A NEW EVAPORATION FORMULA 
111 
Table 38— Sample of observation equations from Solution \\, Lake Michigan-Huron 
Unit is 0.001 foot of depth on the lake. 
Date 
Terms of the equation 
+e (ft) 
+ ['( S W°)]«> 
-r, (fti) 
-r, (ft) 
-r, (ft) 
-u (ft) 
-r»(ft) 
-MS.) 
+!-« 
1911 
May 2 
+ 6 (ft) 
+ 7 (ft) 
-16 (ft,) 
-39 (ft) 
-15 (ft) 
+0 (ft) 
+2 (ft) 
+3 (ft) 
- 13 = n 
3 
+ 6 
+ 1 
+ 14 
-16 
-39 
-2 
+ 2 
+2 
+ 161=1), 
4 
+ 10 
- 2 
+ 17 
+ 14 
-16 
-5 
+ 2 
+ 1 
- 8 = »i 
6 
+ 12 
- 6 
+ 12 
+ 17 
+ 14 
-6 


+ 2 = r 4 
6 
+ 13 
- 6 
+ 7 
+ 12 
+ 17 

-5 
+ 2 
+ 23 = r. 
7 
+ 18 
- 4 
+ 11 
+ 7 
+ 12 
+3 
-7 
+5 
- 17 = », 
8 
+20 
- 4 
+ 12 
+ 11 
+ 7 
+3 
-6 
+ 4 
+ 30 = e 7 
9 
+ 13 
-10 
- 2 
+ 12 
+ 11 
+ 2 
-2 
+ 1 
- 11 = 1-, 
10 
+ 14 
- 4 
-11 
- 2 
+ 12 
+ 2 
+3 
-3 
— 6 = v, 
The coefficient of E h el inn — J, can not be verified from previous sample 
computations. Its method of computation should be clear, however, from observ- 
ing the method of computation of w shown in the last column but one in 
Table 11. If from such a column of values, one subtracts 2.4 and multiplies the 
difference by e, the result will be the coefficient of E t , which now is either + or — 
because the w used in (28) refers to any wind velocity. 
FINAL NORMAL EQUATIONS FOR SOLUTION V x , LAKE MICHIGAN-HURON 
An observation equation like those shown in Table 38 was written for each 
day of the 28 months July to October 1909 and May to October of 1910, 1911, 1912, 
and 1913. In some cases two or more days were combined into one equation, and 
in others certain equations were rejected, in accordance with the rules adopted for 
combinations and rejections already stated on pages 51 and 73. In all, there 
were 845 days of observation representing 787 observation equations used in 
Solution Vu 
The final normal equations formed from the 787 observation equations like 
and including those in Table 38 are shown in Table 39. 
Table 39 — Final normal equations for Solution Vi, Lake Michigan-Huron 
1031ft: -94169 = 0) 
4077ft. - 8213 = 
836ft +37093 = 
131ft. -19665 = 01 
+ 244008ft -44512ft + 28088ft + 10385ft + 4661ft + 830ft - 
- 44512ft +71090ft - 7721ft, - 21652ft, - 10603ft3 + 1822ft 4 - 
+ 28088ft - 7721ft +291773ft, + 151238ft, - 9144ft, - 7389ft 4 + 
+ 10385ft -21652ft +151238fti +296816ft, +165826ft, - 5231ft 4 - 
+ 4661ft -10603ft — 9144ft t + 165826ft, +291696ft, +12663ft 4 - 
+ 830ft + 1822ft - 7389fti - 6231ft, + 12663ft, + 8622ft 4 - 
- 2020ft- 845ft + 4333fti - 702ft,- 8663ft,- 837ft 4 +19458ft 6 + 97ft, + 4294 = 
- 1031ft + 4077ft - 836fti - 131ft, + 3402ft, + 788ft 4 + 97ft 6 +33162ft 6 + 1106 = 
2020ft 6 - 
845ft 6 + 
4333ft 5 - 
702ft 6 - 
8663ft 6 + 3402ft 6 -33669 = 
837ft 6 + 788ft« - 2686 = 
(31) 
The solution of the normal equations, (31), gives the following values for the 
unknowns: 
#, = +0.493 ±0.035 
£, = +0.466 ±0.065 
#,= - 0.291 ±0.039 
#, = + 0.253 ±0.049 
R,= - 0.035 ±0.044 
# 4 = +0.121 ±0.188 
R>=- 0.086 ±0.1 17 
fl,= -0.081 ±0.089 
(32) 
