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422 lines
23 KiB
HTML
<!DOCTYPE html>
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<html lang="en">
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<head>
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<meta charset="UTF-8">
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<title>📝 2.5d rotations wtih shear transformation</title>
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<meta name="description" content="2.5d rotations wtih shear transformation">
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<title>MathJax example</title>
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<script>
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MathJax = {
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tex: { macros: {cis: "\\mathop{\\rm{cis}}\\nolimits"} },
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<link rel="stylesheet" href="//cdnjs.cloudflare.com/ajax/libs/highlight.js/10.4.1/styles/default.min.css">
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<link rel="stylesheet" href="style.css">
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<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css" integrity="sha384-JcKb8q3iqJ61gNV9KGb8thSsNjpSL0n8PARn9HuZOnIxN0hoP+VmmDGMN5t9UJ0Z" crossorigin="anonymous">
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<link rel="stylesheet" type="text/css" href="https://tikzjax.com/v1/fonts.css">
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</head>
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<body>
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<center>
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<h1>2.5d rotations with matrix transformations</h1>
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<h3><s>(WACE) Mathematics Specialist ATAR</s><b style="color: red;">**</b></h3>
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</center>
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<div class="card bg-danger">
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<div class="card-body" style="color: white;">
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<center><h4><b>⚠WARNING⚠</b></h4></center>
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<h5>At this point this article is probably outside of the curriculum. The transformations are in the curriculum, but most of this article is on a real math problem I've experienced which of course requires a bit of programming.</h5>
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</div>
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</div>
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<a class="link" style="left:1%; top: 1%;" href="https://npc-strider.github.io/maths">🔗 Back to MATHS home page</a><br>
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<a class="link" style="left:1%; top: 1%;" href="https://npc-strider.github.io">🔗 Back to home page</a><br>
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Warning: This page requires javascript to render the math.
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<hr><br>
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<div class="card">
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<div class="card-body">
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<center><b>Introduction - What is '2.5d'?</b></center>
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'2.5d' is not an actual dimension, but refers to techniques in computer graphics which aim to create the presence of 3d depth, but in a 2d level or environment.<br><br>
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<center>
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<script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw (-2.5,0) rectangle (2.5,3);
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\draw [fill=gray] (-2.5,0) rectangle (2.5,-3);
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\node [label=left:{[-2.5, 0]}] at (-2.5,0) {\textbullet};
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\node [label=right:{[2.5, 0]}] at (2.5,0) {\textbullet};
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\node [label=right:{[2.5, 3]}] at (2.5,3) {\textbullet};
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\node [label=right:{[2.5, -3]}] at (2.5,-3) {\textbullet};
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\node [label=left:{[-2.5, 3]}] at (-2.5,3) {\textbullet};
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\node [label=left:{[-2.5, -3]}] at (-2.5,-3) {\textbullet};
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\end{tikzpicture}
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</script>
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</center>
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This box is a simplified example of 2.5d graphics.<br>
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The gray on the front-facing side and the white top side is an example of shading, and creates the illusion of a 3d environment as a result of basic lighting. The camera is placed at a 45 degree angle, as we see equal proportions of the top and front faces of the box<br>
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However, this is not real 3d graphics, as we're representing this box within a 2d space with only 2d position vectors.<br>
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Therefore, this is 2.5d graphics.
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</div>
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</div>
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<br>
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<div class="card">
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<div class="card-body">
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<center><b>The problem</b></center>
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I was making a mod for a game that added in rockets. The rockets took a ballistic trajectory to reach a target point<br>
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Of course, because this is a 2.5d game I couldn't simply model the flight with the actual equations, because there is no concept of 'height' or 'altitude' in the game<br>
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<br>
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<center>
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<script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[->] (0, 0) -- (6, 0) node[right] {$\vec{i}$};
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\draw[->] (0, 0) -- (0, 6) node[above] {$\vec{j}$};
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\node [label=below:{$\mathbf{S}[1, 1.5]$}] at (1,1.5) {\textbullet};
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\node [label=below:{$\mathbf{T}[5, 1.5]$}] at (5,1.5) {\textbullet};
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\draw[scale=1, domain=1:5, smooth, variable=\x, red] plot ({\x}, {-(\x-1)*(\x-5)+1.5});
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\draw[->, red, dashed] (1,1.5) -- (5,1.5);
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\end{tikzpicture}
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</script>
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</center>
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This parabola works fine for targets on the same \(\vec{j}\) as the silo<br>
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What if we want a trajectory like this?
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<center>
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<script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[->] (0, 0) -- (6, 0) node[right] {$\vec{i}$};
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\draw[->] (0, 0) -- (0, 6) node[above] {$\vec{j}$};
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\node [label=below:{$\mathbf{S}[1, 2]$}] at (1,2) {\textbullet};
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\node [label=above:{$\mathbf{T}[5, 4]$}] at (5,4) {\textbullet};
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\draw[->, red, dashed] (1,2) -- (5,4);
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\end{tikzpicture}
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</script>
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</center>
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We can try rotating the parabola, but that results in a trajectory which looks very unrealistic.
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<center>
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<script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[scale=1, domain=0:4.47213595, smooth, variable=\x, blue, dashed] plot ({\x}, {-(\x)*(\x-4.47213595)});
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\draw[->] (0, 0) -- (6, 0) node[right] {$\vec{i}$};
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\draw[->] (0, 0) -- (0, 6) node[above] {$\vec{j}$};
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\node [label=below:{$\mathbf{S}[1, 2]$}] at (1,2) {\textbullet};
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\node [label=above:{$\mathbf{T}[5, 4]$}] at (5,4) {\textbullet};
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\draw[shift={(1,2)}, scale=1, domain=0:4.5, smooth, variable=\x, red, rotate=30] plot ({\x}, {-(\x)*(\x-4.47213595)});
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\draw[->, red, dashed] (1,2) -- (5,4);
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\end{tikzpicture}
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</script>
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</center>
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The blue dashed parabola represents the pre-transformed trajectory (we are only given the distance \(d\) from target \(\mathbf{T}\) to silo \(\mathbf{S}\))<br>
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<div class="alert alert-warning" role="alert">
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<center>How this transformation is achieved [Advanced]</center>
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<br>
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Our transformation requires a shift. We can't just add two matrices of different dimensions so we need to somehow represent the shift as a matrix multiplication. For this we need a <b>homogeneous coordinate</b><br>
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We represent a 2d vector as a 3d vector, but the \(\vec{k}\) component is a constant \(1\). This constant allows the addition of the shift through matrix multiplication. <br>
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\[\mathbf{X}=\begin{bmatrix}
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x_0 & x_1 & x_2 & x_3 & x_4 & x_5 & \dots \\
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y_0 & y_1 & y_2 & y_3 & y_4 & y_5 & \dots \\
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1 & 1 & 1 & 1 & 1 & 1 & \dots
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\end{bmatrix}\]
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In this case, we applied the following transformations to <b>rotate</b> then <b>translate</b> \(\mathbf{T}\) into 'inaccurate' trajectory, \(\mathbf{T'}\)
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\begin{align}
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\mathbf{X'} &= \begin{bmatrix}
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1 & 0 & \Delta{x} \\
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0 & 1 & \Delta{y} \\
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0 & 0 & 1 \\
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\end{bmatrix}\begin{bmatrix}
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\cos(\theta) & -\sin(\theta) & 0 \\
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\sin(\theta) & \cos(\theta) & 0 \\
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0 & 0 & 1 \\
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\end{bmatrix}\mathbf{X}
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\\\\
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&= \begin{bmatrix}
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1 & 0 & 1 \\
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0 & 1 & 2 \\
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0 & 0 & 1 \\
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\end{bmatrix}\begin{bmatrix}
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\cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) & 0 \\
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\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) & 0 \\
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0 & 0 & 1 \\
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\end{bmatrix}\mathbf{X}
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\end{align}
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However, in reality I would just use a for loop and iterate through each pair of points.<br>
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<pre><code class="python">import math
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import numpy as np
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def rotate(theta):
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return np.array([[cos(theta), -sin(theta)],
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[sin(theta), cos(theta)]])
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T = [[0,0], ... ,[4,0]] #Our starting list of points T
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T_ = []
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C = [1,2]
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R = rotate(math.pi/6) #Rotation matrix when theta=pi/6. Store to R to save function calls
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for P in T:
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P = np.array(P) #convert python list (A point P, like [0,0]) to numpy array
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P_ = np.add( np.matmul(R,P), C ) #multiply with rotation matrix R (matmul), then add C
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T_.append(P_) #Add our transformed point P_ to our transformed list, T_
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end
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# T_ is now our transformed matrix.
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</code></pre>
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</div>
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</div>
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</div>
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<br>
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<div class="card">
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<div class="card-body">
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<center><b>Shear matrices</b></center>
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\begin{align}
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\text{Parallel to the y-axis, Vertical shear}\quad
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\mathbf{X'} &= \begin{bmatrix}
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1 & 0\\
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m & 1
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\end{bmatrix}\mathbf{X}\\
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&= \begin{bmatrix}
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x\\
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mx+y
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\end{bmatrix} \\\\
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\text{Parallel to the x-axis, Horizontal shear}\quad
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\mathbf{X'} &= \begin{bmatrix}
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1 & m\\
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0 & 1
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\end{bmatrix}\mathbf{X} \\
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&= \begin{bmatrix}
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x+my\\
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y
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\end{bmatrix}\\\\
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\text{Where}\quad\mathbf{X} &= \begin{bmatrix}
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x\\
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y
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\end{bmatrix}\\
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\end{align}
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</div>
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</div>
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<br><hr><br>
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<div class="card">
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<div class="card-body">
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<center><b>The solution: shear matrices</b></center><br>
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A shear matrix is better explained visually.
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<center><script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[->] (-6, 1) -- (-1, 1) node[right] {$\vec{i}$};
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\draw[->] (-6, 1) -- (-6, 6) node[above] {$\vec{j}$};
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\draw[->] (2, 1) -- (7, 1) node[right] {$\vec{i}$};
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\draw[->] (2, 1) -- (2, 6) node[above] {$\vec{j}$};
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\draw[->] (-0.7,3) -- (1.7,3);
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\draw (-6,1) rectangle (-2,3);
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\draw (6,3) -- (2,1) -- (2,3) -- (6,5) -- cycle;
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\end{tikzpicture}
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</script></center>
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In this case we applied a vertical shear (parallel to the y-axis). Our x-values are the same, but our y-values have been transformed in a way that they match the line \(y'=mx+y\) (see the matrix representation).<br>
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It has almost all of the properties required for our 2.5d rotation - we're only distorting the matrix on one axis and preserving the other. The only issue is that our transformed object will appear longer.<br>
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<br>
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In my case, I do not need to know the angle - I can use the gradient between the silo \(\mathbf{S}\) and target \(\mathbf{T}\).
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\[m=\frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x}\]
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\(m\) can also be obtained with a trig ratio, yielding the following shear transformation that involve angles rather than gradients.
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\begin{align}
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\text{Parallel to the y-axis, Vertical shear}\quad
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\mathbf{X'} &= \begin{bmatrix}
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1 & 0\\
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\sin(\theta) & 1
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\end{bmatrix}\mathbf{X}\\\\
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\text{Parallel to the x-axis, Horizontal shear}\quad
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\mathbf{X'} &= \begin{bmatrix}
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1 & \sin(\theta)\\
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0 & 1
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\end{bmatrix}\mathbf{X}\\\\
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\end{align}
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To resolve our issue with the transformed object being 'lengthened', we need to scale it back.<br>
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<center><script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[->] (-6, 1) -- (-1, 1) node[right] {$\vec{i}$};
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\draw[->] (-6, 1) -- (-6, 6) node[above] {$\vec{j}$};
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\draw[->] (2, 1) -- (7, 1) node[right] {$\vec{i}$};
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\draw[->] (2, 1) -- (2, 6) node[above] {$\vec{j}$};
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\draw[->] (-0.7,3) -- (1.7,3);
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\draw (-6,1) rectangle (-2,3);
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\draw (6,3) -- (2,1) -- (2,3) -- (6,5) -- cycle;
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\node [label=below:{$\mathbf{P_1}$}] at (-6,1) {\textbullet};
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\node [label=below:{$\mathbf{P_2}$}] at (-2,1) {\textbullet};
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\node [label=below:{$\mathbf{P'_1}$}] at (2,1) {\textbullet};
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\node [label=right:{$\mathbf{P'_2}$}] at (6,3) {\textbullet};
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\end{tikzpicture}
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</script></center>
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Let's choose two points, \(\mathbf{P_1}\) and \(\mathbf{P_2}\).<br>
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We obtain the original distance, \(d\) and the transformed distance, \(d'\)
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\begin{align}
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d &= \|\mathbf{P_1}-\mathbf{P_2}\| \\
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&= \sqrt{(\mathbf{P_1}_x-\mathbf{P_2}_x)^2+(\mathbf{P_1}_y-\mathbf{P_2}_y)^2} \\
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d' &= \|\mathbf{P'_1}-\mathbf{P'_2}\| \\
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&= \sqrt{(\mathbf{P'_1}_x-\mathbf{P'_2}_x)^2+(\mathbf{P'_1}_y-\mathbf{P'_2}_y)^2}\\\\
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\text{ where }\quad\mathbf{P'_1},\mathbf{P'_2} &= \begin{bmatrix}
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1 & 0\\
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\frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x} & 1
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\end{bmatrix}\mathbf{P_1},\mathbf{P_2}\\
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&= \begin{bmatrix}
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1 & 0\\
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\sin(\theta) & 1
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\end{bmatrix}\mathbf{P_1},\mathbf{P_2}\\
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\end{align}
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Note that \(d\) is also the distance from the silo \(\mathbf{S}\) and target \(\mathbf{T}\)<br>
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\(\mathbf{P_1}\) and \(\mathbf{P_2}\) are not the same as \(\mathbf{S}\) and \(\mathbf{T}\), because we've created these points with \(\begin{bmatrix}0\\0\end{bmatrix}\) as the origin, <b>not</b> \(\mathbf{S}\). We will translate all of the points later so that our starting point is \(\mathbf{S}\) after rotating.<br>
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However, the distances are the same, so we might as well use \(\mathbf{S}\) and \(\mathbf{T}\) to calculate \(d\).
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\[d = \|\mathbf{P_1}-\mathbf{P_2}\| = \|\mathbf{T}-\mathbf{S}\|\]
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Our scale factor is simply the ratio between the two.
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\begin{align}
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s &= \frac{d}{d'} \\
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&= \frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}
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\end{align}
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Putting it all together, our 2.5d rotation is now:
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\begin{align}
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\mathbf{X'} &= s\begin{bmatrix}
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1 & 0\\
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m & 0
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\end{bmatrix}\mathbf{X}\\
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&= \frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}\cdot\begin{bmatrix}
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1 & 0\\
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\frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x} & 1
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\end{bmatrix}\mathbf{X}\\
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\text{ or }\quad &= \frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}\cdot\begin{bmatrix}
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1 & 0\\
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\sin(\theta) & 1
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\end{bmatrix}\mathbf{X}\\\\
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\end{align}
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So we need a 'pre transformation' on two points, \(\mathbf{P_1}\) and \(\mathbf{P_2}\) in order to obtain our scale factor
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<center><script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[->] (-6, 1) -- (-1, 1) node[right] {$\vec{i}$};
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\draw[->] (-6, 1) -- (-6, 6) node[above] {$\vec{j}$};
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\draw[->] (2, 1) -- (7, 1) node[right] {$\vec{i}$};
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\draw[->] (2, 1) -- (2, 6) node[above] {$\vec{j}$};
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\draw[->] (-0.7,3) -- (1.7,3);
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\draw (-6,1) rectangle (-2,3);
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\draw (6,3) -- (2,1) -- (2,3) -- (6,5) -- cycle;
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\node [label=below:{$\mathbf{P_1}$}] at (-6,1) {\textbullet};
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\node [label=below:{$\mathbf{P_2}$}] at (-2,1) {\textbullet};
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\node [label=below:{$\mathbf{P'_1}$}] at (2,1) {\textbullet};
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\node [label=right:{$\mathbf{P'_2}$}] at (6,3) {\textbullet};
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\end{tikzpicture}
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</script></center><br>
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<center><script type="text/tikz">
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\begin{tikzpicture}[thick,scale=1, every node/.style={scale=1.9,inner sep=0pt}]
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\draw[->] (-6, 1) -- (-1, 1) node[right] {$\vec{i}$};
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\draw[->] (-6, 1) -- (-6, 6) node[above] {$\vec{j}$};
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\draw[->] (2, 1) -- (7, 1) node[right] {$\vec{i}$};
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\draw[->] (2, 1) -- (2, 6) node[above] {$\vec{j}$};
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\draw[->] (-0.7,3) -- (1.7,3);
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\draw (-6,1) rectangle (-2,3);
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\draw (5.36656315,2.68328157) -- (2,1) -- (2,3) -- (5.36656315,4.47213595) -- cycle;
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\node [label=below:{$\mathbf{P_1}$}] at (-6,1) {\textbullet};
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\node [label=below:{$\mathbf{P_2}$}] at (-2,1) {\textbullet};
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\node [label=below:{$\mathbf{P''_1}$}] at (2,1) {\textbullet};
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\node [label=right:{$\mathbf{P''_2}$}] at (5.36656315,2.68328157) {\textbullet};
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\end{tikzpicture}
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</script></center>
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The limitation of this technique is that we need to perform the rotation about \(\begin{bmatrix}0\\0\end{bmatrix}\), the origin. After this we can translate the rotated matrix to match the silo and target location.<br>
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You can use homogeneous coordinates to achieve this (see the yellow box earlier), or programatically do it.<br>
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<br>
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We also have to solve one last issue with this method: it only works within the domain \(\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)\).<br>
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This is because the gradient can only describe so much information.<br>
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For instance, the line \(f(x)=1\cdot x\).<br>
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We cannot tell if the line is going from the top-right to bottom-left, or bottom-left to top-right because it describes both situations.<br>
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We need to adjust the signs on the shear matrix according to the quadrant our target is in.
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<pre><code class="python">import math
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import numpy as np
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def quad(A,B): #A is our origin/fix point, B is our other point.
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if B[0] > A[0] and B[1] > A[1]:
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return 1 #First quadrant
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elif B[0] < A[0] and B[1] > A[1]:
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return 2 #Second quadrant
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elif B[0] < A[0] and B[1] < A[1]:
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return 3 #Third quadrant
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else:
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return 4 #Fourth quadrant
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#By elimination. We could do another if but it's unnecessary
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# elif B[0] < A[0] and B[1] > A[1]:
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def trajectory(distance, n):
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# Here we create a list of points. n is the amount we create
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# (higher n results in greater precision)
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# Not going to include it here - uses bezier interpolation and stuff to create
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# the illusion of a ballistic trajectory with gravity (not a simple quadratic function)
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#
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return points
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S = np.array([10,4]) #Silo position vector
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T = np.array([5204,954]) #Target position vector
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d = np.linalg.norm(T - S) #Get the distance: d = ((Tx-Sx)^2+(Ty-Sy)^2)^(0.5)
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m = (T[1]-S[1])/(T[0]-S[0]) #Gradient of line between target and silo.
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X = trajectory(d, 50) #50 is a constant for n. only affects trajectory precision
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#X now contains a 'flat trajectory', a list of points which have the correct
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#distance but the wrong angle and offset.
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#It does not intersect with the target T.
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def shear(m, X):
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q = quad(S,T) #Silo and target is passed to the quadrant finding function.
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#q is the quadrant
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if q == 1 or q == 4:
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return np.array([[1,0],
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[m,1]]) #shear matrix.
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else:
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return np.array([[-1,0],
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[m,-1]]) #shear matrix, but for quadrants 2 and 3 (negative x).
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# AN easier way to do this would be to just compare S[0] > T[0]. Forget about the quadrant stuff.
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# because we only need to know if it's in the negative or positive x.
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SHEAR = shear(m, X) #we get the shear matrix, which the function returns
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P_ = np.matmul(SHEAR, X[(0,-1),]) #first, take a slice of the array X to obtain the first (P1)
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# and last (P2) points. Then multiply this new 2x2 matrix
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# containing P1 and P2 by the shear matrix
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# to get P_, which contains P'1 and P'2
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d_ = np.linalg.norm(P_[1] - P_[0]) #adjusted distance
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s = d/d_ #scale factor
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X_ = s * np.matmul(SHEAR, X) + S # shear, then scale every point by s.
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# Then add S, the silo position, to each position to yield the
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# correct positions
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# X_ (X') is now a numpy array containing the shifted position vectors
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# that have been transformed through the 2.5d rotation.
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# The trajectory will still look 'realistic', not curving to the side but still
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# 'visually correct' and will be functionally correct in that it
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# starts at silo S and ends at target T.
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</code></pre>
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</div>
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</div>
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<br>
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<a class="link" style="left:1%; bottom: 1%;" href="https://npc-strider.github.io/maths">🔗 Back to MATHS home page</a><br>
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<a class="link" style="left:1%; bottom: 1%;" href="https://npc-strider.github.io">🔗 Back to home page</a><br>
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