User Tools
Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
|
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 00:37] Brian Moore [Generalized Solution] |
||
|---|---|---|---|
| Line 220: | Line 220: | ||
| $$ | $$ | ||
| - | 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} | ||
| Line 246: | Line 246: | ||
| 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. |
| $$ | $$ | ||
