[Shrike]

I decided to do a 'real' FS calc, based on visuals.  I chose the scene involving the Lucifer blasting the Orion.  It's a fairly crude calc, and it's not all neatly laid out.  But anyhow...

Assumption:  Orion's armor is homogenous elemental molybdenum.
Atomic weight 95.94
Density @ 293 K 10.2 g/cm3
Atomic volume 9.4 cm3/mol
melting point 2890.2 K
boiling point 4923 K
Heat of fusion 32.0 kJ/mol
Heat of vaporization 598.0 kJ/mol
Specific heat 0.25 J/gK

The Calcs:
1 m3 = 1e6 cm3
1e6 cm3/9.4 cm3/mol = 106,383 mol
1 m3 = 10.2e6 g/m3
10.2e6 g/m3 x 4923 k = 50,215e6 gk/m3
50,215e6 gk x 0.25 j/gk = 12,554e6 j/m3
106,383 mol x (32 + 598)kj/m = 67,021e6 j/m3
Total = 7.9575e10 j/m3

Beam radius:  ~40m
Hull thickness (per side):  ~20m
Beam footprint: 5024m2
Volume of hull affected: 200960m3
1.6e16
1.6e16/4.18e15 = 3.83 MT
Note that this assumes a beam strike perpendicular to the hull, the actual beam hits at 45' or so, so it's low-end.

[an0n]

So what? The Lucifers beams were 3.83 MegaTons? If that's right then it kinda ****s when compared to cyclop and helios power.

[Shrike]

Well, that's a low-end estimate... a loose value of about 1 MT/s.

Now, to make that jive with the low gigaton levels of Harbingers, we have to postulate that FS armor is significantly stronger than normal molybdenum.  The term 'Collapsed Molybdenum' should give us a good hint.  If that refers to collapsed electron shells, it could theoretically take extreme amounts of energy before melting/vaporising.

[an0n]

Uh.....Wouldn't something with a collapsed electron shell just be a bunch of neutrons that could potentially emit high levels of beta radiation if heated?

[Sandwich]

Shrike, you, sir, are bored.

[an0n]

Most bored people will turn to rendering or modelling, but not Shrike. He sits and does math.

[Shrike]

No, at the time I did these (last week) I didn't feel like modelling.

[Fineus]

Tells you something about his personality doesn't it

He's the kinda person who should be a President but never will be.

[Zeronet]

I'd be modelling but after downloading 30 megs of GMAX only for its zip to be corrupt, i dunno if i can take the chance again.

[Galemp]

Originally posted by Thunder
He's the kinda person who should be a President but never will be.

Yeah, he can't be twisted by greed and avarice.

[EdrickV]

Originally posted by Zeronet
I'd be modelling but after downloading 30 megs of GMAX only for its zip to be corrupt, i dunno if i can take the chance again.

Download GetRight. And you could try Wings3D. The run time engine thing it needs is about 23 megs, but the Windows install file for the program itself is about 913k. (That said, I ought to finish my download of GMAX so I can really compare it to Wings3D.)

http://www.GetRight.com

[Styxx]

Originally posted by GalacticEmperor
Yeah, he can't be twisted by greed and avarice.

Well, all he thinks about is... oh, wait, forget that.

[Alikchi]

Originally posted by Styxx


Well, all he thinks about is... oh, wait, forget that.

Yeah, he'd be a good President..of the U.S. anyways.

[CP5670]

Most bored people will turn to rendering or modelling, but not Shrike. He sits and does math.

I do that all the time! Here's an interesting problem, for example:

Solve the following:

a
ò ³Ö(x²-a²) dx
-a

Here's the indefinite antiderivative of that: (a bit messy, as can be seen)

3/5x ³Ö(x²-a²) - 2/5 a²x ³Ö( [ (a²-x²)/(a²(x²-a²)) ]² ) ²F¹( { ½, 2/3 }, 3/2, x²/a² )
(²F¹ is the Gauss hypergeometric function)

Plugging in the values, this reduces to:

2/5a^(5/3) Ö(p) G(1/3) / G(5/6)
(G(x) is the gamma function)

Or approximately,

»1.68261852639054581 a^(5/3)

Problem solved! w00t! What a beautiful solution too!!


I'm working on this one right now: (should have the answer in a few days)

Find the rate at which the ratio of the surface area to the volume of an n-dimensional hypersphere changes with the number of dimensions.

This basically boils down to finding:

d(ò...ò Ö(1 + (¶f/¶v1)² + (¶f/¶v2)² ... (¶f/¶vn)² ) dv1 dv2 ... dvn / ò...ò f(v1,v2...vn) dv1 dv2 ... dvn )/dn

where f(v1,v2...vn)=Ö(v1²+v2²...vn²) , v1 through vn are the independent variables (each representing one dimension) and n is the number of dimensions.

It's easier to convert it into "n-spherical" coordinates before solving (otherwise the integrals become pretty messy), so that is what I have been doing.

[Galemp]

And I thought I was cool doing derivitaves and summations...

Good ol' Sigma. Where would we be without it.

[LtNarol]

dear lord, you people have no lives...and as for collapsed Molybdenum, well, who says that the atoms would lack electrons?

[CP5670]

Good ol' Sigma. Where would we be without it.

s S

I hate greek letters...

a b g d e z h q i k l m n o p r s t u f c x y w

bah...

[Galemp]

I love uppercase Greek, but I get confused in lowercase letters sometimes.

[CP5670]

Some of those are the same as the Latin letters, which makes it a bit easier.

A B G D E Z H Q I K L M N O P R S T U F C X Y W

I don't like them because they are used by mathematicians simply to look smart.

And then there are these two things that aren't even Greek letters:
Ã(x) (weierstrass P function symbol - just look at this doodle thing...what the heck?!)
¶z/¶x (partial derivative symbol - what is this, a mirrored 6?)

Let me try this two-part integral sign (another weird symbol); see how it looks...
ó
õ
not bad...

[Styxx]

Too much math... eyes burning!