NOMOGRAPHIC SOLUTIONS FOR THE STANDARD CASE 57 
% g 27 
Y ee 26 
25 
‘a Xx # 
x ae 
: fou b=0 23 
8 2 
~ 2! 
“ 20 
ef 19 
18 
15 
a7 
-. 
cae 16 
16 aco 
17 pe 14 
EXAMPLE SHOWN BY DASHED LINE < 
B = 0.112 3 a 
2 Y = 12.2 2 
I OL, = 0.465, &,= 0.910 cd - 
20 10 
2 ae? .09 
22 08 
23 3 .07 
cS, 
by .06 
25 
26 05 
27 J 
28 .0¢ 
2 
30 .03 
31 
32 
35 yo 02 
{ 01 
a rs 
20 0 
a 6% 
TVicure 2 
This equation, unfortunately, has too many variables 
for nomographic solution in a single step. We first 
define a quantity C, such that 
C= (3.281)? (914) (150) (n — b) (— 2 + 8a) 
; hy f mc a? . 
Obtain the simple product hif,,,, which is a charac- 
teristic of the set. Then use the nomogram of Figure 
5 to obtain the values of C for the selected ranges of 
k and a. Then we can determine H from the nomo- 
gram of Figure 6, for each value of S and C. Finally, 
from a nomogram (not shown here), representing 
the equation 
: ho = H — (1 — a)hy 
we determine hz. Actually, for much of the range, 
a~landh,~d. 
