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cs:hydrophones:pinger_bearing:start [2017/10/07 00:08]
Brian Moore General Case; Derivation and Solution
cs:hydrophones:pinger_bearing:start [2017/10/07 20:14] (current)
Brian Moore [Generalized Derivation]
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 \hat{P} = \frac{\vec{P}}{|\vec{P}|} = \frac{(P_x,​P_y,​P_z)}{|\vec{P}|} = (\hat{i},​\hat{j},​\hat{k}) \\ \hat{P} = \frac{\vec{P}}{|\vec{P}|} = \frac{(P_x,​P_y,​P_z)}{|\vec{P}|} = (\hat{i},​\hat{j},​\hat{k}) \\
 |\hat{P}| = 1\\ |\hat{P}| = 1\\
-proj_\vec{P}\vec{H_n} = \vec{H_n} \cdot \hat{P} = \frac{|\vec{H_n}||\vec{P}|}{|\vec{P}|}\cos(\theta) = |\vec{H_n}|\cos(\theta) = d_n \\+proj_\vec{P}\vec{H_n} = \vec{H_n} \cdot \hat{P} = |\vec{H_n}||\hat{P}|\cos(\theta) = |\vec{H_n}|\cos(\theta) = d_n \\
 $$ $$
  
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 $$ $$
  
-Recall ​that Matrix multiplication ​consists of systematically ​taking dot products.+Therefore 
 + 
 +$$  
 +\cos{\theta} = d_x/x 
 +$$ 
 + 
 +This trend continues for the other two vectors: 
 + 
 +$$ 
 +(0,y,0) \cdot \hat{P} = |y|\cos{\phi} = y\cos{\phi} = d_y, \quad \cos{\phi} = d_y/y \\ 
 +(0,0,z) \cdot \hat{P} = |y|\cos{\psi} = y\cos{\psi} = d_z, \quad \cos{\psi} = d_z/z \\ 
 +$$ 
 + 
 +So what we want to find is a bearing vector $\hat{P}=(\hat{i},​\hat{j},​\hat{k})$ ​that satisfies having all three of these specific angles $\psi$, $\phi$, and $\theta$ between itself and the three hydrophone vectors simultaneously. ​ To solve this problem, we need to consider the results of dot products simultaneously. 
 + 
 +Matrix multiplication ​between two vectors is identical to taking ​their dot product
  
 $$ $$
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 \end{bmatrix} \end{bmatrix}
  
-+= H_{nx}\hat{i} + H_{ny}\hat{j} + H_{nz}\hat{k} 
 + 
 +=
 \begin{bmatrix} \begin{bmatrix}
 d_n d_n
 \end{bmatrix} \end{bmatrix}
 $$ $$
 +
 +Matrix Multiplication between two-dimensional matrices is merely a systematic way of calculating dot products between all relevant vectors. ​ The product matrix elements each correspond to a dot product, and their position denotes which vectors were involved. ​ Row vector 3 doted with column vector 1 becomes element $(3,​1)$. ​ Matrices are also involved in solving simultaneous equations.
  
 ==== Generalized Solution ==== ==== Generalized Solution ====
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 $$ $$
-H^{-1}H\hat{P}^T = H^{-1}D \\+inv(H)H\hat{P}^T = inv(H)D \\
  
-\hat{P} = H^{-1}D \\+\hat{P}^T inv(H)D \\
  
 \begin{bmatrix} \begin{bmatrix}
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 We can see based on this structure that we need at least three hydrophones measurements ($d_1$, $d_2$, and $d_3$) to solve for the three unknown elements of unit vector $\hat{P}$. ​ Furthermore,​ the three position vectors of the hydrophones ($\vec{H_1}$,​ $\vec{H_2}$,​ and $\vec{H_3}$) must constitute an invertible, full-rank matrix. ​ In this case that simply means the three hydrophones cannot all lie on the same plane.  ​ We can see based on this structure that we need at least three hydrophones measurements ($d_1$, $d_2$, and $d_3$) to solve for the three unknown elements of unit vector $\hat{P}$. ​ Furthermore,​ the three position vectors of the hydrophones ($\vec{H_1}$,​ $\vec{H_2}$,​ and $\vec{H_3}$) must constitute an invertible, full-rank matrix. ​ In this case that simply means the three hydrophones cannot all lie on the same plane.  ​
  
-As a sanity-check,​ this formulation should re-produce the degenerate case where all hydrophones lie perfectly along the $x$, $y$, and $z$ axes.+As a sanity-check,​ this formulation should re-produce ​the solution to the degenerate case demonstrated above where all hydrophones lie perfectly along the $x$, $y$, and $z$ axes.
  
 $$ $$