102 U. S. COAST AND GEODETIC SURVEY 
such as to make the constituent symmetrically situated in regard to 
the maxima and minima of M,., the amplitude necessary to account 
for the mean duration of rise of the tide may be readily calculated. 
293. Let DR=duration of rise of tide in hours as obtained from 
the lunitidal intervals, 
a=Hourly speed of M,.=28.°984. 
M,.=Amplitude of Mp. 
M.°=Epoch of M3. 
M,=Amplitude of M,. 
M,°=Epoch of Mg. 
Then, for M, to be symmetrically situated with respect to the maxima. 
and minima of M, 
M.°=2 M.°+90° (427) 
in which the upper or lower sign is to be used according to whether 
a(DR) is greater or less, respectively, than 180°. Multiples of 360° 
may be added or rejected to obtain the result as a positive angle less 
than 360°. 
The equations of the constituents M, and M, may be written 
Yi=M, cos (at+a) (428) 
Yo= My, cos (2at+ B) (429) 
and the resultant curve 
y=M, cos (at-+a)+M, cos (2at-+ 8) (430) 
294. Values of ¢ which will render (428) a maximum must satisfy 
the derived equation 
aM, sin (at-+a)=0 (431) 
and for a maximum of (430) ¢ must satisfy the derived equation 
aM, sin (at-+a)+2aM, sin (2at+ 8)=0 (432) 
For a maximum of (428) 
jae (433) 
in which n is any integer. 
295. Let = the acceleration in the high waters of M, due to the 
presence of M,. With the M, wave symmetrically situated with 
respect to the M, wave, . will also equal the retardation in the low 
2, : 
water of M., due to the presence of My, and = will equal the total 
amount by which the duration of rise of the tide has been diminished 
by M,;. If the duration of rise has been increased, @ will be negative. 
Then, for a maximum of (430) 
__ 2nz—a—d 
rm a 
t (434) 
and this value of ¢ must satisfy equation (432). 
