Ah cheers. The link works fine now.

This looks like an interesting development indeed, especially given the details supplied. It would be interesting to know the exact specifics of power consumed in order to vaporise the copper from the warhead.

It is stated that, "despite the high charge, the electrical load on the battery is no more than that caused by starting the engine on a cold morning." I would have thought that it would take considerable power to vaporise copper given the following:

Latent Heat - The latent heat of a substance is the amount of heat required to change the state of unit mass of a substance from solid to liquid or from liquid to gas/vapor without any change in temperature. When solid changes to liquid, the heat provided is called the latent heat of fusion (this heat will be given out when liquid changes to solid). When liquid changes to gas/vapor then heat provided is called the latent heat of vaporisation (this heat is given out when gas changes to liquid). The quantity of heat required to change the state of m g of a substance having latent heat L is Q= mL. (heat = mass/latent heat of vaporisation)

The heat of vaporisation of copper is 300.3 kJ/mol. (Check out

http://environmentalchemistry.com for loads of info).

The atomic mass of copper is 63.540 g/mol. Therefore the heat energy required in order to vaporise 1g of copper from its liquid state is:

Q = 63.540 x 300.3

gives 19,081 kJ

In 1Kg we have 1000g, therefore in order to vaporise 1Kg of molten copper we need to use 19,081,062 kJ of energy or 19081062 x 1000 gives 19,081,062,000 Joules of energy required. (19 billion Joules of energy).

This approximates to (using an online energy converter at

http://cactus2000.vi...ice/massenr.htm) to 5300 kW or 5300000 Watts.

Given the equations P = I2R (I squared R) and P = VI

P = Power (Watts)

I = Current (Amps)

R = Resistance (Ohms)

V = Voltage (Volts)

In order to achieve the remarkably high figure of 5300000 Watts it would be preferable to achieve this by maximising the voltage used in the energy discharge. By minimising the amount of current flow you effectively bypass the ill affects imposed by P = I2R, where resistance impedes power given an increase in current.

The supply voltage curve on a starter motor might typically range from (12V..0A - 6V..1500A). This means that a maximum current of 1500A or a maximum voltage of 12V could be supplied by the starter motor alone. The necessary voltage levels to provide the necessary power output could be:

Given P = VI

5300000 = VI

V = 5300000/I

Say that we can use the starter motor top output of 1500A. Then,

V = 5300000/1500

gives 3533 Volts. Hoever, given the high current I certainly wouldn't want to touch the interior. Therefore it would be a good idea to raise the voltage as high as posible and minimise the current.

Anyhow, it all looks possible using just the energy consumption stipulated (starter motor on a cold morning), but I wonder what the frequency of attack was on the vehicle? I wonder how quicky the system re-charges after an attack. After all there must be a re-charging time for the capacitor.

I wonder if part of the specification requires the operator to stand on the tank turret whilst shouting, "please cease fire whilst we re-charge. We should be done in a minute or two! Thanks for your co-operation!"

Anyhow, looks interesting. It would be a bit harder to do if the copper wasn't already molten though.

To be honest, I don't think it would quite work out either as personal body armour because as well as frying speeding bullets, it would probably fry you first!

Anyhow, sorry for all the math, which is probably wrong anyway (you are welcome to check it), but I just wondered if this was one of those flights of fancy or a real invention. Looks like it could be a reality!

:su