= Solution
In the ideal <semidiurnal tide> model, successive high tides occur after the <Earth>–<Moon> relative angle advances by $\pi$. Since the motions have the same sense, subtract their <angular speeds>:
$$
\boxed{\Delta t=\frac{\pi}{\Omega_E-\Omega_M}
=\frac1{2(T_E^{-1}-T_M^{-1})}.}
$$
The lunar <orbital period> is sidereal, so the consistent terrestrial <rotation> period is $T_E=23.9345\,\mathrm h$. With $T_M=27.3\times24\,\mathrm h$, this gives \b[about 12 h 25 min]. The sidereal terrestrial period is distinct from the 24-hour solar day.<Earth Fact Sheet by National Aeronautics and Space Administration>[Orbital parameters, sidereal rotation period 23.9345 h and length of day 24 h]{ref}
If the <Earth>'s <rotation> is approximated as one revolution in 24 hours, the same calculation instead gives \b[12 h 27 min]. That small difference is the day convention, not a change of sign in the relative <angular speed>. Actual coastal <tide> times also depend on ocean dynamics and the solar <tide>.
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