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Next: Exercises Up: Terrestrial Ocean Tides Previous: Proudman Equations

Hemispherical Ocean Tides

Consider a hemispherical ocean for which $ \phi_-=0$ and $ \phi_+=\pi$ . In general, there are many linearly independent solutions to the eigenvalue problem (12.235) and (12.236), subject to the orthonormality constraint (12.244), that correspond to a given value of $ \lambda_r$ . It is convenient to differentiate these solutions by means of two indices, $ n$ and $ m$ . Thus, $ {\mit\Phi}_r\rightarrow {\mit\Phi}_n^{\,m}$ and $ \lambda_r\rightarrow\lambda_n^{\,m}$ , where the index $ n$ ranges from $ 1$ to $ \infty $ . Moreover, for a given value of $ n$ , the index $ m$ ranges from 0 to $ n$ . Now,

$\displaystyle \int_0^\pi \cos(m\,\phi)\,\cos(m'\,\phi)\,d\phi =\left\{ \begin{a...
...}&m=m'=0\\ [0.5ex] \pi/2&&m=m'\neq 0\\ [0.5ex] 0&&m\neq m' \end{array} \right..$ (12.307)

Hence, it follows that

$\displaystyle {\mit\Phi}_n^{\,m}(\theta,\phi) = a_n^{\,m}\,\skew{5}\hat{P}_n^{\,m}(\cos\theta)\,\cos(m\,\phi),$ (12.308)

and

$\displaystyle \lambda_n^{\,m} = n\,(n+1),$ (12.309)

where

$\displaystyle a_n^{\,m} =\left\{\begin{array}{lll} (4/\lambda_n^{\,m})^{1/2}&\m...
...space{0.5cm}}& m>0\\ [0.5ex] (2/\lambda_n^{\,m})^{1/2}&&m=0 \end{array}\right..$ (12.310)

Likewise, the solutions to the eigenvalue problem (12.248) and (12.249), subject to the orthonormality constraint (12.250), is such that $ {\mit\Psi}_r\rightarrow {\mit\Psi}_n^{\,m}$ and $ \mu_r\rightarrow\mu_n^{\,m}$ , where

$\displaystyle {\mit\Psi}_n^{\,m}(\theta,\phi) = a_n^{\,m}\,\skew{5}\hat{P}_n^{\,m}(\cos\theta)\,\sin(m\,\phi),$ (12.311)

and

$\displaystyle \mu_n^{\,m} = n\,(n+1).$ (12.312)

We also have (see Section 12.19)

$\displaystyle \beta_{r,\,s}$ $\displaystyle \rightarrow \beta_{n,\,n'}^{\,m,\,m'}= -\int_{\mit\Omega}\frac{\c...
...}\, \frac{\partial{\mit\Phi}_{n'}^{\,m'}}{\partial\theta}\right) d{\mit\Omega},$ (12.313)
$\displaystyle \beta_{-r,\,s}$ $\displaystyle \rightarrow \beta_{-n,\,n'}^{\,-m,\,m'}= -\int_{\mit\Omega}\cos\t...
...hi}\, \frac{\partial{\mit\Phi}_{n'}^{\,m'}}{\partial\phi}\right) d{\mit\Omega},$ (12.314)
$\displaystyle \beta_{r,\,-s}$ $\displaystyle \rightarrow \beta_{n,\,-n'}^{\,m,\,-m'}= \int_{\mit\Omega}\cos\th...
...hi}\, \frac{\partial{\mit\Psi}_{n'}^{\,m'}}{\partial\phi}\right) d{\mit\Omega},$ (12.315)
$\displaystyle \beta_{-r,\,-s}$ $\displaystyle \rightarrow \beta_{-n,\,-n'}^{\,-m,\,-m'}= -\int_{\mit\Omega}\fra...
...}\, \frac{\partial{\mit\Psi}_{n'}^{\,m'}}{\partial\theta}\right) d{\mit\Omega},$ (12.316)

which yields

$\displaystyle \beta_{n,\,n'}^{\,m,\,m'}$ $\displaystyle = \int_0^\pi\int_{-1}^{1}\left( \frac{\partial{\mit\Phi}_n^{\,m}}...
...}\, \frac{\partial{\mit\Phi}_{n'}^{\,m'}}{\partial\mu}\right) \mu\,d\mu\,d\phi,$ (12.317)
$\displaystyle \beta_{-n,\,n'}^{\,-m,\,m'}$ $\displaystyle = -\int_0^\pi\int_{-1}^{1}\left[(1-\mu^{\,2})\, \frac{\partial{\m...
...}\, \frac{\partial{\mit\Phi}_{n'}^{\,m'}}{\partial\phi}\right]\mu\,d\mu\,d\phi,$ (12.318)
$\displaystyle \beta_{n,\,-n'}^{\,m,\,-m'}$ $\displaystyle = \int_0^\pi\int_{-1}^{1}\left[(1-\mu^{\,2})\, \frac{\partial{\mi...
...\, \frac{\partial{\mit\Psi}_{n'}^{\,m'}}{\partial\phi}\right] \mu\,d\mu\,d\phi,$ (12.319)
$\displaystyle \beta_{-n,\,-n'}^{\,-m,\,-m'}$ $\displaystyle = \int_0^\pi\int_{-1}^{1}\left( \frac{\partial{\mit\Psi}_n^{\,m}}...
...}\, \frac{\partial{\mit\Psi}_{n'}^{\,m'}}{\partial\mu}\right) \mu\,d\mu\,d\phi,$ (12.320)

where $ \mu=\cos\theta$ . However,

$\displaystyle \int_0^\pi \sin(m\,\phi)\,\cos(m'\,\phi)\,d\phi = \left\{ \begin{...
...5cm}} &\mbox{$m+m'$\ odd}\\ [0.5ex] 0&&\mbox{$m+m'$\ even} \end{array}\right. .$ (12.321)

Hence, if $ m+m'$ is even then the $ \beta_{\pm n,\,\pm n'}^{\,\pm m,\,\pm m'}$ are zero. Otherwise, we obtain

$\displaystyle \frac{\beta_{n,\,n'}^{\,m,\,m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}$ $\displaystyle = \frac{2}{m^{\,2}-m'^{\,2}}\int_{-1}^1 \left(m^{\,2}\,P_n^{\,m} ...
...,m'}}{d\mu} + m'^{\,2}\,\frac{dP_n^{\,m}}{d\mu}\,P_{n'}^{\,m'}\right)\mu\,d\mu,$ (12.322)
$\displaystyle \frac{\beta_{-n,\,n'}^{\,-m,\,m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}$ $\displaystyle =- \frac{2\,m}{m^{\,2}-m'^{\,2}}\int_{-1}^1 \left[(1-\mu^{\,2})\,...
...d\mu} + \frac{m'^{\,2}}{1-\mu^{\,2}}\,P_n^{\,m}\,P_{n'}^{\,m'}\right]\mu\,d\mu,$ (12.323)
$\displaystyle \frac{\beta_{-n,\,-n'}^{\,-m,\,-m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}$ $\displaystyle =\frac{2\,m\,m'}{m^{\,2}-m'^{\,2}}\int_{-1}^1 \left(P_n^{\,m} \,\...
...dP_{n'}^{\,m'}}{d\mu} + \frac{dP_n^{\,m}}{d\mu}\,P_{n'}^{\,m'}\right)\mu\,d\mu,$ (12.324)

as well as

$\displaystyle \beta_{n,\,-n'}^{\,m,\,-m'}=-\beta_{-n',\,n}^{\,-m',\,m},$ (12.325)

where

$\displaystyle c_n^{\,m}= a_n^{\,m}\,b_n^{\,m}.$ (12.326)

Let

$\displaystyle I_{n,\,n'}^{\,m,\,m'} = \int_{-1}^1 P_n^{\,m}(\mu)\,P_{n'}^{\,m'}(\mu)\,d\mu.$ (12.327)

It follows, from symmetry, that $ I_{n,\,n'}^{\,m,\,m'}=0$ when $ n+n'-m-m'$ is odd. When $ n+n'-m-m'$ is even (Wong 1998),

\begin{multline}
I_{n,\,n'}^{\,m,\,m'} = \sum_{p=0,p_{\rm max}}\sum_{p'=0,p'_{\r...
...)/2]\,{\mit\Gamma}[(m+m'+2p+2p'+2)/2]}{{\mit\Gamma}[(n+n'+3)/2]},
\end{multline}

where $ {\mit\Gamma}(x)$ denotes a gamma function (Abramowitz and Stegun 1965),

$\displaystyle \alpha_{n,p}^{\,m} = \frac{(-1)^{m+p}\,(n+m)!}{2^{\,m+2p}\,(m+p)!\,p!\,(n-m-2p)!},$ (12.328)

and

$\displaystyle p_{\rm max} = \left\{ \begin{array}{lll} (n-m)/2&\mbox{\hspace{0....
...\ [0.5ex] (n-m-1)/2&\mbox{\hspace{0.5cm}}&\mbox{$n-m$\ odd} \end{array}\right.,$ (12.329)

with an analogous definition for $ p'_{\rm max}$ .

It can be shown that (Longuet-Higgins and Pond 1970)

$\displaystyle \frac{\beta_{n,\,n'}^{\,m,\,m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}$ $\displaystyle = \left[\frac{n'\,(n'+1)+m'\,(m'+1)}{m'+1}-2\,m'\,\frac{n\,(n+1)-n'\,(n'+1)+m'}{m^{\,2}+m'^{\,2}}\right]I_{n,\,n'}^{\,m,\,m'}$    
  $\displaystyle \phantom{=} + \left(\frac{1}{m'+1}\right)I_{n,\,n'}^{\,m,\,m'+2},$ (12.330)
$\displaystyle \frac{\beta_{n,\,-n'}^{\,m,\,-m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}$ $\displaystyle =\frac{2\,m}{(m'^{\,2}-m^{\,2})\,(2\,n'+1)} \left[(n'-1)\,(n'+1)\,(n'+m')\,I_{n,\,n'-1}^{\,m,\,m'} \right.$    
  $\displaystyle \phantom{\frac{\beta_{n,\,-n'}^{\,m,\,-m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}=} \left.+n'\,(n'+2)\,(n'-m'+1)\,I_{n,\,n'+1}^{\,m,\,m'} \right],$ (12.331)
$\displaystyle \frac{\beta_{-n,\,-n'}^{\,-m,\,-m'}}{c_n^{\,m}\,c_{n'}^{\,m'}}$ $\displaystyle =\frac{2\,m\,m'}{m'^{\,2}-m^{\,2}}\,I_{n,\,n'}^{\,m,\,m'}.$ (12.332)

According to Equation (12.306), we can also write $ {\mit\Gamma}_{n,r}^{\,m}\rightarrow {\mit\Gamma}_{n,n'}^{\,m,m'}$ , where

$\displaystyle {\mit\Gamma}_{n,n'}^{\,m,m'}=(a_n^{\,m})^{-1}\left(\delta_{n,n'}^{\,m,m'}+{\rm i}\,\kappa_{n,n'}^{\,m,m'}\right),$ (12.333)

and

$\displaystyle \delta_{n,\,n'}^{\,m,\,m'} = \delta_{n\,n'}\,\delta_{m\,m'}.$ (12.334)

Here, $ \kappa_{n,\,n'}^{\,m,\,m'}=0$ if $ m+m'$ is even, and

$\displaystyle \kappa_{n,\,n'}^{\,m,\,m'} = \left(\frac{2\,m}{m^{\,2}-m'^{\,2}}\right)\lambda_{n'}^{\,m'}\,c_n^{\,m}\,c_{n'}^{\,m'}\,I_{n,\,n'}^{\,m,\,m'},$ (12.335)

otherwise. It follows from Equation (12.305) that $ \gamma_{r,\,s}\rightarrow \gamma_{n,\,n'}^{\,m,\,m'}$ , where

$\displaystyle \gamma_{n,\,n'}^{\,m,\,m'} = (1+k_n'-h_n')\,\alpha_n\,(a_n^{\,m})...
...{\,m,\,m'}- {\rm i}\,\epsilon_{n,\,n'}^{\,m,\,m'} + \theta_{n,\,n'}^{\,m,\,m'}.$ (12.336)

Here, $ \epsilon_{n,\,n'}^{\,m,\,m'}=0$ if $ m+m'$ is even, and

$\displaystyle \epsilon_{n,\,n'}^{\,m,\,m'} =(1+k_n'-h_n')\,\alpha_n\,(a_n^{\,m}...
... (1+k_{n'}'-h_{n'}')\,\alpha_{n'}\,(a_{n'}^{\,m'})^{-2}\,\kappa_{n',n}^{\,m',m}$ (12.337)

otherwise. Now, if $ m$ and $ m'$ are both even then

$\displaystyle \theta_{n,\,n'}^{\,m,\,m'}= \sum_{n''=1,\infty}\sum_{m''=0,n''}^{...
...\,(a_{n''}^{\,m''})^{-2}\,\kappa_{n'',n}^{\,m'',m}\,\kappa_{n'',n'}^{\,m'',m'};$ (12.338)

if $ m$ and $ m'$ are both odd then

$\displaystyle \theta_{n,\,n'}^{\,m,\,m'}= \sum_{n''=1,\infty}\sum_{m''=0,n''}^{...
...\,(a_{n''}^{\,m''})^{-2}\,\kappa_{n'',n}^{\,m'',m}\,\kappa_{n'',n'}^{\,m'',m'};$ (12.339)

and if $ m+m'$ is odd then $ \theta_{n,\,n'}^{\,m,\,m'}= 0$ .

If we let $ p_r\rightarrow {\rm i}^{\,n}\,p_n^{\,m}$ and $ p_{-r}\rightarrow {\rm i}^{\,n+1}\,p_{-n}^{\,-m}$ then Equations (12.311) and (12.312) become

\begin{multline}
\sum_{n'=1,\infty}\sum_{m'=0,n'}\left(\left[\left(\beta^{\,-1}...
...n}^{\,m_j,\,m}+ \skew{3}\hat{\kappa}_{2,\,n}^{\,m_j,\,m}\right),
\end{multline}

and

$\displaystyle \sum_{n'=1,\infty}\sum_{m'=0,n'}\left(\left[-f_j^{\,2}\,\skew{3}\...
...-m'} + f_j\,\skew{3}\hat{\beta}_{-n,\,n'}^{\,-m,\,m'}\,p_{n'}^{\,m'}\right)= 0,$ (12.340)

respectively. Here,

$\displaystyle \skew{3}\hat{\delta}_{n,\,n'}^{\,m,\,m'}$ $\displaystyle = {\rm i}^{\,n+n'}\,\delta_{n,\,n'}^{\,m,\,m'},$ (12.341)
$\displaystyle \skew{3}\hat{\beta}_{n,\,n'}^{\,m,\,m'}$ $\displaystyle = {\rm i}^{\,n+n'+1}\,\beta_{n,\,n'}^{\,m,\,m'},$ (12.342)
$\displaystyle \skew{3}\hat{\beta}_{n,\,-n'}^{\,m,\,-m'}$ $\displaystyle = {\rm i}^{\,n+n'}\,\beta_{n,\,-n'}^{\,m,\,-m'},$ (12.343)
$\displaystyle \skew{3}\hat{\beta}_{-n,\,n'}^{\,-m,\,m'}$ $\displaystyle = {\rm i}^{\,n+n'}\,\beta_{-n,\,n'}^{\,-m,\,m'},$ (12.344)
$\displaystyle \skew{3}\hat{\beta}_{-n,\,-n'}^{\,-m,\,-m'}$ $\displaystyle = {\rm i}^{\,n+n'+1}\,\beta_{-n,\,-n'}^{\,-m,\,-m'},$ (12.345)
$\displaystyle \skew{3}\hat{\epsilon}_{n,\,n'}^{\,m,\,m'}$ $\displaystyle = {\rm i}^{\,n+n'+1}\, \epsilon_{n,\,n'}^{\,m,\,m'},$ (12.346)
$\displaystyle \skew{3}\hat{\theta}_{n,\,n'}^{\,m,\,m'}$ $\displaystyle = {\rm i}^{\,n+n'}\, \theta_{n,\,n'}^{\,m,\,m'},$ (12.347)
$\displaystyle \skew{3}\hat{\kappa}_{n,\,n'}^{\,m,\,m'}$ $\displaystyle = {\rm i}^{\,n+n'+1}\, \kappa_{n,\,n'}^{\,m,\,m'}.$ (12.348)

By symmetry, $ \delta_{n,\,n'}^{\,m,\,m'}$ and $ \theta_{n,\,n'}^{\,m,\,m'}$ are only non-zero when $ m+m'$ is even, and $ n+n'$ is even; $ \beta_{n,\,n'}^{\,m,\,m'}$ , $ \beta_{-n,\,-n'}^{\,-m,\,-m'}$ , $ \epsilon_{n,\,n'}^{\,m,\,m'}$ , and $ \kappa_{n,\,n'}^{\,m,\,m'}$ are only non-zero when $ m+m'$ is odd, and $ n+n'$ is odd; and $ \beta_{n,\,-n'}^{\,m,\,-m'}$ and $ \beta_{-n,\,n'}^{\,-m,\,m'}$ are only non-zero when $ m+m'$ is odd, and $ n+n'$ is even. It follows that all quantities appearing in Equations (12.347) and (12.348) are real. Once we have solved these equations to obtain the $ p_n^{\,m}$ (which is a relatively straightforward numerical task), we can reconstruct the tidal elevation as follows:

$\displaystyle \zeta(\theta,\phi,t) = \zeta_c(\theta,\phi)\,\cos(\sigma_j\,t) + \zeta_s(\theta,\phi)\,\sin(\sigma_j\,t),$ (12.349)

where

$\displaystyle \zeta_c(\theta,\phi)$ $\displaystyle = \sum_{n=1,\infty}^{n\,{\rm even}}\sum_{m=0,n}\zeta_n^{\,m}\,{\mit\Phi}_n^m(\theta,\phi)\,\cos(n\,\pi/2),$ (12.350)
$\displaystyle \zeta_s(\theta,\phi)$ $\displaystyle = -\sum_{n=1,\infty}^{n\,{\rm odd}}\sum_{m=0,n}\zeta_n^{\,m}\,{\mit\Phi}_n^m(\theta,\phi)\,\sin(n\,\pi/2),$ (12.351)

and

$\displaystyle \zeta_n^{\,m} = -\lambda_n^{\,m}\,p_n^{\,m}.$ (12.352)

Thus, the tidal amplitude at a given point on the ocean is

$\displaystyle \vert\zeta\vert(\theta,\phi) = \left[\zeta_c^{\,2}(\theta,\phi)+\zeta_s^{\,2}(\theta,\phi)\right]^{1/2}.$ (12.353)

It is easily demonstrated that

$\displaystyle \frac{{\mit\Delta}t(\theta,\phi)}{T} = \frac{{\mit\Delta}\varphi(\theta,\phi)}{2\pi},$ (12.354)

where

$\displaystyle {\mit\Delta\varphi}(\theta,\phi) =m_j\,\frac{\pi}{2} + \tan^{-1}\left[\frac{\zeta_s(\theta,\phi)}{\zeta_c(\theta,\phi)}\right].$ (12.355)

Here, $ T=2\pi/\sigma_j$ is the oscillation period of the harmonic of the tide generating potential under consideration, $ {\mit\Delta}t$ is the time-lag between the peak tide at a given point on the ocean and the maximal tide generating potential at $ \phi=\pi/2$ , and $ {\mit \Delta }\varphi $ is the corresponding phase-lag.

Figures 12.4 and 12.5 show the amplitude and phase-lag of the ($ j=3$ ) $ M_f$ long-period tide in a hemispherical ocean of mean depth $ d=3.8\,{\rm km}$ (which corresponds to $ \beta=23.1$ ), calculated assuming that $ \rho/\skew{3}\bar{\rho}=5.5$ and $ \mu/(\rho\,g\,a)=0.35$ . The calculation includes all azimuthal harmonics up to $ n_{\rm max}=20$ . Note that only a quarter of the ocean is shown, because the amplitude is symmetric about the meridians $ \vartheta=0$ and $ \phi=\pi/2$ , whereas the phase-lag is symmetric about the meridian $ \vartheta=0$ , and antisymmetric about the meridian $ \phi=\pi/2$ . Here, $ \vartheta\equiv\pi/2-\theta$ . Given that $ \skew{5}\tilde{\zeta}_3=4.25\times 10^{-2}\,{\rm m}$ (see Table 12.3), the maximum amplitude of the tide is about $ 3.5\,{\rm cm}$ , and occurs at the poles. Moreover, it is clear from a comparison with Figure 12.1 that the tide is direct (i.e., it is in phase with the equilibrium tide). In fact, the $ M_f$ tide in a hemispherical ocean is about four times larger in amplitude than that in a global ocean (i.e., an ocean that covers the whole surface of the Earth) of the same depth. Otherwise, the two tides have fairly similar properties.

Figures 12.6 and 12.7 show the amplitude and phase-lag of the ($ j=6$ ) $ K_1$ diurnal tide in a hemispherical ocean of mean depth $ d=3.8\,{\rm km}$ (which corresponds to $ \beta=23.1$ ), calculated assuming that $ \rho/\skew{3}\bar{\rho}=5.5$ and $ \mu/(\rho\,g\,a)=0.35$ . The calculation includes all azimuthal harmonics up to $ n_{\rm max}=20$ . Given that $ \skew{5}\tilde{\zeta}_6=-10.36\times 10^{-2}\,{\rm m}$ (see Table 12.3), the maximum amplitude of the tide is about $ 15.8\,{\rm cm}$ , and occurs at mid-latitudes. This is very different to the case of a global ocean, where the amplitude of the $ K_1$ tide is identically zero everywhere. (See Figure 12.2.) Note that the tidal phase-lag only exhibits a fairly weak dependence on the azimuthal angle, $ \phi$ . In fact, the $ K_1$ tide in a hemispherical ocean essentially oscillates in anti-phase with the equilibrium tide at the ocean's central longitudinal meridian ( $ \phi=\pi/2$ ), except close to the poles, where it oscillates in phase. Again, this is very different to the case of a global ocean, where the tidal maximum lies on a meridian of longitude, and rotates steadily around the Earth from east to west.

Figures 12.8 and 12.9 show the amplitude and phase-lag of the ($ j=8$ ) $ M_2$ semi-diurnal tide in a hemispherical ocean of mean depth $ d=3.8\,{\rm km}$ (which corresponds to $ \beta=23.1$ ), calculated assuming that $ \rho/\skew{3}\bar{\rho}=5.5$ and $ \mu/(\rho\,g\,a)=0.35$ . The calculation includes all azimuthal harmonics up to $ n_{\rm max}=20$ . Given that $ \skew{5}\tilde{\zeta}_8=8.95\times 10^{-2}\,{\rm m}$ (see Table 12.3), the maximum amplitude of the tide is about $ 60.7\,{\rm cm}$ , and occurs at the poles. This is very different to the case of a global ocean, where the amplitude of the $ M_f$ tide is zero at the poles. (See Figure 12.3.) Another major difference is that, in a hemispherical ocean, the tidal maxima circulate around points of zero tidal amplitude--such points are known as amphidromic points. Of course, in the case of a global ocean, the tidal maxima lie on opposite meridians of longitude, and rotate steadily around the Earth from east to west. The systems of tidal waves, circulating around amphidromic points, that is evident in Figure 12.9, are known as amphidromic systems, and are one of the the major features of tides in real oceans (Cartwright 1999). Generally speaking, the sense of circulation of amphidromic systems in real oceans is counter-clockwise (seen from above) in the Earth's northern hemisphere, and clockwise in the southern hemisphere.

In conclusion, our investigation of tides in a hemispherical ocean suggests that the impedance of the flow of tidal waves around the Earth, due to the presence of the continents, is likely to have a comparatively little effect on long-period tides, but a very significant effect on diurnal and semi-diurnal tides. In particular, for semi-diurnal tides, the impedance gives rise to the formation of amphidromic systems.

Figure: Relative amplitude, $ \vert\zeta\vert/\skew{5}\tilde{\zeta}_3$ , of the ($ j=3$ ) $ M_f$ long-period tide in a hemispherical ocean.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter12/ampmf.eps}}
\end{figure}

Figure: Phase-lag, $ {\mit \Delta }\varphi $ , of the ($ j=3$ ) $ M_f$ long-period tide in a hemispherical ocean.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter12/phasemf.eps}}
\end{figure}

Figure: Relative amplitude, $ \vert\zeta\vert/\vert\skew{5}\tilde{\zeta}_6\vert$ , of the ($ j=6$ ) $ K_1$ diurnal tide in a hemispherical ocean.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter12/ampk1.eps}}
\end{figure}

Figure: Phase-lag, $ {\mit \Delta }\varphi $ , of the ($ j=6$ ) $ K_1$ diurnal tide in a hemispherical ocean.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter12/phasek1.eps}}
\end{figure}

Figure: Relative amplitude, $ \vert\zeta\vert/\skew{5}\tilde{\zeta}_8$ , of the ($ j=8$ ) $ M_2$ semi-diurnal tide in a hemispherical ocean.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter12/ampm2.eps}}
\end{figure}

Figure: Phase-lag, $ {\mit \Delta }\varphi $ , of the ($ j=8$ ) $ M_2$ semi-diurnal tide in a hemispherical ocean.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter12/phasem2.eps}}
\end{figure}


next up previous
Next: Exercises Up: Terrestrial Ocean Tides Previous: Proudman Equations
Richard Fitzpatrick 2016-03-31