190 T. H. Havelock 
With 6 and d infinite, this gives, for a semi-infinite stream with an upper 
free surface, 
2 PN on a Ze r 
bily = 1~3(°) sind w6( 2) (1 8sin 0) 
1 [2 1 /2\4 
+35 ai (3 sin 0 — Ssin®4) + 515 (”) (6—41 sin? + 46 sin40)+.... (63) 
With a = b = id, we obtain 
Bf Th malt) @ —sin?6) 
d 
m* (2p\* ae = 4 
+7as0(z) (64—97 sin? 6+ 66 sin*0)+.... (64) 
For numerical comparison, we take the same case as before, and obtain 
the following: 
0 = 10°; 2p/d = 0-2 
a/d 0-3 0-4 0-5 0-6 0-7 
L/Lp 0-924 0-951 0-969 0-983 0-994 
Similarly, for the moment, we have 
M/My = 1+¢,p+c, p?+¢cs p?+c,ptt+..., (65) 
¢, = —2r)sin8, 
Cy = 3r2sin?@—4r,(1+4sin?O)—7;, 
C3 = — 473 sin? 6 +797,(3 sin 6 + 2 sin? 0) 
+ 47,(sin 6 — 8 sin? @) + 2797;(3 sin 0 — 4 sin? 9), (66) 
C, = Sr§ sin? 6 —42737, sin? 0 — 2797,(sin? 6 — 3 sin? 0) 
er dr3(1+ 14sin? 6 — 12 sin* 0) + 47,(1—8 sin? 6) 
— 3r27r{(5 sin? 6 — 8sin*@)+7,7}(1+ 10 sin? 0 — 20 sin’ 6) 
+72(1+ 6 sin? 6 — 16 sin4 @) + 473(1—4sin26). 
With } and d infinite, we obtain 
1/2p)\.. 1 (2p\? ube 1 (22)\ee : 
M/M, = 1=5 (7) sino-5,(2*) bape (7sin 6— 12 sin?@) 
1 
2p i i 
g( 2 Jasin 0+ 8sin*@)+.. (67) 
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