216 Long Waves in Canals and Standing Waves in Closed Basins 



canal which are shown in Figs. 91 and 92. Comparing with Fig. 90, we can 

 see the changes caused by the closing of the canal. The inside amphidromy 

 in front of this barrier is developed regularly on all sides and it looks as if 

 the wave penetrates into the canal at the left side and leaves the amphidromy 

 at the right side of the canal, turning around its centre. The outside amphi- 

 dromy is identical with the one shown in Fig. 90. The amplitudes are always 

 largest at the bay shores and attain a maximum in the two corners of the bay. 



In the outward sections of the canal, the currents (Fig. 92) run always 

 parallel to the sides of the canal, and their distribution here is identical with 

 the currents produced by the superposition of two Kelvin waves. In the inside 

 section of the canal, the currents are parallel to the wall — as is required — 

 but in the middle they are not any longer alternating currents, but have 

 become currents which rotate in the direction of the wave. The current 

 figures are ellipses, and their eccentricity decreases towards the middle. Just 

 in front of the closed end of the canal the major axis of the ellipse is parallel 

 to the transverse barrier, whereas it is parallel to the longitudinal walls 

 towards the opening. At the same time, the transverse velocity decreases and 

 vanishes completely in the outer section of the canal. The current distribu- 

 tion for four phases (t =0, 1 J, 3 and A\ h) can be found in Defant (1923a, 

 p. 57). 



The free oscillations of a rotating basin could be easily computed, in case 

 the basin has a circular or an elliptical shape (see Goldstein, 1929, p. 213). 

 Rayleigh (1903, 1909) has given a solution for a rectangular basin of uni- 

 form depth, for the case where the rotating velocity is relatively small which 

 does not apply to the rotating earth. Taylor arrived at a general solution 

 following the same method which he used for the reflection of a Kelvin wave 

 in a canal closed at one side. An appropriate second solution of the equa- 

 tions of motion is added to the superposition of two Kelvin waves, in such 

 a manner that the longitudinal current is always zero at the transverse bar- 

 riers closing the canal at both ends. This will give the condition for the natural 

 period of the rotating rectangular basin. 



We now assume that the superposition of two Kelvin waves moving in opposite directions 

 has the form: 



ax ax 



m = S \ cosh ay cos — cos at— sinhavsin — sin at ) , 



(VI. 125) 



The origin of the co-ordinate system is laid in the middle of the basin; let its width be n, ex- 

 tending from — \n ^ y ^ + \n, its length 2/ and — / ^ x =% + /. The second solution is supposed 

 to have the form (VI. 119). As it is now required that in two points x the longitudinal current 

 equals (VI. 125), and when w 2 > k 2 , we have to put v 1 and v 2 : 



v x — 2j C„sinh S n x sin ny and v 2 = 2j C n coshS n xcosny . (VI. 126) 



even « odd n 



