The Forces on a Circular Cylinder S53) 
in v1. Using the notation of (21), and separating out the real and 
imaginary parts, we obtain, after some reduction, 
— X = 4r®ec?a (x a)? eI [1 — 2r, (koa)? — (r2 — 3737 + 5*)(Koa)? 
— (4r3 — ro? — nr, + £7, — 4r4s" + S?) (koa)® 
SP (ola == clalay —= GIRL ae eee a a ae AW a APES Sing 7 
— (10r? — 3r, — 3re + 335) 5? + st} (oa)® + ...], (29) 
Y = 4reca (xa)? [—4re + ryre (K9a)? + H(rarg — 3rPre + 125") (Koa)” 
+ (2rfre — 4rs8 — yrs + aly Paha + $728" — 2ryres*) (Koa)” 
ae {eils® = sia No — folate sia lone — BIA = TAIRA eg 
+ (Sryr2 — shire — 31s dre +ils sla + yx) S 
= ress} (koa)® + --], (30) 
with r,, 5 given by (22). 
The first term in (29) is the expression for the wave resistance of a 
circular cylinder which was obtained by Lamb. The first term in (30) is, 
after putting in the value of r, from (22), the first approximation for the 
vertical force which I obtained by the method of images in the paper 
already quoted. 
5—It is of interest to obtain the wave resistance, which should be equal 
to —X, from considerations of energy applied to the regular waves 
behind the cylinder. The current function ¥ is given by the imaginary 
part of the expression (12). Putting y= (+ 7, we obtain at once the 
complete expression for the surface elevation as 
n= ia? | f@ efra—«f de + ig? [ ee KS LE r& (i) ef de, (31) 
0 0 B= Ky = UB 
where the imaginary part is to be taken, f(«) is given by (11), and uy is to 
be made zero ultimately. This expression separates into two parts, a 
local disturbance 4, which decreases with increasing distance from the 
cylinder, and a system of regular waves 7, to the rear, that is, for negative 
values of x. The latter part is found, by methods familiar in these 
problems, to be given by 
Ne = — Ark a2f* (Ko) emt, (32) 
the imaginary part to be taken. 
If his the amplitude of the regular waves at a great distance behind the 
cylinder, the wave resistance R is given by 
R = 4geh". (33) 
426 
