Ship Resistance. 495 
clearly the difference in question. Numerically, if & is the position of 
the maximum, at 3¢) the value of p from (16) has fallen to 3/5 of the 
maximum, while from (18) it has fallen to 1/7 of the same value. 
The poles of the function in (18) are +(1+7%)a; thus from (8) we have 
Eeixé 
x= 29" | Teal} (Efe +0) (Ea 0} heasy 
+21 | See eee CN tee esa 
{E-a(1+1)} {E+a(1+2)} {§—«(1—%)} |-ea-o 
= a e7* sin ax. 
The wave-making resistance, R, is given by 
AgpotR = xr? A2e~ 2 gin? ax. (19) 
We have now the oscillating factor sin?««. There will be, for instance, a 
hump on the resistance curve when 2e« = 7, that is, when the half wave- 
length is equal to 2x. It may be noticed that this is nearly, but not exactly, 
the distance between the maximum and minimum of y; from (18) it follows 
that the latter distance is 2«{/(4/3), or approximately 2°15a. 
We also have R exactly zero in the hollows in the resistance curve, a result 
which follows from the numerical equality of the positive and negative 
pressures at equal distances from the origin. We can make the negative 
pressures less by considering an unsymmetrical distribution. 
7. Let the pressure be 
en: 
P= Ba pe+ aa ep 
In this case the graph would be as in fig. 1, with the curve B for positive & 
and the curve C for negative values. 
If the poles of (20) are a1 +75, and ag+72be, we have 
aq +a2 = 0, 
ay? + by? + ag? + bo? + 4araq = 0, 
2 {ay (a2? + bo?) + a2 (a? +0,7)} = B 
(ay? + by) (aig? + bs?) = 4at. 
In forming the function y .by the previous method we have two parts. 
The part for the pole a,+ 7; is 
2a Eet«t 
E—(a—%i)} {E—(a2+tba)} {E—(a2—wh2)} 
(a ate 2b) eik2,g—Kdj 
et * (G—my— (G2 —b?) + Libs (a —aa) (22) 
(21) 
a,+ib 
100 
