A NEW EVAPORATION FORMULA 105 
clear. In order to clarify the method of computing the r's, a numerical example, 
shown in Table 37, is taken from the least-square solution designated as V x per- 
taining to Lake Michigan-Huron. 
Attention is invited to Table 37, in connection with the definitions of r,, r 2 , r„ 
. . . r 6 on page 104. In order to estimate the evaporation from land, Ei, it 
is necessary to assume values for E x and E t , and first estimate the evaporation 
from the water (lake) surface, E w , on the basis of these assumed values, the assumed 
straight-line evaporation law and the observed weather elements e and w. The 
best values of E x and E 2 obtainable from previous least-square solutions were 
Ei= +0.581 and E 2 = +0.624. These were accordingly used in Solution V x in 
estimating the evaporation. The sum of the products eE x and el — rr-: — jE t 
are shown in column 4. This is the estimated evaporation, E w , from the lake sur- 
face in units of 0.001 foot of depth per day. Using the ratio ^-r = 0.62 obtained on 
pages 16 to 18, and the ratio of land to lake area for Lake Michigan-Huron shown 
92 493 
in Table 7, page 17, . ' ^ . =2.04, the estimated evaporation from the land surface, 
E h in units of 0.001 foot of depth on the lake area is E W X 0.62X2.4 = 1.26 E„ 
This is shown in column 13, Table 37. 
Similarly, in order to estimate the run-off from the land into the lake on any 
day, it is necessary to assume the run-off constants, R 1} R 2 , R 3 , . . . R*. This is 
more difficult than assuming values of Ei and E 2 . The estimated constant part of 
the run-off into the lake each day in 0.001 foot of depth is shown in column 5 as 
7 C = 6. The value originally estimated was 8 (page 19). The method of revising 
the original value, 8, to get the final value, 6, has yet to be stated. 1 The estimated 
variable part of the run-off into the lake each day is shown in columns 6 to 11 
inclusive, based upon the following estimated values, (29), or R x , R 2 , R 3 , . . . R e . 
#i = +0.01 fl«=+0.80i 
R 2 =+0.07 R>= +0.20 (29) 
tf 3 =+0.17 R,= +0.27 i 
The quantity r h by definition, is the total change in storage in the ground 
within the drainage area of the lake on the current day, and it is expressed in the 
same units as the absolute term of the equation, I. Now, from equation (28) it 
should be evident that R x is that fraction of r x which is delivered to the lake on the 
current day. From this, and the definitions of r h r 2) r 3 , . . . r 6 , it can be shown 
that, if one assume that a change in storage on the current day, r,, is followed by 
no change in storage for a long time thereafter, 
100 R x per cent of r x is delivered to the lake on the current day, or 0th day; 
100 R 2 per cent of r x is delivered to the lake on the first following day; 
100 R 3 per cent of r x is delivered to the lake on the second following Jay; 
5.0 Ri per cent of r x is delivered to the lake on each of the third and fourth following 
days; 
2.5 Ri per cent of r x is delivered to the lake on each of the fifth to eighth following days, 
inclusive; 
1.25 R 6 per cent of n is delivered to the lake on each of the ninth to sixteenth following 
days, inclusive. 
■See page 117. 
