OF AN OPEN CHANNEL BY GANGUILLET AND KUTTER’S FORMULA. 3) 
‘sample 1. To find the standard section for Q=50 eubie feet per 
second, when 8 =: 30°, n=0°03 and J=='/1000- In Plate I, we read 
Example 2. In the above example, if the depth of the channel is 
to be 25 feet, find the corresponding bottom breadth. 
We have 
d, 2°5 
=P Son =(0765. 
Therefore, from the hyperbolas, 2 22, so that 
b,=322 x 1°75 = 5:63 feet, 
the corresponding bottom breadth. 
Example 3. Find the quantity of discharge of a channel whose 
G=30°, n=0:03, J=rgb5, bı=5°5 feet and d,=3 feet. 
We have 
ars 
59x 1°87 
Therefore, by drawing a line from the origin of the hyperbolas through 
= 0'292. 
to 
: a b 
the point m==F—0'292 at x==1, we read ——2'72. 
b 
b, dd 
Hence b = 972 7272 
of discharge is read off from the curve, i.e., Q=74 cub. ft. per second. 
—=2'02 feet, and the corresponding quantity 
Example 4. Find the mean velocity of flow for the channel in the 
preceding example. 
The sectional area=(bı+d, cot 30°) x d, 
a= (5ion. +3 21273) x3 
== 32°1 square feet. 
& 
(4 
Therefore, the mean velocity of tlow—=— 357 
Im 
—%3 feet per second. 
The above examples are for 030° throughout, but the processes of 
working are quite similar for other values of @ in Plates II to IV. 
In preparing the present paper, the author desires to acknowledge 
the kind assistance of a student (now a graduate) of the Agricultural 
College, who wishes to remain incognito and who worked out a mass of 
numerical calculations. His results were checked, corrected and plotted 
into curves by Mr. Naosaburö Kusakabe, Kdzakushi, and by the author. 
Tokyo Imperial University, 
Tokyo, Japan, Dec., 1909. 
