OF AN OPEN CHANNEL BY GANGUILLET AND KUTTER S FORMULA. 7 
Then the line ST drawn parallel to the side line DC gives with EB 
and BT the required cross section. 
For, by the construction, PN= nd MN=d, the depth. 
Op 
OA 
Also enti oa G ri Therefore MS=öp, and the relation 
Op 
is satisfied, 
Hence, EBTS is the section with given depth d, and discharge Q. 
The construction is similar for deeper sections than the standard. 
The above is a graphical solution of the problem. 
OA d, 2nk + ja 
dD 
OA 
Now from (2)’ and (8), ), so that 7, must 
Op enter +3 ns 2) 
always lie between 14 N '/, of the depth of the standard section. 
Dencting the ratio — en by r, let us examine what sort of value 
0 
it would attain for various values of nk and d. 
r . . . 
The valnes of nk in onr ease are given by the following schedule: 
Values of nk 
ES 0090 , 0250 0030 
vw | 0-85 106 | 1:97 
a 086 | 1075 | 1-29 
a veil 1:33 
nn ODE Sa 1.50 
en 111 | 1:39 | 1-67 
Taking the least and the greatest values of nk, the respective values 
of r are; 
