C—la,

34

The problem is to determine the values of R audi for any particular tide. If the observations arc summed continuously from day 1 oto day v n. then v = n 2 ft,,c = R sin (« + /12-si) ~_, „ ,„_ „ sin 12-.-,/ ? cos(tt-£ +n. 1*6.) + .... (2) where for the purpose of illustration one term only of (I) is considered, and the sum of the observations from v= n t —Ito v = n., is the difference between the total sums from v oin each case, or v — n 2 V as » 2 [» =B //,, — 1 P,=SSt.A t . i/ = 2A,, i/ — SAt, v v = ?i| v = o y = o vO si „ ; 2 ., / cos tt-l+ (», + »,) 1 2-5 ,+.... (4j Put j, _ aiu(»o, ii, +I) 12-5 i sin 12-5 i \°) ■N = (Mj + w 2 ) 12-5 i (6) t hen P, = R .J cos (. f - £ + N) + (7) = A cos it + B sin it + .... (8) where A = R ,| cos (£ — N) (9) and B = R J sin (i - N) (10) Now, if the 25 values of P, are submitted to analysis, we have t = 24 i P, cos 28°-8« = F'=mA+it 1 B +PA, + gßy + .... (11) t = 0 < = 24 i P, sin 28°-B<= G 1 =n, A+ rß+ sAv+« B y + .... (12) 1 = 0 m A-f-w, B = F' -(pA v +?B, + ....)= F. (13) » a A+rß=Gl-(8 A y + « B y + ....) =G. (14) A = T F - "» G. n 5) mr — n l n 2 mr — n 1 n 2 V l "^ B = '" Q _ _^— F . (16) mr — n l u 2 mr — n l n 2 V *"/ Cotfc-N) = J (17) R- | v.V + (18) For the M 2 tide, i = 28°-9841042 per mean solar hour, and the maximum value of JF is required. £ is a maximum when sin (n. 2 —n l + 1) 12-5. = I—that is, when (n 2 —n Y +1) 12-5» = 90°; hence the maximum value is given by n 2 —n x + 1 = 39. Hence, if the values ol n 2 and n l —l are so chosen as to make n 2 —n l + 1 = 39, then the corresponding value of jf will be a maximum and give the best value of the tide M 2. As it is desirable to use all the available observations, they are used in cont-nuous sections I hrotighout the year. Thus, beginning with », =o, n 2 =38 for firsi seel ion, then n t =38 and « 2 =76 for second section; w x = 76 and n 2 = 114 for third section, and so on. Schedule for M 2 Tide from lAst of Sums. ii. 0 h. 1 h. 23 h. 24 h. 38 2275 1906 .. 2701 2524 76 3711 3705 .. 3792 3737 114 6050 5733 .. 6423 6268 304 15001 15008 .. 15110 14998 342 17635 17426 .. 17667 17664 Example for the third section: n 2 = 114, n t = 76, n t — n L + 1= 39. Here the sums of the hourly readings are evidently the differences between the total sums opposite n = 76 and « = 114, and are— Oh. Ni. 2 h. 24 k. 25 h. P t = 2439 2028 1675 .. 2631 2531 Then F, - + 506T8, G x = — 72827. and for a first approximation the corrections due to the other tides are omitted and F assumed =F l and G = G l . Hence A -i- 425-15 B = = ~~ ' 74866 = COt 306 °' 82 C - N - 306°-82 (n, + n.,) I*s. = N = 77 -25 '.'.£- 24 -07 A 2 + B 2 = 503233 y/A' + B 2 = 709-39 Q . = -00501935 o | V \ - 4- IS - X = - Q T X = 3-5607 ft.