A NEW METHOD OF ESTIMATING STREAM-FLOW 163 
The solution of the final normal equations for Solution M, (69) gives the fol- 
lowing values for the unknowns: 
tf=+66 fl<=-0.091 R* =+0.054 
R t = + 0.023 R„=- .630 R, a = + .832 
ff,= - .020 # 6 =- .950 R, b = + .453 
K,= - .013 R 7 =- .430 72,, = 4-1. 586 
(70) 
Combining (70) with (66) according to (61), there is obtained for the values of 
the unknowns: 
S e =+127 (±4.0) fi' 4 =+0.079( = 0.035) R\ = +0.92 ( = 0.03) 
R\= + 0.28 ( = .05) R\=+ .50 (= .17) R', a = +1. 15 (= .15) 
R' 2 = + .21 ( = .06) R' t =+ .10 (= .11) fl / l6 =+0.51(= .15) 
#' 3 = + .14 (= .06) R'-,= + .40 (= .09) fl'„=+2.17( = .21) 
• (71) 
The probable errors shown in parentheses were not computed rigorously from 
the normal equations, (69), and from the residuals, v, of the 242 observation equa- 
tions of Solution M, but were estimated on the basis of probable errors computed 
rigorously from normal equations like (69) in Solution L and the residuals of 183 
observation equations. 
In interpreting the above constants, (71), the units in which the r's were 
expressed should be borne in mind. There was a shift in decimal place to the left 
between n and r 6 and again between r th and no, hence, to get the above constants 
into their true relationship with reference to each other, R' t to R', b inclusive should 
be divided by 10 and R' 10 by 100. This will become clearer from the following 
paragraphs. 
METHOD OF CONVERTING CONSTANTS OF STREAM-FLOW FORMULAS 
TO PERCENTAGES 
From such constants, (71), it is possible to determine what percentage of the 
change in storage each day reaches the stream on a current or subsequent day and, 
from this, what percentage of the total change in storage on a day reaches the stream 
throughout the period covered by (71). For example, the derived fraction, R\, of 
the change in storage, n, reaches the point of measurement in the stream on the 
current day. Since the product, R\ n, is in the same units as D, viz, 0.001 c.f.s., 
(-RVO/IOOO c.f.s. is that part of the stream-flow above (or below) S c the current 
day which is due to the change in storage, n, on the current day. To convert dis- 
charge on Stream A in c.f.s. to units of 0.01 inch of depth per day on the watershed 
R' r 
one must multiply by 10.7, hence — — X10.7 of the change in storage the current 
R' r 
day reaches the stream that day, in 0.01 inch of depth. Therefore — ^X10.7X 
J 1000 
— or 1.07 R\ is the percentage of the total change in storage the current day, n, 
n 
which is delivered on that day. 
Now, assume that on a specific day there is rain or net melting sufficient to pro- 
duce a gain in storage, and that thereafter for 256 days there is no change of 
storage on any day, the rainfall or net melting on each day being just sufficient to 
equal the evaporation plus stream flow. Then, according to the theory developed 
