The Initial Wave Resistance of « Mowing Surface Pressure. 246 
This case is the simplest of the type which allows of exact evaluation of (15), 
and for which the integral pressure is finite. To derive the corresponding 
pressure system, we make use of Euler’s integrals of the form 
| Kr Teco cog (Ax sin «) dk = A~"T'(n) cos nz, (17) 
0 
r~> 0, n> 0, —in<ca<hr. 
Using the Fourier integral theorem, combined with (16) and (17 ), we find 
mF (x) = | Ge—aycten cos Kx dk 
0 
= DG) (7? +2?) 8 cos (f tan7! 2/7) — oT (2) (72 +.22)—58 cos (5 tan~} x7). 
(18) 
The two terms of this expression are easily graphed when expressed in 
terms of the angle tan7!(x/r); two numerical cases are shown later. 
We can now find the resistance R for the system (18), travelling with velocity 
c= /(g/xo). Substituting (17) in (15), and writing « = w®, git =, ct = Ps 
we have 
ao 
—trgpR = —K | (u® — Kut) e~?™ sin pu? cos qu du 
0 
o 
+ Ko? | (wu! —Kyu*) e-?" cos pu? sin qudu. (19) 
0 
The integrals involved can all be derived, by differentiation with respect to 
the parameters, from 
| e- &-P) cos qu. du = }[m](p—ip) en" 0-0, (20) 
0 
Carrying out these operations, we obtain finally 
—7PgoR =_ Koll? (Ar? + 07¢?)—5/4 e—gPrl2 (412+ ct?) 
x [o8 { —42 A sin (FO—$) +48 97A2 sin (20— $)—337'A’ sin (420—¢) 
+ pe QAt sin (426 —$)} + 0%” (3 sin ($0—p)— 397A sin (Z0—9) 
vee sin ($0—$)} +498 9A? cos ($0—g)—198 9A cos (1,0—$) 
34 7A* cos (126+ 6)—74_97A® cos (120—¢) 
ene (F0—$) + £9°A? cos (3 0—) — lg gPA cos (1.0—4)} ] 
(21) 
where 
q=9t; A= (4 +4c%?2)-4; O= tan“'(ct/2r); = get?/4 (47? + 2?), 
6. Before working out numerical examples, it is convenient to record the 
asymptotic expansion suitable for large values of et/2r, From (21), by 
111 
