442 
Proceedings of the Royal Society of Edinburgh. [Sess. 
Hence 
(P-V o)/(Po-^o) = e ~^ .... (8). 
In one of Shaw’s tests* it was found that for £ = 0, k(p 0 — w 0 ) = ^x 29 ; 
and for t = 3600, k(p — w 0 ) = § X 9. Hence 
whence 
e36oo^ = 29/9: 
/x = (log 29 - log 9)/3600 log e 
j= -000325 ; 
T = l//x = 3080 sec - 
For rough purposes we shall hereafter take t = 3000. 
Case 2. Suppose that the air-chamber and the outer air have come to a 
common pressure m 0 , and that the outer air then suddenly rises to the 
pressure ts 0 + to 0 . Then we have 
P={ra„ + /*(w 0 + to 0 ) 1 dte^ t }e~^ t 
.= {cr 0 + (uT 0 + 8cr 0 )(e^ - l)}e-^ 
= C7 0 + Scx 0 - Sc r 0 e-M«. 
Hence, if y denote x*y 0 + <5x*y 0 — jp, to which the microbarograph reading is 
proportional, we have 
y = Bzs 0 e~^ (9). 
Case 3. Suppose the pressure in the outer air to begin to rise with a 
uniform time gradient y ; so that V5 = cx 0 + yt. We have 
Hence 
rt 
pe^ = c? 0 + /X dt (d 0 + y t)e^ 
Jo 
= Z3 0 erf + y {tt& - (&*- l)//x}. 
p = TZ 0 + y {(t + .... ( 10 ). 
Therefore, if we put t=1//a, and y—.js—p, we get 
y = yr( l-e-t/r) (11); 
tr •*-* ( 12 >- 
Hence we have the following figure (3),f where the graphs of y — xz—p 
and y 1 = w — d 0 are OC and OD, the latter being the tangent to the former 
at the origin, as might be expected. 
Thus, at the start, the microbarograph gives the correct gradient ; but 
after a considerable time, if the outside gradient remains constant, the 
* Quart. Journ. Roy. Met. Soc., xxxi., p. 43 (1905). 
t This figure and the corresponding ones which follow are not drawn to scale, but are 
diagrammatic merely. 
