Example 6:

Suppose we have the same transformer problem, but have only 3" of available depth and a sheet thickness of 1/4":

fR = 120 Hz;
h = 3";
t = 0.250"
Try d = 0.500"; then:

e = 0.250 + 0.8 x 0.500
   = 0.65" and we find;

p = 0.61%
A = 31968 sq in/hole, and
n = l/A = 0.03 holes/sq in.

The hole spacing is;
b = (32.19/cos 30�)1/2 = 6.1".

Example 7:

If we repeat Example 6 with 1" holes, we have: e = 0.250 = 0.8 x 1 = 1.05" from which we find: p = 1.0%;

a = 78.54 sq in/hole;
n = 0.013 holes/sq in; and
b = 9.5".

All these variations of perforation pattern in Examples 5, 6 and 7 lead to the same resonance frequency of 120 Hz, to match the dominant frequency of the transformer noise.