THE DAMPING OF 
by-step with successive approximations to a solution satisfying 
both the condition on the surface of the spheroid and that on 
the free surface of the water. We could then, at any stage, 
obtain the fluid pressure on the spheroid and hence the resultant 
vertical force. Of this force, the part in phase with the accelera- 
tion represents a change in virtual inertia, the part in phase with 
the velocity is connected directly with the loss of energy in the 
wave motion. At present we are concerned only with the latter 
part of the force, and we adopt the simpler procedure of obtaining 
directly the energy loss, taking only two terms in a successive 
approximation to the velocity potential. 
For the first term we take the exact solution ¢o for the motion 
of the spheroid in an infinite liquid. Taking, momentarily, the 
origin O at the centre of the spheroid, and the usual spheroidal 
co-ordinates 
x=aeul; y=ae(l — pC — 1)1/2 cosa; 
z=ae(l — p7)'2 (2 — 1)2 sinw 
we have the known solution, as given in Lamb’s Hydrodynamics, 
p. 142, with a slight change of notation. 
$p = — $aV(QZ—1)(1 +k) Pi (uw) Q!(Q)sinweos ar . (9) 
where () = e—! = (a? — b?)-1/2, and ky is the virtual inertia 
coefficient for motion perpendicular to the axis. For the next 
step we add a potential ¢, such that ¢y + ¢, satisfies the con- 
dition at the free surface. For this purpose it is convenient to 
express gy as the potential of an equivalent dipole distribution. 
Using the general formula 
e 
(a2 e2 — h)112s Ps (h/ae) 
; Ze. .O\s i 
PS (2) Qa(Oe =a(s5 T iss) [@—A)?+y2+Z2]!/2 ae 
(10) 
and taking the particular case in (9), we see that do is the potential 
of a line distribution of vertical dipoles along the axis of the 
spheroid between the two foci and of moment per unit length 
1 1 — é?2 
——(1 + k) (@ e? — h?) Veos ct 
4 @ cD) 
We could now obtain ¢, by integrating (1) with respect to h. 
For our purpose we proceed directly to the energy loss from 
(7) and (8). We require the integral 
ae 
I (a2 e2 — h?) ei Kohcoso dh (12) 
=H 
and this has the value 
Ave @ 112 J5/9 (Kg a e cos 8) 
2 (ko ae cos §)3/2 
(13) 
Collecting thé various factors from (7), (8), (11), and (13), we 
obtain the energy loss for heaving, which we shall denote by a 
suffix H, namely 
7/2 
Ey =472p on3a2b4(1 +k,)2V20-240F | Jan (Koae cos 8) |p 
(ko ae cos §)3 
(14) 
3. Suppose the spheroid to be making rotational oscillations 
about the transverse axis with angular velocity cos af. The 
velocity potential for an infinite liquid’is 
do =3 e(G—-)[lL+2QZ—-12k] 
Q PI (u) Q!(Q sin w cos ot 
in which k’ is the virtual inertia coefficient for rotation. 
a5) 
Using 
HEAVE AND PITCH 3 
(10), we find that ¢o is the potential of a line distribution of 
vertical dipoles along the axis between the foci and of moment 
per unit length 
1/4 (9 (43 — 1) [1 + 2B — 1)? kK JQUN@ ce? — h?)coscr . (16) 
For (8) we require the integral 
ae 
| h (a e? — hh?) eikohcos0 dh 
ne 
(17) 
and this has the value 
m\ 12, Js) (Kg a e cos 8) 
4 a4 et (7) ea 2 
From (7), (8), (16), and (18) we obtain the energy loss for pitching, 
which we denote by a suffix P, 
7/2 
32/9 (ko a e-cos 6) 40 
(Ko ae cos A)3 
Ep = 4 7? p ok} at bf e2 
[1 + @ @ = 1)? kh’? e—2 kof 2 
(9) 
4. We obtain now the corresponding expressions by the two- 
dimensional strip method, denoting them by an additional 
suffix S. 
For a circular cylinder of radius r with its axis at depth f, 
making heaving oscillations Vcosat, we have the known 
expression for the energy. loss to the same approximation, 
E=2 7? pox r4 V2 e-2hof (20) 
per unit length of the cylinder. 
For the spheroid, this result is assumed to hold for each thin 
disc of width dh; in fact, we might picture the method by 
assuming thin partitions transverse to the axis, separating the 
elementary discs and making the fluid motion purely two- 
dimensional for each disc. Integrating along the axis, we obtain 
by this method 
Eys = 2 7? p ok§ vt etn | b4 (1 — h2/a2)2 dh 
= 2p o Ke a bt V2 e—2 Kof 
iG (21) 
For pitching oscillations by this method, we simply substitute 
hQ for V; hence 
a 
Eps = 2 72 p ok? (2 entect | b4 h2 (1 — h2/a?)? dh 
Ee 
oe 7 p o K2 a3 b4 Q)2 e—? Kof 
105 
(22) 
5. The particular point in question is the ratio of the damping 
coefficients obtained by the two methods, which we take as equal 
to the ratio of the corresponding energy loss. 
From (14) and (21) we have for heaving 
7/2 
Ex 15 J3,, (ky a 8 cos A) 
pes, 1 e,)2 3/2 \Ko , (3 
aa © oils i) | (ko ae cos )3 ao > ES) 
0 
From (19) and (22) we have for pitching 
70/2 
Ep 105 9 J2)5 (Kp ae cos 8) 
—- = 2 2 — 1/2 k’/2 5 2A Oa Sane 
= gz Kore fi+@@-1) “P| Gala BeOS: dé 
(24) 
As the length of the spheroid is increased, with a given breadth, 
613 
