1072 Proceedings of Royal Society of Edinburgh. [sess. 
the theorem of (97) proved. Go back to the problem of § 41 
above: but instead of taking e= *9, as in§§ 46-61, take e = 1 — 10 -4 * 
and c= 1/(2/+ 1). By (86) and (87) of § 45 we have the following 
solution 
d= tf+ $ (100) 
where 
J= { \ sin (./ + 1)0 tan" 1 - £ cos ( / + \)6 log 1 + 2 >/e cos + _e 1 ( , 01 
< 1 - e " 1 - 2 Je cos -ktf + e ) 
and 
^=1 • 
3 2 ; + 1 + 2 / 
e cos 0 e 2 cos 20 , .. 
1 + 9* + Be^cos jO 
Je cos J0 + e 
( 102 ). 
Fig. 29 has been calculated by putting 0— — . and 
A j + h 1 
taking j = 20. The explanation is that, as we shall see by (78) of 
§ 43 above, (100), (101), (102), express the water disturbance due 
to an infinite row of forcives at consecutive distances each equal 
to (20|r) A ; the expression for each forcive being 
cbafeir 
(103), 
b 2 + (x - naf 
where n is zero or any positive or negative integer ; and by (79) 
we have 
20-5. lO" 4 . A 
b = 
27 T 
(104). 
Thus we see that the pressure at O due to each of the forcives next 
to O, on the two sides, is l/{ 1 +(27r.l0 4 ) 2 } of the pressure due to 
the forcive whose centre is O. Thus we see that the pressures due 
to all the forcives, except the last mentioned, may be neglected 
through several wave-lengths on each side of O : and we conclude 
that (100), (101), (102) express, to a very high degree of approxi- 
mation, the disturbance produced in the water by the single 
travelling forcive whose centre is at O. 
§79. To prove (97) take 0=180° in (100), (101), (102); we 
thus find 
( - l)>d(180->*y { tar' 1 -£ + £ + « - e)~> ■ } (105). 
Instead now of taking e= 1 - 10 -4 , as we took in our calculations 
