59d SCIENTIFIC RESULTS OF ZIEGLER POLAR EXPEDITION 



Substituting the values for K°i and 0°, from the table of harmonic constants already given, 

 we obtain : 



Cape Flora Teplitz Bay 

 h m h in 



D,HWI 2 39.8« 2 34.9a (30) 



TlDS INBQUAI,ITIBS AND RANGES 



An inequality in the interval, range, or height of tide is a systematic departure of the 

 same from the mean value at a given station. The inequality having a period of a half synodic 

 month is the phase inequality ; that having an anomalistic month is the parallax inequality ; 

 that which has the period of a tropical month causes the two high waters or two low waters of 

 the day to differ in height, and is called the diurnal inequality. 



The age of an inequality is the amount of time by which it follows its astronomical cause. 

 The ages of the principal inequalities are given by the expressions : 



Age of phase inequality =0.984 (S°, — M°j) hours (31) 



Ageof parallax inequality =1.837 (M°j — N°j) hours (32) 



Age of diurnal inequality = 0.91 1 (K°i — 0°i) hours (33) 



Substituting the values of the epochs or kappas given in the table of harmonic constants, 

 we obtain : 



Cape Flora Teplitz Bay 



h h 



Age of phase inequality 53.6 50.4 (34) 



Age of parallax inequality 61.9 42.6 (35) 



Age of diurnal inequality — 15-9 —21.5 (36) 



The mean range of tide, as given by the direct summation of high and low waters, usually 

 requires to be corrected for the longitude of the moon's ascending node, there being whole 

 series of years during which the mean annual range is greater than an average for the lunar 

 cycle, followed by another series of years having a smaller mean annual range than the average. 



If we put Mn for the corrected mean range or rise and fall of tides, and Mn' for the uncor- 

 rected mean range, we may find the corrected range from the equation 



Mn = Mn' X F (Mn) (37) 



The values of F (Mn) are obtained from Table 14,* using I and (K, + Oj) -^ M, as argu- 

 ments. In the present case these arguments are 



Cape Flora Teplitz Bay 



o o 



I 18.38 18.33 



Ratio Ratio 



(Kj + OJ -f- M, 0.68 0.28 



Entering Table 14* with these arguments, we find 



For Cape Flora Mn = 0.99x0.976=:: 0.966 feet (38) , 



For Teplitz Bay Mn^ 1.17x0.972 = 1. 138 feet (39) 



* 



See note, p. 588. 



