Imperial Star Destroyer turbolasers
Heavy turbolasers
On Mr. Anderson's site (st-v-sw.net) I found this interesting piece of txt from RotS novelization:
In the novelization of Revenge of the Sith, our introduction to the story comes with the following:
" The skies of Coruscant blaze with war.
The artificial daylight spread by the capital's orbital mirrors is sliced by intersecting flames of ion drives and punctuated by starburst explosions; contrails of debris raining into the atmosphere become tangled ribbons of cloud. The nightside sky is an infinite lattice of shining hairlines that interlock planetoids and track erratic spirals of glowing gnats. Beings watching from rooftops of Coruscant's endless cityscape can find it beautiful.
From the inside, it's different. 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."
Now, first thing we need is to obtain "small town" to blow up.
http://en.wikipedia.org/wiki/Town
If we go with US definition, "small town" might be around 500 to 2 000 people. To define area, being relatively lazy, I will use this list beacouse it already has defined values in square kilometers. By using that list, we can assume "small town" as being 10 to 40 square kilometers. Assuming circle, that gives us radius of 1.784 to 3.57 kilometers. Only small town we saw in Star Wars is Mos Eisley, which is closer to former size.
Using this calculator, we can deduce energy required to destroy such town in form of widespread destruction from air blast radius to be 0.015 to 0.12 megatons. However, that is absolutely tiny low-end estimate, and not very reasonable given what has been said in quote.
More reasonable estimate is by using fireball radius.
I will go with these values:
Fireball radius (minimum): 35 megatons
Fireball radius (airburst): 22 megatons
Fireball radius (ground-contact airburst): 10 megatons
Since we are talking about "vaporizing town", 10 megaton value for heavy turbolasers is reasonable, especially when we compare only "small town" seen in Star Wars canon - Mos Eisley - to US "small town" I used.
I decided to use this calculator to check effects. If we want to destroy New York in radius from 1700 to 3600 meters (1.0625 to 2.25 miles), it would give us 330 kilotons to 3.2 megatons.
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Before scene in which TIE bombers drop bombs on Falcon's asteroid hideout (and after scene in which Luke proclaims to Yoda "I won't fail you"), we see Star Destroyer destroying asteroids with its medium turbolasers. Asteroids are instantly vaporized, but first we have to determine their size.
Here is Imperial Star Destroyer. It is 1600 meters long - at 276 mm that gives us 5.797 meters per milimeter size ratio (values are for original picture; picture seen is automatically resized by webpage). Trench is therefore 14.5 meters tall normal or 23 meters tall at widest part. Width of TL bolts is 1/4th of trench width, or 3.625 meters. Largest asteroid vaporized is around 3 to 4 times bolt width, or 11.5 to 14.5 meters in diameter. That gives us volume of that asteroid to be 796.32 to 1596 cubic meters.
Asteroids are most probably rock, but might be silicon too. Heat capacity of silicon is 711 J(kg x K) and density is 2.329 g/cm^3. However, I can't find other values so I will use Wong's calculator. Resulting figures are 5.8 to 22.8 kilotons.
CONCLUSION: Heavy turbolasers on ISD are 0.33 to 3.2 megatons , and medium turbolasers are 5.8 to 22.8 kilotons.
I would put likely value for heavy TL as 1.8 megatons, and medium TL as 4 kilotons.
Rate of fire of heavy TL is 1 shot every 10 seconds for all heavy TL combined, and medium TL 2-3 shots per second combined. So we have average of 44.6 kilotons to 388.4 kilotons per second for combined firepower of ISD.
NOTE: Some say that turbolasers firing on asteroids were light ones. That is completely wrong - not only too high power and too small rate of fire, but also fact that these turbolasers were most likely positioned on other places of hull.
On my site
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