lb1000br,
I have been thinking about that too, and I agree. I think the shock front is acting as a bug shield and deflecting most glancing blows. The drop would have to either be a direct hit or be large enough to penetrate or deflect the shock wave sufficiently to apply a force on the bullet. Or maybe the drop would vaporize as the wave passes through it and you have an angry little puff of expanding water vapor along side the jacket. Those are just ideas -- I don't know enough about gas dynamics even to make an educated guess.
If you say water is 62.4 lbs per cubic foot, 0.008 grams would imply a drop diamater of 0.038" or a little less than 1 mm. Sounds reasonable.
If all momentum is transferred to the bullet there would be two possible effects: lateral velocity and yawing/destabilization. I'll not touch the latter as it is way out of my expertise, but lateral velocity should be vanishingly small with a single hit, something like a millionth of a meter per second. If you do energy transfer methods with phase change and the like, it should come out roughly the same order of magnitude.
So a single average drop is not capable of deflecting a bullet significantly, or at least that's the way it looks to me.
If you take the same 0.038" diameter drop and figure an inch of rain per hour, that is 1.2 such drops per square inch per second. If you say that the bullet's effective cross section (due to shock wave shielding) is 0.1", then the "impact circle" is 0.138" diameter. 1.2 of those circles has an area of 0.015 square inches. So the probability of a hit is 0.015 square inches per square inch or 1.5% -- in 1" of rain per hour.
If you shoot 200 bullets in that rain, 3 of them will hit a single drop on their way to a distant target. The effect of those hits is probably not measurable.
So that's my scrutiny, for what it's worth.
Scott
I have been thinking about that too, and I agree. I think the shock front is acting as a bug shield and deflecting most glancing blows. The drop would have to either be a direct hit or be large enough to penetrate or deflect the shock wave sufficiently to apply a force on the bullet. Or maybe the drop would vaporize as the wave passes through it and you have an angry little puff of expanding water vapor along side the jacket. Those are just ideas -- I don't know enough about gas dynamics even to make an educated guess.
If you say water is 62.4 lbs per cubic foot, 0.008 grams would imply a drop diamater of 0.038" or a little less than 1 mm. Sounds reasonable.
If all momentum is transferred to the bullet there would be two possible effects: lateral velocity and yawing/destabilization. I'll not touch the latter as it is way out of my expertise, but lateral velocity should be vanishingly small with a single hit, something like a millionth of a meter per second. If you do energy transfer methods with phase change and the like, it should come out roughly the same order of magnitude.
So a single average drop is not capable of deflecting a bullet significantly, or at least that's the way it looks to me.
If you take the same 0.038" diameter drop and figure an inch of rain per hour, that is 1.2 such drops per square inch per second. If you say that the bullet's effective cross section (due to shock wave shielding) is 0.1", then the "impact circle" is 0.138" diameter. 1.2 of those circles has an area of 0.015 square inches. So the probability of a hit is 0.015 square inches per square inch or 1.5% -- in 1" of rain per hour.
If you shoot 200 bullets in that rain, 3 of them will hit a single drop on their way to a distant target. The effect of those hits is probably not measurable.
So that's my scrutiny, for what it's worth.
Scott