110 M. Girard on Navigable Canals. 
Or, making the whole number of boats which ascend 
= N; and the whole number of descending boats = M ; 
the sum of the quantities 
yy +, Se.p2' +2" 424, &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=2 (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, 
TT’ 
> N+(M—K) 
- Y east et 
N+(M—K) 
fe AM 
is & <N}(M—R) 
By recurring to our general formula ee | 
Y=2(N+(™M ~—K))—(T’ oF") 
we may remark that the first term of the second member is 
at its mintmum 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 pe 
go singly. 
In this first hypothesis the formula becomes 
Y=Na—(T’—T’) 
as we have already see 
The term (N+ (MK) of the general Surana 2 arrives, 
on the contrary, at its maximum value when K =1, since the 
nmr of convoys ascending or descending cannot be less 
than o . 
In shis second hypothesis we have 
Y=ax (N+(M-1))— “(tT ; : 
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. 
e last ee ie becomes 
2(2N =N-(P 
