2023-08-05 01:05:35 +08:00
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<title>📝 2.5d rotations wtih shear transformation</title>
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2023-08-05 01:51:30 +08:00
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2023-08-05 01:05:35 +08:00
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<body>
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<b
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>PLEASE DO NOT USE ANY CONTENT FROM HERE! ALL UNMAINTAINED AND THERE'S
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PROBABLY NOTHING USEFUL HERE! CONTENT IS NOW UN-INDEXED TO PREVENT
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CONFUSION, SO YOU CAN ONLY ACCESS THIS PART OF THE SITE THROUGH LINKS.
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CONTENT IS ONLY KEPT HERE FOR HISTORY OF THE SITE -2023</b
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>
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<center>
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<h1>2.5d rotations with matrix transformations</h1>
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<h3>
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<s>(WACE) Mathematics Specialist ATAR</s><b style="color: red">**</b>
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</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>
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<h4><b>⚠WARNING⚠</b></h4>
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</center>
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<h5>
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At this point this article is probably outside of the curriculum. The
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transformations are in the curriculum, but most of this article is on
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a real math problem I've experienced which of course requires a bit of
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programming.
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</h5>
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</div>
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</div>
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<a
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class="link"
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style="left: 1%; top: 1%"
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href="https://peter-tanner.github.io/legacy_site/maths"
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>🔗 Back to MATHS home page</a
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><br />
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<a
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class="link"
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style="left: 1%; top: 1%"
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href="https://peter-tanner.github.io/legacy_site"
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>🔗 Back to home page</a
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><br />
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Warning: This page requires javascript to render the math. This website runs
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better on a chromium browser. Untested on Firefox.
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<b>TikZ graphics may not render on other platforms!</b>
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<hr />
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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>Introduction - What is '2.5d'?</b></center>
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'2.5d' is not an actual dimension, but refers to techniques in computer
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graphics which aim to create the presence of 3d depth, but in a 2d level
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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
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of shading, and creates the illusion of a 3d environment as a result of
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basic lighting. The camera is placed at a 45 degree angle, as we see
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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
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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
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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
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flight with the actual equations, because there is no concept of
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'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
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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
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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
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are only given the distance \(d\) from target \(\mathbf{T}\) to silo
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\(\mathbf{S}\))<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
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different dimensions so we need to somehow represent the shift as a
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matrix multiplication. For this we need a <b>homogeneous coordinate</b
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><br />
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We represent a 2d vector as a 3d vector, but the \(\vec{k}\) component
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is a constant \(1\). This constant allows the addition of the shift
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through matrix multiplication. <br />
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\[\mathbf{X}=\begin{bmatrix} x_0 & x_1 & x_2 & x_3 & x_4 & x_5 & \dots
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\\ y_0 & y_1 & y_2 & y_3 & y_4 & y_5 & \dots \\ 1 & 1 & 1 & 1 & 1 & 1
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& \dots \end{bmatrix}\] In this case, we applied the following
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transformations to <b>rotate</b> then <b>translate</b> \(\mathbf{T}\)
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into 'inaccurate' trajectory, \(\mathbf{T'}\) \begin{align}
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\mathbf{X'} &= \begin{bmatrix} 1 & 0 & \Delta{x} \\ 0 & 1 & \Delta{y}
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\\ 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} \cos(\theta) &
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-\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \\
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\end{bmatrix}\mathbf{X} \\\\ &= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2
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\\ 0 & 0 & 1 \\ \end{bmatrix}\begin{bmatrix} \cos(\frac{\pi}{6}) &
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-\sin(\frac{\pi}{6}) & 0 \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6})
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& 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\mathbf{X} \end{align} However, in
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reality I would just use a for loop and iterate through each pair of
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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} \text{Parallel to the y-axis, Vertical shear}\quad
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\mathbf{X'} &= \begin{bmatrix} 1 & 0\\ m & 1 \end{bmatrix}\mathbf{X}\\
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&= \begin{bmatrix} x\\ mx+y \end{bmatrix} \\\\ \text{Parallel to the
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x-axis, Horizontal shear}\quad \mathbf{X'} &= \begin{bmatrix} 1 & m\\ 0
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& 1 \end{bmatrix}\mathbf{X} \\ &= \begin{bmatrix} x+my\\ y
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\end{bmatrix}\\\\ \text{Where}\quad\mathbf{X} &= \begin{bmatrix} x\\ y
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\end{bmatrix}\\ \end{align}
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</div>
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</div>
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<br />
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<hr />
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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 solution: shear matrices</b></center>
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<br />
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A shear matrix is better explained visually.
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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[->] (-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>
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</center>
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In this case we applied a vertical shear (parallel to the y-axis). Our
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x-values are the same, but our y-values have been transformed in a way
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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 -
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we're only distorting the matrix on one axis and preserving the other.
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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
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between the silo \(\mathbf{S}\) and target \(\mathbf{T}\).
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\[m=\frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x}\] \(m\)
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can also be obtained with a trig ratio, yielding the following shear
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transformation that involve angles rather than gradients. \begin{align}
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\text{Parallel to the y-axis, Vertical shear}\quad \mathbf{X'} &=
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\begin{bmatrix} 1 & 0\\ \sin(\theta) & 1 \end{bmatrix}\mathbf{X}\\\\
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\text{Parallel to the x-axis, Horizontal shear}\quad \mathbf{X'} &=
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\begin{bmatrix} 1 & \sin(\theta)\\ 0 & 1 \end{bmatrix}\mathbf{X}\\\\
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\end{align} To resolve our issue with the transformed object being
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'lengthened', we need to scale it back.<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[->] (-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>
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</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,
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\(d'\) \begin{align} 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} 1 &
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0\\ \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}\\ &= \begin{bmatrix} 1 & 0\\
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\sin(\theta) & 1 \end{bmatrix}\mathbf{P_1},\mathbf{P_2}\\ \end{align}
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Note that \(d\) is also the distance from the silo \(\mathbf{S}\) and
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target \(\mathbf{T}\)<br />
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\(\mathbf{P_1}\) and \(\mathbf{P_2}\) are not the same as \(\mathbf{S}\)
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and \(\mathbf{T}\), because we've created these points with
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\(\begin{bmatrix}0\\0\end{bmatrix}\) as the origin,
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<b>not</b> \(\mathbf{S}\). We will translate all of the points later so
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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
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\(\mathbf{S}\) and \(\mathbf{T}\) to calculate \(d\). \[d =
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\|\mathbf{P_1}-\mathbf{P_2}\| = \|\mathbf{T}-\mathbf{S}\|\] Our scale
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factor is simply the ratio between the two. \begin{align} s &=
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\frac{d}{d'} \\ &=
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\frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}
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\end{align} Putting it all together, our 2.5d rotation is now:
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\begin{align} \mathbf{X'} &= s\begin{bmatrix} 1 & 0\\ 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\\ \frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x} & 1
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\end{bmatrix}\mathbf{X}\\ \text{ or }\quad &=
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\frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}\cdot\begin{bmatrix}
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1 & 0\\ \sin(\theta) & 1 \end{bmatrix}\mathbf{X}\\\\ \end{align} So we
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need a 'pre transformation' on two points, \(\mathbf{P_1}\) and
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\(\mathbf{P_2}\) in order to obtain our scale factor
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<center>
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<script type="text/tikz">
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|
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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>
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</center>
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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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|
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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>
|
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|
</center>
|
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The limitation of this technique is that we need to perform the rotation
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about \(\begin{bmatrix}0\\0\end{bmatrix}\), the origin. After this we
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can translate the rotated matrix to match the silo and target
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location.<br />
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You can use homogeneous coordinates to achieve this (see the yellow box
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|
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
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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,
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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
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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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|
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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
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|
class="link"
|
|
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|
style="left: 1%; bottom: 1%"
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href="https://peter-tanner.github.io/legacy_site/maths"
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>🔗 Back to MATHS home page</a
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|
><br />
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<a
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class="link"
|
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|
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|
style="left: 1%; bottom: 1%"
|
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|
href="https://peter-tanner.github.io/legacy_site"
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|
|
>🔗 Back to home page</a
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><br />
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</body>
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</html>
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