NonCredibleDefense

Oh I'm going to do the math on this:

assuming the warship is being fired from a naval gun of truly massive proportions based on the AGS Mark 5, firing a saboted warship based on the LRLAP ( since they were made for the Zumwalt class and cost a million a round, making them the obvious best choice.)

And why not, let's also make the warship a Zumwaltz class, for flavor synergy.

The LRLAP weighd 225lbs and had an effective range of 150km, so thats ~ 100kg at 150km, and it would travel at 825 mps, let's say ours needs to go 800, were in no rush.

Projectile Mass: A Zumwalt weighs, rounded down, 15,000 long tons, which is 15,240,704 kg Let's say 15,000,000 kilograms. They emptied it, unloaded the ammunition, decommissioned the AGS, so it weighs a little less. Zumwalts are 610 feet long, let's make that 1000 total to account for an aerodynamic sabot and charge, so 300m. The bore length was 378" and the shell 88", so that gives us a ratio of 4.29:1

Our barrel is 1287m long, let's just say 1200m.

Saboted Zumwalt: 15,000,000Kg

Barrel length of gun: 1200m

Distance: We're gonna park this gun on Gotland, middle of lake NATO. Well need some space, and that's right in the middle. It's 1,168.31 km as the crow flies, so we'll round up to 1200 km.

So, our 1.2Km long nuclear cannon would send our saboted 15,000,000 ton Zumwalt 1200km @ 800 mps, requiring 1200 Kilotons of energy, rounded up. The boat would be in flight for about 25 minutes.

My math is quite accurate, please do not double check it.

at least 3

For these purposes I am, of course, assuming that air resistance doesn't exist. Which would probably increase this answer by a lot.

Narva bay to Moscow = 713 km Fully loaded Arleigh Burke destroyer displacement = 8,432,800 kg

v = launch velocity d = horizontal distance = 713000 m θ = angle we're launching the ship at = 45 degrees g = acceleration (from gravity) = 9.81 ms^-1

Range of a projectile equation:

d = (v^2 sin(2θ)) / g

Rearrange to find v:

dg = v^2 sin(2θ) dg / sin(2θ) = v^2 v = (dg / sin(2θ))^0.5

Plug our numbers in:

v = (713000*9.81 / sin(2*45))^0.5 v = (6994530 / 1)^0.5 v = 2644.72 ms^-1

So we need to launch at 2,645 metres per second (9,522 kph, 5,917 mph). To get the energy, we use the kinetic energy equation:

e = 0.5 m v^2 e = 0.5*8432800*2644.72^2 e = 29,491,794,808,885.76 joules e = 29.5 terajoules

For comparison, the nuclear bomb dropped on Hiroshima exploded with about 60 terajoules of energy. So once you account for air resistance you're probably looking at a nuke of energy.

So let’s you want to yeet this over 1000km from a NATO country to Moscow.

You’d need at least 3000 m/s of velocity to do this (a ton more since this without air resistance).

A fletcher class destroyer is around 2000 tons.

So you’d need more than 8 terajoules or 2 kilotons of TNT.

Probably depends entirely on the aerodynamics involved, if we're assuming it's approaching as an aircraft. This is kind of an intermediate range, and it has shitty ballistics, so the energy to just get it off the ground will be dwarfed.

Nuclear trebuchet