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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 14:53] Brian Moore [Generalized Derivation] |
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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. | ||
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| + | 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. |
| $$ | $$ | ||
