Cheezy the Wiz
Socialist In A Hurry
These observations beg to differ.
From this site, concerning yields of Photon Torpedoes and Star Wars shield strength.
Photon torpedo yield: Warhead
It is difficult to estimate photon torpedo yields because there has never been a substantive quantification of their output. However, the TM indicates that a photon torpedo carries 1.5kg of antimatter which presumably reacts with an equal amount of matter. This allows us to determine that the upper limit for photon torpedo yield is 2.7E17 joules (64.3 megatons), since Einstein's Theory of General Relativity predicts that E=mc²
Naturally, Federation cultists use this figure as a benchmark, assuming (for example) that if a GCS fires 10 photon torpedoes at a ship, then the target vessel will be hit with 643 megatons of energy. However, this is incorrect. The 64.3 megaton figure is an upper limit, and not necessarily a realistic estimate. Upper limits are extremely generous methods of estimation- for example, the upper limit for a Death Star blast, based on the Alderaan explosion, is well over 1E40 joules!. There are two principal assumptions incorporated into this particular upper limit:
The antimatter within the torpedo will react with matter at 100% efficiency. In other words, not one solitary atom or subatomic particle of antimatter will escape and be hurled out into space by the explosion.
All of the reaction products will contribute to the destructive effect of the torpedo. In other words, every single particle and erg of energy in the entire blast will directly affect the target vessel, with no useless particles and no wasted energy.
Assumption #1: 100% Efficiency
The 100%-efficiency assumption is common among Federation cultists, and they make this assumption so often that they don't even seem to realize they are doing it! One will often hear a Federation cultist saying that the yield of a photon torpedo "is 64 megatons as dictated by the laws of physics" even though the true of laws of physics only dictate that the yield of a photon torpedo will be "equal to or less than 64 megatons". This is not a minor difference of semantics- a theoretical maximum is just that: theoretical. The true yield of a photon torpedo will be lower than this because no process in ever 100% efficient, and it may in fact be much lower. The principal reason for efficiency loss is the fact that the atoms of antimatter and matter do not react simultaneously- there is a reaction process which involves a finite timescale.
How do we know that the reaction isn't instantaneous? Besides the obvious fact that everything in the universe occurs over finite timescales, the TM states that the matter and antimatter are broken up into "many thousand minute packets". Therefore, we can determine that individual packets may be as small as 1 gram apiece. However, even a 1-gram packet will still contain more than 3E23 atoms of deuterium or anti-deuterium (in other words, three hundred billion trillion atoms). When the blocking field between matter and antimatter packets is removed, the atoms in each packet will not experience simultaneous reaction. Although the time delays between individual reactions will seem infinitesimally small to us, they are nonetheless finite and real.
OK, the reaction isn't instantaneous. What difference does that make? Most reasonable people are willing to accept that the reaction cannot possibly be instantaneous, but many have difficulty understanding why this would make a difference. The problem is as follows: the atoms at the packet boundaries will be the first to react (starting with electrons and positrons and moving on to nucleons and antinucleons), producing highly energetic charged pions and gamma rays. These initial reactions will rapidly superheat surrounding (as-yet unreacted) portions of the matter and antimatter, imparting large amounts of kinetic energy to surrounding matter. Successive annihilation reactions will impart more kinetic energy to surrounding atoms, until a point is reached where unreacted atoms inside individual packets have so much energy that they can streak out of the reaction area without contacting anything.
Why would energetic particles escape? Normal matter (not the stuff of black holes or pulsars) is mostly empty space. This is a fact that physicists have been trying to explain to laypeople for decades, with little success. There is an intuitive tendency, among the average layperson, to cling tenaciously to the fact that ordinary matter "feels" solid. It is easy for a physicist to state that the atomic radius of a deuterium atom is roughly 0.46 angstroms while its nuclear radius is less than 2 femtometres. However, these sterile figures are often difficult for laypeople to intuitively understand. It may be instructive to examine the nature of matter from an everyday perspective. Toronto's SkyDome sports stadium facility is a good example of a cavernous man-made building. To those who have never seen this impressive structure, the following facts may help readers imagine its size. SkyDome is so large that the Roman Coliseum, St. Paul's Cathedral, or a 31-storey apartment building could fit inside, even with the roof closed. Eight Boeing 747 jumbo jets can fit on the playing field. More than 60,000 people can be seated. The scoreboard alone is more than three storeys tall. So, if SkyDome were a deuterium atom, then how big would its nucleus be? It would be smaller than a marble. Deuterium atoms are not unique in their abundance of empty internal space. Even a far more massive atom, such as Au-197 (Gold, with 197 nucleons), has an atomic radius of 1.44 angstroms and a nuclear radius of 6.23 fm. If Toronto's SkyDome were a gold atom, its nucleus would still be smaller than a marble. Hopefully, these example will illuminate the oft-repeated physicists' statement that ordinary matter is mostly empty space.
OK, so there's more empty space in an atom than I thought. What difference does that make? Because matter is mostly empty space, subatomic particles or matter and antimatter will not necessarily strike one another. In fact, the only reason that they will collide with any great regularity is the electrostatic attraction that will draw oppositely-charged electrons and positrons (or protons and antiprotons) together. With normal matter, there is electrostatic repulsion rather than attraction, so protons almost never strike other protons, except under extreme conditions (ie- nuclear fusion). However, if an individual particle is moving quickly enough, the magnetic attraction that should draw it to another particle will be insufficient to alter its velocity enough to cause collision. A reaction between a gram of matter and antimatter is nothing at all like collisions in particle accelerators, because the particles will be moving in random directions rather than being shot directly at one another. As an example, take a proton which passes near an antiproton. Even if it passes to within 0.1 angstroms (less than one tenth the width of a typical atom), its electric potential energy will only be -150 eV. It is therefore clear that if this proton and antiproton were to possess, for example, 1 MeV of kinetic energy each, they could very well hurtle by one another without being drawn together by electromagnetic forces. At that speed, an antiproton might very well be able to escape. In fact, current theoretical studies of antiproton-based propulsion at NASA suggest that most antiprotons will not react until they have thermalized (slowed down), and that 90% of antiprotons at 500keV or higher will fail to react with matter.
In conclusion, it is quite clear that the matter/antimatter annihilation process is much more complex than the simplistic "put two pieces together and they will completely annihilate" assumption usually put forward by Federation cultists. Unfortunately, we have no way of estimating what the efficiency would be- we can only point out that it cannot possibly be 100%. The DS9 TM seems to support this, by stating that standard photon torpedo yield is roughly 18.5 isotons (a unit of unknown meaning) while 25 isotons is the theoretical maximum. That would suggest an approximate efficiency of 74%.
Assumption #2: All Reaction Products are Dangerous
On the surface, this appears to be an obvious, and eminently reasonable assumption. However, the situation is more complex than some people seem to think (simple solutions and simple explanations are often wrong- this is a maxim of life in general). In reality, much of the energy of a photon torpedo will be wasted.
The first, and most obvious example of wasted energy is the random release of energy in all directions. A simple geometric observation leads to the conclusion that a minimum of 50% of the torpedo's energy will be wasted by radiating away uselessly into space, away from the target vessel. Some Federation cultists have proposed that the weapon may be designed to focus its energies toward the target, much like a proton torpedo. However, we know from numerous episodes (such as Q Who) that a photon torpedo detonation can be extremely dangerous even to the ship which launched the torpedo. This is canon evidence that a torpedo does in fact release its energy in all directions, rather than focusing it all forwards.
A second example of wasted energy is the charged pion production in a photon torpedo. A proton/antiproton annihilation releases roughly 70% of its energy in the form of charged pions, while releasing the other 30% as energetic gamma rays. A charged pion will decay within nanoseconds into an electron and several neutrinos. As a result, much of the charged pion's energy will be useless. However, the charged pions will still travel a considerable distance (several metres to several hundred metres) before decaying, since they are often hurled away from a matter/annihilation at great speed. Therefore, if a photon torpedo detonates directly against a defensive shield, it can be assumed that the charged pions are just as dangerous as the gamma rays. But if it detonates at a significant distance (several hundred metres) away from the target vessel, then a significant portion of the energy carried away by the charged pions may be useless.
There are some disclaimers which must be applied to the above statement- charged pions, like any charged particle, react with matter because of electromagnetic interactions. Therefore, it is inevitable that they will impart some energy to the photon torpedo casing as they pass through it, but not all of their energy because they will be moving too quickly to be entirely stopped.
In conclusion, it is clear that a photon torpedo can deliver an absolute maximum of 32 megatons to its target, and the realistic figure will probably be somewhat lower. This energy will strike the target ship in several forms. The energetic gamma rays from the matter/antimatter annihilation will strike the defensive shields directly, but they will also tend to superheat the photon torpedo's casing, which is composed of several hundred kilograms of matter. This superheated matter, as well as energetic charged particles from the matter/antimatter annihilation itself, will strike the defensive shields at high velocities. The superheated matter will also emit large amounts of so-called "thermal radiation" (low-frequency electromagnetic radiation caused by the high temperatures), which will strike the shields and create a brilliant, blinding flash. The effect is very similar to that of a large nuclear fusion explosion.
The overall impact of a photon torpedo on its target is therefore an amount of energy, in the form of superheated matter, gamma radiation, thermal radiation, and highly energetic subatomic particles, which is less than or equal to 32 megatons in quantity for a direct impact, and as little as 10 megatons in quantity for a medium-proximity blast (decreasing with increasing distance, based on the radius beyond which charged pions decay into useless neutrinos). If we use the 74% efficiency estimate derived from the DS9 TM, we can determine that a photon torpedo should deliver roughly 24 megatons for a direct impact and as little as 7 megatons for a medium-proximity blast.
Photon torpedo yield: Kinetic Energy
Many Federation cultists claim that the theoretical high-sublight speed of a photon torpedo can be used to calculate kinetic energy, and that this kinetic energy should be considered in any photon torpedo yield calculations. Typically, they claim that a torpedo travels at 0.9c or even higher, so it carries hundreds or perhaps even thousands of megatons of kinetic energy in addition to its warhead yield. As usual, there are numerous flaws with this line of reasoning:
If the high-sublight speed and kinetic energy formulas are taken to their logical conclusion, then the warhead yield is an insignificant fraction of the total energy of the torpedo. This is a ludicrous conclusion; one would have to question why the Federation would put warheads on their torpedoes at all, since they require complex and damage-prone antimatter loading and containment systems.
Federation sublight drives employ a low-level variant of warp drive to propel starships at high relativistic speeds without corresponding expenditures of energy. Their kinetic energy is therefore also low (they cannot get a large amount of kinetic energy with a small expenditure of energy- this would be a violation of Conservation of Energy), most likely because they have somehow reduced the "effective mass" of their starships. This technology is also used on photon torpedoes, in the form of a "warp sustainer" engine described in the TM. Therefore, the kinetic energy of a photon torpedo is not related to its speed, since it is propelled by a spatial distortion rather than conventional impulse reaction physics.
The formula for the maximum speed of a photon torpedo is v=vi+0.75vi/c, where vi is the launch velocity, and there is a 75% boost available for torpedoes fired at low sublight (ref TM pg. 129). However, maximum speed is a nonsensical concept in space, where continued acceleration is always possible provided that fuel is available. This is consistent with a space-warp propulsion system rather than a conventional impulse reaction propulsion system, and it is yet another piece of evidence that the KE=½mv² formula cannot be applied to photon torpedoes.
Photon torpedo speeds are greatly exaggerated. Photon torpedoes invariably accelerate to a velocity of a few kilometres per second relative to the launching platform, rather than reaching high sublight speeds regardless of launch-platform speed as some believe. They can travel at superluminal speeds when launched from a ship at warp, but when launched from low sublight platforms or immobile platforms such as DS9, they invariably remain at low sublight speeds. The velocity increase (relative to the launching platform) is consistently in the range of 1-10 km/s, not 100,000 km/s. This can be clearly seen in "Way of the Warrior" and "A Call to Arms".
The low-sublight DS9 torpedoes seen in "Way of the Warrior" and "A Call to Arms" were just as effective as torpedoes launched from mobile platforms such as starships. If the kinetic energy of a photon torpedo were commensurate with the launch platform's velocity, the low-speed DS9 torpedoes would have been ineffective compared to torpedoes launched from a mobile starship.
Without knowing the rate of acceleration, it is impossible to determine how fast a photon torpedo is traveling by the time it reaches its target. If a torpedo accelerates at the same rate as a GCS (10 km/s²) and it strikes its target in 5 seconds, it will only be traveling at 50 km/s. This would explain the relatively low speeds seen in the episodes while being consistent with the high speed claims of the TM. Given enough time, thanks to its space-distortion drive system, a photon torpedo may indeed be capable of achieving high relativistic speeds. However, those speeds will be tactically useless, because of the long time required to achieve them.
In conclusion, because photon torpedoes use a warp sustainer engine rather than conventional impulse reaction engines to achieve its speed, they do not possess the kind of kinetic energy that one would normally expect from a torpedo travelling at high relativistic velocities. Furthermore, their observed velocities in actual combat are far lower than the claims being made on their behalf. We can only assume that this must be yet another propaganda campaign on the part of the Federation.
Other photon torpedo operating characteristics
According to the range figures from the TM, photon torpedo operating range is from 15km to 4.5 million km. The minimum range of photon torpedoes is most likely a safety restriction, much like the safeties on 20th century submarine torpedoes.
Photon torpedoes can penetrate and travel through the ground quite easily, as seen in several TNG episodes when they "burrowed" underground. This allows them to be quite destructive against planetary surfaces, although this feature is of little use in combat against shielded starships.
Since no photon torpedo explosion has ever destroyed a ship without assistance from a warp core breach (which is inevitable in the event of a starship's destruction), we can conclude that photon torpedoes are not capable of vaporizing large starships as is often claimed. Instead, they appear to do sufficient damage to destabilize the warp core. "Booby Trap" was one of the few incidents in which an powerless starship (ie- no functioning warp core) was hit with photon torpedoes. In that incident, it required 5 torpedoes to destroy the ship, even though it was unshielded.
Bird of Prey torpedoes are weaker than capital ship torpedoes. In ST3, ST5, ST6, and ST Generations, BOP torpedoes were shown to exhibit poor effectiveness against a shielded or even an unshielded starship. The E-D survived more than a dozen torpedo hits in ST Generations even though its shields were compromised, and the E-A survived well over a dozen torpedo hits in ST6, including several hits after its shields collapsed. Low BOP torpedo yields are consistent with the Klingon commander's statement in ST3 that "he outguns us ten to one".
The TM describes two types of photon torpedoes: an older, lower-yield weapon used in the early part of the 23rd century, and a higher-yield weapon used from 2271 up until the development of quantum torpedoes in DS9. The original Enterprise was destroyed in 2285, which means that older torpedoes were used in TOS and newer torpedoes were used in the movies and TNG. The torpedoes fired in ST2, ST3, ST5, etc. were therefore similar to TNG-era torpedoes.
Photon torpedo tactical effectiveness
Photon torpedoes are roughly equivalent to high-yield nuclear fusion weapons. We know that shielded Federation starships can withstand nuclear explosions (albeit with damage and radiation exposure to some crewmembers) from "Balance of Terror". This indicates that a single hit from a large nuclear weapon or photon torpedo could be expected to penetrate a TOS Federation starship's shields, but several would be required to destroy it. This is consistent with the pattern of battle in TOS, where a single volley of photon torpedoes (only 2 torpedoes with the old TOS launchers) heavily damaged a fully shielded Klingon warship in "Elaan of Troyius".
Photon torpedo accuracy is excellent against targets which are either immobile or travelling in a predictable path- a photon torpedo struck a 1.5 metre target from 80,000km away in "The Changeling". This is not surprising- active guidance systems like those in 20th century cruise missiles correct and refine their courses as they approach their targets so that the range is irrelevant to the accuracy (a cruise missile has the same target accuracy from 10km away as it does from 400km away). The same would probably be true for photon torpedoes, which use active guidance systems.
In general terms, photon torpedoes have much longer ranges than phasers and much higher power, but they are not very maneuverable and are therefore not effective against highly maneuverable targets (otherwise, why would phasers be used at all?).
Erroneous claims from Federation cultists:
They claim that photon torpedoes executed 80 degree "snap" turns in "Shattered Mirror", but large-radius turns can appear to be small-radius turns if they occur at a great distance. If a torpedo executed a 90 degree turn over a 2 metre radius in "Shattered Mirror", it would have appeared from the camera's distance that the torpedo ricocheted off a solid object. There were large sweeping turns executed in "Message in a Bottle" and Star Trek VI: The Undiscovered Country, and those turns do not indicate particularly great maneuverability.
They claim that "Scorpion" established the ability of a 5 million isoton explosion to affect everything within a 5 light year radius. However, this is merely another unscientific and nonsensical Federation cultist claim. Even if the weapon released infinite energy, it could not have propelled the nanoprobes into space any faster than c (or even at c). This incident is teeming with flaws and confusing statements. It was stated that the super-yield torpedo would require 50 trillion nanoprobes, but a 5-light-year radius sphere has an internal volume of 4.4E50 cubic metres. Therefore, this theoretical torpedo would disperse its 50 trillion nanoprobes so that there was one nanoprobe for every 8.9E36 cubic metres of space! An 8.9E36 cubic metre volume of space is equivalent to a cube that is over 2 billion km long on every side! At such high dispersion, they would be lucky to hit a planet with a probe, let alone a small ship.
They claim that an 83 isoton weapon was capable of severely damaging Sisko's already-damaged Jem'Hadar fighter at a distance of 50km. However, they were at a distance of less than 1 km away from the weapon when it detonated, not 50km. They were trapped inside the Dominion White drug depot asteroid's shield and took the daring move of going to warp just as the bomb detonated and the shield went off.
They point out that full-power torpedoes destroyed a freighter in "The Ultimate Computer". But we must keep in mind that an unarmed freighter is not exactly a difficult test for a weapon system (not to mention the fact that its detonating warp core undoubtedly assisted).
They point out that 5 photon torpedoes obliterated an ancient Promellian warship in "Booby Trap". But we must also remember that the Promellian warship had been subject to gamma radiation bombardment for over a thousand years (remembering that gamma radiation makes metal more brittle with time). If a photon torpedo carried enough energy to totally obliterate an enemy ship without assistance from its warp core, why then did the Enterprise require 5 photon torpedoes to destroy the Promellian warship? The answer is that the Promellian warship, unlike most targets, did not have a functioning warp core that could detonate and aid in its own destruction, so the Enterprise was forced to use 5 torpedoes. The same holds true for any other situation where a photon torpedo destroyed a vessel- in every case, significant energy was contributed to the explosion by the vessel's warp core. The ancient, brittle Promellian vessel was different because it did not contribute to its own destruction.
Quantum torpedoes
Quantum torpedoes are newer than photon torpedoes, but the fundamental differences and operating principles are as yet unknown. The DS9 TM states that quantum torpedo yield is roughly 50 isotons, while photon torpedoes are limited to a theoretical maximum of 25 isotons.
Since the theoretical maximum yield from a photon torpedo is roughly 2.7E17 joules, this would suggest that a quantum torpedo's yield is roughly 5.4E17 joules (129 megatons). We don't know whether quantum torpedoes will suffer from the same process inefficiencies that affect photon torpedoes, so the most conservative approach would be to assume that their full yield will be released as destructive energy. Of this, less than half (64 megatons) would be directed against the target. This represents an effective upper limit to the energy that can be delivered to a target vessel by a quantum torpedo.
The new quantum torpedoes appear on paper to be much more dangerous than the old photon torpedoes, but they are not yet being used exclusively throughout Starfleet. Either they are very expensive and difficult to produce, hence the slow deployment rate, or they have some weaknesses with respect to photon torpedoes, making the decision more difficult than it would seem. Even though their yield appears to be superior, there may be extenuating circumstances; perhaps their range or accuracy are decreased. The quantum torpedoes in STFC were all fired at close range, and they didn't even try to use one against the fleeing Borg sphere, which was heading directly toward Earth and which could not have been more than a few thousand kilometers ahead of the E-E. They did not fire upon it until after it entered a stable orbit, and they were able to approach to extreme close range.
Conclusion
Photon torpedoes and quantum torpedoes both release energy equal to large nuclear fusion weapons. Star Wars shields have been designed to withstand such weapons- in fact, starfighters routinely fire directed-energy nuclear weapons at capital ships with little or no effect. The ROTJ novelization specifically describes a thermonuclear explosion immediately outside Admiral Ackbar's bridge window in the opening minutes of the Battle of Endor, which was undoubtedly caused by a fighter-launched missile since the capital ships of the two fleets had not yet engaged in battle. This explosion had negligible effect upon Ackbar's vessel, which is of course what one would expect. We project that our Star Destroyers should be able to withstand roughly 1000 photon torpedoes or 370 quantum torpedoes before losing shields. This estimate is based on the ISD shield strength of 1E20 joules determined in the Imperial shield analysis.
However, those torpedoes are almost certain to strike our large Star Destroyers repeatedly in spite of their poor maneuverability, because Star Destroyers have slow turning rates (and are very large targets). Also, we project that our TIE fighters will suffer heavy (but acceptable) losses if the Federation uses proximity-fused photon torpedoes against them.
It is our belief that the historically short combat lifespan of Federation vessels is due to inherent design flaws in those vessels, primarily their use of metastable power generation technology which is prone to catastrophic failure due to system damage. While their starships have been designed and optimised for very short-duration confrontations, all Star Wars Imperial warships were designed for prolonged survivability in combat. It is this design philosophy that gives us our primary advantage over the Federation- we can simply outlast their vessels by continuing to absorb and release punishment after their vessels succumb.
