ST-v-SW.Net :: To Vaporize a Small Town

In the novelization of Revenge of the Sith, our introduction to the story comes with the following:

The part that's been of interest for folks like me has always been the mention of "turbolaser bolts powerful enough to vaporize a small town". Though vague (since to be otherwise would be dramatic death), the statement suggests a range of possibilities that one might be able to calculate, which is of course what Vs. geeks live for.

Given that these bolts have to be visible from the surface (or at least rooftops) per the description, it's clear that these are the largest bolts being fired. After all, even the big bolts from RoTJ attributed to heavy turbolaser cannons are barely visible from a distance of just a few kilometers in open space, and even during the battle of Coruscant being described by the novel weapons fire between capital ships was invisible past a few kilometers. Observe the following cropped DVD caps of heavy weapons fire between an ISD and a Mon Calamari vessel, the aforementioned "big bolts of RotJ" that my opponents have declared to be "heavy turbolaser" bolts:

It's apparent that the ISD and its adversary are only a few kilometers distant, and yet the bolts are barely visible. Indeed, before the advent of the DVD versions arguments over this scene were frequent, since most people couldn't see the two continuing bolts visible in the last shot. (I'd even wager that at a glance you missed the green bolt just passing over the Mon Cal ship in the first shot.) At a few dozen kilometers, one wouldn't imagine being able to see anything at all.

The situation is similar over Coruscant. The opening scene as the fighters 'fall' over the side of a Venator shows us that a battle is afoot, and most of the ships that are visible have only flashes around them . . . the distant bolts aren't visible at all against the planet. Where weapons fire occurs against the blackness of space the situation is similar. Observe these two battles in progress . . .

The first features a Republic cruiser on the right (her nose visible) fighting a Separatist vessel. The blue bolts of the Republic ship have almost a whitish inner glow at close range, but this feature is invisible further away. Similarly, the returning red bolts of the Separatist ship are scarcely visible at all. In the second image, a similar battle features flashes on the hulls of the two vessels, but no bolts are seen between them whatsoever.

Even in modern-day Earth urban environments, with what one would expect to be a small amount of light pollution compared to that of Coruscant, you won't see anything fainter than a 3 or so in our goofy magnitude system. It's not a leap to imagine that from the cityscape of Coruscant (even on the rooftops above most of the lights and, in some cases, the low-level cloud cover) the situation would be far worse, especially when looking for the dim hairlines of a turbolaser bolt. Thus, it's clear that any bolts visible from the surface must've been the biggest and most powerful.

That said, the rocket booster of Sputnik 1 was visible as a magnitude 1 object compared to the polished half-meter of Sputnik itself, which was only mag 6. That's on par with the brightness of Saturn. It was a 28m by 3m cylinder that reached a height that isn't available in my research, but in any case it is said to have reached orbit and hung out there for a couple of months, suggesting that it maybe hit 100km or some similar LEO figure. For a painted-white, high-albedo, 28x3 vehicle to show as mag 1 still suggests that the turbolaser bolts, only about as bright as the reflected sunlight on the dingy ship hulls in the RotS battle when seen from afar, would've had to be quite large to be noticed.

Now, it's worth noting that this isn't the first time I've tried calculating this. Back at the STrek-v-SWars forums we discussed the lack of any formal definition for words like "town", and especially "small town". I ended up using the "city" I grew up in (population 30,000ish), assuming the diameter of the town to be based on the maximum extent of the city limits, and then calculating the yield required to vaporize a person at the edge of town from a hit dead-center.

But as time has passed the notion of calling my old haunt a "small town" just seemed less and less proper, and of course the old STrek-v-SWars forums are long gone and thus I don't have the calculation anymore. Finally, I was curious to see what a more fair estimation would be.

II. Preliminaries

A. Defining "Small Town"

For starters, we need a definition of "small town". Interestingly, there is no real denotation of this term. It's just one of those things like obscenity that you just sort of know when you see. It is based on several factors, including the people, the land area, and the relationship between them (i.e. the population density). A tall building housing 10,000 people is not a small town, after all. But buildings count, too . . . a million people standing in the desert are not a city any more than the tall building was a town.

Ideally we'd have a specifically identified small town to work from. And so our options here are two-fold.

1. Star Wars Itself

On the one hand, we could try to establish what a small town might be in the Star Wars universe. For instance, in the novelization to A New Hope we see that Anchorhead and Mos Eisley are "towns", with the latter being an explicitly larger town (a hub of commerce and a spaceport). Mos Espa is also described as a town in the script to The Phantom Menace, though at times it's also identified as one of three cities Darth Maul sees upon arrival on Tatooine. Dex's Diner is described in the Attack of the Clones script as being in a "tough part of town", though that's obviously a colloquial description given that the entire planet is basically one vast cityscape.

Since we have no small town mentioned, we could just drop the matter there, and indeed that's what was done for the first version of this page. However, we do get a very good overview shot of Mos Eisley, explicitly described as a larger town (larger, at least, than Anchorhead).

And as luck would have it, we can ballpark its size. How so? Well first, let's zoom in a bit for a tighter view:

Now, let's see if we can find anything recognizable. After all, we see several scenes of Luke and company driving through town on their way to the cantina, so surely something might catch our eye. For instance, there's that very interesting leaning tower, looking like a crashed ship, sitting cattywompus in the dirt that Luke's speeder passes by:

Coincidence? Unfortunately, yes. You'll note that we can see the angle of the object and its shadow, which are largely consistent with the later, in-town scene. However, we ought not be able to see the shadow, because there's a dome-shaped building just behind it in the in-town scene. That dome would be in the way of the shadow given the distant view's logical vantage point.

However, we can at least say with some confidence that the leaning tower in the distant view and the leaning tower in the in-town view are comparable in size. After all, this is not Coruscant . . . the buildings of Mos Eisley are occasionally a handful of stories tall including the top dome, but are generally only about two stories in height. Individual buildings are clearly visible in the distant view, and unless we wish to presume there are huge buildings always just out of sight in the in-town shots, then the individual buildings we see must be the same as those in the distant view.

Judging by sun and shadow angles on the crashed ship and the ground below, the vessel reaches a height of probably about 25 meters. If the leaning tower of the distant view really is comparable in size, that suggests . . . given that the tower is approximately 18px tall in the circled image above . . . that the entire town of Mos Eisley is only about a kilometer in width, if not less.

Other suspicious shapes exist in the distant view, but none seem to correspond to what we want from the in-town shot. However, all are of similar proportion, limiting Mos Eisley to a kilometer or two in size.

2. Small Town America

Given the American origin of Star Wars it seems fair to try to use a U.S. consensus of small-town-hood to determine the meaning of the word as written. Not having seen the results of a vote on the matter lately, however, and not quite being ready to conduct a formal survey of people, we'll just roll with an informal look at what we've got from the internet.

Wikipedia, for instance, features a page on the term 'town', wherein they discuss the fact that there's no definition for it, and indeed that the term is applied differently in different states. Most of the variation is legal, and in some cases laws mean that a legal town can be an enormous place of almost 200,000 people. That's not exactly the mental image most people get when they think of towns, though. It certainly isn't what I think of.

Googling the term "small town", meanwhile, points us toward smaller concentrations of people. There's a book by the excellent author Bill Bryson, for instance, called The Lost Continent: Travels in Small-Town America. We can consider him a bit of an authority on language given his book on the etymology of U.S. words and word usage Made in America (my own copy of it is rather dog-eared from continual re-reading). The Amazon.com page shows a handful of city and town names. The ones I can't identify (unlike "New York", for example) are presumably some of the small towns in question . . . Bryson City, Cripple Creek, Warm Springs, Idaho Falls, and New Salem. Googling these, we find:

So what we have available from Bill Bryson via Amazon suggests that small towns are towns of around 1500 or less, and about 1.5 miles wide on the high side. But frankly, my own personal view is that such a definition is too small . . . those are tiny, blink-and-you-miss-them places, like Mos Eisley.

Returning to the Wiki article, right there at the start of it is something interesting. Pictured on the current version of the page is the downtown area of Bastrop, Texas, with the caption "Main street in Bastrop, Texas, a small town".

That place has an estimated population of about 7300, and a land area of 7.3 square miles (18.9 square kilometers). Now we're talking.

But let's stop and double-check ourselves. Earlier we went with the idea of Star Wars as being American and, as such, we should seek an American definition. But where exactly is the author Stover from? The best info I could find in a quick-and-dirty search is that he was educated at Drake University in Des Moines, Iowa (which, coincidentally, is where Bryson set out from for his book). Assuming he's from that area (since I can't seem to find his actual birthplace), then Googling for "small town iowa" might be useful to see what Iowans think. And as it happens, a sociologist at Iowa State defines a small town as being a location of 500 to 10,000 inhabitants.

Bastrop has 7300, so it'll do fairly well as a Wiki-consensus example of a small town. And so even on a "backstage" or "author's intent" level . . . which I shy away from for analysis purposes . . . the notion holds.

But Bastrop has one of those large subdivisions with an obnoxious assortment of Hawaiian street names right beside a large golf course. This type of community is a sort of suburb, but a distant "money suburb" where people who make money in a larger municipality's business district commute from. Sure enough, Bastrop is about 20 miles from Austin, TX, making it a fairly short commute.

While I'm not keen on using a unique type of golf community suburbiaville like that and would prefer a full-fledged small town away from larger cities, it is the example given on Wikipedia (which, though imperfect, is at least impartial and hence objective in this case), and seems to fit the general connotation of the terms. And, we get to imagine its silly Hawaiian-themed streetsigns melting if we use it, so without further delay let's blow up Bastrop.

B. Bastrop Must Die

If we plug Bastrop, Texas into Google Maps, we get a look at our target. Google shows us the city's limits of incorporation. At maximum extent, the strangely-Alaska-shaped town is about 6.2km wide (from the southwest to the southeast tip). Mathematically the town, at 18.9km�, would be a circle of about 4.9 kilometers, or a square of 4.34km on a side. But we want to be sure to vape the place, so let's calculate our yield based on a diameter of six kilometers.

C. Defining Vaporization

Of course, we could just stop here and make use of any number of websites which detail effects of nuclear explosions. For instance, this PBS site notes that a one megaton surface blast will flatten nearly everything within 2.7 kilometers (1.7 miles), meaning that between the heat and shockwave everything is destroyed.

The issue, though, is the shockwave versus the idea of trying to "vaporize a small town". Let's stop and ask ourselves what that means in this context. Sure, a shockwave can blast a house to bits, but that's not really thermal vaporization so much as extremely efficient mechanical separation of constituent pieces (and pieces of pieces).

1. Hiroshima

Do we need a crater the size of Bastrop and a giant cloud that used to be Bastrop and the soil below in the atmosphere, or will it suffice to thermally vaporize everything on the surface? Or is it sufficient to vaporize the place akin to the way everyone uses the term to describe what happened to Hiroshima (1, 2, 3, 4), which (for the most part) was simply blasted and then burned down?

Ideally, the text would be explicit in suggesting thermal vaporization. However, in the context of weapons of mass destruction, thermal vaporization is seldom what is referred to.

Given the layman's meaning of the term "vaporize" in the context of Hiroshima, it's not very likely that the author of the RotS novelization was attempting to give a strict scientific statement of thermal vaporization. After all, it's physically impossible for a ~15kT atomic bomb to thermally vaporize a city of 350,000. The 'vaporization' is going to be a result of thermal effects, related incendiary effects, and, of course, blast. Damage maps of Hiroshima suggest that the fire damage due to incendiary effects was limited to about 1,800 meters. In other words, if we scaled based on the connotation of the term, then the turbolaser bolt hitting Bastrop could wind up sub-megaton.

But, despite having been accused of being a literalism Nazi by my opponents, I'd still rather err on the side of taking the statement at something closer to a literalist, denotative value. Certainly Hiroshima wasn't even close to vaporized in a scientific sense.

2. The Thermal Problem

For instance, even some of the wood structures of Hiroshima survived the initial thermal pulse from the atomic bomb. It was the blast that knocked them down and the incendiary effects that ignited them, laying waste to them.

How did they survive? Well, the radiant thermal pulse of a nuke is quick enough that despite the ridiculous energy involved, very little is truly vaporized that way. The surface of a nearby object, such as the wall of a house, has its first few millimeters heated to ridiculous temperature. This surface layer will ablate, but in those few seconds or fractions of a second there simply isn't time for the heat to go much further. Not only do the ablative effects provide a bit of protection, but also the heat simply won't conduct fast enough to provide deeper effects.

Larger nukes have longer-duration thermal effects (seconds instead of milliseconds), but the underlying thermal problem is the same.

To be sure, sufficient thermal energy could overcome this. However, a bomb so big that it's instantly vaporizing a house is probably going to be large enough that it's blasting the house away at the same time, and not before. The main problem is the atmosphere . . . there are simply limits to thermal radiation in an oxygen-nitrogen atmosphere. As the local atmosphere is ionized in the first moments of a nuclear blast, for instance, a layer of "smog" forms around the detonating bomb, shielding the outside world from some of the effects within. And when the blast shockwave separates from the initial fireball, the atmosphere is so rapidly compressed and heated by the absurdly-fast shockwave that it ionizes and becomes incandescent. While extremely hot itself, this shockwave shell is opaque, and the far hotter interior is thus temporarily invisible. It is only when the shockfront cools to 3000 degrees or so that radiant energy from the interior starts to pass through it again.

For a 20-kiloton nuclear weapon this all happens quite quickly . . . the shockfront hits a 'mere' 3000 degrees while the shockwave is only 220 meters across, though its moving at a speed of multiple kilometers per second. Maximum thermal effects thus begin around 150 milliseconds into the event. But even a one-megaton bomb takes longer for this . . . almost a full second elapses. By this time the blast wave's shockfront has travelled at least a couple of kilometers, if not more, meaning that by the time of maximum thermal effect for even a single megaton bomb, most of Bastrop would've been blasted into debris.

3. Bolts and Bombs

Obviously, a turbolaser bolt is not a nuclear weapon. As a result, there will be a little variation in how things work. The types of radiation will vary, for instance . . . we know a blaster bolt is composed of radioactive particles. Witness the note from the Episode III novelization, wherein we're told that "the bolts flared between their blades until their galvening faded and the particles of the packeted beams dispersed into radioactive fog" (Ch. 20).

Presumably, then, turbolaser bolts are also packeted "beams" of radioactive particles. The exact type of radiation is unclear, as is the method of detonation for a turbolaser bolt. We know, for instance, that they can be detonated in a flak burst, but how you make the radioactive particles go up in a blast is uncertain at best, absurd at worst. What radiation is emitted in such a burst is also unclear . . . do the particles just happen to be radioactive but this is largely nullified when the bolt detonates, or does the radiation only get worse? Alas, we cannot know.

However, whether we're talking about a bomb or a bolt, most of the basic principles of high-energy interaction with the atmosphere will remain largely the same. Thus, it's clear that we can't think of this matter realistically in the absence of acknowledgement of the atmosphere.

But then it's so much easier that way, so let's start by pondering the vaporization of a human being at the city limits as if the atmosphere did not exist, and then see what happens from there.

III. Running the Numbers

A. Burning Flesh

Ignoring, just for the sake of ease, such denser constructions as bones and teeth, then we already know the energy required to vaporize your average 60 kilogram ugly bag of mostly water from a starting temperature of 37�C . . . about 162 megajoules, or the equivalent of 39 kilograms of TNT. And again, for the below, we'll also ignore atmospheric effects on the weapon detonation.

So, we want 162 megajoules deposited against a human body at the maximum radius of the town, or three kilometers as decided earlier.

The frontal area of a human being is roughly 1.8 meters times .5 meters, or .9m�. However, since not everyone would be facing the center of the blast or standing, we'll use an average value of .7 meters squared.

(It's worth stopping for a moment here to compare. We're talking about 162 megajoules over .7 square meters, or 231MJ/m�. That's also 23.1kJ/cm�. The Sublette Nuclear Weapons FAQ notes that 100 calories (not the food kind) per square centimeter are enough to cause flesh to flash into steam, "flaying exposed body areas to the bone". 100 cal/cm� is just 0.42kJ/cm�, or over 50 times less than we're using.)

Since we already know the desired amount of energy to throw at this .7m� target, all we have to do is work backward via the inverse square law to know the total energy of the turbolaser. Using S / 4r� = I (where S is the energy at the center of the expanding sphere and I is the intensity), or for our purposes S = I (4r�), let's ponder the data. The intensity at 3000 meters is 162MJ divided by .7m�, or 231.4 MJ/m�. The square of 3000 meters is 9,000,000 m�. That figure times 4 equals 113,097,335.53, and we thus multiply this figure times 231.4 MJ to determine the energy at the center. That value is:

That figure represents an enormous amount of energy. Deposited in one second, it would be the equivalent of 261,707,000,000,000 (261.7 trillion) 100 watt light bulbs.

However, compared to more energetic objects it seems rather less impressive. After all, a ton of TNT releases an estimated 4.184 gigajoules, and this is the figure used when calculating the explosive tonnage of an item. Converting our result from earlier into tons, we find that the energy required to vaporize a person at three kilometers is just 6.25 megatons. To be sure, this is nothing to sneeze at, and it seems more than a little absurd to find such a value disappointing.

B. Detonating Wood

A 6.25 megaton nuke would, going by a rough guesstimate based on the concepts above, blast Bastrop off the map via the shockwave well before the rest of the world saw the majority of the nuke's thermal effects after the shockfront became transparent to thermal radiation. Indeed, the blast radius for a 20psi overpressure from a 6.25 megaton nuke, using Sublette's equation (6250^0.33 x .28), comes out to five kilometers. This is well beyond the three kilometers we need for Bastrop, and at 20psi even the heaviest concrete structures would be wrecked, if not demolished to dust and rubble altogether. Within the dome of the shockfront would be a new form of hell as the remaining debris was exposed to the many thousands of degrees of the fireball and the superheated wind left in the passing of the shockwave.

An average wood-construction house weighs in at around 120,000lbs (plus or minus a bit depending on construction era), not including things like foundation (1, 2). Living wood might feature 60 percent of its weight as water, but wood as prepared for construction has probably stabilized at or near 20 percent.

Given that the house would've already been subjected to violent disassembly and would now probably be little more than dried splinters and the dust of gypsum and charcoal from the walls and wood frame, we can pretty much call the remaining 80% of the house's mass . . . some 43,500kg . . . vaporized for most purposes. What little remains in any solid form will be burned by fire.

C. Original or Extra Crispy?

It's worth reiterating here that we've used an intensity value 50 times greater than what would cause flesh to flash into steam down to the bone in order to get our initial 6.25 megaton figure. We also ignored atmospheric effects altogether. Coupled with the fact that we're also vaporizing a much larger town of far sturdier structures with that figure by way of the massive overpressure, it seems our generosity has been extreme.

If we seek a 20psi overpressure across all of Bastrop's three kilometer radius, we would have the following equation:

Y is the yield, and in doing the math we find Y to be about 1325 kilotons, or 1.325 megatons.

As noted earlier, even a 1 megaton blast will feature a blast wave that destroys the first two kilometers or so of the town's radius before significant thermal effects come into play. The extra third-of-a-megaton will only enhance that, meaning that the figure above is probably much more accurate than the human-centric, zero-atmosphere figure.

D. Remembering Mos Eisley

Recall that initially we were discussing Mos Eisley, a larger town on Tatooine. While Bastrop was some six kilometers in width, Mos Eisley appeared to be just one or two at maximum. If we assumed that Mos Eisley was 1.5 kilometers wide (possibly double the true value, but no matter), and again employed our human-centric vaporization standard, we would find that the energy release at the center of town would be some sixteen times less than what we calculated for Bastrop. In other words, the value for vaporizing humans in Mos Eisley would give us a 'mere' 400 kilotons per turbolaser bolt.

Getting a blast wave overpressure of 20psi across the whole town could be achieved with a 20 kiloton bolt, though the shockwave isn't powerful enough with such a small bomb to encase the town within an opaque shockfront. Thus a higher value is preferred for our purposes.

While there's no real way to calculate anything based on the above . . . we don't know any details of wall thickness, coolant efficacy, and so on . . . we can certainly presume that the destruction of some of the buildings would be a more energy-intensive affair than, say, blowing up a 20th Century wood-frame house. The towering spires of Coruscant are also built of durasteel and something called permacrete, suggesting that such materials as are used in Mos Eisley are probably pretty normal for robust buildings.

The most likely direct destruction mechanism for such a structure would be blast, just as in the case of Hiroshima. Mos Eisley doesn't seem a prime contender for firestorm, so that could be largely ignored. Thus, the higher the overpressure the better, again suggesting that we should aim more toward the 400 kiloton value.

IV. Conclusion

Using a canonically-identified larger town of robust construction in Star Wars, we find that a bolt powerful enough to vaporize a small town could be as powerful as about 400 kilotons, give or take.

Giving some wiggle room and using a larger modern American town, we find that a bolt powerful enough to vaporize a smallish town could be as powerful as about 1.5 megatons, give or take.

In fairness, the canonically-derived value is probably the most accurate. The temptation is to go with that figure, or at least split the difference. However, in the interests of bias avoidance, plus allowing for some Imperial advancement in the intervening two decades, one might wish to split the difference to a megaton yet also leave wiggle room for Imperial advancement in the intervening two decades.

Thus, we can peg those few heaviest shipboard weapons of Star Wars at about 1.5 megatons, high-end.

Appendix A

Let's see what happens if we go for mad overkill . . . vaporization at 60 MPH:

A 1969 Ford Mustang Mach One weighs in at around 3571 lbs., or about 1620kg. Assuming 90 percent of that weight was in the form of iron just for calculation's sake (or 1450kg), and assuming that it takes about 7.6MJ/kg to vaporize iron, then the Mach One would need 11,020 megajoules to be vaporized. That's about the same as the eleven tonnes of water from the house example, but it has to strike a much smaller area. The Mach One is 4.76 meters long, 1.82m wide, and 1.27m tall. Assuming the energy is to be deposited on four square meters, then, then the required intensity would be 2,755 MJ/m�, which is just freaking nuts. That's almost 12 times the intensity required to vaporize a person at three kilometers, and 600 times the intensity required to make exposed flesh flash into steam. It would require a strike of about 75 megatons.

Appendix B

Depending on various factors, a house might simply collapse if all the water in the wood could be vaporized, since it would then become charcoal. The carbon of charcoal is one of the most difficult things to vaporize properly, but generally lacks structural integrity. That's where blast comes in.

Color is also important in a house, not to mention the possibility of white paint serving as a sort of ghetto ablative armor. As seen here in a most interesting article, a white house using 1940s/50s construction methods easily survived a nuclear explosion's radiant intensity of 12 calories per square centimeter, or about 50 joules per square centimeter. That's 500kJ/m�, or about half a megajoule per square meter . . . some four hundred times less than what we calculated with for vaporizing humans at range. Even if four hundred times more intensity wasn't enough to do more to the house, as seen on the page, that's more than enough to start igniting fires, and our value is certainly enough to burn down anything that remains.

Appendix C

To be sure, durasteel is often described in less-than-impressive terms in the novelizations . . . even in Revenge of the Sith's novelization (ch. 20) we're told of durasteel melting away rapidly in the lava flow, suggesting (assuming Hawaii-normal lava temperatures) a melting point no more than a few hundred degrees less than that of modern steel. Even so, however, that's still around a couple of thousand degrees Fahrenheit.

To Vaporize a Small Town

I. Introduction