## 2 Materials and Methods  
The lunisolar tides deform the Earth up to 60 cm twice a day, moreover the weight of the ocean tides gives a periodic load on the Earth's surface strongly dependent on bathymetry [23]. Although the displacements are relatively large, the associated changes in strain at the Earth's surface are tiny, extremely difficult to measure accurately, and even more tricky to model. The main difference between liquid tides and solid tides is in the phase: rocks react quickly to solicitations, while fluid masses need a characteristic time span to move, so they are affected by the tide with a certain phase shift. We model tidal stress according to the following method. Considering two massive celestial bodies with a spherical distributed mass density one realizes that the gravitational and centrifugal forces are balanced whenever their volumes are deformed by tides. \[\vec{F}_{\mathcal{M}}(P)=\vec{F}_{G}(P)+\vec{F}_{C}(P) \tag{1}\]  
from which the tidal acceleration is immediately obtained  
\[|\vec{a}_{\mathcal{M}}(R)|=\frac{GM}{(R-r)^{2}}-\frac{GM}{R^{2}}\simeq\frac{2 MGr }{R^{3}} \tag{2}\]  
in the last step \(R\gg r\) is assumed. The gravitational potential due to celestial body with mass \(M\), at a point \(P\) at a distance \(r\) from the center of the planet, can be expanded in a series of powers of \(r/R\), where \(R\) is the distance from the center of the Earth to the celestial body. \[W(P)=V(P)+\vec{F}_{G}(P)\cdot\vec{r}(P)+const \tag{3}\]  
\[W(\vec{R})=-\frac{GM}{|\vec{R}+\vec{r}|}+\frac{GM}{R^{3}}\vec{R}\cdot\vec{r}- \frac{GM}{R} \tag{4}\]  
if \(l=|\vec{R}+\vec{r}|\)  
\[W(\vec{R})=-\frac{GM}{R}\left(\frac{R}{l}-\frac{\vec{R}\cdot\vec{r}}{R^{2}}-1\right) \tag{5}\]by expanding in series of Legendre polynomials we obtain  
\[W(R,\Psi)=\frac{GM}{R}\sum_{n=2}^{\infty}\Big{(}\frac{r}{R}\Big{)}^{n}P_{n}(\cos\Psi) \tag{6}\]  
Since \(r/R\) for the Moon is \(\sim\)1/60 while for the Sun is \(\sim\)1/23,000, the contributions of successive terms to the potential rapidly decrease. For the Moon, \(\sim\)98% of the total tidal potential, and for the Sun, the higher orders are completely negligible for our purpose. Moreover, from the ratios of masses and their mean distances, it follows that the solar tidal perturbation is \(\sim\)0.459 times the lunar tides. So, we can write that the total tidal potential is given by the sum of lunar and solar perturbations in the following form  
\[W(R,\Psi)\simeq\frac{GM}{R}\bigg{(}\frac{r}{R}\bigg{)}^{2}P_{2}(\cos\Psi) \tag{7}\]  
where \(\Psi\) is the zenith of the body with respect to P and \(P_{2}\) is the second degree Legendre polynomial. \[cos\Psi=\cos\theta\cos\delta+\sin\theta\sin\delta\cos(\phi-\alpha) \tag{8}\]  
\(\theta\) is the colatitude and \(\phi\) is the easterly longitude of P, \(\delta\) is the codeclination, and \(\alpha\) is the right-ascension of the body. We can write the potential so that three different contributions are highlighted  
\[W(R,\Psi_{1},\Psi_{2},\Psi_{3})=\frac{3 GMr^{2}}{4R^{3}}[\Psi_{1}+\Psi_{2}+\Psi _{3}] \tag{9}\]  
with  
\[\Psi_{1}=3\bigg{(}\sin^{2}\Bigl{(}\frac{\pi}{2}-\theta\Bigr{)}-\frac{1}{3} \bigg{)}\bigg{(}\sin^{2}\Bigl{(}\frac{\pi}{2}-\delta\Bigr{)}-\frac{1}{3}\bigg{)} \tag{10}\]  
\[|\vec{a}_{\mathcal{M}}(R)|=\frac{GM}{(R-r)^{2}}-\frac{GM}{R^{2}}\simeq\frac{2 MGr }{R^{3}} \tag{2}\]0  
\[|\vec{a}_{\mathcal{M}}(R)|=\frac{GM}{(R-r)^{2}}-\frac{GM}{R^{2}}\simeq\frac{2 MGr }{R^{3}} \tag{2}\]1  
For the sake of simplicity  
\[|\vec{a}_{\mathcal{M}}(R)|=\frac{GM}{(R-r)^{2}}-\frac{GM}{R^{2}}\simeq\frac{2 MGr }{R^{3}} \tag{2}\]2  
is called Doodson's parameter. The vertical displacement is obtained by dividing \(W\) for the local value of the gravitational acceleration  
\[|\vec{a}_{\mathcal{M}}(R)|=\frac{GM}{(R-r)^{2}}-\frac{GM}{R^{2}}\simeq\frac{2 MGr }{R^{3}} \tag{2}\]3  
The three terms represent the zonal, tesseral, and sectoral tides respectively. The zonal and sectoral contributes are responsible for tides with a half the Moon's revolution period, while the tesseral one for planet's rotation period tides.