+
The solution: shear matrices
+ A shear matrix is better explained visually.
+
+ 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).
+ 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.
+
+ In my case, I do not need to know the angle - I can use the gradient between the silo \(\mathbf{S}\) and target \(\mathbf{T}\).
+ \[m=\frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x}\]
+ \(m\) can also be obtained with a trig ratio, yielding the following shear transformation that involve angles rather than gradients.
+ \begin{align}
+ \text{Parallel to the y-axis, Vertical shear}\quad
+ \mathbf{X'} &= \begin{bmatrix}
+ 1 & 0\\
+ \sin(\theta) & 1
+ \end{bmatrix}\mathbf{X}\\\\
+ \text{Parallel to the x-axis, Horizontal shear}\quad
+ \mathbf{X'} &= \begin{bmatrix}
+ 1 & \sin(\theta)\\
+ 0 & 1
+ \end{bmatrix}\mathbf{X}\\\\
+ \end{align}
+ To resolve our issue with the transformed object being 'lengthened', we need to scale it back.
+
+
+ Let's choose two points, \(\mathbf{P_1}\) and \(\mathbf{P_2}\).
+ We obtain the original distance, \(d\) and the transformed distance, \(d'\)
+ \begin{align}
+ d &= \|\mathbf{P_1}-\mathbf{P_2}\| \\
+ &= \sqrt{(\mathbf{P_1}_x-\mathbf{P_2}_x)^2+(\mathbf{P_1}_y-\mathbf{P_2}_y)^2} \\
+ d' &= \|\mathbf{P'_1}-\mathbf{P'_2}\| \\
+ &= \sqrt{(\mathbf{P'_1}_x-\mathbf{P'_2}_x)^2+(\mathbf{P'_1}_y-\mathbf{P'_2}_y)^2}\\\\
+ \text{ where }\quad\mathbf{P'_1},\mathbf{P'_2} &= \begin{bmatrix}
+ 1 & 0\\
+ \frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x} & 1
+ \end{bmatrix}\mathbf{P_1},\mathbf{P_2}\\
+ &= \begin{bmatrix}
+ 1 & 0\\
+ \sin(\theta) & 1
+ \end{bmatrix}\mathbf{P_1},\mathbf{P_2}\\
+
+ \end{align}
+ Note that \(d\) is also the distance from the silo \(\mathbf{S}\) and target \(\mathbf{T}\)
+ \(\mathbf{P_1}\) 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,
not \(\mathbf{S}\). We will translate all of the points later so that our starting point is \(\mathbf{S}\) after rotating.
+ However, the distances are the same, so we might as well use \(\mathbf{S}\) and \(\mathbf{T}\) to calculate \(d\).
+ \[d = \|\mathbf{P_1}-\mathbf{P_2}\| = \|\mathbf{T}-\mathbf{S}\|\]
+ Our scale factor is simply the ratio between the two.
+ \begin{align}
+ s &= \frac{d}{d'} \\
+ &= \frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}
+ \end{align}
+ Putting it all together, our 2.5d rotation is now:
+ \begin{align}
+ \mathbf{X'} &= s\begin{bmatrix}
+ 1 & 0\\
+ m & 0
+ \end{bmatrix}\mathbf{X}\\
+ &= \frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}\cdot\begin{bmatrix}
+ 1 & 0\\
+ \frac{\mathbf{S}_y-\mathbf{T}_y}{\mathbf{S}_x-\mathbf{T}_x} & 1
+ \end{bmatrix}\mathbf{X}\\
+ \text{ or }\quad &= \frac{\|\mathbf{T}-\mathbf{S}\|}{\|\mathbf{P'_1}-\mathbf{P'_2}\|}\cdot\begin{bmatrix}
+ 1 & 0\\
+ \sin(\theta) & 1
+ \end{bmatrix}\mathbf{X}\\\\
+ \end{align}
+ So we need a 'pre transformation' on two points, \(\mathbf{P_1}\) and \(\mathbf{P_2}\) in order to obtain our scale factor
+
+
+ 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.
+ You can use homogeneous coordinates to achieve this (see the yellow box earlier), or programatically do it.
+
+ We also have to solve one last issue with this method: it only works within the domain \(\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)\).
+ This is because the gradient can only describe so much information.
+ For instance, the line \(f(x)=1\cdot x\).
+ 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.
+ We need to adjust the signs on the shear matrix according to the quadrant our target is in.
+
import math
+import numpy as np
+
+def quad(A,B): #A is our origin/fix point, B is our other point.
+ if B[0] > A[0] and B[1] > A[1]:
+ return 1 #First quadrant
+ elif B[0] < A[0] and B[1] > A[1]:
+ return 2 #Second quadrant
+ elif B[0] < A[0] and B[1] < A[1]:
+ return 3 #Third quadrant
+ else:
+ return 4 #Fourth quadrant
+ #By elimination. We could do another if but it's unnecessary
+ # elif B[0] < A[0] and B[1] > A[1]:
+
+def trajectory(distance, n):
+ # Here we create a list of points. n is the amount we create
+ # (higher n results in greater precision)
+ # Not going to include it here - uses bezier interpolation and stuff to create
+ # the illusion of a ballistic trajectory with gravity (not a simple quadratic function)
+ #
+ return points
+
+S = np.array([10,4]) #Silo position vector
+T = np.array([5204,954]) #Target position vector
+
+d = np.linalg.norm(T - S) #Get the distance: d = ((Tx-Sx)^2+(Ty-Sy)^2)^(0.5)
+m = (T[1]-S[1])/(T[0]-S[0]) #Gradient of line between target and silo.
+
+
+X = trajectory(d, 50) #50 is a constant for n. only affects trajectory precision
+ #X now contains a 'flat trajectory', a list of points which have the correct
+ #distance but the wrong angle and offset.
+ #It does not intersect with the target T.
+
+def shear(m, X):
+ q = quad(S,T) #Silo and target is passed to the quadrant finding function.
+ #q is the quadrant
+ if q == 1 or q == 4:
+ return np.array([[1,0],
+ [m,1]]) #shear matrix.
+ else:
+ return np.array([[-1,0],
+ [m,-1]]) #shear matrix, but for quadrants 2 and 3 (negative x).
+ # AN easier way to do this would be to just compare S[0] > T[0]. Forget about the quadrant stuff.
+ # because we only need to know if it's in the negative or positive x.
+
+
+
+SHEAR = shear(m, X) #we get the shear matrix, which the function returns
+
+P_ = np.matmul(SHEAR, X[(0,-1),]) #first, take a slice of the array X to obtain the first (P1)
+ # and last (P2) points. Then multiply this new 2x2 matrix
+ # containing P1 and P2 by the shear matrix
+ # to get P_, which contains P'1 and P'2
+
+d_ = np.linalg.norm(P_[1] - P_[0]) #adjusted distance
+s = d/d_ #scale factor
+
+X_ = s * np.matmul(SHEAR, X) + S # shear, then scale every point by s.
+ # Then add S, the silo position, to each position to yield the
+ # correct positions
+# X_ (X') is now a numpy array containing the shifted position vectors
+# that have been transformed through the 2.5d rotation.
+# The trajectory will still look 'realistic', not curving to the side but still
+# 'visually correct' and will be functionally correct in that it
+# starts at silo S and ends at target T.
+
+