0\cdot2} + {\color{red}0\cdot3} + {\color{red}0\cdot4} = 1\) and the same applies for the other 3 elements of the vector.
+ ベクトルは完全に元のままです。これは積の定義から明らかです。結果となるベクトルの1つ目の要素は行列の1行目にベクトルの各要素を掛けたものです。行列の一行目は1つめの要素を除いて0なので、得られる値は\({\color{red}1}\cdot1 + {\color{red}0}\cdot2 + {\color{red}0}\cdot3 + {\color{red}0}\cdot4 = 1\)となり、残りの3つの要素についても同様です。
</p>
<note>
@@ -360,8 +360,8 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
</p>
<p>
- Let's try scaling the vector \(\color{red}{\bar{v}} = (3,2)\). We will scale the vector along the x-axis by <code>0.5</code>, thus making it twice as narrow; and we'll scale the vector by <code>2</code> along the y-axis, making it twice as high. Let's see what it looks like if we scale the vector by <code>(0.5,2)</code> as \(\color{blue}{\bar{s}}\):
- \(\color{red}{\bar{v}} = (3, 2)\)を拡大してみましょう。x軸方向に<code>0.5</code>倍、つまり半分に切り詰め、y軸方向に<code>2</code>倍、つまり倍に引き伸ばします。このベクトルを<code>(0.5, 2)</code>により拡大したベクトル\(\color{blue}{\bar{s}}\)は以下のようになります:
+ Let's try scaling the vector \({\color{red}\bar{v}} = (3,2)\). We will scale the vector along the x-axis by <code>0.5</code>, thus making it twice as narrow; and we'll scale the vector by <code>2</code> along the y-axis, making it twice as high. Let's see what it looks like if we scale the vector by <code>(0.5,2)</code> as \({\color{blue}\bar{s}}\):
+ \({\color{red}\bar{v}} = (3, 2)\)を拡大してみましょう。x軸方向に<code>0.5</code>倍、つまり半分に切り詰め、y軸方向に<code>2</code>倍、つまり倍に引き伸ばします。このベクトルを<code>(0.5, 2)</code>により拡大したベクトル\({\color{blue}\bar{s}}\)は以下のようになります:
</p>
@@ -373,10 +373,10 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
</p>
<p>
- Let's start building a transformation matrix that does the scaling for us. We saw from the identity matrix that each of the diagonal elements were multiplied with its corresponding vector element. What if we were to change the <code>1</code>s in the identity matrix to <code>3</code>s? In that case, we would be multiplying each of the vector elements by a value of <code>3</code> and thus effectively uniformly scale the vector by 3. If we represent the scaling variables as \( (\color{red}{S_1}, \color{green}{S_2}, \color{blue}{S_3}) \) we can define a scaling matrix on any vector \((x,y,z)\) as:
- それでは拡大を行う変換行列を作成してみましょう。先程単位行列を見た際、対角線上の各要素がそれに対応するベクトルの要素と掛け合わせられるのを確認しました。単位行列の対角成分を<code>1</code>ではなく<code>3</code>にしたらどうなるでしょう。この場合、ベクトルの各要素に<code>3</code>が掛けられ、ベクトルが一様に3倍になります。各軸に対する拡大率をそれぞれ\( (\color{red}{S_1}, \color{green}{S_2}, \color{blue}{S_3})\)とした場合、ベクトル\((x, y, z)\)に対する拡大行列は:
+ Let's start building a transformation matrix that does the scaling for us. We saw from the identity matrix that each of the diagonal elements were multiplied with its corresponding vector element. What if we were to change the <code>1</code>s in the identity matrix to <code>3</code>s? In that case, we would be multiplying each of the vector elements by a value of <code>3</code> and thus effectively uniformly scale the vector by 3. If we represent the scaling variables as \( ({\color{red}S_1}, {\color{green}S_2}, {\color{blue}S_3}) \) we can define a scaling matrix on any vector \((x,y,z)\) as:
+ それでは拡大を行う変換行列を作成してみましょう。先程単位行列を見た際、対角線上の各要素がそれに対応するベクトルの要素と掛け合わせられるのを確認しました。単位行列の対角成分を<code>1</code>ではなく<code>3</code>にしたらどうなるでしょう。この場合、ベクトルの各要素に<code>3</code>が掛けられ、ベクトルが一様に3倍になります。各軸に対する拡大率をそれぞれ\( ({\color{red}S_1}, {\color{green}S_2}, {\color{blue}S_3})\)とした場合、ベクトル\((x, y, z)\)に対する拡大行列は:
- \[\begin{bmatrix} \color{red}{S_1} & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}{S_2} & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}{S_3} & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} \color{red}{S_1} \cdot x \\ \color{green}{S_2} \cdot y \\ \color{blue}{S_3} \cdot z \\ 1 \end{pmatrix} \]
+ \[\begin{bmatrix} {\color{red}S_1} & {\color{red}0} & {\color{red}0} & {\color{red}0} \\ {\color{green}0} & {\color{green}S_2} & {\color{green}0} & {\color{green}0} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}S_3} & {\color{blue}0} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} {\color{red}S_1} \cdot x \\ {\color{green}S_2} \cdot y \\ {\color{blue}S_3} \cdot z \\ 1 \end{pmatrix} \]
Note that we keep the 4th scaling value <code>1</code>. The <code>w</code> component is used for other purposes as we'll see later on.
となります。4つ目の要素は<code>1</code>であることに注意して下さい。この<code>w</code>要素は後程別の用途に利用します。
@@ -390,10 +390,10 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
</p>
<p>
- Just like the scaling matrix there are several locations on a 4-by-4 matrix that we can use to perform certain operations and for translation those are the top-3 values of the 4th column. If we represent the translation vector as \((\color{red}{T_x},\color{green}{T_y},\color{blue}{T_z})\) we can define the translation matrix by:
- 拡大行列と同様、4x4の行列のある部分を利用することで平行移動を実現できます。その部分とは4番目の列の上3つの要素です。平行移動ベクトルを\((\color{red}{T_x},\color{green}{T_y},\color{blue}{T_z})\)で表わすと、平行移動行列は:
+ Just like the scaling matrix there are several locations on a 4-by-4 matrix that we can use to perform certain operations and for translation those are the top-3 values of the 4th column. If we represent the translation vector as \(({\color{red}T_x},{\color{green}T_y},{\color{blue}T_z})\) we can define the translation matrix by:
+ 拡大行列と同様、4x4の行列のある部分を利用することで平行移動を実現できます。その部分とは4番目の列の上3つの要素です。平行移動ベクトルを\(({\color{red}T_x},{\color{green}T_y},{\color{blue}T_z})\)で表わすと、平行移動行列は:
- \[\begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}{T_x} \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}{T_y} \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}{T_z} \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} x + \color{red}{T_x} \\ y + \color{green}{T_y} \\ z + \color{blue}{T_z} \\ 1 \end{pmatrix} \]
+ \[\begin{bmatrix} {\color{red}1} & {\color{red}0} & {\color{red}0} & {\color{red}T_x} \\ {\color{green}0} & {\color{green}1} & {\color{green}0} & {\color{green}T_y} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}1} & {\color{blue}T_z} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} x + {\color{red}T_x} \\ y + {\color{green}T_y} \\ z + {\color{blue}T_z} \\ 1 \end{pmatrix} \]
This works because all of the translation values are multiplied by the vector's <code>w</code> column and added to the vector's original values (remember the matrix-multiplication rules). This wouldn't have been possible with a 3-by-3 matrix.
のようになります。平行移動の値はベクトルの<code>w</code>要素に掛けられ、ベクトルの元の値に足し合せられます（行列の積の規則を思い出して下さい）。これは3x3の行列では実現できません。
@@ -437,8 +437,8 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
ここで<code>\(\pi\)</code>はおよそ<code>3.14159265359</code>です。
</note>
- Rotating half a circle rotates us 360/2 = 180 degrees and rotating 1/5th to the right means we rotate 360/5 = 72 degrees to the right. This is demonstrated for a basic 2D vector where \(\color{red}{\bar{v}}\) is rotated 72 degrees to the right, or clockwise, from \(\color{green}{\bar{k}}\):
- 半円分の回転は360/2 = 180度の回転を意味し、右に1/5回転させるのは360/5 = 72度時計回りに回転させることを意味します。以下にこの回転を2次元ベクトルで行ったものを図示します。\(\color{red}{\bar{v}}\)は\(\color{green}{\bar{k}}\)を時計回りに72度回転させたものです:
+ Rotating half a circle rotates us 360/2 = 180 degrees and rotating 1/5th to the right means we rotate 360/5 = 72 degrees to the right. This is demonstrated for a basic 2D vector where \({\color{red}\bar{v}}\) is rotated 72 degrees to the right, or clockwise, from \({\color{green}\bar{k}}\):
+ 半円分の回転は360/2 = 180度の回転を意味し、右に1/5回転させるのは360/5 = 72度時計回りに回転させることを意味します。以下にこの回転を2次元ベクトルで行ったものを図示します。\({\color{red}\bar{v}}\)は\({\color{green}\bar{k}}\)を時計回りに72度回転させたものです:
</p>
<img src="/img/getting-started/vectors_angle.png" class="clean" />
@@ -462,29 +462,29 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
Rotation around the X-axis:
x軸周りの回転:
- \[\begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}{\cos \theta} & - \color{green}{\sin \theta} & \color{green}0 \\ \color{blue}0 & \color{blue}{\sin \theta} & \color{blue}{\cos \theta} & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} x \\ \color{green}{\cos \theta} \cdot y - \color{green}{\sin \theta} \cdot z \\ \color{blue}{\sin \theta} \cdot y + \color{blue}{\cos \theta} \cdot z \\ 1 \end{pmatrix}\]
+ \[\begin{bmatrix} {\color{red}1} & {\color{red}0} & {\color{red}0} & {\color{red}0} \\ {\color{green}0} & {\color{green}\cos \theta} & - {\color{green}\sin \theta} & {\color{green}0} \\ {\color{blue}0} & {\color{blue}\sin \theta} & {\color{blue}\cos \theta} & {\color{blue}0} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} x \\ {\color{green}\cos \theta} \cdot y - {\color{green}\sin \theta} \cdot z \\ {\color{blue}\sin \theta} \cdot y + {\color{blue}\cos \theta} \cdot z \\ 1 \end{pmatrix}\]
</p>
<p>
Rotation around the Y-axis:
y軸周りの回転:
- \[\begin{bmatrix} \color{red}{\cos \theta} & \color{red}0 & \color{red}{\sin \theta} & \color{red}0 \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}0 \\ - \color{blue}{\sin \theta} & \color{blue}0 & \color{blue}{\cos \theta} & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} \color{red}{\cos \theta} \cdot x + \color{red}{\sin \theta} \cdot z \\ y \\ - \color{blue}{\sin \theta} \cdot x + \color{blue}{\cos \theta} \cdot z \\ 1 \end{pmatrix} \]
+ \[\begin{bmatrix} {\color{red}\cos \theta} & {\color{red}0} & {\color{red}\sin \theta} & {\color{red}0} \\ {\color{green}0} & {\color{green}1} & {\color{green}0} & {\color{green}0} \\ - {\color{blue}\sin \theta} & {\color{blue}0} & {\color{blue}\cos \theta} & {\color{blue}0} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} {\color{red}\cos \theta} \cdot x + {\color{red}\sin \theta} \cdot z \\ y \\ - {\color{blue}\sin \theta} \cdot x + {\color{blue}\cos \theta} \cdot z \\ 1 \end{pmatrix} \]
</p>
<p>
Rotation around the Z-axis:
z軸周りの回転:
- \[\begin{bmatrix} \color{red}{\cos \theta} & - \color{red}{\sin \theta} & \color{red}0 & \color{red}0 \\ \color{green}{\sin \theta} & \color{green}{\cos \theta} & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} \color{red}{\cos \theta} \cdot x - \color{red}{\sin \theta} \cdot y \\ \color{green}{\sin \theta} \cdot x + \color{green}{\cos \theta} \cdot y \\ z \\ 1 \end{pmatrix} \]
+ \[\begin{bmatrix} {\color{red}\cos \theta} & - {\color{red}\sin \theta} & {\color{red}0} & {\color{red}0} \\ {\color{green}\sin \theta} & {\color{green}\cos \theta} & {\color{green}0} & {\color{green}0} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}1} & {\color{blue}0} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} {\color{red}\cos \theta} \cdot x - {\color{red}\sin \theta} \cdot y \\ {\color{green}\sin \theta} \cdot x + {\color{green}\cos \theta} \cdot y \\ z \\ 1 \end{pmatrix} \]
</p>
<p>
- Using the rotation matrices we can transform our position vectors around one of the three unit axes. To rotate around an arbitrary 3D axis we can combine all 3 them by first rotating around the X-axis, then Y and then Z for example. However, this quickly introduces a problem called <def>Gimbal lock</def>. We won't discuss the details, but a better solution is to rotate around an arbitrary unit axis e.g. <code>(0.662,0.2,0.722)</code> (note that this is a unit vector) right away instead of combining the rotation matrices. Such a (verbose) matrix exists and is given below with \((\color{red}{R_x}, \color{green}{R_y}, \color{blue}{R_z})\) as the arbitrary rotation axis:
- これらの回転行列により、位置ベクトルを座標軸に沿って回転させることができます。任意の回転軸に沿って回転させるにはこの3つの回転を組み合わせればいいのですが、これでは<def>ジンバルロック</def>と呼ばれる問題が生じます。この問題には深く立ち入りません。よりよい方法は任意の単位ベクトルに沿って回転させる回転行列を利用することです。そのような直接的な行列は確かに存在し、\((\color{red}{R_x}, \color{green}{R_y}, \color{blue}{R_z})\)を回転軸としたとき以下のように表わされます。
+ Using the rotation matrices we can transform our position vectors around one of the three unit axes. To rotate around an arbitrary 3D axis we can combine all 3 them by first rotating around the X-axis, then Y and then Z for example. However, this quickly introduces a problem called <def>Gimbal lock</def>. We won't discuss the details, but a better solution is to rotate around an arbitrary unit axis e.g. <code>(0.662,0.2,0.722)</code> (note that this is a unit vector) right away instead of combining the rotation matrices. Such a (verbose) matrix exists and is given below with \(({\color{red}R_x}, {\color{green}R_y}, {\color{blue}R_z})\) as the arbitrary rotation axis:
+ これらの回転行列により、位置ベクトルを座標軸に沿って回転させることができます。任意の回転軸に沿って回転させるにはこの3つの回転を組み合わせればいいのですが、これでは<def>ジンバルロック</def>と呼ばれる問題が生じます。この問題には深く立ち入りません。よりよい方法は任意の単位ベクトルに沿って回転させる回転行列を利用することです。そのような直接的な行列は確かに存在し、\(({\color{red}R_x}, {\color{green}R_y}, {\color{blue}R_z})\)を回転軸としたとき以下のように表わされます。
- \[\begin{bmatrix} \cos \theta + \color{red}{R_x}^2(1 - \cos \theta) & \color{red}{R_x}\color{green}{R_y}(1 - \cos \theta) - \color{blue}{R_z} \sin \theta & \color{red}{R_x}\color{blue}{R_z}(1 - \cos \theta) + \color{green}{R_y} \sin \theta & 0 \\ \color{green}{R_y}\color{red}{R_x} (1 - \cos \theta) + \color{blue}{R_z} \sin \theta & \cos \theta + \color{green}{R_y}^2(1 - \cos \theta) & \color{green}{R_y}\color{blue}{R_z}(1 - \cos \theta) - \color{red}{R_x} \sin \theta & 0 \\ \color{blue}{R_z}\color{red}{R_x}(1 - \cos \theta) - \color{green}{R_y} \sin \theta & \color{blue}{R_z}\color{green}{R_y}(1 - \cos \theta) + \color{red}{R_x} \sin \theta & \cos \theta + \color{blue}{R_z}^2(1 - \cos \theta) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\]
+ \[\begin{bmatrix} \cos \theta + {\color{red}R_x}^2(1 - \cos \theta) & {\color{red}R_x}{\color{green}R_y}(1 - \cos \theta) - {\color{blue}R_z} \sin \theta & {\color{red}R_x}{\color{blue}R_z}(1 - \cos \theta) + {\color{green}R_y} \sin \theta & 0 \\ {\color{green}R_y}{\color{red}R_x} (1 - \cos \theta) + {\color{blue}R_z} \sin \theta & \cos \theta + {\color{green}R_y}^2(1 - \cos \theta) & {\color{green}R_y}{\color{blue}R_z}(1 - \cos \theta) - {\color{red}R_x} \sin \theta & 0 \\ {\color{blue}R_z}{\color{red}R_x}(1 - \cos \theta) - {\color{green}R_y} \sin \theta & {\color{blue}R_z}{\color{green}R_y}(1 - \cos \theta) + {\color{red}R_x} \sin \theta & \cos \theta + {\color{blue}R_z}^2(1 - \cos \theta) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\]
A mathematical discussion of generating such a matrix is out of the scope of this chapter. Keep in mind that even this matrix does not completely prevent gimbal lock (although it gets a lot harder). To truly prevent Gimbal locks we have to represent rotations using <def>quaternions</def>, that are not only safer, but also more computationally friendly. However, a discussion of quaternions is out of this chapter's scope.
@@ -497,7 +497,7 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
The true power from using matrices for transformations is that we can combine multiple transformations in a single matrix thanks to matrix-matrix multiplication. Let's see if we can generate a transformation matrix that combines several transformations. Say we have a vector <code>(x,y,z)</code> and we want to scale it by 2 and then translate it by <code>(1,2,3)</code>. We need a translation and a scaling matrix for our required steps. The resulting transformation matrix would then look like:
座標変換に行列を利用する真のパワーは、行列の積のおかげで複数の変換をひとつの行列により表わすことができることです。いくつかの変換を組み合わせた変換行列の例を見てみましょう。ここにベクトル<code>(x, y, z)</code>があり、二倍に拡大した後<code>(1, 2, 3)</code>だけ平行移動させたいとしましょう。この操作のために平行移動の行列と拡大の行列が必要です。その結果変換行列は以下のようになります:
- \[Trans . Scale = \begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}1 \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}2 \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}3 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} . \begin{bmatrix} \color{red}2 & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}2 & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}2 & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} = \begin{bmatrix} \color{red}2 & \color{red}0 & \color{red}0 & \color{red}1 \\ \color{green}0 & \color{green}2 & \color{green}0 & \color{green}2 \\ \color{blue}0 & \color{blue}0 & \color{blue}2 & \color{blue}3 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \]
+ \[Trans . Scale = \begin{bmatrix} {\color{red}1} & {\color{red}0} & {\color{red}0} & {\color{red}1} \\ {\color{green}0} & {\color{green}1} & {\color{green}0} & {\color{green}2} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}1} & {\color{blue}3} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} . \begin{bmatrix} {\color{red}2} & {\color{red}0} & {\color{red}0} & {\color{red}0} \\ {\color{green}0} & {\color{green}2} & {\color{green}0} & {\color{green}0} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}2} & {\color{blue}0} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} = \begin{bmatrix} {\color{red}2} & {\color{red}0} & {\color{red}0} & {\color{red}1} \\ {\color{green}0} & {\color{green}2} & {\color{green}0} & {\color{green}2} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}2} & {\color{blue}3} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} \]
Note that we first do a translation and then a scale transformation when multiplying matrices. Matrix multiplication is not commutative, which means their order is important. When multiplying matrices the right-most matrix is first multiplied with the vector so you should read the multiplications from right to left. It is advised to first do scaling operations, then rotations and lastly translations when combining matrices otherwise they may (negatively) affect each other. For example, if you would first do a translation and then scale, the translation vector would also scale!
行列の積において初めに平行移動をしてその後拡大をしていることに注意して下さい。行列の積は非可換なので順番が大切です。行列の積において、ベクトルに対してまず右側の行列から掛けられるので、行列の積は右から左に見ていかなければいけません。行列を組み合わせる際、まず拡大の操作を行い、その後平行移動の操作に移って下さい。そうしないとお互いに間違った効果が生じてしまいます。例えば平行移動を行った後に拡大すると、平行移動のベクトル自体に拡大の影響がでてしいます。
@@ -507,7 +507,7 @@ Now it also makes sense as to why those single numbers are called scalars. A sca
Running the final transformation matrix on our vector results in the following vector:
組み合わせた結果の変換行列をベクトルに掛けると以下のような結果が得られます:
- \[\begin{bmatrix} \color{red}2 & \color{red}0 & \color{red}0 & \color{red}1 \\ \color{green}0 & \color{green}2 & \color{green}0 & \color{green}2 \\ \color{blue}0 & \color{blue}0 & \color{blue}2 & \color{blue}3 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} . \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = \begin{bmatrix} \color{red}2x + \color{red}1 \\ \color{green}2y + \color{green}2 \\ \color{blue}2z + \color{blue}3 \\ 1 \end{bmatrix} \]
+ \[\begin{bmatrix} {\color{red}2} & {\color{red}0} & {\color{red}0} & {\color{red}1} \\ {\color{green}0} & {\color{green}2} & {\color{green}0} & {\color{green}2} \\ {\color{blue}0} & {\color{blue}0} & {\color{blue}2} & {\color{blue}3} \\ {\color{purple}0} & {\color{purple}0} & {\color{purple}0} & {\color{purple}1} \end{bmatrix} . \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = \begin{bmatrix} {\color{red}2}x + {\color{red}1} \\ {\color{green}2}y + {\color{green}2} \\ {\color{blue}2}z + {\color{blue}3} \\ 1 \end{bmatrix} \]
Great! The vector is first scaled by two and then translated by <code>(1,2,3)</code>.
ええやん。ベクトルはまず2倍に拡大され、<code>(1, 2, 3)</code>だけ平行いどうしました。
