Once again, no one is arguing the presence of a pressure wave, but what it's effects are. Certainly 500 or 1000 PSI would be harmful; Thing is, as the pressure wave travels in a 3-dimensional parabolic shape, the pressure attenuates at an exponential rate. Even excluding the shock-absorbing capabilities of soft tissue and presuming pure liquid, that dissipation is still geometric. Double the volume of the wave, halve the pressure. Double it again from the
volume (not diameter or circumference) of the last increase, halve the pressure again (1/2, 1/4, 1/8, etc.) and so on.
Someone with better math skills can probably articulate this more effectively than I can, but this is concept rather than exact, so we'll use simple linear math for it. Now remember that spherical volume is calculated at a factor of 4/3*pie*radius cubed, (roughly 4.19*radius cubed) so a semisphere (hemishpere) is half of that (4.19*Radius cubed/2).
For the purpose of these illustrations with my rudimentary geometry skills 15 years since I studied, we'll use the calculation of a true hemisphere rather than a half ellipsoid with cylindrical base or other shapes that make volume calculations more difficult.
So, if calculating using a .45 cal RN {(4.19*.225cubed)/2}, we have a pressure wave point of origin (POO) resulting from a displacement of .05CI. Suppose the pressure at POO is 1,000 PSI. Increase the area of our pressure wave by a factor of 2 (a .9" semisphere diameter) and now you have a
volume of .19CI. Assuming a linear dissipation, we have to take the 1000 PSI we began with and divide it by ~4 (.19/.05). So only .225" from the POO, our pressure is reduced to 250 PSI. By the time you get an inch away from the POO (1.225" radius semisphere with a volume of 3.85CI), pressure is down to ~13 PSI.
As I said, my math is not as good as some of our members who could accurately calculate with a correct elliptiod (and superscript that I can't figure out how to get). The point remains.
Some variables to consider that require better math skills and a lot more specific details than I (we?) have at hand:
Causing pressure to be greater than the simple equation:
-The math I did is very simple and assumes a static pressure; The velocity and magnuitude of the pressure wave will affect the pressure as it travels outward, resulting in a pressure differential on either side of the wave with greater pressure on the outside.
-The pressure will not be equal across the arc of the wave; it will be greatest at the front.
Causing the
loss of pressure to be more dramatic than simple math indicates:
-As with the pressure, the magnitude and velocity of the wave will decrease as it travels; Drop a boulder of, say, 50 pounds from a height of 5 feet into a 50,000 gallon pond, you get a big splash and fast moving wave right next to it, but very small and slow wave at the shoreline.
-Pressure will continue to be generated as the bullet travels deeper, but at a decreasing magnitude as velocity is shed; Countering this, some of that pressure wave will dissipate
behind the hemisphere as the bullet penetrates deeper, causing the dissipation to become more spherical, and thus increasing the loss.
-Unlike shooting into a bucket with a homogenous liquid medium, the human body is a conglomeration of different tissues that are separated by membranes and other fascia, as well as just space. They're also all contained within a flexible, expandable unit and able to compress/expand and absorb impact themselves; Every part of the human body is designed to absorb and dissipate impact and pressure to prevent injury (The exception being the brain, which is contained rigidly and does not tolerate much pressure). Now, Imagine the hydraulic shock of shooting a water ballon filled with ballistic gelatin. Then imagine what the effect will be on a gelatin-filled water balloon placed
next to the one that is impacted, with a couple sheets of clingwrap between them and a 1/8" layer of shaving cream between those sheets. Now imagine the pressure that is generated in a thin tube made of, say, latex, filled with water when it is wrapped around the balloon that
isn't struck. Can you see where I'm going with this?