Nuclear weapons calculator, weapon shape number?

[adam_grif]

So I was messing around with the nuclear effects calculator linked from Atomic Rockets. It's working fine for spherical bursts, but I wanted to crunch numbers for shaped charges. The "weapon shape" value is apparently the ticket, but I don't know what to put in there.

After searching, I've discovered that the 12.566 value is the approximation of 4pi, which is part of the formula for determining intensity at a distance. However, since my math-fu is exceedingly weak, I have no clue how to determine what it would be for a nuclear shaped charge in weaponized form (so the angle that most of the shit is fired out of is less than the 22.5 for the Orion drive charges).

So:

- What number should I put in?

and

- What is a reasonable angle for a weaponized shaped charge?

[Kuroneko]

If charge fires in a cone with opening half-angle α, then the corresponding solid angle is Ω = 2π(1-cos α) steradians. This uses the angle between the central axis of the cone and the edge, rather than the full vertex angle between the edges of the cone. So, for example, α = 90° corresponds to a hemisphere (Ω = 2π).

[adam_grif]

Pretty sure I'm doing something wrong.

I put in alpha of 5 (for a five degree cone), and ended up with ~0.0239. Might just be that windows calculator is shite.

[Kuroneko]

I put in alpha of 5 (for a five degree cone), and ended up with ~0.0239. Might just be that windows calculator is shite.

That's right. Note that an opening half-angle of 5° means that the vertex angle in a cross-section of the cone is actually 10°. That's a pretty well-focused beam.

[adam_grif]

I have no clue what a real shaped charge is capable of achieving. Ideally you'd get it down to between 1 and 2.5 degrees if possible, but I get the feeling that there will be some sort of hard limit around the 10 degree mark or above.

Can you explain what an "opening half angle" is and how that differs from a vertex angle?

[Kuroneko]

Code: Select all

A   C   B If α is the angle between
.\  |  /  the central axis and the edge
  \ |α/   of the cone, then the full
   \|/    opening angle is 2α.
    V

By "vertex angle", I mean AVB, and by "opening half-angle" I mean CVB.

[adam_grif]

So for the initial project orion example with 22.5 degrees, is that a 22.5 vertex or opening half angle? I'll have to adjust my calculations accordingly.

[Singular Intellect]

Forgive my ignorance, but doesn't that sheer destructive power of a nuclear yield put a damper on any effort to shape the direction of energy release?

[adam_grif]

Forgive my ignorance, but doesn't that sheer destructive power of a nuclear yield put a damper on any effort to shape the direction of energy release?

No.


The following are taken from the Atomic Rocket website:

The nuclear device is encased in an shell of x-ray opaque material (uranium) with a hole in the top. This forces the x-rays to to exit from the hole. Where they run full tilt into a large mass of beryllium oxide. The beryllium transforms the nuclear fury of x-rays into a nuclear fury of heat. Perched on top of the beryllium is the propellant: a thick plate of tungsten. The blast of heat turns the tungsten plate into a star-core hot spindle shaped plume of tungsten plasma. The x-ray opaque material and the beryllium oxide also vaporize a few microseconds later, as the tungsten plasma jet hits the Orion drive pusher plate. With the reference design of nuclear pulse unit, the plume is confined to a cone of about 22.5 degrees. About 85% of the nuclear devices's energy is directed into the desired direction.

About this time the representatatives of the military (who were funding this project) noticed that if you could make the plume a little faster and with a narrower cone, it would no longer be a propulsion system component. It would be a directed energy weapon. Thus was born project Casaba-Howitzer.