wf/nA = 48
Wf = 7.68 oz
(5) For Rf less than 20 feet,
Vs = Vo a 6,258 ft/sec.
(6) Using Equation (28) to determine the number of fragments
having a weight greater than 1.5 oz:
Nf = NT(1 - CL)
From Figure 22, for given casing geometry:
(B2NT)/W, = 0.275
N T = [0.275(220.7)(16)1/0.09 = 10,790
Determine the design Confidence Level corresponding to
Wf = 1.5 oz.
From Figure 21, CL = 0.955
Therefore, Nf = 10,790(1 - 0.955)
= 486 fragments.
Presented area of fragment, in2
Vent area, ft2
Fragment distribution constant
Charge depth, ft
Diameter of hole in roof, ft
Average inside diameter of casing, in
- Gurney's energy constant
Factor to account for effect of casing on effective charge weight