[tech] Determining the Power of an Imperator Star Destroyer Acknowledgments

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[TECH] Determining the Power of an Imperator Star Destroyer
First, a thank you to Rob Wilson and Stuart Mackey, both of whom had worked so hard to help make this essay a better one. We should also thank Commander Thelea, for her Imperial Fleet stuff, Kyle, for his Firepower Distribution for ISD stuff and Commander Wilkens, for his Wilkens Calcs.
I feel I should also thank Mike Griffiths for the BTM CD stuff.
Some other things people reading this a year from now needs to know. My E-mail is . If you have any difficulty understanding this at all, please E-mail me. Ditto if you feel you see an error.

The Canon Analysis With No Holds Barred
There are four canon films of SW, less than ten hours of first-order evidence. Well, there is absolutely no problem. Since the canon films do provide very good evidence.
TPM features no Star Destroyers, nor anything particularly useful about starship weaponry. Starships only shoot at starships when they shoot at all. And starships shooting starships, thanks to shields and armor, are no fun. Let's head over to ANH.

ANH: The Canon that Forever Seals a Trekkie's Fate
We see Imperator Star Destroyers, all of the first subclass in ANH, Devastator, Conquest, and one other. We have the Death Star casually blowing up Alderaan and remaining unscathed. And we have Dodonna's quote about the Death Star "having more firepower than half the Starfleet." There is no evidence to show that he is exaggerating in any way or it is very hard to be figurative about this kind of quote.
DS1's infamous superlaser blast of history produced 1E38J-4.3E38J (with upper-limit estimates winding all the way up in the 1E40 range using the fastest chunks), but let's use this range. This is where most people would be nice to the Trek side and round it down, but I want to try a new approach, so I won't. Anyway, since the Death Star requires 24 hours to charge its superlaser and there are 86400 seconds (the Galactic Standard Time, according to Saxton, seems identical to ours for seconds, minutes, hours and days) in a day, the time-averaged firepower is anywhere from 1.24E33W-4.975E33W.
You can find the calcs for the 1E38 and 4.3E38 at .
Therefore, half of Imperial Starfleet firepower=(approx) 1.24E33W-4.975E33W, with a small Starfleet, this would become a pretty large number.
Personally, I am a big fan of Thelea's 378 million hyperspace capable+306 million non-hyperspace capable ship Starfleet. And using it helps prevent the numbers from ballooning to even higher heights. I'll leave her to explain her rationalization for it below (references).
I've decided to be realistic, from 1.24E33W to 4.98E33W it is, divided by half the Starfleet, or 324 million vessels.
4.975E33W/3.42E8 ships=1.45E25W for the average ship in Starfleet.
1.24E33W/3.42E8 ships=3.64E24W for the average ship in Starfleet.
The average Imp starship therefore, assuming Thelea is right has 3.64E24W-1.46E25W of time averaged firepower. Actually, given the numbers involved, which may include superlaser armed ships, but certainly also many corvettes, and even non hyperspace vessels, so this is at best fair, if not slightly on the low side of the ledger. In fact, the Star Destroyer is within less than one percent of the top firepower arc. Let’s round it up, that should do it!
ISD Firepower=(approx)3.64E24W-1.46E25W, if not somewhat more. I believe I've made a real effort to make this as easy to understand as possible, but if you are still having difficulties, please come and talk to me. My E-mail is .
And here is the typical Trekkie attack on it, so I had better deal with it.
Trekkie: But Dodonna's just talking about the surface turbolasers!

Me: Do you have any proof that he is talking about the surface turbolasers? I don't think so. The superlaser is the main reason for having to attack the station. Chances are he'll mention something about what it can do. If you're briefing an attack on bombing some artillery unit, you'll mention why you had to do it, like, "The enemy artillery here is menacing our troops along Highway 3 and 5", and in this case, the reason for bombing the Death Star is because "it has more firepower than half the Starfleet". If you judge whether you brief something by whether it is a threat to your forces, the defense turbolasers weren't very useful either. The SWEGWT also seems to be against it, mentioning, "the resulting energy beam had more firepower than half the Imperial starfleet and could instantly reduce a world to asteroid fragments and space dust." Chances are, Dodonna is talking about the superlaser.

Trekkie: But he said "more".

Me: Nitpicking! It can't be that much more! There are many gradations between "half" (50%)and "the entire Starfleet" (100%). If it was over 2/3rds, a more appropriate choice of words is "over 2/3rds of Starfleet". If it was over 3/4ths, he'd have said "over three-quarters of the Starfleet". If you want to nitpick, why don't you nitpick about the fact neither the 1E38J nor the 4.3E38J calculation factored in any thermal energy that would increase the value further, but only kinetic energy, or the fact that most of the ships mentioned in "half the Starfleet" would not have a hundredth of the firepower of an Imperator Star Destroyer and thus the division is conservative for the Star Destroyer's part? Remember that you guys don't like Thelea's estimate and think it is hopelessly high? Dividing the same firepower by fewer ships merely means the number goes up.

Trekkie: But what about other superlasers?

Me: You guys have been trying to convince us that there is no other superlasers! The other superlasers are not nearly as powerful, and given the conservative nature of these calculations, even if half of the sum is taken by other superlasers, that still leaves a substantial sum for the Star Destroyers and other ships.

Trekkie: But you wish to be conservative, right?

Me: Tell me that Trekkies were conservative in the old days of "The Outrageous Okona", or more modern "The Die is Cast". Trekkies would scream long as a Warsie suggests a firepower 1W above what they feel it should be. If I ain't going to get a consensus anyway, why not wish big?

There, now you don't have to say that, because I've said it already. Other questions? No? Move on.

TESB: A Distraction for Fanatic Trekkies, Or Is It?
In TESB, we get asteroids getting blown apart with the tiny weenie turbolasers on the Star Destroyer's ventral flank. The values say >30000TW. Trek fans attacks on it have taken basically two forms 1) "The asteroid isn’t that big!" or 2) "The asteroid was disintegrated!"
But who cares, it is a lower limit from a tiny gun on the ventral surface of a Star Destroyer firing short duration bursts. But we know for sure, even if we accept the "disintegrated, not vaporized" argument, thanks to Cmdwilkens...
ISD LTL/CIWS firepower>1.76E15W
Oh, Wong’s statement on how it didn’t even matter whether it was vaped or shattered is equally interesting.
(note from MW: this idea deserves further explanation: speed is everything. Thermal conductivity through the asteroid's mass is insufficient to account for the effects we saw, because the rock simply cannot conduct that much heat that quickly, even if it's pure iron. The effect would be more of an explosive effect, with a tiny area being superheated and a concussive shock wave moving out and shattering the asteroid. However, in order to shatter these asteroids so quickly, the fragments would have had to move through the rest of the asteroid at more than 600 m/s! This would require extremely rapid large-scale deformation of material, and the mechanics of solid material deformation happens to be an area which I've studied in depth. Deformation involves work, as defined by the stress-strain curve of the material, and that work becomes energy in the resulting deformed matter; this effect is known as work heating. The question of whether the asteroid was heated or shattered is therefore moot, because the act of shattering it at such great speed would create so much work-heating that the resulting material would be superheated anyway). (2001)

Commander Wilkens performed a set of calculations, and came up with some pretty interesting conclusions. Basically, he managed to use the firepower of the asteroid killing to work out a BDZ power level (base, land crust only or 30%, 1/3rd meter melt, no inefficiency = between 24-65 minutes, depending on how big the asteroids are). Still, the number is unlikely to be higher than 1.46E25W. Still, it is a lower limit. Not yet my kind of BDZ (see section below on it), but it clearly proves that something of the sort is easily possible.

It is pretty easy to see how he did it, but in his recent posts, despite referring to them, he had not posted up any precise formulas. Therefore, in the "Appendix: The Reconstruction of the Wilkens Calcs", I will attempt to reconstruct his logic pattern. It isn't all that hard to do - I think I can handle doing it myself, but he's the first maker of the calcs, so I believe we should give him credit. Any help offered by him, will of course, be welcome..

ROTJ: Not Applicable In This Case
In ROTJ, we see a lot of starship fighting, but not a whole lot of ground bombardment (none, in fact). Without "real" objects (like trees, rocks, planets) to provide a comparison, attempting to evaluate a turbolaser based on its effects on armored, shielded starships is no good.

This ends the canon analysis, with one Thelea speculation (certainly in canon there is no evidence to support either the 25000 ISD or 2 mil "Heavy Destroyer" (minor semantic gap here) claims) thrown in to keep the firepower number from ballooning into even higher heights (heights that you seem to support by claiming Imps have got only 10 million capships) and no superlasers.
ANH Value: 3.64E24W-1.45E25W (dialogue based with a visual of the firepower of the DS, an estimate, not a lower limit, but it is on the low side already)
TESB Value>6.1E19W (Wilkens Calcs, a real conservative estimate)

Official Source Analysis
Based on the Law of Conservation of Mass and Energy, you can't spend energy you don't have. Any number that exceeds the likely output of the power generation devices onboard a Star Destroyer, therefore is out. We have three major ways to determine the power of the total power resources on an ISD. Unfortunately, none of them are amenable to reducing that immense number shown to you above. Which just goes to show that everybody but Trekkies understand the truth - even SW people that don't understand an Executor is 11 miles and not 5 miles long understand the immense power meant for ISDs, and by extension all capships in the SW world by George Lucas. Well, KJA also doesn't understand, I figure. Sometimes I wonder whether KJA is a Pro-Trek plant in the SW universe.

Determining an ISD's Power Generation Capability


The DS1 Reactor: The Upper Limit of a Star Destroyer's Strength
This is admittedly from Mike Wong, but no one has ever criticized this before in my memory. Therefore, let's use it. The DS1 hypermatter reactor is 16km in diameter. An ISD's dome is perhaps 140m in diameter. I hadn't seen SWICS for myself, but I did remember seeing pictures of DS innards and the solar ionization reactor dome for the ISD can be seen in many places. These numbers look OK to me, whether he measured them or the book had a number written for it I just don't know.
Besides, it is pretty damn hard to lie about these things. The books are probably everywhere in America (though I can't find a single copy, no matter how expensive in Hong Kong). Besides, the result is roughly consistent with other results of the ISD's power output.
Well, the problem is pretty simple. To unleash an Alderaan sized beam every 24 hours (one day), the DS must produce at least 1.2E33W in its reactor. The reactor is nearly 15 million times the volume of the ISD's reactor. Assuming the same power density m^3 for m^3, the Star Destroyer may produce as much as 8E26W.
Two caveats here apply. The Star Destroyer probably won't be able to channel all its power to the guns, and no one said a hypermatter reactor is every bit the same as an ISD. It is probably at least somewhat more efficient. Plus, the entire energy transfer process would hardly be perfect and inefficiencies will of course be involved. Still, we can use these to determine limits to the ISD's firepower.
As such, ISD firepower<8E26W.

The "Miniature Sun" Description: The Middle Value
This one is dirt simple. As a "miniature sun", the solar ionization reactor is assumed to produce comparable power to a typical small star. A small star like our Sun, which happens to produce 3.827E26W of energy. This is from Wong, but this is also common knowledge.
As such, ISD Firepower<3.827E26W. Again, the efficiency and channeling caveats apply.

The Star Wars Technical Journal Hyperdrive for ISD Descriptions: An Indication of How Much Power can go to Individual Systems
A single hyperjump, and every source I've seen seems tight on this, for an ISD would consume as much energy as many planetary civilizations will consume in their entire lifetimes. Assuming that the "planetary civilization" has similar overall energy consumption to the United States and a lifespan of 10,000 years (a pretty young civ in the SW galaxy). The civ only has 20 billion people, and that's like one city in SW, considering Coruscant has 500 trillion-1 quadrillion!
Some Poeple have claimed that a Planetary civilization, part of a Galactic Republic 25,000 years old, would produce less power than The US (a planetary civilisation that hasn't even managed to leave it's own solar system let alone span a Galaxy).... I'll leave it to you to spot the desperation and idiocy behind that!
Reasonable conclusion would say that they led similar lifestyles to us, using more ultramodern, power drawing tech and thus drawing a lot more power. Besides, even without all of this, I can argue with equal validity that they'll consume more simply because they're more advanced. For humans, our consumption increases as tech increases. Conservation and efficiency improvement efforts can only so far to reduce energy consumption.
Still, this number using these assumptions would wind up at around 6E25 joules. If we assume it takes roughly 3 seconds to do a hyperjump (in canon they jump mighty fast), then the reactor can produce at least 2E25W.
Is that inconsistent with the above values? No, it really isn't. Total power generation does not mean you can dump it all into one system. It is perfectly conceivable that the Star Destroyer can produce 3.827E26W on its reactor, but only channel 2E25W into the hyperdrive...or to any other system. By system, I mean parts of the ship, including...weaponry.
As such, ISD Firepower<=2E25W. Even at 50 percent efficiency, it'll still be <=1E25W. Without hard evidence, there is absolutely no justification for claiming a lesser efficiency, especially given the canon stuff I showed above and the fact that an Imperial Naval Ship is not the same as a Federation Ship.

In my memory, I’ve never seen Pro-Trek debaters analyze these values. I know I hadn’t been in ASVS or any other ST-SW debate for a long time, but while there have been 2 attacks on BDZ, one wonders why they never try to attack these. I will not speculate on the reasons for this, but since they’re not under attack, I guess these aren’t such bad values to use.

In Conclusion, Based on Official Power Generation, ISD Firepower<=2E25W

Along with Canon, there would be a lower limit of about 6.1E19W (TESB) and an upper limit of 1.45E25W.
We got a mighty good bracket now, don't we? Do you have any difficulties? If so, please talk to me.

Some Other Official Evidence
This stuff is often lower than the numbers above, but then no one said the Star Destroyer was firing with it's full weaponry or didn’t have other reasons. In any case, if worst comes to worst, all we have to say is “canon rules!” and the battle is over. Still, let’s analyze it.

Many people have rolled over on Darksaber, a less than well-written novel by Kevin James Anderson. So I’ll just limit myself to a few words. Let’s put it this way, (Fleet) Admiral Dalla orders a full-power bombardment, but what we see is not a full power bombardment. Indeed the power shown in the bombardment is less than an ISD's LTL's would produce (as shown above) let alone an SSD's full weapons compliment.
Thankfully, the canon has a tendency to crush the official. Other official evidence is also against it. So it is out.
Trekkie: But you can't ignore it!

Me: Yes I can, especially when it isn't even self-consistent with itself, is contradicted by canon, is contradicted by the BTM CD, and so on and so on!

The ISD Forger
Everyone knows what Forger did. It is apparently in a book called "Shield of Lies" and a description is also at Again, it is one of the less attacked things Wong has written. Anyway, the shots have to be a one-time fire, which means optimally a heavy HTL gun. There is no reason to use the LTL unless they could also achieve the result. Let's try it once with each gun firing once. The coordination must be perfect for optimum effect, therefore, it'll be hard to use a full ship broadside.
Let's be conservative and assume that 3 VHTL turrets, totalling 6 guns had fired, hitting three spots. Each explosion was evaluated at a minimum of 730,000TJ. Kyle had estimated that the ISD-I's (like Forger) VHTL guns have fifty times the firepower of a LTL. Since I'm being conservative and assuming that both guns on the turret fired in each case, each gun only fired 730,000TJ/2=3.65E17J, which is equal to 50x. Therefore x=7.3E15J. Therefore, each LTL, in this scenario, has 7300TJ of kinetic energy. Kyle then estimates that the overall firepower of the ship will be distributed in such a manner that the total turbolaser firepower is 1025.5 times the power of a single LTL. Therefore, 7.3E15*1025.5=7.49E18J (actually a tad lower, but I rounded it off).
On the high end of the scale, we'll assume 3 light turbolasers fired (I didn't say CIWS yet, but I might). Each light turbolaser must produce 730,000TJ, with one on each target. Scaled like before, it means the ISD would produce 7.49E20J as a whole. Two seconds per shot, so divide by two...
First I wish to thank Mike Griffiths for so graciously providing these values. A KDY W-165, however is credited, by him, with up to 2E25J per shot. This comes in fact, quite close to the maximal firepower of the ISD as estimated by canon. The turbolaser cannon is reputedly capable of dueling with or even destroying Star Destroyers on equal terms, therefore, the Star Destroyer's total weaponry should have comparable or at least within the same order of magnitude strength.
Assuming a 2sec/shot cycle, the power of this weapon is 1E25W.
Since Mike Griffiths uses the lower 1E38J as his Alderaan power level, we'll use it also. Using the 1E38J estimate and Dodonna's quote, the average Imperator-I Star Destroyer carries 3.46E24W of firepower. If we take 4.3E38J as the power of the ANH detonation, then it is equally simple, multiply everything by four point three, but the ratio is still the same.
The KDY W-165 Planetary Defender, therefore, has 2.89 times the firepower of a Star Destroyer. This would seem to explain why Star Destroyers come in threes and sixes on major attacks.. One Star Destroyer duelling with the Planetary Defender may get disabled too fast. The Star Destroyer's firepower can't be too far away from the Planetary Defender, or they wouldn't even try an attack - they'll get snuffed. If the Star Destroyer is far more powerful, nobody would buy a Planetary Defender for it can fire all day and not hurt the Star Destroyer a whit. Therefore, I believe this is a reasonable value.

Slave Ship
After so much work, who needs Slave Ship. Suffice it to say, though, that it is possible to have a recoil equivalent to multi-giga explosions without having the bolt weigh ridiculous amounts. Need I say more? Most of the energy in the turbolaser is probably thermal energy, not kinetic, and thus recoil limitations do little to limit the power of the turbolaser bolt.

The last and the worst of the estimates. It has been one of the most rolled over numbers. As you can see from all the other examples, I am not going to even try to decide what an ISD can do based on what I think a BDZ is. In fact, the firepower values seen greatly outstrip what most conservative BDZ firepower specifies. I will instead use the value derived so far to work out what it might mean for the BDZ itself. Mathematically speaking, instead of putting (ISD Firepower)=(Estimated BDZ Destruction)/(Estimated BDZ duration), I am going for (Estimated BDZ Destruction)=(ISD Firepower)/(Estimated BDZ Duration) and alternatively (Estimated BDZ Duration)=(Estimated BDZ Destruction)/(ISD Firepower). Which is in more than one way better and perhaps even a new approach to the problem. I've given my best answers for the ISD Firepower. Now toy around with either different durations or different destruction levels to see it.
However, if I may say one thing that would hurt debaters from both sides who may just be getting reconciled, it is that in truth, I don't think you can just slag the 30% of the planet that is "land" and ignore the other 70% that's water.
Both sides: Why not?
What I notice, is a tendency these days for people to think that the Caasmas BDZ was made by slagging merely the land, unless of course there is no significant water in Caasmas, which is pretty unlikely as a whole. Carbon-based creatures require water, after all. Life is unlikely to flourish in planets without water, at least life as we know it.
Both sides: And what does that have to do with it?
Let's assume Caamas is about Earth sized and has a similar proportion of water. If we slag 1 meter of only the 30% that's land, and assuming nothing else goes wrong, it'll take around 30% of the 2E24J specified by Wong, and that makes it 6E23 joules.
Both sides: So?
And so, Caamas is gonna get slagged, but it'll quickly solidify all over again! All of 2E24J is gonna raise the temperature of the Earth's seawater by a lousy 0.4 degrees Kelvin. 6E23J is gonna be around 30% of that, or zero point one two Kelvins. With such a high temperature differential (the rocks at thousands of Kelvins and the surrounding water at around 300), heat transfer would be fairly rapid. I'm not a mathematician or scientist who could deduce exactly how fast all this stuff would cool, but it is probable that several decades later, like Caasmas, it would be easily survivable, with temperatures cooling to relative normality and almost certainly we won't see a molten planet or firestorms any more. We'll see a solid planet, a little hotter than it might otherwise be, but the soil would be solid.
Both sides: And? Look, we've just had an agreement.
Sorry pal I currently believe, to guarantee Caamas stays with its firestorms and others for years requires that all the seawater on the planet has to be vaped off.
Both sides: And how much energy would that take?
Well, the quickie calcs suggest several hundred times the 2E24J quote. By the time we finish with the seawater, the amount needed to melt a meter off the rest of the crust would be nothing. By the time you add inefficiencies, I'll say something close to 1E27J for the first meter would be quite reasonable, lesser amounts for the others (much closer to 2.2E24J). Saxton also suggests that the estimate would wind up to be hundreds or thousands of times more than his statement. Here is an example of the reasons!
This is not an artificial raising, nor would it take too much time. Look at all the other evidence! There is plenty of power! All this 1E27J "bullshit" would be accomplishable in about a few minutes, which may also fulfill the secrecy and speed requirements, in case you're interested! Even if we assume an instant start ability in SW starships, just getting out of the shock of a sudden bombardment and running to the ship would take a few minutes!

Both sides: But you've gone against the Imperial Sourcebook and the SWTJ?

The Imperial Sourcebook never said melting only 1/3rd of a meter, or one meter for that matter. It used "reduce a civilized world to slag". It said nothing about melting only the crust that isn't covered by seawater. Nor does it give any time limits, which Trekkies often like to think "It might mean hours" but it could equally mean "it might mean minutes". Besides, some planets, like Mon Cal or Helska or even Naboo are fairly well covered with water. Good luck getting at all the facilities without boiling off some seawater.
As for the SWTJ, it gave a time frame (1h 1 sec to 23 hours 59 mins 59 seconds), but no exact depth of damage either. If we have time after dealing with one meter in short order, we can work on another meter. And another, and vape each one a little more thoroughly until the a time over 1 hour is reached. Nobody said anything on depth, did they?
Besides, the firepower originally derived was based on canon firepower divided by what most people consider as an excessively large estimate of Imperial Fleet numbers. Canon>official, as we all know!
Both sides: But this is not conservative!
No one said this is a particularly conservative paper. History tells us that whenever SW fans make a "conservative" estimate, Trek fans scream anyway.
Trekkies: How about blowing off the atmosphere? Wouldn't that reduce the vapor pressure? Many official BDZs involve fling off all or most of the atmosphere!
Hmm, good point. Let's see...there is 5.1E14 square meters of land, at least according to Wong. (L*W*1) gives the same number as L*W. At sea level, each cubic meter of air has a mass of 1.224kg. Because density in the atmosphere does not fall perfectly uniformly, but I do not know the exact formulas, I will use a great many gradations, as much as possible, while using averaging the densities in those small gradations. It is not a perfect method, but the best one available to me, and will hopefully be approximately correct. And we're only going for order of magnitude stuff here after all. The data came from .

After we got mass of the atmosphere, then we can talk about how much effort would it take to overcome GPE. The calcs are in the "Appendix: BDZ Atmospheric Vent Calcs" section, for they're quite tedious and this way no one is obliged to be tortured by them. You read the Appendix, you do so at your own risk.(BTW, if someone knows the density formula, give it to me and they'll be thanked.)

I'm back. It turns out that the idea wasn't so great after all. It takes over 1.9E26J to blow off Earth's entire atmosphere (actually only the mass in the first seven kilometers) to escape velocity to ensure it never comes back. And that is assuming that turbolasers would spend all their energy on blowing the atmosphere off in a perfectly outbound vector, with nothing else getting in the way, rather than wrecking ground or water, an unlikely prospect because they're aimed down. Worse, the energy requirement increases rapidly with increasing timeframes, for gravity would begin dragging the air back down quickly. This is based on a one time infusion of energy and perfect efficiency, the former condition like a superlaser hit. If it took multiple infusions over many time periods, the total energy requirement would be...well, it is probably beyond me to do the calcs, but one thing is for sure - it'll increase. These numbers are every bit as conservative as the original BDZ calcs in their own way.
Even if we ludicriously assume it'll take one hour and it'll still take only 1.9E26J total, the time averaged firepower would still be >5.4E22W! Which is still a more than eighty-eight point five times the power to merely slag the crust in Wong's perfect BDZ land. According to Kyle's Division, as we've mentioned before, a LTL is 1/1025.5 of the total firepower, which means that a LTL in this case has over 5.27E19W time averaged firepower. 5.27E19W means that on average, the light turbolaser would be spitting out 12.57 gigatons equivalent of energy. If we assume an VHTL to be fifty times that (also from Kyle's Division), then each VHTL gun can produce a time averaged firepower of 2.63E21W, or 628.47GT per second averaged. If we assume Kyle's right and the VHTLs fire at 4 seconds/round, then we get over 2 teratons per shot!

From all this evidence, as well as any other already given, a Star Destroyer has got plenty of destructive power, and can easily do most things required by a BDZ. In fact, we've fixed the power of the ISD within one order of magnitude of 1E25W.
All we have to do is divide it. According to Kyle, an ISD LTL got 1/1025.5 of the total firepower of the ISD. Therefore, a LTL got 1.95E22W time averaged. If you're screaming that this number doesn't jive with the BTM CD, which suggests, even after doing the 4.3x thing, a mere 2.15E18 joules as an upper limit. But it is really no problem.
Shunt a little more of the fraction into the HTLs than before. All a matter of distribution. The heavy turrets don't have fixed firepowers! The BTM CD says nothing about it, to be certain, and therefore distribution is, as far as I know, to the discretion of every person, which is why Kyle and Andras Otto Schneider's divisions are a little different from each other.

Appendix: BDZ Atmospheric Vent Calcs
The formula chosen to approximate the process is listed below:


Where D1 is the density at the lower altitude of the sample, D2 is the density at the higher altitude of the sample. 5.1E14 is the surface area of the sphere in square meters, which when multiplied by the altitude differential between A2 and A1 (the differential in meters) should give the volume of the air in the sample in cubic meters. Volume times density should give mass. Density is in kg^m3 format, and mass is in kg, because kg and meters are the SI units to calculate joules required to do anything. Answers are rounded off the three significant digits, for the source data is also accurate only to three digits.
The values are as follows:

0-121m: (1.224+1.209)/2*5.1E14*(121-0)=7.507E16kg

121-365m: (1.209+1.181)/2*5.1E14*(365-121)=1.487E17kg

365-487m: (1.181+1.167)/2*5.1E14*(487-365)=7.305E16kg

487-609m: (1.167+1.154)/2*5.1E14*(609-487)=7.221E16kg

609-741m: (1.154+1.14)/2*5.1E14*(741-609)=7.722E16kg

741-853m: (1.14+1.127)/2*5.1E14*(853-741)=6.475E16kg

853-975m: (1.127+1.113)/2*5.1E14*(975-853)=6.969E16kg

975-1219m: (1.113+1.087)/2*5.1E14*(1219-975)=1.369E17kg

1219-1341m: (1.087+1.074)/2*5.1E14*(1341-1219)=6.723E16kg

1341-1463m: (1.074+1.061)/2*5.1E14*(1463-1341)=6.642E16kg

1463-1584m: (1.061+1.048)/2*5.1E14*(1584-1463)=6.507E16kg

1584-1706m: (1.048+1.035)/2*5.1E14*(1706-1584)=6.48E16kg

1706-2072m: (1.035+0.998)/2*5.1E14*(2072-1706)=1.897E17kg

2072-2194m: (0.998+0.986)/2*5.1E14*(2194-2072)=6.071E16kg

2194-2560m: (0.986+0.95)/2*5.1E14*(2560-2194)=1.867E17kg

2560-2926m: (0.95+0.915)/2*5.1E14*(2926-2560)=1.741E17kg

2926-3048m: (0.915+0.904)/2*5.1E14*(3048-2926)=5.659E16kg

3048-3413m: (0.904+0.87)/2*5.1E14*(3413-3048)=1.612E17kg

3413-3779m: (0.87+0.838)/2*5.1E14*(3779-3048)=3.184E17kg

3779-3901m: (0.838+0.827)/2*5.1E14*(3901-3779)=5.18E16kg

3901-4267m: (0.827+0.796)/2*5.1E14*(4267-3901)=1.515E17kg

4267-4389m: (0.796+0.785)/2*5.1E14*(4389-4267)=4.918E16kg

4389-4632m: (0.785+0.765)/2*5.1E14*(4632-4389)=9.605E16kg

4632-4754m: (0.765+0.755)/2*5.1E14*(4754-4632)=4.729E16kg

4754-5120m: (0.755+0.726)/2*5.1E14*(5120-4754)=1.382E17kg

5120-5486m: (0.726+0.698)/2*5.1E14*(5486-5120)=1.329E17kg

5486-5974m: (0.698+0.661)/2*5.1E14*(5974-5486)=1.691E17kg

5974-6096m: (0.661+0.652)/2*5.1E14*(6096-5974)=4.085E16kg

6096-6217m: (0.652+0.644)/2*5.1E14*(6217-6096)=3.999E16kg

6217-6339m: (0.644+0.635)/2*5.1E14*(6339-6217)=3.979E16kg

6339-6461m: (0.635+0.626)/2*5.1E14*(6461-6339)=4.019E16kg

6461-6827m: (0.626+0.601)/2*5.1E14*(6827-6461)=1.145E17kg
TOTAL FOR FIRST 6827m of AIR>3.1E18kg
Let's stop it as this point. The density is less than half of what it started out as. I think we've got over half the atmosphere into the mass now. Let's leave it at that. Besides, you aren't the only guys that are getting bored by this.
The mass ain't exactly a lot, but it ain't anything to sneeze at either. Let's see if we can get an inkling of how much energy assuming, as usual, 100% efficiency it would take to blow this all off to escape velocity.
The escape velocity for Earth is: (1.12E4m/s)
The mass to be flung out is: >3.1E18kg
0.5(3.1E18)(11200m/s)^2>1.94E26J is required to fling out all of the atmosphere.
I know I'm supposed to use calculus to determine GPE because gravity is not perfectly constant and thus escape velocity is also not perfectly constant! But let's be serious, I don't know Calculus, and any attempts to use something you don't know often results in ridiculous errors (look at Stilgar). Besides, we're only talking around six or seven kilometers of altitude here to cause variation. The real answer may be fractionally lower, but...

Appendix: The Reconstruction of the Wilkens Calcs
It really isn't that hard. Wilkens seems to assume that the little cannons firing in TESB would be 22500TW apiece, using the 40m asteroids. By adding in the fact that there are over 100 light turbolasers (120 actually) and adding the various medium, heavy and very heavy turbolasers, scaling the larger guns in ratio, we can derive a firepower value adequately powerful to destroy 30% of the crust to 1/3rd meter depth in 24-65 (the latter meaning the asteroids were smaller than he thought) minutes. The energy requirement for melting the whole crust to 1 meter is 2.2E24J and assuming it takes one hour, the power output would be just over 6.11E20W.
The power requirement for this less powerful BDZ would be 2.2E23J. If it takes 24 minutes, the job would require a firepower of 1.53E20W and the latter requiring 5.641E19W.
Firepower Requirement: 5.641E19-1.53E20W.
The light turbolasers cannot be scaled. They're all the same. We can either assume they're CIWS Quad Laser Cannons or LTLs. Let's assume that they're CIWS cannons, because blasting little asteroids out of the path seems a likely occupation for them. There are 120 of them, at least according to Kyle, too small to see in most depictions. Kyle believes that each is roughly 40% of a light turbolaser, and 1025.5 light turbolasers is the equivalent of the entire ship's firepower. Because I cannot get Cmdrwilkens to come and inform me which distribution ratios he decides to use, let's use Kyle's which are quite well documented.
Thus x=56250TW=5.625E16W
If 1025.5x=ISD Firepower then...
ISD Firepower=5.625E16*1025.5>5.768E19W
Kyle assumes that the largest VHTL guns on an ISD-I is only 50 times as powerful as the light turbolasers or 125 times as powerful as the little CIWS cannons. But what if we shuffle it somewhat, and make it so that one VHTL turret is 125 times as powerful as the LTLs? Different people have different power distribution assumptions, and while it is not a major problem in cases where you use the total firepower and judging how much power should go to each gun, it can be a damn big problem when you're using one gun to judge the entire ship's firepower.
Anyway, in that case, the 50x*12=600x slice in Kyle's distribution needs to be altered. In this case, we need to alter it so it is 125*12=1500x. There are now 1925.5x units of firepower. Remember that x is 5.625E16W.
ISD Firepower=1925.5(5.625E16W)=1.083E20W.
As we can see, by using the power demonstrated by the little turbolasers as a base, then factoring in medium, heavy and very heavy turbolasers, we can find a good estimate for firepower. This is honestly not the strongest approach in the universe, for people can disagree on how the power is distributed, but it makes for an adequate support theory.

Reference: The Thelea Ship Calcs
You can get a copy here but for some mysterious reason, there is a newer, more comprehensive set which is not the version on Dalton's web. (This has since been fixed. The rest of this section has been snipped except for the commentary. --Ed.)
My Commentary: Personally, I love these numbers. Somehow they make sense to me. They almost certainly make sense to Commander Thelea and to at least some other Warsies, but to be honest, it is a pretty fringe theory even among Warsies, so don't be too surprised if fellow Warsies don't believe this stuff.

Reference: Kyle's ISD-I Firepower Distribution Copied below
Shortly after Wong realized he made a mistake of assuming the rear two turrets are turbolasers rather than ion cannon (a favor which Wong incorrectly attributed to me, AFAIK Kyle noticed it first, making a web page on it, and in terms of recent mentionings on ASVS, Andras Otto Schneider detected it first. One of the two should get credit, not me), Kyle still feels he hasn't counted in all the turrets yet, so he comes in with his calcs. Here, Kyle's Firepower Distribution Post is posted as a reference, since I make use of it from time to time in this essay, and I don't want anybody saying that he can't find what the hell I'm talking about when I mention Kyle and his distributions.
I personally think Kyle is way too gentle to Trekkies. And therefore, I don't agree with him for his firepower assessment for the ISD (meaning I don't agree with his ray shielding assessment too, for shielding is often tied in with firepower), but I feel his distribution was quite reasonable so I use it. Here it is below as the finish to this essay.
I'm sure some people have noticed some minor errors in Mike Wongs BDZ calculations. Nothing horrible but he uses an incorrect number of HTLs and doesn't include any MTLs. (Correction, I now see he recently corrected the HTL problem).
So I'm going to redo them using the information I've gathered on the weaponry of an Imperator class while working on my website. I hope no one is dense enough to think this is an attack on Mike Wongs work, it's merely a continuation of it with several new variable factored in.


For those who haven't visited my website (get your ass over there:) I'm going to post some basic figures from it, to set the stage for this.
Light TL = 10 MT

Medium TL = 150 MT /2 Seconds

Heavy TL = 500 MT /4 Seconds

Very Heavy TL = 2000 MT /4 Seconds

Quad LC = 2 MT *2 Seconds
Now I'm not not going to use these to set up a relative scale of numbers, with the LTL being the baseline.
LTL = x

MTL = 7.5x

HTL = 12.5x

VHTL = 50x

QLC = .4x
Now I factor in the number of weapons.
120 LTLs = 120x

21 MTLs = 157.5x

8 HTLs = 100x

12 VHTLs = 600x

120 QLC = 48x

Total = 1025.5x

A one hour BDZ requires a sustained bombardment of 6.1e20 Watts.
So, 1025.5 = 6.1e20 Watts
Therefore, x = 5.95e17 W or 142 Megatons
Now I use the relative individual firepower numbers to determine the power of each Energy weapons on an Imperator.
LTL = 142 MT

MTL = 2.13 GT /2 Seconds

HTL = 7.1 GT /4 Seconds

VHTL = 28.4 GT /4 Seconds

QLC = 28.4 MT *2 Seconds or 7.1 Megatons per Laser Cannon shot

Now I'm going to redo it with another variable added. According to some sources an X-Wings quad laser blast is equal to one Proton Torpedo.

So I'll now use a scale of;
LTL = x

MTL = 7.5x

HTL = 12.5x

VHTL = 50x

QLC = .4x

FPT = .2x

Now I'll add in the number of weapons mounted.
120 LTLs = 120x

21 MTLs = 157.5x

8 HTLs = 100x

12 VHTLs = 600x

120 QLC = 48x

120 FPT = 24x

Total = 1049.5x
A one hour BDZ requires a sustained bombardment of 6.1e20 Watts.
So, 1049.5.5 = 6.1e20 Watts
Therefore, x = 5.81e17 W or 139 Megatons
Once again I'll now determine the individual power of each weapon.
LTL = 139 MT

MTL = 2.09 GT /2 Scones

HTL = 6.95 GT /4 Seconds

VHTL = 27.8 GT /4 Seconds

QLC = 27.8 MT *2 Seconds or 6.95 Megatons per Laser Cannon shot

FPT = 27.8 MT /2 Seconds

Guess what? I'm going to do it again, but this time using the WEG quote that anti-ship torpedo's are 20 times the strength of anti-fighter.
So I'll now use a scale of;
LTL = x

MTL = 7.5x

HTL = 12.5x

VHTL = 50x

QLC = .4x

CPT = 4x

Now I'll add in the number of weapons mounted.
120 LTLs = 120x

21 MTLs = 157.5x

8 HTLs = 100x

12 VHTLs = 600x

120 QLC = 48x

120 CPT = 480x

Total = 1505.5x
A one hour BDZ requires a sustained bombardment of 6.1e20 Watts.
So, 1505.5 = 6.1e20 Watts
Therefore, x = 4.05e17 W or 97 Megatons
Once again I'll now determine the individual power of each weapon.
LTL = 97 MT

MTL = 1.46 GT /2 Scones

HTL = 4.85 GT /4 Seconds

VHTL = 19.4 GT /4 Seconds

QLC = 19.4 MT *2 Seconds or 4.85 Megatons per Laser Cannon shot

CPT = 776 MT /2 Seconds



AIM: KyleJK82

WWW: The Imperial Navy Yard

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