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ewm90
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« on: February 22, 2008, 02:25:39 AM »

What are the yield of fire power compared to conventional weapons?

I seen 3 photon torps in "ent" take out the inside of a one story building. I have also seen 3 quondam torp in "Movie" take out a borg sphere.

Which suggests photon torps may be the strength of a artillery round.

quondam may be the strength of of a 3 tun bomb.
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« Reply #1 on: February 22, 2008, 02:42:51 AM »

Photon torpedoes are antimatter weapons, suggesting destructive power being equivalent to around one thousand nuclear weapons (they do seem to regard nuclear weapons as primitive in star trek).  Quantum torpedoes are roughly twice as effective than photon torpedoes against shields, but I'm not sure against normal material objects.  I think the way they're portrayed tends to involve a lot of writer's license though Cheesy
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ewm90
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« Reply #2 on: February 22, 2008, 03:48:52 AM »

check out "Storm Front" Part 8 skip to 06:20 on the time bar. YouTube - "Storm Front" Part 8

The torp dose not nook like a nook at all. Nooks start with a flash then the mushroom cloud shows its ugly head.

Heres usas nuke test looks nothing like in ent.

YouTube - Nuke Test 2


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« Reply #3 on: February 22, 2008, 06:09:13 AM »

Ewm a Photon Torpedo isn't literally a nuke...it just has the destructive power of a nuke. I think >.>
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ewm90
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« Reply #4 on: February 22, 2008, 07:48:14 AM »

It did not look like that much power is what I was trying to say in the clip. and when the borg used there granitic torps in 1st contact they jut took out one building at a same and some time not even the building was distorted.
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« Reply #5 on: February 22, 2008, 02:22:56 PM »

Yea but that could be because they were the first mark torps, whilst in Kirks time they use mark VI and in TNG who knows Smiley just one possibility
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« Reply #6 on: February 22, 2008, 03:39:08 PM »

What about the gravimetric topedos in the movie 1st contact?
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« Reply #7 on: February 22, 2008, 05:19:50 PM »

What about the gravimetric torpedo's in the movie 1st contact?
You have to remember a few things. First there is yield calculation, second there is a special effects budget to take into account, third there is the issue of will we have a story if we actually use the weapons to anywhere near their power level. Then there is also the issue of Star Trek ignores most of physics for the purpose of helping out the writers, (hence why it is pointless to point out yield calculations, and other general math in Star Trek v Star Wars, because those calculations assume that Star Trek obeys physics, which it does not). Otherwise if you are really looking for the destructive power of a Photon Torpedo, look at how much damage they do against Asteroids etc.
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« Reply #8 on: February 22, 2008, 10:11:48 PM »

I think we're severely underestimating the power of antimatter annihilation relative to nuclear fission and fusion. In fisson or fusion, a tiny percentage of the available mass is converted in to energy (the mass defect). Even a small amount of mass results in the release of an inconceivable amount of energy; this is simply because to calculate the energy released, we multiply the mass loss by the square of the speed of light.

Does everyone realise that in an antimatter reaction, all of the mass, both matter and antimatter, is converted to photons? If the warhead in a photon torpedo had the same mass as that in a nuke, it would result in a shockingly huge detonation, and would make a nuclear explosion look feeble by comparison. Presumably the detonation is controlled by making the warhead tiny.

But there is one problem in star trek battles: the warp engine, which controls an antimatter reaction big enough to warp space. A warp core breach would be orders of magnitude bigger than that depicted in the show. Not only that, but a warp core breach would occur everytime a ship is destroyed (you cannot keep the matter and antimatter separate once the antimatter confinement fails). The resulting detonation would wipe out an entire fleet of starhips, ending the battle quickly. And would probably take the nearest solar system along with it.

Just my "two cents". 
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« Reply #9 on: February 22, 2008, 10:39:42 PM »

Take the equation e=mc^2
When an electron crashes into a position, two photons emerge.  In other words, there is a 100% conversion of matter into energy in an antimatter reaction.  So let's plug stuff in:  Let's say that 1 gramme of matter combines with 1 gramme of antimatter, that would give us 2 grammes for the purpose of calculating energy.  So, approximating c as 3*10^8 m/s, we can calculate that that reaction will produce 1.8*10^14 joules of energy, a jewel being kg*(m^2)/(s^2).  Now, let's take 5.13*10^21 atoms of uranium 235 (about 2 grammes), and let them undertake a fission reaction.  This will give us 1.64*10^11 joules.  That's 0.0912% the amount of energy as an equivalent mass antimatter reaction.  In that case, if little boy was an antimatter bomb, we certainly wouldn't have needed to drop a second bomb, as there wouldn't even have been a japan to drop a second bomb on!

Quantum torps are based upon the principle of zero-point energy, whereby a certain fundamental energy exists even when a system has reached a point of zero energy (no movement, 0 degrees kelvin, etc).  I'm not really sure how this is supposed to be turned into a weapon though, but it's the principle behind the idea of a truly infinitely-running perpetual motion machine.   
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« Reply #10 on: February 22, 2008, 10:50:11 PM »


Quantum torps are based upon the principle of zero-point energy, whereby a certain fundamental energy exists even when a system has reached a point of zero energy (no movement, 0 degrees kelvin, etc).  I'm not really sure how this is supposed to be turned into a weapon though, but it's the principle behind the idea of a truly infinitely-running perpetual motion machine.   


i thought quantum torps used quantum shielding to direct the full force of the antimatter detonation towards the target or it's shields, like a shaped charge. photon torps waste 75% of it's energy because most of the explosion is directed outwards, where as a quantum torp will hit it's target with roughly 80% of it's total energy output. so a quantum torp is about 3 time as powerful as a standard photon torp.

or am i wrong?
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« Reply #11 on: February 22, 2008, 11:04:21 PM »

Actually, that makes a whole hell of a lot more sense now, forget about my zero-point energy crap (although that is a real theory); but still, what exactly does "quantum shielding" mean?
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« Reply #12 on: February 22, 2008, 11:18:03 PM »

The writers are just using some licence when they think up things like quantum shields, they really don't know what that is, and they're probably unconcerned.

Needless to say, if we were ever capable of creating an antimatter weapon, we'd never need to upgrade it using hard to access zero point energy. We'd just have to use a bigger warhead. A few more nanograms.

Thanks for the calculation Mortographer, I didn't think it was necessary but numbers can help us to appreciate the scale. I'd like to point out that possibly you meant Joules rather than Jewels, and grams rather than grammes?  Sorry to be an arse.
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« Reply #13 on: February 22, 2008, 11:20:37 PM »

Actually, that makes a whole hell of a lot more sense now, forget about my zero-point energy crap (although that is a real theory); but still, what exactly does "quantum shielding" mean?

not a clue.  Azn
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« Reply #14 on: February 22, 2008, 11:28:13 PM »

Btw zero point energy is only a theory in the sense that gravity is a theory - it is very real. You can use schrodinger's equation to calculate vibrational energy levels, and the ground state has non-zero energy. Therefore a system at zero Kelvin will have residual energy. Whether you can access it, that's another matter.
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