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Rh overcome by the momentum of the latter, it is manifest that the jacket resistance can never exceed the mean recoil force as due to the gases: thus we have a limit to the possible velocity of the air blast.

As to the first of these, we have the combined area of the air passages, about 5 square inches, or volume per second at 200 ft. per second = $200 × 5/144$ = 7 cubic ft. = 0.54 lbs., equivalent in heat capacity to 0.13 lbs. of water. Now, the heat units to be taken up = 27, hence the air will be increased in temperature $27/0.13$ = 208 Fahrenheit. This result is concordant. Evidently no velocity much less than 200 ft. a second will pass the volume necessary.

Next as to the recoil calculation. If we take the resistance of the jacket as calculable on the basis of skin-friction we may fairly assume the single surface coefficient as .005, and at 200 ft, per second this gives 0.3 lbs. per square foot, or, on a total surface of 6 square feet, the resistance is 1.8 lbs.; this is the mean force which must be applied by the exhaust blast of the powder gases to maintain an air current of 200 ft. per second velocity.

It has been established by experiment that the mean recoil as due to the powder gases in the service cartridge is 0.28 that of the projectile; now the latter already calculated amounts to 2 lbs, per shot per second, or 20 lbs, at 10 shots per second. Thus the mean force due to the momentum of the powder gases is .028×20 = 5.6 lbs. But the air leaving the jacket muzzle retains approximately its mean velocity, and this represents by its momentum a force of 0.54 × 200/32.2 = 3.35 lbs., so the account becomes: