DSK wrote in message ...

To put this in perspective, this is roughly equivalent to stopping ten
rounds from a .50 cal machine gun. Destructive, nyet?

Your analogy is extremely misleading, since the area to which this
force would be applied is hundreds of times smaller. This leads to
hull-pressures that are hundreds of times larger

That's not quite right, the force due to deceleration isn't linear.
Comparing a braking distance of 2' to 4', the energy is less than
half... about a third would be closer...

Thats just plain wrong, the force (which is what Jordan was talking
about) due to decelaration IS linear with breaking distance (given a
certain initial velocity). The formula for force due to a constant
deceleration (in terms of initial velocity and 'breaking distance')
is:
F = m*v^2/(2d)