Phaser Output Errors

[keiku]

I have found a few errors pertaining to the phaser output of the Enterprise D.

Legacy
“In this episode the Enterprise-D must use a phaser array to tunnel its way through 1,600 metres of rock…”

This tunnel as stated by DITL appears to be 1,600 meters down and 70 meters in diameter. Using the formula π r2 we get 6,157,521.601 m3 of rock. The mass of the rock is 14,303,922,679.2066 kilograms based on a density of 2,323 kg/m-3. Fairly close to what DITL lists. DITL states it takes 13 seconds to complete the operation. If we assume the rock is common sandstone, as I did, and use the same formula as used on DITL, I get a result of 1.74842348 petawatts of power. Close but not identical to DITL (1.731 petawatts).

Now I also tried to calculate this in another way. I’ve found many sites listing 18,000 J as the energy required to vaporize one gram of normal earth rock. (18,000 J * g) or (18,000 kJ * kg) Using this as a guide, I took:
18,000 kJ * 14,303,922,679.2066 kg (the amount of rock vaporized) and got 257.470608 petajoules or 19.80543 petawatts (the energy divided by the 13 seconds it took the Enterprise to vaporize the rock). This is per emitter.

Q Who?
In the episode, “Q Who?” just as was stated on the DITL site, The Enterprise D vaporizes an area of the Borg cube that is a roughly hemispherical region about 600 meters in diameter.

Using the formula ) / 2 to get the volume of the hemispherical section destroyed, we get 56,548,667.765 m3

Like stated on DITL, we do not know the Borg ships composition but we assume it to be tritanium (tri not ti) and according to the Star Trek The Next Generation Technical Manual, it takes 7.2 x 1012 J to vaporize 3 m3 of tritanium. Now to make this easier to calculate I divided 7.2 x 1012 by 3 to get 2.4 x 1012 to vaporize 1 m3 of tritanium. So then, to vaporize this area of the Borg ship we use the following equation:

56,548,667.765 x (2.4 x 1012) = 1.35716802 x 1020 or 135,716.8026 petajoules then multiply this by 0.01 to factor in the actual amount of tritanium the Borg ship had to get 1,357.16802 petajoules.

This is quite different then the 6.8 x 1019 J listed on DITL. DITL found the number in the following manner:

F = (0.5 x 4/3 x pi x r3) / (3 x 2)
= 56,556,000 / 6
= 9,426,000 times larger.
The energy required is thus:
E = 9,426,000 x 7.2 x 1012
= 6.8 x 1019 Joules

Also after multiplying the resulting Joules by 0.01 to factor in the actual amount of tritanium the Borg ship had, the result would be 68,000 petajoules, not terajoules as stated on the site. If fired for one second this would result in 68,000 petawatts.

Conclusion
This results are showing the power or energy for one phaser emitter firing, not all phasers on the ship. So we can conclude from this the actual power of each of the emitters is in the petawatts, not terawatts. Likely, this is between a low of 1.74842348 petawatts and a high of 68,000 petawatts the middle number is 257.470608 petajoules or 19.80542 petawatts and a setup up from this is 1,357.16802 PW. This also leads us to conclude that the figures given on the specification pages for the total output of all phasers firing at once is incorrect. (50,000 TW for the Galaxy – half of what DITL actually calculated) It must be substantially higher. Total power output for the MK 1 Galaxy class (assuming 2,150 emitters – which can be counted on the ship model) could be quite high.

That said, I think the power output of the Galaxy class is just ridiculous! Just as an article on DITL showed, it would need far more antimatter to have such power intensive systems. I calculate the maximum energy output of the warp core (assuming all fuel used at once– which it should not be for firing phasers) is 327,398.4 exajoules. This result is based on a combination of information from the technical manual and the Enterprise D blueprints. I find it highly unlikely they would use up all their fuel just to fire their phasers for a few seconds.

If my math is off or I didn’t make something clear, let me know, but I think I have it fairly close. I have not checked the shield outputs yet, that will be my next task. I want to have the most accurate results I can.

[Mikey]

You're making a lot of assumptions about the materials involved, no? I mean, assuming that "space rock x" is identical to Arizona sandstone is something of a stretch to my way of thinking.

[keiku]

You're making a lot of assumptions about the materials involved, no? I mean, assuming that "space rock x" is identical to Arizona sandstone is something of a stretch to my way of thinking.

That is true. We really have no way of knowing what type of rock it was. Perhaps a geologist would know, but that is not my area of expertise. But I have nothing else to go on.

[McAvoy]

It could be something entirely different than what we see on Earth. It could be as dense as granite or as dense as balsa wood for all we know.

[keiku]

Yeah, I mostly am trying to make sense of what is written on DITL and double check the math. There are multiple examples in my post and the original DITL to make the case for the actual output of the Enterprise D / Galaxy Class MK 1. Of course we don't know exactly, which is unfortunate.

[Mikey]

First, I forgot myself - welcome to the forum. Second, I applaud your dedication. Third... assuming that the rock in question was at least marginally rock-like as we know it, I'd say that the calcs you provided are close enough to the DITL calcs to be significantly within the margin of error created by the unknown specific properties of the rock in question. As far as the Borg ship... unfortunately, we know little enough about the actual ratio of mass to empty volume - and the specific densities of the different masses involved in the interior construction - to make any calculation little more than a guess and a prayer.

[keiku]

It does seem fairly impossible to accurately determine the actual phaser output based upon available data. However I still believe that the methods used by DITL are sound, but the completed calculations are not entirely correct. Like I showed in my post, the output of the phasers is in the petawatts, not terawatts, and moreover this is for only one emitter out of 2,150 across the surface of the ship. So to give the Galaxy a maximum phaser output of 50,000 TW is incorrect. By extension all the phaser outputs listed on DITL are incorrect. Again we don't know exactly what these outputs really are, but I think my calculations get us closer to the truth.

[Captain Seafort]

However I still believe that the methods used by DITL are sound

Unfortunately they're not, at least for phasers/disruptors - the calculations assume that phasers are DET weapons, which pretty every scene involving phaser "vaporisation" demonstrates that they're not.

this is for only one emitter out of 2,150 across the surface of the ship

In almost every scene involving a GCS in ship-to-ship combat we see two points of light converging on the firing point from each end of the array.

So to give the Galaxy a maximum phaser output of 50,000 TW is incorrect.

True. It's actually about 400TW, assuming warp nine power (20 PW, per Deja Q) to the Enterprise's deflector dish weapon, which was expected to overwhelm the Borg ship in BoBW while normal phasers caused only a two per cent drop in power. This fits well with the TW-range plasma the ship produces (The Masterpiece Society) and the fact that such a weapon would be roughly tactically equivalent to a torpedo (somewhere <1 Mt) against shields, which are presumably unaffected by the normal chain reaction mechanism.

Welcome to DITL.

[Graham Kennedy]

That article is really, really old. I could probably do with rewriting it entirely, but I can't really be bothered. Besides, the numbers you get from things like that are highly dependent on what assumptions you plug in in the first place (rather like speeds and such), so I'm far from convinced these days that you really could do something like that sensibly. I've thought more than once about just deleting it.