Mr. Oragahn wrote:MT wrote:
For whatever it's worth, I don't think that most Trek ships could handle a multiple-teraton torpedo detonate within several kilometers of them. Gigatons applied over a period of seconds via phasers (or an exploding planetoid)? Sure. Teratons applied over a period of microseconds? Nope. This is how I explain the knife-fight ranges we often see in fleet battles: by closing in to within several kilometers of your adversary, you make it impossible for them to one-hit-kill you without one-hit-killing themselves as collateral damage.
With the levels of energy you're talking about, any short range detonation would have all ships' shields in the close vicinity to flare up every single time a torpedo would blow up against another ship's shield.
My position is that torpedoes in "knife fight range" situations are set to ~10 gigatons. That fits in well with the 2 gigaton/second phaser firepower we see in episodes like Masks. Under such assumptions, shield flares would only occur if hundreds of megatons were deposited upon the shields in short order. In short, the benchmark for "what causes a shield flare" has been moved up.
On the contrary, the number of torpedoes carried by the Enterprise do not number more than 100. If it had hundreds of thousands of these munitions, perhaps it would be plausible to place them in the kilotons. However, if we assume that (say) 1% of the planet needs to be covered in 5 psi overpressure, then each photon torpedo needs to cover 50,000 square kilometers.
Let's not forget phasers. They can be set to a very large angle and have been seen to dissolve matter with far more efficiency than doing it through traditional mechanical work, material heating.
That would be the case IF we assumed that "disappearing" took less energy than blowing something up. I, for one, take the opposite view. The only real way we can have this sort "disappearing" happen in real life is if either a) everything gets turned into neutrinos, or b) phasers on high settings literally make matter go "out of phase," temporarily or permanently, in the weird "no physical interaction" sense that most sci-fi franchises use the phrase to describe. Given that neither effect has been observed in particle accelerators, we can presume that either a) the sorts of energies (per individual particle, measured in electron volts) modern particle accelerators are capable of generating pale in comparison to the sorts of energies phasers are capable of generating, or b) the physics of Star Trek are so radically different from the physics of the real world that any attempt at calculating
anything for the Star Trek universe is pointless.
Ludicrously powerful or disconnected from our physics entirely. Take your pick.
It's not exactly hard to imagine what a large angle shot from orbit could do, after a prolongated firing, to an entire region when one remembers what personnal phasers on certain settings can do to people and their gear.
TNG era phasers are capable of "disappearing" about 400,000,000 tons/second at 10% power. That's about 4,000,000,000 tons per second at 100% power, assuming that "disappearing" scales linearly.
Over
here they guestimate that Los Angeles weighs 12,000,000,000 tons. So the phasers of the Enterprise-D could indeed "disappear" Los Angeles in three seconds or so.
But wait. While 10 exawatts is a "fair guesstimate" for TNG phaser firepower (under the assumption that "disappearing" water takes no more energy than vaporizing it), we know for a fact that energy weapons two hundred years earlier were limited to a power of five terawatts. If we take the geometric mean of the two figures (assuming that the "rate of progression" in energy weapon power is roughly constant), TOS phaser power would be in the vicinity of 7 petawatts - which could only disappear about 3,000,000 tons per second (incidentally, these numbers would indicate that laser weapons in the 2040's would have outputs on the order of 3.5 gigawatts, which fits in reasonably well with current levels of technological progression). Disappearing Los Angeles during the TOS era would have taken close to an hour. Multiply that by every particularly populous city on the planet, and we're talking about an operation that would take several days. And then there's all the small towns scattered across the planet that you have to destroy as well. And there's the agricultural base that needs to be destroyed in order to really ensure that this civilization
ends. And somewhere in some base your scanners haven't detected yet - or one that your computers have mistakenly identified as a "low priority target" - there's a group of people that could be capable of mounting a counteroffensive. Maybe it's under a mountain. Maybe it's in the middle of nowhere, far from population centers. Maybe it's at the bottom of the ocean. Wherever it is, it has time to figure out a way to stop you. You can't give them that time, especially not if they're actively hostile to the Federation.
Once we start including farmland in the equation, the numbers required to pull off General Order 24 skyrocket. Suddenly, instead of destroying the 5,000,000 square kilometers of cities on an earth-like planet, you have to destroy that AND the 20,000,000 square kilometers of farmland. Instead of taking days to phaser civilization to dust, it will take weeks - which is unacceptable. The only way to reduce the amount of time to carry out General Order 24 to a tactically potent level is to assume that the torpedoes are doing most of the heavy lifting - and since there are only about a hundred of them, each of them must have a yield well into the gigatons.
The asylum is the only habitation that we are made aware of in the episode, but there could easily be other domed colonies elsewhere on the planet.
I read the script several months ago and if there is one thing sure it's that there's practically no reason to believe the asylum to be one amongst many other inhabited zones. If anything, evidence of other inhabited zones is quite necessary.
Given McCoy's concern with leaving a "margin of safety" for the people below the shield, we have basically two options: either there are people on the side of the planet opposite the asylum, in which case the fact that blasting through the force field would kill every living thing on the planet becomes all the more impressive, or there are not, in which case McCoy is worried that a weapons strike on one side of the planet will destroy the asylum on the other side, which would arguably be even more impressive. Take your pick.
Given that we haven't seen overloaded shield generators explode in Star Trek, we have to choose option C: that the firepower of the Enterprise is so great that blasting through the shields on one side of the planet would cause lethal effects on the other side of the planet, propagated atmospherically or seismically, not by coring through the planet.
We haven't seen one single shield pushed so far as to encase an entire world either.
Perhaps there's a very good reason that is never done: because the device that you rely on for protection is now a bomb that will kill you all if the shield fails.
It's even worse if one single ship managing to poke a single tiny whole into the entire bigger-than-a-planet shell makes the shielding device blow up. It makes the shield extremely dangerous, absolutely unwise to use it under such circumstances.
However, considering the delusional nutcase who was running the entire circus down there and how he tried to trick other people, lied to them and bluffed, it suddenly becomes possible that one person would be mad enough to push a planetary shield to such limits.
Emphasis added.
1) Even on ships, shields are not produced by a single projector, but by a grid of deflectors covering the hull. That's what the grid on the hull of every post-TMP ship is supposed to be: the shield generators. Even if their designation as such by the designers of the ships isn't enough for you, we often have different portions of the shield (aft, fore, port, starboard, etc.) taking damage separately, which - while admittedly not conclusive - certainly indicates that different generators are devoted to protecting different sections of the ship. On your theory, a mere ship requires multiple distinct elements covering each section of its hull, but an entire planet can be covered by a single shield generator.
2) In episodes like "Year of Hell" "Dagger of the Mind," the
absence of planetary shields elicits surprise.
3) The idea that the shield generator, and not the ship's weapons, would be the cause of the asylum's destruction implicitly contradicts McCoy's sentiment of "How can we be powerful enough to wipe out a planet and still be so helpless?"
4) At the end of the day, the italicized "perhaps" is the cornerstone of this entire theory. Perhaps this shield generator
is different from all the others in that it is centralized and dangerously explosive - but there is
no real reason to think that this is the case. From my perspective, the whole "explosive shield generator on Elba II" concept is a contrived theory designed specifically to avoid the implication that, in the Trekverse, capital ships can cause mass extinctions as quickly as they can empty their torpedo bays.
I think I'll have to provide a link to the rather recent thread I partook in that was covering the Elba II case.
Be my guest.
Bad example. Until recently, the tomahawk missile could be mounted with a variable yield W80 thermonuclear warhead. The maximum yield? 150 kilotons, some ten times the yield of Little Boy.
"More advanced" means "smarter and smaller, but - if need be - capable of far superior power."
Not necessarily. The level of sophistication of a weapon is hardly measured
solely on its firepower. In fact, most of the evolutions done today all deal with better guidance, penetration, autonomy and survival, from infantry level to air bombs or even greater, long range missiles. Yields barely enter the equation these days.
Emphasis added.
You cannot deny that a one ton bomb today is significantly more powerful than a one ton bomb from the forties. In addition to better guidance, etc. we have indeed improved the yield.
A primitive weapon of any given size is going to have a dramatically inferior yield to an advanced weapon of the same size.
I wouldn't be so sure of that. The Sphere Builders helped the Xindi build that one, and IIRC they were time travelers.
Indeed, it's quite a fence to jump here to assume that because it was used in the past, "modern" UFP would easily crack its components and even have a joy at upgrading them, considering who helped the Xindi in making it.
Precisely my point.
My argument would be that the warpcore almost never runs at "one hundred percent," and that its "normal" output may indeed be far less than "one percent."
A margin of excess might be considered, but claiming that a power plant is usually used at 1% of less of its capabilities sounds rather odd.
Odd, yes, but arguably necessary to explain the Enterprise's moving neutron stars around.
I, for one, find it odd that he's referring to the "power of the plasma" in the first place, but for some reason, that's what he's doing.
Wouldn't that just be a technician's way to talk about the pwoer carried by plasma through the conduits? Nothing really weird there.
I cannot predict how a 24th century technician would talk about
anything, and that's precisely the point. There is plenty of room for interpretation, especially in light of what the supposedly "terawatt-range" reactor accomplishes later in the very same episode.
Oh, just before we miss that one:
On the other hand, pulse phasers and quantum torpedoes are not the most efficient means of heating something.
They actually are. The widespread angle of a phaser bank would be the best way to homogeneously heat up a surface at large. Anything else would be far less efficient.
Has the Defiant ever displayed the ability to set its pulse phasers to widebeam? I'm not aware of it ever having done so, but if I'm wrong, feel free to correct me.
But I'll humor you for a moment. The Defiant is much smaller than the Enterprise D, but it is also much more combat oriented, so we'll assume that it is capable of putting out the same phaser firepower as a Galaxy class starship's phaser array - roundabout ten exawatts or the equivalent thereof. We will assume the Great Link is about the size of the Baltic Sea, and that it needs to heat the entire thing from 17 degrees Celsius to about 65 degrees Celsius - making the Founders considerably more tolerant of high temperatures than their "solid" counterparts, but not capable of handling literally boiling temperatures. Conveniently, it also cuts the energy from the 1.0e22 J required to heat the Baltic Sea from 4 degrees C to boiling in half.
So 5e21 J / 1e19 W = 500 s.
I don't know about you, but I
seriously doubt that the Defiant could last for a full eight minutes and twenty seconds against seven battlebugs without even trying to return fire for a second. And, of course, there's every reason to think that the pulse phasers on the Defiant are not
more powerful than a Galaxy class starship's phaser array, but
less.
Clearly, Garak is going to be relying on the quantum torpedoes to be supplying the bulk of the firepower in this operation.
In any case, allow me to engage in some rather more precise calculations regarding the firepower of the Defiant based on what Garak expected to be able to do in "Broken Link."
We will assume that the Great Link is comparable in size to the Baltic Sea (which is one of the smallest seas on the planet). So, a surface area of roundabout 3.77e11 square meters, and an average depth of 55 m.
Given the specific heat capacity of water (4.184 kJ/kg*K), the density of water (1000 kg/cubic meter), and our required change in temperature (48 degrees), we can calculate the necessary amount of energy required per cubic meter: 2.0e8 J/cubic meter. Given the depth of Great Link is by assumption 55 meters, we can conclude that 1.1e10 joules need to be applied to every square meter of the Great Link in order to sterilize it.
Now, is this an average energy intensity, or is it a minimum? Let us suppose it were an average. Then there would be regions where the intensity would be so great as to flash-boil a region of the Great Link (these regions would be directly below the Quantum torpedoes) surrounded by regions that had "just enough" to sterilize them, surrounded by regions that were only sterilized to a handful of centimeters. Sure, this situation would leave us with huge shockwaves propagating through the Great Link, but given that
solid homogeneous tissues are resistant to blast effects, it is likely that significant numbers of Founders would survive. This hardly counts as "wiping out every Founder on the planet."
Therefore, the entire 3.77e11 square meter surface of the Great Link must be covered with an energy intensity of 1.1e10 J per meter squared or more.
Now we need to work out how much of the Great Link needs to be sterilized by each torpedo.
Imagine a hexagonal grid overlaid over the surface of the Baltic Sea such that there are one hundred of them over the sea itself. One hundred is probably a good "guesstimate" of the number of torpedoes the Defiant carries at any given time, considering that ships a hundred times its size only carry around twice as many. At any rate, a torpedo would be detonated at the center of each hexagon. Each hexagon has an area of 3.77e9 square meters. The regions at the corners are the spots farthest away from the epicenter of the explosion, about 38.1 km. Since each corner is shared by three hexagons, we may assume that each bomb must cast an intensity of (11/3) gigajoules per square meter at that point 38.1 kilometers away from the epicenter of the detonation.
The next bit involves some trigonometry.
Imagine a right triangle ABC. The right angle is at point B. Leg BC is horizontal, leg AB is vertical.
leg AB = h = the height above the surface of the Great Link that the torpedo is detonated at.
leg BC = 38,100 m
let "theta" = angle ACB
The explosion occurs at point A.
Consider the equation I = Y/(4*pi*R^2)
Where I equals the intensity in J/square meter, Y equals the yield in joules, and R equals the distance from the explosion in meters.
For our purposes, the hypotenuse AC = R.
Interestingly, we can express theta and R in terms of h as follows:
theta = arctan(h/38,100 m)
R = h/sin(theta) = h/sin[arctan(h/38,100 m)]
The previous equation for intensity only applies when the surface in question is at right angles to incident rays. In this case, the rays are striking the surface at angle theta. So:
I*sin(theta) = 1.1e10 J / 3 square meters
rearranging things a bit:
I = 1.1e10 J / (3*sin(theta))
Then we substitute:
Y/(4*pi*R^2) = 1.1e10 J / (3*sin(theta) square meters)
More rearranging and substituting:
Y = pi * [4.4e10 J / (3*sin[arctan(h/38,100 m)] square meters)] * (h/sin[arctan(h/38,100 m)])^2
Basically, what's going on here is two things: as the height increases, theta approaches 90 degrees (which means that the torpedo has to be less powerful to get the required 11/3 gigajoules per square meter), but R approaches infinity (which means that the torpedo has to be more powerful to get the required 11/3 gigajoules per square meter). Which means that there's some h that is neither too far up nor too far down such that Y is at a minimum.
When I first started this calculation, I was all like "once I get the equation, I'll take the derivative, find the zeroes, figure out which one is the minimum, and then I'll be done!" And then it came out as that monster, so I decided to cheat and use my graphing calculator instead. So between about 25 km and 29 km, the value is 1.7e20 J. That's 41 gigatons per quantum torpedo.
This calculation is made under several assumptions, including the following:
1) That the pulse phasers do not contribute to the overall firepower. It is probable that this is not the case. Assuming (again) that all four pulse phasers put together can equal the firepower of a Galaxy class starship's ventral phaser array, and assuming that they can fire for no more than a minute (a reasonable survival time when the Defiant is surrounded by no less than seven battlebugs), they can sterilize (at most) 5.5e10 square meters, leaving 3.2e11 square meters for the torpedoes. Leg BC becomes 35.2 km, giving us a yield of 1.5e20 J, or 36 gigatons. This is, quite literally, as low as we can go.
2) The Defiant has time to fire a hundred quantum torpedoes. It is probable that this is not the case. If we cut the number of torpedoes in half, the minimum yield rises to 3.5e20 J, or 84 gigatons.
3) All one hundred quantum torpedoes detonate at the same time. It is possible that this could be the case, if quantum torpedoes fired early were programmed to move more quickly than quantum torpedoes fired later, but it's not particularly probable. If the torpedoes detonate one at a time, each individual torpedo would have to provide the full 11 gigajoules per square meter all on its own, increasing the yield to 5.2e20 J, or 124 gigatons. Also, there would be substantial displacement of large portions of the Great Link with every detonation, meaning that later torpedoes would have to deal with a greater depth of Founder, increasing the required yield further still.
4) The depth of the Great Link is constant. It is probable that this is not the case. If it were roughly as variable as the depth of the Baltic sea, a non-homogenous spread of torpedoes would be ideal, with maximum torpedo density over maximum depth of Founder-goo. I am unsure how this factor would effect the calculation, but I am fairly confident that it would not lower my figures by an order of magnitude.
5) The energy released by each torpedo is exclusively in the form of thermal radiation. This is, to my understanding of the physics involved, impossible. If we include blast effects (which, as mentioned before, would not be particularly harmful to the homogenous founders), we will have to increase the yield of quantum torpedoes substantially. A 50/50 split between blast and thermal effects like that seen in conventional nuclear weapon, for example, would require us to double the yield.
6) That Founder material on the top of the Great Link will transmit as much heat energy downward as is necessary to sterilize the material below. This is exceedingly improbable. It is far more likely that upper layers would ablate (4e6 J/square meter is enough to cause human flesh to flash into steam, flaying it to the bone), and while this ablation would cause shock waves to run through the Founders below, these shockwaves would likely not be damaging to the homogenous changelings, and most of the heat energy would be carried upwards into the planet's atmosphere. I am unsure of exactly how this would effect my calculations, but it could easily raise the yield of the quantum torpedo by an order of magnitude.
7) That the Great Link is only the size of the Baltic Sea. Given that the "reference range" of seas and oceans available to the makers of Deep Space Nine ranges from the Baltic Sea to the Pacific Ocean, this is likely a low end figure. We could just as easily pick, say, the Red Sea, which has a surface area of 4.38e11 square meters and an average depth of 490 meters. It would require 9.8e10 J/square meter to sterilize. Leg BC would equal 41,100 m. The torpedo yield would rise to 1.8e21 J, or 430 gigatons. If we pick the Mediterranean sea (which conveniently has a volume roughly equivalent to the geometric mean between the volume of the Pacific Ocean and the volume of the Baltic Sea), which has a surface area of 2.5e12 square meters and an average depth of 1500 meters. It would require 3.0e11 J/square meter to sterilize. Leg BC would equal 98,100 m. The torpedo yield would rise to 3.1e22 J, or 7.4 teratons. I will leave it as an exercise for the reader to determine what sorts of yields we would be looking at if, instead of taking oceans from the lower half of the scale, we were to take oceans from the upper half.
8) That the physical and thermodynamic properties of Founder fluid are comparable to those of liquid water. Given what we know about Founders (for instance, their apparent ability to alter their mass/size more or less at will), this is unlikely. It is impossible to tell how this would effect the calculation, but I am confident that it would not lower the yield by an order of magnitude.
So, in the end, the lowest figures we can get for torpedo yield that are consistent with "Broken Link" are in the double-digit gigatons.
If they wanted us to think it's only as deep as a kiddie pool, they wouldn't have made it stretch from horizon to horizon.
Yet if that was the way it were, that thin, stretched over the entire planet or so, it would precisely be the way it would look like.
It would also look precisely like that if it were a hundred kilometers deep.
Besides, Interstellar. Remember the planet with the huge tides? Until they arrived, all we were to see was a vast sea that was less than half a meter deep.
Never seen it. In any case, there is no evidence that the Founder homeworld has tides any greater than Earth does.
Still, I don't think it matters much.
If I get the crux of your calculation, you're using classicaly physics to obtain a firepower figure,
Actually, I'm using it to obtain a firepower figure from "Broken Link," in order to show that "The Die is Cast" fits in well with the rest of Trek.
while the episode has clearly shown waves crossing over vast areas: waves so fast that if they had anything to do with regular physics, they'd have heated up the matter to boiling point and most likely,
But of course, these aren't waves. Given the way that they slow down at a fixed distance from the point of origin, it is far more likely to assume that we're seeing some kind of "stuff" propogating away from the epicenter and slowing down as it does so, or (more likely) is the visible effect on the surface of something going on underground.
because of that too, ejected a lot of matter into space, which would have easily been visible from space.
If the explosions occurred underground, and the phasers were triggering fault-lines, then we wouldn't expect to see any ejecta, we wouldn't expect to see any fireballs, and we wouldn't expect to see any magma showing through the crust. In fact, we would need an explanation for seeing
anything from orbit in the first place.