Turbolasers:

How Turbolasers work:

There precise nature of turbolasers is not known but we do have a general idea. First turbolasers despite their names are not lasers. How do we know this? Well, turbolasers radiate light in all directions in a vacuum, something which a laser wont do. Then there is the fact that turbolasers travel slower than the speed of light and turbolasers barrels have physical holes in the end of them, instead of lens. The final nail in the coffin is in the ROTS novelizations which refers to turbolasers as "galvened particle beams". So it is rather clear that turbolasers are particle weapons of some kind.

An interesting effect of turbolasers and blasters (which appear to be scaled down versions of turbolasers), they will often cause damage before the visible part of the bolt hits. As seen in this picture:

As you can see the skin of Luke's hand is melted before the actual bolt hits. There have been many theories put forward to explain that effect, most notably the theory that turbolasers are actual an invisible light speed bolt and the visible part is just a tracer or something to that effect. This however doesn't make sense, how could Jedi deflect blaster bolts if they deflect the visible part only? Obviously they can't, so I think the best thing to do is just dismiss the pre-bolt damage as a visual effects error.

There is also a "flak effect" sometimes accosiated with turbolasers. This is demonstrated in TESB when the ISDs are chasing the MF and the anti-speeder fire from the AT-ATs. Flak is also refered to in the ROTS novel. How this flak effects works is a bit of a mystery, excpacial considering that the bolts seem to detonate at perdeterined distances. To explain that I popose that turbolasers are a particle weapon whose bolts are made coherent by a magnetic bottle around the bolt and for flak effects the power of the magnetic bottle is dialed down so it is no longer powerful enough to hold togeather the bolt after sometime, leading to the flak effect.

Turbolaser Firepower:

Unfortunatly there a few examples in Star Wars of turbolasers impacting known materials, so it is hard to find firepower. There are however two examples in canon Star Wars that we can use for firepower estimates. They are the asteriod scene from TESB and the ROTS novel that discribes the power of turbolasers.

In TESB we see an ISD blasting asteroids with light turbolasers(LTL) in an effort to avoid damage.

Through some rough scaling those asteroids apear to be about 20m in diameter. Turbolaser Commentaries at StarDestroyer.net written by Brian Young has these calculations:

Volume of asteroid = 4188.79 m³
Mass of asteroid = 32,965,759 kg
Heat Capacity of iron = 447 J/kg·K
Initial temp of asteroid = ~200 K, normal for objects in space
Final temp of asteroid = 1853 K for melting
Energy for vaporisation of 1 kg of iron: 7.6 megajoules

With this we can calculate a rough figure of around 30 terajoules (TJ) to melt the asteroid, and around 250 TJ to vaporise it.

-Brian Young, Turbolaser Commentaries

So LTLs have a yeild of 250TJ or 59.75KT of TNT, depending on which you perfer.

Next is heavy turbolasers (HTL). Unlike LTL we have never seen a HTL fire against something of known properties so we can't use visuals for calculations; however there is a quote from the ROTS novel that is useful.

The gnats are drive-glows of starfighters. The shining hairlines are light-scatter from turbolaser bolts powerful enough to vaporize a small town. The planetoids are capital ships.

-ROTS novel, page 2

It takes 2.7 megajoules to vaporize one kilogram of water that starts at 37 degrees Celsius (body temp). People weigh in at about 60 kilograms, on average. So, let's ballpark a value of 150 megajoules to vaporize a person . . . or at least to ruin their whole damned day hardcore.

* Ballpark Via Population

Now, let's multiply. 15,000 times 150 megajoules is 2,250,000 megajoules, or 2.25 terajoules, or half a kiloton. But, of course, this is a crappy estimate.

Instead, let's take that 150 megajoule figure and do something else with it.

* Ballpark Via Range

Using the inverse square law, we can try to estimate the necessary yield to have 150 megajoules per square meter (just as a rough estimate) at a range of 2 miles.

So, we'll need a 4 mile diameter circle (i.e. 6.4km wide, 3.2km radius). That's about right for the radius of incorporated towns that size in this area and population density, at least. Then, we can fiddle as needed. I'll show the work, but I'm pressed for time so forgive me if I'm sloppy:

Source / (4pi(r^2)) = Intensity at Range

or

S / (4pi(r^2)) = I

or

S = (4pi(r^2)) * I

Replacing, we get:

S = (4pi((3200m)^2)) * (150MJ / m^2)

or

S = (12.56637 * 10240000m^2) * 150MJ / m^2

or

S = 128679628.8 * 150MJ

or

S = 19,301,944,320 MJ

or

S = 4.6 megatons
So, just from the energy and ignoring things like atmosphere, blast, et cetera, this ought to vape a guy at over three klicks, or at least piss him off real bad.
Your tonnage may vary, of course, but I'd say this is a (very) rough upper limit.

-DSG2K, strek-v-swars.net Forum, April 04, 2005

While I haven't had time to do my own calculations this is the generally accepted figure so I will use it for the time being. There is some question over what type of turbolaser this quote refers to. In fact some people have even claimed that it doesn't refer to HTL at all but LTL. That makes no sense, the context of the quote is that is the view from Coruscant and as HTL are the most visible from of turbolasers it is logical to assume that this quote refers to HTL.

There you have it. LTL have a yield of 250TJ(59.75KT) and HTL have a yield of 4.6MT.

Other Weapons:

Coming soon

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