Someone shoots a tungsten spike of length 0.5cm and width 10mm (0.1cm) at
1000 m/s at a target with a personal shield. The target is 1.8m tall and thus
the shield is spherical with a diameter of 1.8m. The spike hits the shield
directly in the back and the shield dissipates the kinetic energy to heat
around the aft hemisphere. Assume this happens on the moon, so there's no
air resistance, and that the shield has the heat capacity of copper, that
its thickness is 0.1 cm, and to provide for the diffusion of gases, it is only
99.2% solid.

How much does the hind/aft shield heat up?

------- 
	Tungsten's density: 19250 kg/m^3
	Radius of spike: 0.1/2 = 0.05 cm = 0.0005m
	Height of spike: 0.5 cm          = 0.005m
	
	Volume of the spike (cone): 1/3 pi * r * r * h
		= 1/3 * pi * 0.0005^2 * 0.005
		= 6.5 * 10^-12 m^3
	
	Total surface area of the shield: 4 * pi * r^2 * solidity
		= 4 * pi * (1.8/2)^2 * 0.992
		= 4 * pi * 0.9^2 * 0.992
		= 10 m^2
	
	Total surface area of where heat is dissipated: 10 / 2 = 5 m^2

	Specific heat capacity of copper at STP: 385 J/Kg-K

	----

	Density of the spike: 19250 * 6.5 * 10^-12
		= 1.25 * 10^-7 kg
		= 0.0004 g
	
	Speed as fraction of c: 1000 / 3e9 m/s = 0.3 * 10^-7

	(So it's not relativistic, and we can use the Newtonian equation)

	Kinetic energy: 0.5m*v^2
		= 0.5 * 1.25 * 10^-7 * 1000^2 
		= 0.0625 J (not much)
	
	--

	Thickness of the shield: 0.1cm = 0.001 m
	Density of copper: 8920 kg/m^3
	Total volume of the hemisphere: 5m^2 * 0.001 = 0.005m^3
	Imaginary mass: 8920 * 0.005 = 44.6 kg
	
	Heat capacity of the aft shield: 385 * 44.6 J/K
		= 17171 J/K
	
	--

	Heat increase from impact: 0.0625 / 17171 = 0.36 * 10^-9 K

The impact of the spike results in an average heating of the shield by 0.36 nK.
This is far too little to be of any concern.

Say we have a mass driver (an automated turret) that shoots an object of any 
size at 0.005c (1.15 * 10^7 m/s). How large does the tungsten spike have to be
in order to heat the shield by 30 K?

KE = 0.5 m*v^2 
KE = 17171 * 30 J
   = 515130 J
v^2 = 1.3225 * 10^14

0.5 * x * 1.3225 * 10^14 = 515130
6.6*10^13 * x = 515130

m = 7.79 * 10^-9 kg

This is less than the spike above. So velocity is good.

-

What is the usual velocity of a single round of the H&K P-90?
http://www.fnmfg.com/products/p90/p90.htm says
	900 rounds per minute
	50 rounds per magazine
	2346 feet per second velocity (715 m/s)
	31 grains projectile weight (0.002 kg)

Each projectile then delivers 0.5 * 0.002 * 715^2 = 511.225 J, and a 50 round
burst will

	511.225 * 50 = 25561 J
	25561 / 17171 = 1.49
	
heat the shield by 1.49 Kelvin.	So if their shields were anything like my
idealized model, it's no wonder guns had no effect[1]. Only specialized weapons
would be able to punch through the shield (assume it overloads after being
heated 30 K), and these weapons probably would have to be fueled by some sort
of nuclear device (decay or fission battery) to get the required energy. It may
be even better to punch a hole with a laser first, then fire a spike at
traditional velocities afterwards.

The P-90 would have an abysmal adjusted hit rate - only one in 1000 shots 
passes the shield. {Still, something like Metal Storm may actually be 
damaging.}

[1] What is strange is that swords do have an effect. Unless they use shields
from Dune, that is -- could make sense that the shield has some sort of
velocity discriminator so the user can still climb ladders, carry objects, etc.