SHIP VIBRATIONS: THE VIRTUAL INERTIA OF A SPHEROID IN SHALLOW WATER 
Another relation, given by the present writer™ is largely 
the basis of the following calculations; it is, in the 
present notation, 
rn) . 0 
PW Wet =4(5 +855) 
1 
(1 — h?)** Ps (h) dh 
[@ =m +P +2 
-1 
This result expresses the spheroidal harmonic as a line 
distribution of multi-poles along the axis of the spheroid 
between the two foci. 
& 
(8) 
Transverse Motion at Right Angles to the Boundary 
8. We begin with the form for an infinite liquid, 
do = P,, (H) Q;, (2) sin w (9) 
For convenience, we omit the time factor and the con- 
stant C. Take parallel axes O'(x’, y’, z’) with the 
origin O’ at the point x = 0, y = 0, z = 2f/ae; and let 
ph’, 6, w’ be spheroidal co-ordinates referred to these new 
axes. It is easily seen that to maintain zero normal 
velocity over the plane z = f/a e, we must take 
o) = — P, (#) Q, @’) sin (10) 
To obtain ¢, we expand (10) in the neighbourhood of 
the spheroid ¢ = ) in the form 
— 3 3S (As cossw +Besins wo) P*(y)P5(Q (11) 
n=0 s=0 
Then ¢, is given by 
b2 = 
x & (A; cossw + By, sin sw) Pr (u) Qi (2) Pr (S0)/Q), (So) 
(12) 
General expressions for the coefficients could be obtained; 
but it is easily seen that in order to calculate the kinetic 
energy we only need the coefficient B}, noting that the 
normal velocity on the spheroid is unaltered and also 
using the orthogonal properties of the functions. From 
(8) we have 
(22)! Pl(k) dk 
Slesarei ey 
=i 
Plu’) QC) sin w! = 5 
d (1 — k*)! Pi (k) dk 
dq Ike —k)?+y?+(z—q)*}? 
(13) 
with g = 2f/ae. If the point (x y z) is (uf w) and the 
point (Kk oq) is (4; ¢,@,) in the same spheroidal co- 
ordinates, we have in (6) the expansion of the inverse 
distance between these two points, with 
k= py 2); 0= (1 — pi)? (G — 1)* cos w; 
qg= (1 — pi)? (G — 1)Fsin & 
586 
Further, the expansion is valid in the neighbourhood of 
the spheroid, since €, > Cp. 
We substitute this expansion for the denominator in 
(13), and B} is the coefficient of the term P! (1) P'(£) sin w. 
Hence, from (6) and (13) we have 
2n+1 a 
n(n + 1)? dq 
[4 PLO) P12) QE G)sin oy de (1) 
24 
raft = 
We may put this into a symmetrical form by noting that 
P! (u,) Q' (¢,) sin, is the value of this spheroidal 
harmonic at the point (k 0 q); hence, from (8) 
1 
(1 — #2) PL(h) dh 
: ed 
Ph (141) Q, (¢;) sin w, = 2 dq aG =hy + rap (15) 
This gives 
B} — ntl ey 
n n2 (n ae 1)? dq 
(1 — h?)k (1 — k?)# P! (h) P! (k) 
I [‘ (aera dhdk (16) 
To calculate the kinetic energy, we see that d4/d¢ on 
the spheroid is P! (u) Q! (4) sinw; and from (9), (11), 
(12) the corresponding term in the value of ¢ on the 
spheroid is 
[Q! (Go) — BLP} (Go) + BL PI (£0) Qh (Co)|/ 
QI (Go) . Ph (w) sin w 
or using (7) 
[l —n(m + 1) BUG — 1) Qk) Qi (£0)] 
Qi) (Go) Ph () sin @ (17) 
It is obvious that the kinetic energy is increased by the 
factor within square brackets in (17); hence from (16) 
and (17), the relative increase in kinetic energy, or in 
the inertia coefficient is given by 
8 T/Ty = —2n+1)D,/ ' 
2n (nm + 1) (& — 1) Q) (So) Qi (Lo) (18) 
with 
(i =) — kt P,P, (&) 
ee a) [ [(k — h)? + a}? ana 
re (19) 
Transverse Motion Parallel to the Boundary 
9. When the vibrations are parallel to the boundary, we 
begin with 
$o = Pr (#4) Q; (2) cos w (20) 
