OF AN OPEN CHANNEL BY GANGUILLET AND KUTTER’S FORMULA. 
CE Er an 
GER) JS= a a yi 
Since b,=b+LJ+JS, from the relations (2)’, (9) & (11), we get 
after reduction, 
bı___ a—(2—m)ar 
{Sea 
or, 
(12) (a—1) m?+xm-——(24—1) rm—rx—a (1—2r)=0, 
k’’ 
Transforming the origin to (r, r), (12) becomes 
where Vi 
o 
(12)’ (@—1)m,? +x,m;—a(1—r)*=0, 
where m =m—r, x,—=x—r. 
This is a hyperbola with asymptotes m=0 & (a—1) m, +x,=0, 
as in Fig. 2. 
as 
The family of curves (12) all pass through the point Ar(t,1), as 
is evident from the nature of our problem. If # be a constant, so also 
is a. .Then, the only variable parameter is r and it varies from 14 for 
d=0 to '/, for d= oo, so that the asymptotes displace parallel to them- 
selves by their respective component amounts. 
From (12), we find for a fixed value of m, 
