Phaser Output Errors
Posted: Sat Feb 18, 2012 2:37 am
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.
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.