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## Galaxy Class Total Output |

## True Q |

This sixth season episode features a visit to the Enterprise-D by a young student, Amanda Rodgers. Early on she is visiting the engineering department in the presence of Lieutenant Geordi LaForge and Lieutenant Commander Data, Amanda looks at the Warp Core and the following dialogue ensues :

Amanda : 'It's hard to imagine how much energy is harnessed in there.'

Data : 'Imagination is not necessary; the scale is readily quantifiable. We are presently generating twelve point seven five billion gigawatts per-' [cut off by alarm].

At this point in the episode the Enterprise is in orbit of a planet - she is not using her warp or impulse drives, shields or weapons. The warp core would presumably be 'ticking over' to generate the ships gravity field and maintain life support and so on - rather than running at full power to produce lots of energy energy nobody is using. So we have a figure for the low end of a Galaxy class starships engine output of 12.75 x 10^{6} TeraWatts.

## TNG Technical Manual |

Nowhere in the tech manual is any kind of maximum power output given. However, there are several indirect references which prove useful.

On page 55 a chart of the velocity values of various warp speeds is given. Overlaid on this is the energy required to maintain this speed, given in 'Megajoules per Cochrane' - Cochranes being a measure of the warp field. Strictly speaking this is an energy value rather than a power value - ideally it should read 'Megawatts per Cochrane'. However, we know that warp drive is a non Newtonian system - the Enterprise must continually expend energy in order to maintain a given warp speed. The text also refers several times to the power taken to maintain given warp speeds, and refers to features of the chart with reference to this. It thus seems certain that the graph is intended to give the power needed rather than the energy.

The text also states 'note that the Cochrane value for a given warp factor corresponds to apparent velocity of a spacecraft travelling at that warp factor. For example, a ship travelling at Warp 3 is maintaining a warp field of at least 39 Cochranes and is therefore travelling at 39 times c, the speed of light'.

By multiplying the Cochrane value of each warp factor by the power per Cochrane we can calculate the power required for a Galaxy class ship to cruise at a given warp speed velocity :

| (speed in xc) | (Watts) | (Watts) |

| | ^{8} | ^{8} |

| | ^{9} | ^{10} |

| | ^{10} | ^{11} |

| | ^{11} | ^{13} |

| | ^{11} | ^{14} |

| | ^{12} | ^{15} |

| | ^{13} | ^{16} |

| | ^{14} | ^{17} |

| | ^{14} | ^{18} |

| | ^{15} | ^{18} |

| | ^{15} | ^{18} |

| | ^{17} | ^{20} |

NOTE : The values for Warp 9.9 are extrapolated by projecting the curves slightly. There is therefore a greater degree of error involved in this particular calculation than the others, and it should be approached cautiously.

Page 57 tells us that the Galaxy class can cruise at a top speed of Warp 9.6 for twelve hours. This equates to a maximum designed power output of 4,770,000 TeraWatts.

Hence, we get a figure for the Galaxy class starships maximum power output of 4.770 x 10^{6} TeraWatts.

Now while this is within an order of magnitude of the previous value, it does indicate a substantially reduced output since we are comparing minimum output to maximum output. There are possible answers to this problem : Most importantly, the Enterprise-D actually achieved a speed of Warp 9.7 during Encounter at Farpoint, indicating that the warp core can pull considerably more than the designed maximum out of its warp core when needed. In addition, the technical manual was written in 1991, a date which corresponds to 2368 of Startrek. The 'True Q' episode is set some three months after the beginning of season six of the show, in 2369. It is possible that the Enterprise received a hefty upgrade to its warp core in the interval. In addition, the Enterprise may have been running some kind of very energy intensive scientific experiments at the time of Amanda's visit which required the core to be running at close to the maximum limit.

On page 67 - 69, the Galaxy class fuel arrangements are described. The ship is stated to have a total of 62,500 m^{3} of Deuterium storage, and 3,000 m^{3} of antimatter storage. The density of slush Deuterium is approximately 164 kgm^{-3}, so the total Deuterium storage is 10,250,000 kg.

The density of the antimatter storage is not described, but we can get a good idea of how much there is by looking at how long the Galaxy class can go without refuelling. On page 2 the Galaxy class is described as having 'independent mode exploration capability of seven standard years at nominal warp 6 velocity'. Travelling at warp 6 for seven years would use a total energy of :

E = 1.56 x 10^{15} x (7 x 365.25 x 24 x 60 x 60)

= 3.446 x 10^{23} Joules

Even assuming a 100% conversion efficiency of matter to energy, this would require 1,914,494 kg of antimatter and an equal mass of matter. This allows us to get an absolute minimum density for the storage of antimatter of 638.16 kgm^{-3}, which is nearly four times that of the matter tanks. Given that the ship must also run its antigravity field, inertial dampers and structural integrity field, possibly along with non critical systems, then the true fuel supply must be somewhat above this - we can't really know how much above, though in the Voyager episode 'Revulsion' a hologram on an alien vessel complains that 30% of the ships power was devoted to maintaining its crew, a comment which evokes no real surprise from the Starfleet crew members.

But assuming the minimum supply for a moment, this would be sufficient to sustain the ship for 2.4 days at the maximum cruise of Warp 9.2, or about 20 hours at Warp 9.6. By Warp 9.9 the figure drops to 9.57 minutes, which presents an interesting problem in itself; the original Encyclopedia warp speed chart indicated that the Galaxy class could sustain Warp 9.9 for ten minutes before auto engine shutdown. As with all speeds above Warp 6, it would seem that component wear rather than fuel would be the limiting factor here. This is a fairly minor nit, as one can easily enough assume that compromises can be made with systems such as the Inertial Dampers or Structural Integrity Field - a steady but very high speed will not stress the ships fabric or interior as much as heavy manoeuvring at lower speeds, after all.

Of course, by far the greatest factor in determining the exact mass of antimatter fuel would be the efficiency of the conversion process. Some quotes do exist regarding engine efficiency, but they are not very specific since we rarely know what is being expressed as a percentage of what. To give you a feel for the range of figures, I've drawn up the following table :

| | |

| | ^{-3} |

| | ^{-3} |

| | ^{-3} |

| | ^{-3} |

| | ^{-3} |

| Deuterium supply. | ^{-3} |

Given that there is no real lack of space on board a typical Federation Starship, and certainly not on a Galaxy class, I would suggest that the reason for building the antimatter pods so small is that they must be capable of being ejected into space as rapidly as possible in the event of an emergency, whilst the Deuterium tank is much less of a danger. This, incidentally, indicates that the Federation is capable of generating and controlling enormously powerful magnetic fields with relative ease.

## Impulse Engines |

The TNG Technical manual describes the Impulse drive system in some detail. The opening paragraph (page 75) notes that 'The total IPS [Impulse Propulsion System] consists of two sets of fusion powered engines : the main Impulse Engine and the Saucer Module Impulse Engines'. Each engine is supplied by a group of spherical fusion reactors, totalling 24 in all. Later in the page the TM notes that fuel for the system is provided by the Primary Deuterium Tank.

This indicates that the engines of the Galaxy class do not operate on the Deuterium/Tritium fusion process, but the Deuterium/Deuterium process. This is described by the equation :

^{2}H + ^{2}H ---> ^{3}He + ^{1}n + 3.2 MeV

The total mass of the initial two Deuterium atoms is 4.02726 u, equivalent to 3750.25 MeV. Since the reaction releases 3.2 MeV, it is therefore at most 0.08533% efficient in its conversion of mass to energy. Hence, for each 1 kg of fuel consumed by the Impulse system 7.67 x 10^{13} Joules will be released.

The performance of Impulse engines is a subject of some debate. Page 75 of the TNG TM states that the Ambassador class Starship was designed to accelerate at 10,000 ms^{-2}, but in fact this is seriously low for an Impulse drive system. In ST : TMP we see the Enterprise head out of Earth orbit at full Impulse; 1.8 hours later the ship passes Jupiter, a distance of between 628,000,000 and 928,000,000 km, depending on the orbital arrangement. Even assuming that the ship accelerated continually throughout the trip to give us the lowest possible value, the acceleration was between 29,911 ms^{-2} and 44,200 ms^{-2}. But in fact, the true figure is likely to be considerably higher; as the ship boosts out of orbit we see the Earth shrink dramatically behind it, indicating a sizeable fraction of light speed was reached in a matter of seconds - an acceleration of over a thousand times that of the much later Ambassador class.

But for the moment let's assume the lowest feasible value for acceleration, and further assume that the Constitution was the Ferrari of Starships, so 10,000 ms^{-2} is a standard figure for the TNG era. We know that 'full Impulse' is usually restricted to velocities below 0.25 x c in order to avoid large relativistic distortions. It would thus take a Galaxy class ship 7,500 seconds to accelerate to full Impulse. The force required to produce this acceleration on a 4.96 million metric ton vessel is found via :

F = m x a

= 4.96 x 10^{9} x 10000

= 4.96 x 10^{13} Newtons

We can calculate the distance covered in this time via :

S = 0.5 x a x t^{2}

= 0.5 x 10,000 x 7500^{2}

= 281,250,000,000 m, or 281,250,000 km

And the total work done on the ship by its engines is equal to :

W = F x s

= 4.96 x 10^{13} x 2.8125 x 10^{11}

= 1.395 x 10^{25} Joules

To supply this energy we would need to fuse some 181,877,444 metric tons of Deuterium. Unfortunately, the Primary Deuterium Tank only holds 10,250 metric tons!

Fortunately, we have a get-out. The TM describes the use of 'space-time driver coils', which are described on page 75 as being '...similar to those in standard warp engine nacelles, that would perform a low level continuum distortion without crossing the warp threshold'. The driver coil apparently reduces the effective mass of the ship - or multiplies the force, whichever way around you want to put it. So that the work required is reduced by some driver ratio, r, hence :

W = (F x s) / r

Unfortunately, in order to determine a value for r we would need to know the actual power output of the engines - and we don't. The TM can give us some clues, however. On page 77, it notes that fuel is injected into the reactors as solid pellets of frozen Deuterium, stating 'Pellets can range in size from 0.5 cm to 5 cm, depending on the desired energy output per unit time. A standing pulsed fusion shock front is created by the standard initiators ranged about the forward inner surface of the sphere. The total instantaneous output of the IRC is throttleable from 10^{8} to 10^{11} Megawatts.'

Now it is tempting to say that each reactor can generate up to 10^{11} Megawatts, or 2.4 x 10^{12} MW for the entire system, but in fact this cannot really be so. At the efficiency noted above for the D/D fusion reaction, this would consume 31,290.743 kg of fuel per second - sufficient to exhaust the Primary Deuterium Tanks entire 10,250 metric tons in 5.46 minutes. Obviously this is nonsense.

From the description of the reactor operation above, it seems that the instantaneous power must refer to the rate of energy production while a particular fuel pellet is fusing; a 5 cm diameter pellet of Deuterium would mass 11.407 g, and when fused would release a maximum energy of 8.749 x 10^{11} Joules. In order to reach the maximum instantaneous power of 10^{11} MW, this pellet would be consumed in 8.749 microseconds. Unfortunately, the average power would also need to factor in how many pellets are fused each second, information which is not supplied.

We can only really make an educated guess at this number. We know that the Enterprise-D visited approximately one planetary system every two weeks throughout its life. The average time between refuelling for Starships is commonly pegged at three years; hence, the Enterprise could expect to visit 78 star systems. It has been stated that it is dangerous to engage warp drive within a star system; Kirk waited 1.8 hours before attempting to go to warp to intercept V'Ger in ST : TMP. On the other hand we have seen ships warp in and out of orbit many times in all of the series. It seems that the most we can really say is that it is dangerous to use warp drive whilst within *some* star systems *some* of the time.

On balance, I think it's not unreasonable to suppose that each visit to a star system will involve an average of two hours at impulse - one arriving and one leaving - for 156 hours total over three years. Hence, the ship would spend about 0.6% of its time at impulse speeds. Given the presumably lower fuel requirements for using Impulse as compared to normal Warp 6 cruising, my own personal guess for the power output of the Impulse reactors would be 10^{7} MW each, or 2.4 x 10^{8} MW total. This would equate to 11.43 pellets per reactor per second, or a total of 154,056,456 pellets massing 1,757,322.24 kg - about 17.151% of the total fuel - consumed over the entire three year time between refuelling.

Using this estimate, we would get a figure for the driver coil ratio of :

2.4 x 10^{14} x 7500 = (4.96 x 10^{13} x 2.8125 x 10^{11}) / r

1.8 x 10^{18} = 1.395 x 10^{25} / r

r = 7,750,000

So the driver coils would lower the mass of the ship from 4.96 million metric tons to approximately 640 kg - quite an impressive accomplishment!

Now of course, this figure is wholly dependant on my own assumption for how much power the impulse engine puts out. It's an inverse relationship - the higher you rate the impulse engine power output, the lower you must rate the driver ratio. Alternatively, the higher you rate the acceleration the higher the drive ratio is going to be - and of these two factors, the latter is by far and away the more likely to be off.

Assuming, then, a more realistic figure for the acceleration of 10^{7} ms^{-2}, we would arrive at a drive ratio of 7.75 x 10^{9}, and the effective mass drops to 0.64 kg.

## Conclusion |

The performance of the various power systems on a Galaxy class Starship are a matter of considerable conjecture, and for the most part what I've tried to do here is lay out the assumptions that I generally work by, give the justifications for them, and point out some possible alternatives. I'm not saying that any of this must be The One True Answer - I know for a fact that no less than Rick Sternbach disagrees with the seven year figure for Galaxy class endurance, for example. By laying out the numbers and equations in reasonably simple form I hope that people will be able to use this page as a basis for their own theories.

But some numbers we do know fairly well. To summarize a couple of the more important ones :

Maximum Warp Core Output : 4.77 x 10^{6} TeraWatts - 12.75 x 10^{6} TeraWatts

Average Warp Core Output : 1.56 x 10^{15} TeraWatts

Yellow text = Canon source | Green text = Backstage source | Cyan text = Novel | White text = DITL speculation |

Copyright Graham Kennedy | Page views : 13,662 | Last updated : 1 Jan 1970 |