Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: Shrike on May 01, 2002, 01:14:44 pm
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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.
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So what? The Lucifers beams were 3.83 MegaTons? If that's right then it kinda ****s when compared to cyclop and helios power.
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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.
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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?
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Shrike, you, sir, are bored. :D :lol: :D
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Most bored people will turn to rendering or modelling, but not Shrike. He sits and does math.
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No, at the time I did these (last week) I didn't feel like modelling.
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Tells you something about his personality doesn't it ;)
He's the kinda person who should be a President but never will be.
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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.
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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. :rolleyes:
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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
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Originally posted by GalacticEmperor
Yeah, he can't be twisted by greed and avarice. :rolleyes:
Well, all he thinks about is... oh, wait, forget that. :D
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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. :doubt:
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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! :D 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. ;)
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And I thought I was cool doing derivitaves and summations... :jaw:
Good ol' Sigma. Where would we be without it.
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dear lord, you people have no lives...and as for collapsed Molybdenum, well, who says that the atoms would lack electrons?
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Good ol' Sigma. Where would we be without it.
s S
I hate greek letters... :p :D
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... :p
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I love uppercase Greek, but I get confused in lowercase letters sometimes.
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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. :p
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...
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Too much math... eyes burning! :p
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you can never have too much math... ;7
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Originally posted by CP5670
you can never have too much math... ;7
Yes, you can. The only things you can never have too much are beautiful women, and Spacecrack™. :D
Have you read the book "Flight of the Dragonfly"? You'd like the aliens on it... ;)
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Originally posted by 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. :p
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...
thats the greek alphabet guys not latin...
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Originally posted by CP5670
I do that all the time! Here's an interesting problem, for example:
But all that has no real relevance to FS..... whereas my calculation does. :p
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Originally posted by Shrike
But all that has no real relevance to FS..... whereas my calculation does. :p
And that's why we forgive you. :D
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But all that has no real relevance to FS..... whereas my calculation does. :p
um, um...the expression in the first problem ( ³Ö(x²-a²) ) could be an energy-time function for a reactor in a FS fighter, and with the appropriate values for a, one could determine the work output with that integral, and if the mass is given, the top speed of the fighter could also be calculated! :p :D
As for the other problem, think subspace. ;7
thats the greek alphabet guys not latin...
yeah, as I said, many of the capital Greek letters are the same as Latin ones. ;)
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Originally posted by CP5670
um, um...the expression in the first problem ( ³Ö(x²-a²) ) could be an energy-time function for a reactor in a FS fighter, and with the appropriate values for a, one could determine the work output with that integral, and if the mass is given, the top speed of the fighter could also be calculated! :p :D
As for the other problem, think subspace. ;7
I don't see them actually being used for that, though.
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Is the 3.7 MT figure for the entire beam duration or just a second (for instance) of the strike?
And you're forgetting stress factors from the ships' spaceframe. Chances are the beam would stress it quite badly, if only from thermal expansion.
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Originally posted by wEvil
Is the 3.7 MT figure for the entire beam duration or just a second (for instance) of the strike?
And you're forgetting stress factors from the ships' spaceframe. Chances are the beam would stress it quite badly, if only from thermal expansion.
For the whole beam, I suppose.
And he's getting a low-end estimate, considering the simplest elements possible only. The fact that he considered the hull pure molybdenum indicates that...
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Originally posted by wEvil
Is the 3.7 MT figure for the entire beam duration or just a second (for instance) of the strike?
And you're forgetting stress factors from the ships' spaceframe. Chances are the beam would stress it quite badly, if only from thermal expansion.
Basically what Styxx said. I had a post, but I lost it.
Just one comment though, the energy required to vaporize a large chunk of armor is probably far higher than the energy that would go into heat-stressing the frame.
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Originally posted by EdrickV
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
I have and use DAP.
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Wait, I just checked something; 3.7MT isn't consistent with some of the other values. The FS bible and tech entry says that the original GTM-N1 Harbinger had a 5000MT (5GT) payload.
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Exactly. That's why we have to assume that the armour on the destroyers is much stronger than simple molybdenum...
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BUMP! :bump:
Hey, CP - here's a pic of one of my favorite shirts - mean anything to ya? :D
(http://freespace.volitionwatch.com/qm/pub/sandwich/pictures/what_part_of.jpg)
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LOL, actually the math makes sense to me, but it appears to be a physics formula of some sort. (it's mostly MV calculus, but the variables are the sort commonly seen in mechanics) The F is probably net force, V is linear velocity, w is rotational velocity, R is radius, x is linear displacement and t is obviously time, but I can't really tell what the rest could mean. Notice the betas, rhos and omegas though. :p Actually, now that I look at it, it looks like a variational optimization problem of some sort with those compounded derivatives and the contour integral. :nod:
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Originally posted by CP5670
LOL, actually the math makes sense to me, but it appears to be a physics formula of some sort. (it's mostly MV calculus, but the variables are the sort commonly seen in mechanics) The F is probably net force, V is linear velocity, w is rotational velocity, R is radius, x is linear displacement and t is obviously time, but I can't really tell what the rest could mean. Notice the betas, rhos and omegas though. :p Actually, now that I look at it, it looks like a variational optimization problem of some sort with those compounded derivatives and the contour integral. :nod:
*Reads CP's post*
Drat - Babelfish doesn't do Calculus-to-English.:D
I heard from someone that the appearance of 3 of those "f"s in a row means that the whole thing is nonsense...? :confused:
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LOL, you mean those ò things? That's just the integral sign, so three of them would be a triple integral.
The one thing that is a little annoying is that they have used all the different kinds of notations for derivatives in the same problem: they have the letter subscripts, the ¶f/¶x's, and the time derivatives denoted by the letters with the dots on top. :p Historically, calculus notation has been something of a farce, with everyone using their own system, and all of them lacking in some way. :p
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Originally posted by 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.
that could also explain why they are so low.
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I thought molybdenum had a low shearing value, though?
Adding a kind of reflective coating to the outer hull could possibly deflect much of the energy from beams etc.
Bombs..
By all means, a low-gigaton bomb would completely annihalate an orion. The FS bible and V on this count are totally wrong, unfortunately :( (antimatter less powerful than 1fusion+three salted fission bombs? bah!)
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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.
Use Blender (http://blender.excellentwhale.com/) for heavan's sake!
BTW It does have a mirror. I'll write a tutorial soon for it and others.
To mirror just press 'S' (size) and then 'X' or 'Y'. Sorry 'Z' doesn't work. Then press 'Ctrl' to do even sizing and get it back you size 1.00 on all sides.
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GMAX?!
Last time I downloaded it, it only handles it's own formatting.
Unless the developers have finally created a useful export function or save as.
Great product in theory, but it's meant to create model files that only it can use itself. (Can't import .3ds or .max scene files.)
The makers however, are supposed to be selling patches to let you be able to save as game formats. (At least that's what they said when I filled out the questionaire for them before I could download it.)
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the idea behind GMAX is you learn how to use it better than anything else, so when you go pro and you'll be looking to buy something, you'll buy their stuff, not somebody elses'.
Same with Maya PLE, only saves to its' own format so you can't open it with anything else.
the commercial version of GMAX will cost ya, possibly as much as the commercial version of MAX.
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Well, the gamehouses are supposed to develop or "buy" support for their own file formats if they want the community to be able to use if for mods... I don't think that :v: or Interplay will be getting it POF support in the foreseeable future... :p
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Its part of the new big push.
.GMAX , .XSI all new formats that are so flexible you can base an entire engine around them.
Custom formats are out - why bother developing one that does less than an existing "open" format can? Waste of resources.
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Waste of time is to build code to read complex formats with mostly useless information when you can read a much simpler format that does everything you need. I don't think we'll see that many games with these "standard" file formats that soon...
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well those calcs are intresting but i dunno exactly what they tell us.
Lets try and work out how much energy is needed to to vapourise the hull of the Orion as the lucifers beam does.
Using the volume of molybdenum effected, its density, latent heat capacity, heat of vapourization, boiling point and initial temprature (293 K which is probably wrong since the side that got hit was in the sun, so might be hotter) and a few simpl-ish equations, the energy needed to vapourise the hull is:
2.38e14 J
How the beam of the lucifer suplies this? You'll have to ask :v: , but its about the equivilent of turning a mass of 2.64 grams into pure energy, or the total output of 11000 large nuclear reactors, take your pick.
pete.
oh and dont try and show off with your big equations, your making me jelous
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Uh, all my calculations were right there. So how come yours was so different?