The forces on a submerged spheroid moving in a circular path 302 
Carrying out these operations we obtain for the rotating spheroid at depth f, its 
centre describing a circle of radius h, 
oe | “ (AhOKF + BOK(a?e2— k) Q} dk, (26) 
1 2. P@peesst-a ‘ : 
F= an thP| FER I, (KD) J, {K,(h? + k?)*} sin na sin n(6 — Qt) dk 
0 re 
+ (4702/9) 5 NPI, (Ky) In{k,(h? + k?)*} exp [—K,(f—z2)] sin na cosn(6— Qt), (27) 
1 
_ h-weos(6—Qt) _h—weos(0—t) 
¥ re re 
© K2 eKU—2 
“af (ko) [Jn 4.{K(A? + k?)#} sin (n + 1) 
- mee ie + k?)#} sin (n — 1) a] sin n(@— Qt) dk 
oy YT (ky) [Tn ya{Kn (h2 + k2)H sin (n+ 1) 
—J,_3{Ky(h? + k*)?} sin (n— 1) alexp[—«,,(f—2)] cosn(@—Q#). (28) 
In this k,, = n?2/g, tana = k/h, and 
r? = {osin (0— Qt) —k}? + {h—weos (8— Ot)}* 4 (z oH (29) 
= {osin (0 — Qt) —k}? + {h—w cos (6 — Ot)}*? + (z—f)?. 
By comparison with § 3, we see that for w large we have 
$ © 
ars 2nor(=2) 3 {2nAL, +n3B(Q2/gh) M,,} 
1 
exp [—K,(f—2)] cos {n(0 — Qt— 47) +k, 0-47}, (30) 
with L, = | Tuc (EE oie ain nade 
—ae 
eZ (31) 
a k(a?e? — k?) [J {Ky (h? + k?)#} sin (n +1) a 
—ae 5 
—J,,_1{K,(h? + k?)?} sin (n— 1) a] dk. 
Using this in (8) we obtain the rate of propagation of energy outwards; if R is the 
tangential resistance and G the couple required to maintain the uniform motion, 
this leads to 
2 2Q5 © 
RAO+G0 = 
2 2 
yy (2.4L, +B = 1?) exp [ — 2n?Q2?f/q]. (32) 
1 
7. We may obtain the resistance, the couple and the radial force by calculating 
the resultant fluid pressure on the spheroid. For the wave resistance the only part of 
the pressure which gives a resultant is the term p0¢/0t, and we have 
2a 
R= [vas = par(1 -2)[" Ho of ndudo, (33) 
