I have read this page on closure phase and could not understand the following: \begin{align} \psi_1 & = \phi_1 + e_B - e_C \\ \psi_2 & = \phi_2 - e_B \\ \psi_3 & = \phi_3 - e_C. \end{align} If the error in $\phi_3$ is $e_C$ and the error in $\phi_2$ is $e_B$, then why should the error in $\phi_1$ be connected with the errors of $\phi_2$ and $\phi_3$ in the way given?


It's because the interference phase is the difference between two antennas, it is the difference of a quantity in two different places. To starting out, there are three different shifts $e_A$, $e_B$ and $e_C$, which are the unknown shifts in time at the three locations. This gives three phase errors.

$$ \psi_1 = \phi_1 + e_B - e_C $$ $$ \psi_2 = \phi_2 + e_A - e_B $$ $$ \psi_3 = \phi_3 + e_A - e_C $$

But you just define $e_A=0$, by using the time of the reception at A as your time coordinate. Now you have only two errors and three measurements.

  • $\begingroup$ Especially useful if you can't measure phase directly, as in optical interferometry $\endgroup$ – Martin Beckett Oct 8 '12 at 4:15

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