Continuing my foray into "volumetrics hell" is a bit of a thought experiment conducted using a very rough, spitballed idea of the change in efficiency per cubic meter of volume propelled at warp speed. First, this is rough and it's still sketching out ideas on how to play a bit with things like a ship's fuel efficiency, so note the numbers are going to change. So second, I decided to set a baseline. Working from as official a figure as I could find on fuel consumption, going with 6,000 m³ of fuel (3,000 m³ of deuterium/3,000 m³) for the Enterprise-D, I then worked backward to 3 years of warp 6 till fuel exhaustion to get an overall generation of 1121TW for 3 years. This is a tiny figure in the context of a starship's warp reactor. In terms of volume this works out to about 80cc of fuel per second using the canonical conversion rate of around 93%. Now, the assumption is that the mass of a starship in fact influences how much power is needed to drive it a certain warp factor. So taking the 1121 TW, I then divide it by the Enterprise-D's official mass of 4.5 million metric tons. This gives us a needed output of 635,000 watts per cochrane per ton. Put more simply it works out to about 250 MW per ton to sustain the 392.5 cochranes of Warp 6. I'm assuming for the moment that the energy required to cross a peak transitional threshold remains constant over the years as it's described as being some sort of subspace physics constant. Therefore we then move on to comparative masses. For the Constitution I'm going to go with the more canonical figure of closer to a million tons than the much lower 190,000 metric ton fandom figure. Excelsior and Intrepid I'm using the "official" figure of 2,350,000 tons and 700,000 tons. Overall this means to sustain 392.5c each of the ships require the following output: Galaxy - 1121TW Constitution - 211TW Excelsior - 585TW Intrepid - 175TW Then we divide these output figures by the amount of total internal volume each ship contains. This gives us: Galaxy - 5,171 m³/TW Constitution - 1,038 m³/TW Excelsior - 1,451 m³/TW Intrepid - 3,582 m³/TW Basically for every TW of output each of the ships can move the given amount of payload volume. Basically it's an engine/nacelle efficiency vs. the rest of the ship measure. At around the same mass, a 24th century starship can have a payload 3.5x as voluminous. Now, we can of course assume that advances in warp coils, etc. would actually reduce the power consumption per cochrane per ton, but this is still a rather interesting way to look at the technological advances from the 2260s to the 2380s.

As a side note, the Intrepid/Constitution contrast gives us a nice place to examine the relative usefulness of nacelle as part of hull volume. Specifically of the 210,000 m³ of volume available to a Constitution approximately 25% or 54,000 m³ is dedicated to warp nacelles. (The number is similar for the refit) For the 625,000 m³ Intrepid the ratio is much smaller. 17,000 m³ per nacelle, or 34,000 m³ total which is 5.5% of total volume. We do know that Voyager when landing seems to be much more "back" heavy in that the nacelles are capable of counteracting the weight of the primary hull. The density of the rear half of the ship then becomes at least 50%, and given the relative volume of the hulls, most of the extra mass is probably in the nacelles. If we assume uniform distribution of mass on the Constitution and something of a 15% mass of Voyager in her nacelles, (leaving something like a 50/35 split for her primary hull/secondary hull) we have comparative nacelle masses of: Constitution - 212,500 tons Intrepid - 105,000 tons So per ton of nacelle... Constitution - 4 total tons/ton of nacelle Intrepid - 6.67 total tons/ton of nacelle And...(Rounded to the nearest cubic meter) Constitution - 1 m³/ton of nacelle Intrepid - 6 m³/ton of nacelle And finally on a volume basis(Rounded to the nearest cubic meter): Constitution - 3 payload volume m³/m³ of nacelle Intrepid - 17 payload volume m³/m³ of nacelle

First and foremost, I really dig your analysis and I think you're onto something. I've been thinking along the same lines for quite a while and I'm glad to see I'm not the only one. I see nothing wrong with your analysis written in the second post concerning volumes, mass and densities. You seem to be on pretty solid ground there. (I'd just like to note that the calculations below are rounded to the third order of magnitude. And in some cases even this is not really deserved.) But I would challenge some of your assumptions in the first post about energy usage. First, just so we're on the same page, I assume the 3000kL figure for antimatter comes from the TNG tech manual, page 68. But that's a volume, not a mass. You have to then assume the density to get a mass and from the mass calculate energy released. You do not have to assume the antideuterium storage has the same density as the deuterium storage. So, for example, at standard pressure and 20 Kelvins deuterium is a liquid and has a density of 165.6 g/L. At higher pressures and lower temperatures, its density is more. (Though solid deuterium is less dense than liquid deuterium.) So though the deuterium slush in the engineering hulls' storage tank is likely to be around 165.6 g/L, the antideuterium may be held at a higher pressure and have a higher density. Using the warp power chart on page 55 of TNG:TM, I get a warp 6 power requirement of 1330 TW (392 cochrane x 3.4 TW per cochrane) which is a little more than your 1121TW, 6wf for 3 year estimate. I get that you calculated backwards from 3000 kL and 3 years but I don't get how. Dividing 3000 kL by 3 years gives me 33.7 CC/sec of antimatter. Multiplying by 165.6 g/L gives 5.25 g/sec. Adding the same amount of matter and converting to power at 93% gives me 878 TW. So how'd you get 1121 TW? What did I do wrong? Using the 1330 TW as the power usage for the Ent-D at 6wf and extending it for three years, I get 135e21 joules released for 93% conversion to power. That translates to 1500 tonnes of M/AM, or 752 tonnes of antideuterium. To store that in 3000 kL, the density has to be on the order of 251 g/L. So it has to be under pressure. Indeed, I would argue that there has to be more antimatter than this because this is only enough for the warp drive, not the shields or navigation deflector. The figures may need to be doubled or tripled or more to include those. But my biggest objection is the assumption that warp power usage scales proportionately to mass alone. Warp drive seems to be a geometric effect similar to (but probably distinct from) the Alcubierre drive. As such, I suspect the shape and size of the field has much to do with it's power requirements. And though we can't know what that scaling factor might be, maybe we can make some educated guesses for a first order approximation. We can, for example, place an upper limit to scaling factor by size. Power requirements can not scale directly with the volume of the field. If that were the case then larger vessels would have no advantage over smaller ones: same percentage of volume taken up by fuel, same range;you could go just as far and just as fast in a shuttle as the Enterprise --assuming the coils could take that many cochranes of warp stress. And, of course, the lower limit is no scaling factor: in this case it would take the same power to push the shuttle at warp 6 as it does all the different Enterprises. Naturally, the answer is likely to be between these two extreames. As for mass scaling, if its entirely a geometric effect then mass would have no effect on the power usage. And though I could make arguments in favor of this approach, I do not believe it: if there is no momentum involved in warp travel than there would be no need for navigational deflectors or any threat to warp ramming. Since both are canon, it seems likely that considerable momentum is imparted upon a ship going warp. Thus considerable power must be used to accelerate any mass to warp. But is the mass scaling factor one-to-one? IE, twice the mass, twice the power usage for its acceleration? I don't know. On one hand it makes intuitive sense, even general relativity and quantum mechanics agrees with this requirement. But a symmetric, static warp field decreases the apparent mass of anything within the field. So I don't know that it's 1:1 or some other, more complex relationship. None the less, any scaling factor needs to take into account geometric size differences of the warp field as well as enveloped mass, even for a first approximation. And for further accuracy, shape needs to be factored. Lastly, I'd like to reiterate that I like your analysis. I think it's a fabulous first step. Further, I can think of no objections your second post, concerning density and size of the warp nacelles over the generations; another excellent step towards understanding what's going on.

It'd probably help for me to post up my numbers. (It's rather sloppy that I didn't, and I apologize) My notes say that my assumed mass figure for ²H was 196.7 kg/m³. I can't for the life of me remember where that figure came from. I'll recalculate with standard deuterium densities, I think. I'm actually still debating on whether or not the effects of mass, volume and field geometery on warp propulsion. The thing that kind of troubles me about the power consumption figures in the TNGTM is that the implication throughout on warp drive is that fuel isn't the limiting factor for higher warp factors. This implies that it's coil components or hull stress that starts causing problems at super high warp factors. However, a comparison of the energy requirement chart vs. fuel capacity (even assuming high density slush deuterium at the nearly 200kg/m³ I erroneously had above) suggests that fuel capacity would become an issue significantly faster than supposed other factors would be. Warp 9.6 for can't be sustained for the 12 hour limit because fuel exhaustion would kick in at the 4 hour mark. Also the other problem is that the physical size of the warp core seems extraneous with the fuel figures we have. Even a reaction chamber of only 1 cubic meter would expend all 6,000 m³ of deuterium in an hour and a half at maximum throughput. (Noting that the reaction chamber is supposed at least 2.3 x 2.3 x 2.5m in dimensions, it's probably substantially larger than 1m³ in reaction volume). We also know that the Galaxy-class puts warp reactor output to 75% when they go into Red Alert. And that this readiness state protocols require procedures in case of duty shift changes. In which case we should probably assume the ship can sustain that level of output for multiple duty shifts (4 hr intervals, presumably). But again, even a minimally sized reactor would eat through its entire anti-deuterium supply in a bit over 2 hours at 75% output. Something's not right here. Need to figure out what it is. I'm going to be posting more of an in universe examination, that kind of glosses over the whole tonnage issue soon as a side bar in the procurement thread (I do think size and mass should have some impact on warp propulsion, but that there's probably something in the order of subspace field stress that can shake a ship apart (see: Defiant).

First, I got it wrong. It's not 165.6 g/L at 1atm and 20k, it's 162.4 g/L at 1 atm and 18.5 K. 18.5 K is the triple point at 1 atm, where deuterium can be both solid, liquid and gaseous. If I knew what I was doing I could then calculate the density at many pressures. But I can't. As I think I said in the TOS Rom BOP thread, there is evidence that deuterium can be stored considerably more densely than the "standard" 162.4 g/L. It's called "ultradense deuterium" and is highly speculative at the moment but promises densities of 100 to 150 tonnes per liter. That is not a type-o. The circumstances for this density are, for the moment, not particularly conducive to antimatter storage: it's supposedly found in the pores of certain metals under odd circumstances. And the researcher who claims to have detected it has had his findings questioned. None the less, the theoretical foundations are apparently sound. So it is a possibility. Thus we can make the assumption that up to this rather large limit, we can store as much antideuterium as we need in as small a container as necessary. So, let's calculate M/AM mass requirements from 9.6wf for 12 hours and this will give a lower limit for our needs. For Ent-D 9.6wf needs 4.58e18 watts (2.4e15 W/coch * 1909c). Twelve hours of this is 213e21 joules, at 93% power conversion, equivalent to 2.36 tonnes M/AM, or 1.18 tonnes of antideuterium. Within a 3000kL volume, that's 394 kg/kL. You will note the total amount of fuel-mass here is nearly twice what I calculated earlier for three years at warp six! So, again the numbers don't match. Page 69 gives a normal deuterium tankage of 62,500 kL. Again assuming a density of 162.4 g/L --a questionable assumption since they specifically state a temperature of 13.8 K-- that's 5.28 tonnes of deuterium. I figure that's a lower limit for deuterium storage. If you had the same amount of antideuterium, you'd need 1760 g/L... I enter this data point for thoroughness as I do not expect this tto be accurate. However, I think I proved in the TOS Rom BOP thread that the bussard collectors could gather enough fuel to supplement the deuterium tanks for impulse power while in warp flight if you can fuse hydrogen into deuterium... And maybe even if you can't, but I'll have to double check that. So maybe this isn't a completely useless data point. Of course, that much fuel would run the engines at warp 6 for 31yrs, 10mo. So, I think a valid wild-ass-guess for antideuterium density is between 500 and 1000 kg/kL. That's less then 1% of the above ultradense limit and more than what's needed for the 12h @ 9.6wf requirement. I have no estimate for what 75% of its power output would be. Do you? ... I could never think of a way of estimating it, or even limiting the range. However, I just found on page 64 of TNG:TM that the cross sectional area of the reaction has a radius between 2.1 & 9.3 cm. Thus the "reaction chamber" does not have a volume but an area. Further, page 59 says the magnetic strictures between the injectors and the dilithium crystal "add between 200 & 300 m/sec velocity." Put together, this gives us a volume between 0.554 & 16.3 kL per second of both matter and antimatter. But we don't know at what densities so we can't convert that to mass or energy usage...that I can see. However, given that 9.6wf requires 4.58e18 watts and 93% conversion, that's 54.7 kg and 3.36 kg/kL in the injectors. Not particularly dense, which you'd expect. But, once again that only tells us what's needed for the warp drive. However, page 138 suggests the shields can draw a momentary peak of 3.31 TW (7 generators at 473 GW each) but a more normal maximum and continuous load of 2.688 GW. But even the larger number does not add to fuel consumption much: 39.6 milligrams per second of M/AM. And page 88 seems to place an upper bound to deflector power projection at 2.53 TW (3 projectors with 80% of peak momentary output of 675 GW each). Again, not a huge hit on fuel reserves: 30.2 milligrams per second (remember the assumed 93% energy to power conversion)....And this is supposed to represent maximum output of these systems. So I don't think the manual can be correct. And I agree with your assessment that the manual implies that warp speed limitations are in the materials of the nacelles, not the fuel supply. I will look for your in-universe post.

Specific thoughts on your observations maybe tomorrow. For now: Here's the post in question. http://www.trekbbs.com/showpost.php?p=8758975&postcount=60

@Akimoto -- I very much appreciated your warp nacelle post. I like it because it makes warp energy usage for Ent-D just one of the possible curves it *could* have been, and, indeed, what it was turned into. [LEFT]But it does beg the question if the four nacelle designed are an attempt to allow for enhanced change in warp geometry, allowing for better economy for any given warp factor... At the expense of considerable complexity. [/LEFT]

http://www.trekbbs.com/showthread.php?p=8779888#post8779888 Here's a stab at it, zDarby. Lemme know what you think.