I am pretty damn sure it's wrong. 2GW produces 2 GJ of energy per second. Tungsten has vaporization heat of 774 kJ per mole. This mean that, assuming 100% efficiency, forgetting about melting and solid heat up parts, one could vaporize 2000000/774=2583 moles of tungsten. This gives us 2583×184=475452 grams of tungsten. Tungsten's density is between 17.6 and 19.2 tons per cubic meter depending on its state. This means that 2 GW second long pulse could vaporize, at most, 475452/17600000=0.027 cubic meters. Realistically it would be about an orser of magnitude smaller, perhaps smaller.
I think the calculator has problems with duty cycle on/off ratio moved into infinity.
Not an issue with the duty cycle. Since even at 0.5 it's still penetrating 9km.
But lets do the math!
The formula, as far as I can tell, for the area of a laser beam at distance is:
A = π * (D * tan(θ/2))2
Where D is the distance to the target and θ is the beam divergence angle.
θ is given by:
θ = 1.22 L/RL
Where L is the wavelength of the beam and RL is the radius of the lens.
So:
θ = 1.22 * 0.0000004/1 = 1.22 * 0.0000004 = 0.000000488
A = π * (10,000 * tan(0.000000488/2))2
A = π * (10,000 * tan(0.000000244))2
A = π * (10,000 *2.44e-7)2
A = π * (0.00244)2
A = π *5.9536e-6
A = 0.0000187m^3
Assuming the Tungsten starts at 25C it must first be heated to 3422C. That's a 3,397 degree increase. At 24.27 joules per degree per mole that's 82,445J per mole.
Then it needs to melt which as per it's heat of fusion takes an additional 35,300J per mole.
This is followed by an additional 2,508C climb in temperature to it's boiling point which requires 60,869J per mole.
Finally it boils away for another 774,000J. Bringing the total energy required up to 952,614J/mol.
I specified 2GW for 1.2 seconds which gives a total energy of 2,400,000,000J. Divided by the heating energy gives 2,519 moles.
Tungsten's molar mass is 183.85 grams which multiplied by 2,519 moles give a total mass of 463,187 grams.
Using the 19.25g/cm^3 density, since that's what it would have been at while solid armor, gives a total volume of 24,062cm^3 or 0.024062m^3.
So now we know the volume of Tungsten vaporized and the area of the beam we can calculate the penetration (D) of the beam.
0.024062 = D * 0.0000187
D = 0.024062/0.0000187 = 1,287m = 1.2km.
I'm not quite sure what they did wrong to get such a wildly different answer but they are still right in that it's kilometer range penetration.
It's important to remember even through it's over a kilometer long the hole is only like 2.4 microns in radius.