SOME MEASUREMENTS OF ATMOSPHERIC TURBULENCE. 
9 
From each side of (l) subtract 0 t 2 ) and divide through by t 2 —t v Then 
0 (h x , 0 — 0 {h u t 2 ) _ 0 (h 2 , t a )-0(h u t 2 ) 
t 2 ty t 2 ty 
( 2 ) 
Now the left side of (2) is the finite difference ratio — at the height hy, which is 
ot 
what we want to find in terms of the spacial distribution of 0. Expand the right- 
hand side of (2) in powers of h 2 —h x by the well-known theorem in the calculus. It 
follows that, if subscripts indicate the time and height 
m 
*8t Ai.feJJi 
(h 2 hy) “ 
2 ! 
-f higher terms. 
( 3 ) 
The difference ratio on the left of this equation is centred at the same height h x as 
the differential coefficients on the right of the same, but at a time ^{t 2 —t x ) previous. 
This slight misfit in centering will not matter, because t 2 —t 1 will be of the order of 
one minute or less, whereas we are next going to take the average of each term in 
(3) over a much longer time, say 6 hours. The subscripts may now be omitted as 
unnecessary. Let a bar over a symbol, or group of symbols, denote the mean value 
over this longer period. Let a dash denote the instantaneous deviation from this 
mean, so that, for instance, we have for every fluctuating quantity a formula such as 
00 = /00\ /00 V 
dh \dh) \dh) 
Now the mean of any dashed quantity vanishes.(5) 
Again the mean of the product of any dashed quantity into any barred quantity 
also vanishes.(6) 
We shall further suppose that the mean of h 2 —h x vanishes,.(7) 
that is to say that there is no mean vertical displacement. 
Then, in the first term on the right of (3) 
00 
00\ . /00V 
dh] V dh 
{(K-hJ + fa-hy)'} 
00V 
dh 
( h 2 h x ) .(8) 
because of (6) and (7). 
/'00V 
Now y—j(h 2 —h x )\ after being divided by t 2 —ty and by the “ standard deviations” 
/00V 
of j an d of (h 2 —h x )', becomes equal to the correlation between dOfdh and 
VOL. CCXXI.-A. C 
