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In this episode a fleet of Romulan and Cardassian vessels attack a class-M planet with a phaser, disruptor and photon torpedo barrage. In a briefing prior to the attack, the Romulan commander of the mission talks about what the fleet expects to achieve - he talks about destroying the planet down to its core over a period of six hours. When the fleet arrives, a group of ships commence a bombardment of the surface lasting approximately five seconds. A tactical officer on the lead ship reports that 30% of the planets crust has been vaporized in this initial barrage.

We are forced to make assumptions in order to determine the energy required for this feat. The surface area of the planet Earth (the only class-M planet currently known!) is 510 million square kilometres and has a crust some 40 kilometres thick. Assuming that the founders homeworld is of equal size, then the initial bombardment vaporized some 6,800,000,000 cubic kilometres of rock, or 6.8 x 1018 cubic metres. Taking a reasonable value for the density of rock as 2,300 kilos/cubic metre, this equates to 1.564 x 1022 kilos vaporized. With a specific heat for rock of approximately 720 J/Kg/K, the energy required to bring this mass to a boiling point of some 2,500 K from a normal 293 K can be calculated by the formula :

E = 1.564 x 1022 x 720 x (2500-293)
= 1.564 x 1022 x 720 x 2207
= 2.485 x 1028 Joules.

The fleet comprised approximately twenty Starships, a mixture of Romulan Warbirds and Cardassian uprated Keldon/Galor class warships. The Romulan Warbird is an approximate match for the Federation Galaxy class starship, while the Cardassian vessel is typically regarded as considerably inferior. However, since these ships were apparently considerably uprated I will assume that they were also an approximate match for the Romulan vessels.

We see two shots of the bombardment. The first shows the lead elements of the fleet open fire, a total of 4 Cardassian and 3 Romulan vessels. . A total of seven phaser bursts and five photon torpedoes are fired. The scene then cuts to the viewscreen of the Romulan flagship. Ten phaser strikes and five photon torpedoes are fired in this view.

The total power output of the fleet bombardment can be calculated as :

P = 4.97 x 1027 Watts, or :
= 4.97 x 1015 TeraWatts

Although we only see a small part of the fleet on screen, it seems likely that the entire fleet participated in this opening bombardment. The average output of each ship was therefore :

P = 2.485 x 1014 TeraWatts

We can only guess at how much of the destruction was contributed by phasers as opposed to photon torpedoes. If we assumed that only 1% of the overall energy was contributed by the beam weapons, then the beam weapons contributed an overall total of

E = 2.485 x 1027 Joules

So the average power of the phaser armament of a ship can be calculated via :

P = 2.485 x 1027 / (20x5)
P = 2.485 x 1025 Watts
P = 2.485 x 1013 TeraWatts

This figure is far higher than previous estimates of the output of a Galaxy class starship. One answer to this problem is to assume that the phasers contributed a very small percentage of the overall energy of the attack - assuming that a ships phasers could fire at a power of some 1 x 105 TeraWatts, in line with the phaser output we have calculated for a Galaxy class, then they would have contributed only 0.000000004% of the overall energy!

This may indicate that photon torpedoes are vastly more powerful than phasers. However, the energy output of a photon torpedo can be calculated from the mass of its warhead via E = mc2 and is not remotely high enough to account for the "missing" energy.

Alternately, the Founders planet may have had significant stocks of antimatter located on the surface which added to the destruction once the containment pods were breached. It is also possible that the fleet employed some form of advanced torpedo warhead in the attack, possibly a less powerful version of the Genesis Device seen in Startrek II : The Wrath of Khan.

In any case, whatever the reason I tend to discount this figure - it's simply too high to fit in with the bulk of the evidence.

In "Unification, Part II", a Warbird decloaked and fired disruptor beams at the three stolen Vulcan ships - appearing to vaporize each with a one-half second discharge.

Where size is concerned, we can scale the ships next to a Romulan Warbird. Warbirds, at slightly over 1,200 metres, have a forward section approximately one-quarter of overall length, i.e. 300m. Scaled to this forward section, the cargo vessels are of equal length and, as a rough guess, perhaps one-quarter the height of said section. The Enterprise-A is said to weigh some 1,000,000 metric tons; that being the case, the Vulcan ships of comparable length and greater volume are certainly no less than that figure. 1 x 106 tons is 1,002,000,000 kg. To be conservative, let us assume it is primarily composed of a metal with similar properties to Aluminum (900 J/kg K specific heat and boiling point of 2740 K).

Assuming a relatively high 305 K initial temp, we have the following:

E = Mass x specific heat x Temperature increase
= 1.002 x 109 x 900 x 2435
= 2.195883 x 1015
= 2,195,883,000,000,000 joules

or about 2,195 terajoules

I captured this scene at a rate of twelve frames per second, and each shot lasted a total of six frames - hence each shot lasted exactly one half second. The lower-limit power of the Warbirdâs disruptor beams is, therefore, no less than 4,390 Terawatts.

This compares unfavorably with the 100,000 TW figure for the Type X phaser arrays of the Galaxy class starship, but of course there is no reason to suppose that the warbird would fire on maximum power against defenceless targets. In addition, we know that at least some Federation vessels are composed of materials which are exceptionally difficult to heat - see the entry under the "Materials" entry for details.

Had the Vulcan ships been comprised of Tritanium instead of Aluminium, the blast would have had to be some 450 times more powerful, or 1,975,500 Terawatts! Obviously this is also out of line with the Warbird being an approximate match for a Galaxy class starship.

On the other hand, it is not improbable that a civilian interstellar vessel would have a significantly weaker hull than a Starship, whilst still being much stronger than present day materials. If we take the 78,000 Terawatt figure generated under the "Pegasus" entry below as fact, then the Vulcan ships would have a hull some eighteen times tougher than Aluminium. This is certainly feasible, but we are in the realms of manipulating the data to fit the conclusion here, so look at this one with caution.


In this episode a Romulan Warbird melts a significant portion of a large asteroid in order to seal the Enterprise-D inside a large chamber in the interior. The asteroid is described by the Encyclopedia (2nd edition, page 24) as 'moon sized'. No clear indication of the overall size is possible from the episode itself, but the when the Enterprise is on its way out of the asteroid Lieutenant Worf reports "We have passed through two kilometres of the asteroid. Now within one kilometre of the surface.", indicating that the fissure is three kilometres deep. The fissure must have a diameter of at least six hundred metres in order to accomodate the Enterprise-D. The Warbird melts sufficient rock to fill the entire fissure.

The time scale is uncertain. We see Admiral Pressman and Commander Riker on the Starship Pegasus when the attack begins; they beam to the Enterprise, then the scene cuts and shows them arriving on the bridge to see the rock cooling. Twenty five seconds of screen time elapse, but some twenty or so extra seconds would be required for the two officers to reach the bridge from the transporter room. The total time of the Warbirds barrage is therefore approximately forty five seconds.

The volume of asteroidal material melted can be calculated by the equation :

V = pi x average radius of fissure2 x depth of fissure
= 3.142 x 3002 x 3000
= 3.142 x 3002 x 3000
= 848,340,000 cubic metres.

Assuming that the asteroid is rock the density, boiling point and specific heat capacity should be approximately 2,300 kg/cubic metre, 2500 K, and 720 J/Kg/K respectively. The energy required to melt this volume can be calculated thus :

E = 8.4834 x 108 x 2300 x 2500 x 720
= 3.512 x 1018 Joules

The average power output of the Warbird was therefore equal to :

P = 3.512 x 1018 / 45
= 7.805 x 1016 Watts

Or 78,047 Terawatts. This fits in exactly with the Warbird having slightly less firepower than the 100,000 TW output calculated earlier for the Galaxy class.

Last updated : ## 1998.
This page is Copyright Graham Kennedy 1998.

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