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Nova Cannon Firepower Calculation
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Old 21 Jan 2008, 02:13   #1 (permalink)
Shas'O
 
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Default Nova Cannon Firepower Calculation

I was bored, and inspired by FT's Exterminatus thread.

The result is truly terrifying.

Please comment on any errors or oversights you see.

[hr]

I started by scaling from this image.

The Retribution-Class Battleship is generally accepted to be 7.5km long. Minus the Nova Cannon, the ship measures 668 pixels, giving a conversion factor of 0.011227544 km/pixel.

The barrel of the Nova Cannon measures, on the low side, 6 pixels high, or ~67.37m. Conservatively, I went with 60m.

The main 16" guns on the Iowa-Class Battleship have a 50:1 shell length:caliber measure; I chose 20:1, again quite conservative but giving us 1,200m long shells.

Assuming cylindrical shape, that makes gives the shell a volume of1.357x10[sup]7[/sup] cubic meters.

Using the density of iron (7870 kg/m[sup]3[/sup]), we get 1.0679x10[sup]11[/sup]kg for the shell's mass.

Assuming that "close to the speed of light" from the Battleship Gothic rulebook means 0.75c (this is likely conservative), and plugging it into Mike Wong's Relativity Calculator, we get a Newtonian kinetic energy for the shell of 2.698x10[sup]25[/sup] J, or a Relativistic KE of 2.709x10[sup]25[/sup] joules.

Using the Newtonian measure and dividing by the energy density of TNT (4.2x10[sup]6[/sup] J/kg) we get 6.45x10[sup]18[/sup] kg.

Divide this by 1000kg and we get 6.45x10[sup]15[/sup] tonnes, or 6.45 petatonnes.

Let that sink in for a second.

To give an idea of the implications of petaton-range firepower, the readout for an equivalent nuclear warhead from the SDN nuclear weapons calculator (after dividing by 1x10[sup]6[/sup] to get the MT equivalent) gives us the following:

[table]
[tr]
[td]Thermal radiation radius (3rd degree burns):
Air blast radius (widespread destruction):
Air blast radius (near-total fatalities):
Ionizing radiation radius (500 rem):
Fireball duration:
Fireball radius (minimum):
Fireball radius (airburst):
Fireball radius (ground-contact airburst):[/td]
[td]
122677.3 kilometres
12469.1 kilometres
4724 kilometres
228.1 kilometres
116819.2 seconds
3627 kilometres
4433.1 kilometres
5843.6 kilometres
[/td][/tr][/table]

Remember that these are conservative equivalent effects due to kinetic energy alone, disregarding any properties of the Nova Cannon shell's warhead.

Also note that those numbers are based on an equivalent nuclear explosion, rather than the actual effects of the kinetic impact which would not have 100% efficiency.

Regardless, 10[sup]15[/sup] is very large number :sadnshocked:.
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Old 21 Jan 2008, 02:19   #2 (permalink)
Shas'El
 
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Default Re: Nova Cannon Firepower Calculation

You have time to kill your brain?


I envy you ;D :P
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Old 21 Jan 2008, 02:23   #3 (permalink)
Shas'El
 
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Default Re: Nova Cannon Firepower Calculation

So a Nova cannon could give everyone on a flat earth 3rd degree burns?
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brave kid, i don't know if i could do that. whichever girl gets you for life is the luckiest in the world.
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Old 21 Jan 2008, 02:28   #4 (permalink)
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Default Re: Nova Cannon Firepower Calculation

Quote:
Originally Posted by Scottish Gecko is back!
You have time to kill your brain?

I envy you ;D :P
The way of a science student; occasional nightmare all-nighters of work/study and a huge mess of "do nothing" time :funny:.

Quote:
Originally Posted by T pok
So a Nova cannon could give everyone on a flat earth 3rd degree burns?
Three times over. It'd probably happen on a normal round Earth too, given gravity/the atmosphere restraining the shockwave (I think).
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Old 21 Jan 2008, 04:40   #5 (permalink)
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Default Re: Nova Cannon Firepower Calculation

Of course, that assumes that there is a way to convert 100% of the projectile's kinetic energy into explosive force (here being redirected kinetic energy) and heat. That assumption ends up being the big problem with using the fireball calculator or with converting kinetic energy to an equivalent mass of conventional explosives. If you fired a Nova Cannon at a ship, for example, the projectile might just punch through one side and out the other. Or that a large part of the kinetic energy would go into crack formation rather than material acceleration or heat.

It is pretty staggering, though. I wonder if GW really thought through all the implications of weapons firing masses at close to light speed?
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Old 21 Jan 2008, 04:53   #6 (permalink)
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Default Re: Nova Cannon Firepower Calculation

Question: did you assume a solid iron shell?

Care to give us a liberal amount?
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Old 21 Jan 2008, 04:55   #7 (permalink)
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Default Re: Nova Cannon Firepower Calculation

Quote:
Of course, that assumes that there is a way to convert 100% of the projectile's kinetic energy into explosive force (here being redirected kinetic energy) and heat.
I realize the fact that 100% efficiency isn't going to happen, but this is a calc derived from, at it's heart, one line in a rulebook and pixel scaling in MS Paint - even a ballpark is pretty good .

Quote:
That assumption ends up being the big problem with using the fireball calculator or with converting kinetic energy to an equivalent mass of conventional explosives.
I really just ran it through the nuke calculator to give a rough idea of what an insanely large number 10e15 is... most people probably haven't ever heard of "petatons" before.

Quote:
If you fired a Nova Cannon at a ship, for example, the projectile might just punch through one side and out the other.
One freaky implication of this calc is the durability of Imperial starships skyrocketing. There's a fluff piece of a Chaos cruiser taking a direct hit from a Nova Cannon and not being destroyed, only crippled.

Quote:
Or that a large part of the kinetic energy would go into crack formation rather than material acceleration or heat.
I think I understand what you're saying, but any chance you could dumb it down to Bachelor's level?

Quote:
It is pretty staggering, though. I wonder if GW really thought through all the implications of weapons firing masses at close to light speed?
This is a company who makes warheads our of depleted deuterium.

Probably not :P.

Quote:
Originally Posted by The Immortal Black Mage
Question: did you assume a solid iron shell?
Yes.

Quote:
Care to give us a liberal amount?
Wouldn't know... the shell could be smaller, but it could also be larger. It could travel at 0.9c rather than 0.75c. It's sketchy, moreso even than a "lower limit" calculation. Hazarding a guess... the number could potentially be double or more.
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Old 21 Jan 2008, 05:18   #8 (permalink)
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Default Re: Nova Cannon Firepower Calculation

Sweet. I'd love to have one of those...
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Old 21 Jan 2008, 05:33   #9 (permalink)
Ethereal
 
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Default Re: Nova Cannon Firepower Calculation

Well, I am certainly not an expert on this type of physics, so take this with a little salt.

But if you fire a rifle at a brick wall, you do not just get an explosion at the surface. The kinetic energy of the bullet goes into a lot of different processes. Some of it goes to heat due to friction. Some of it goes into kinetic energy of air waves, producing sound. Some of it goes into waves propagating through the brick wall, which will be gradually dissipated as heat. A lot of it goes into breaking chemical and structural bonds in the bricks and cement, forming thousands of tiny fractures and a few large cracks. And then more kinetic energy goes into these fragments, many of them flying away. And due to the elastic properties of the wall, the bullet may actually keep a lot of its kinetic energy, being flung off in some other direction. Total energy for the system is conserved, but there are a lot of places for it to go.

So explosives are sort of an inherently different problem. With an explosive, you have an energy release across a broad area in the form of heat and kinetic energy of large fragments and air molecules. This is what that equation calculates in terms of distance. The energy in a bomb comes from stored chemical energy in the case of conventional explosives or more elemental stored energy in the case of nuclear weapons.

So the assumption is that you can make the "energy" in a moving particle behave like the energy released by an explosion. This is an area I know almost nothing about. But I know that collision physics makes this a much harder problem than a simple conversion of energy from one state to another. I think the ballpark calculation is fine, but I have no idea how to go from kinetic energy to its effects on the target. Most of the physics we get deals with statics (which is why we can just use conservation of energy), but this problem really falls into the realm of dynamics.
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Old 21 Jan 2008, 14:38   #10 (permalink)
Shas'O
 
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Default Re: Nova Cannon Firepower Calculation

Ah, makes good sense... I recall covering energy transformation now.
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