110 M. Girard on Navigable Canais. 
Or, making the whole number of boats which ascend 
= N; and the whole number of descending boats = M ; 
the sum of the quantities 
y ty +y"' +, &e.p2'+2"42'"4, &e =Y, 
And preserving the denominations 'T’ and T” for the sum 
of the drafts of water of the ascending and the descending 
boats respectively, we shall have in a more simple form, af- 
ter dividing by S: 
Y=a (N+(M—K))—(T’—-T’). 
Therefore, the total expense of water from an upper basin 
of a lock, through which a certain number K of convoys 
shall alternately pass, will be positive, null or negative, ac- 
cording as we have, 
ae Vy 
~ N+(M-—K) 
BiiKee 
~ N+(M—K) 
‘I A ect aD . 
* <N+(M—K) 
By recurring to our general formula 
Y=a(N+(M—K))—(T’—T’) 
we may remark that the first term of the second member is 
at its minimum value when M—K=o, that is, when the de- 
scending boats are equal to the number of convoys which 
they form, or, which amounts to the same thing, when they 
go singly. 
In this first hypothesis the formula becomes 
Y=Ne—(T’—-T’) 
as we have already seen. 
The term « (N-+-(M—K)) of the general formula arrives, 
on the contrary, at its maximum value when K =1, since the 
number of convoys ascending or descending cannot be less 
than one. 
In this second hypothesis we have 
=x (N+(M—1))—(T’—T); 
an equation which appliesto the particular case where all the 
boats which pass the lock should form but two convoys, one 
ascending during a certain time, and the other descending 
during another period of time. 
The last equation becomes 
Y=r (2N — 1) —(T’—T’) 
