Figure 1-25. Free Bolt Velocity lor 0.160 Second.

Integrating under the curve of fig. 1 23 from tr=r 0.0009 to t=0.0091 gives a total area of 0.1310 (lb.-sec:. '). Using equation 1 -A to evaluate Mr gives:

Therefore the required weight is:

Accordingly, the weight of the recoiling parts will be taken as 26 pounds for the remainder of the analvsis.

2. Determination of driving spring design data

The maximum velocity of free recoil is found in the same way as for plain blowback by using the equation expressing the maximum free recoil momentum to solve for the velocity.

v _MpVp -| 4700 M0_ WpVt,+47Q0 W, rf 2 Mr " 2 Wr

For the 26-pound bolt and the conditions of the example:

v _.29X2750+ .070X 470U_o, / ft. \ Vr<'~ 2X2G WJ

Again assuming that this velocity is imparted instantaneously to the bolt, the initial bolt energy is given by equation 1-6:

0 0

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