Example 2.

We have 16 gauge steel sheet, with 0.066" holes on 0.125' staggered centers. Then:

b = 0.125ff
d = 0.066ff
t = 0.0625'f

A = (0.125)³ x 0.87
   = 13.6 x 10^-3 sq in
n = 1/A = 73.6 holes/sq in

a = b - d
  = 0.125 -0.066
  = 0.059"

P = [7r(0.066)²/4 x 13.6 x 10^-3] x 100
  = 26.16%

TI = [73.6 x (0.066)²/0.0625 x (0.059)²]
  = 1474;

or:

TI = [0.04 x 25.16/Tr x (0.0625) x (0.059)²]
   = 1472.

Again, the agreement between the two values of TI is very good. But notice that the 10-kHz-attenuation has increased to 2.9 dB, much more than the attenuation of Example 1, despite the fact that the open area for this example is 53% greater than in the earlier case!