Okay so... I'm going to do this in costs because that's the main data point I have. A ship 135m long has about a 60 billion credits of eezo in the core, a ship with a double eezo mass core (120 billion) has twice the endurance, but doesn't seem to be terribly faster. Fighters have 10 million credit cores, speed unknown. A dreadnought's core should definitely >24 Trillion (the linear scale) but <~32 Trillion (at an estimate). Ships don't go faster that 15Ly/day as both technological limit and a cost limit. Most ships need to do at least 5Ly a day to get anywhere I set this a being 5000 times less than a 15ly core to make civilian ships reasonable. Most ships have a base endurance of 50hours. Ships larger than a 1km start becoming prohibitively expensive. 2km is the sort of thing only a non-capitalist (Geth, Reapers) system could pull off as the cost is pretty much insane (note that cost~mass so at some point you're describing a ship that's 99% eezo.
Other than that... we have piddly canon data (there's some very questionable SR-2 stats) and UberJJK seems to have made bigerships cheaper via some effect so the extra money can be tossed at FTL cores. It maybe that Dreadnoughts due to their size just can't afford 15Ly cores, and 15Ly cores maybe reserved for lighter vessels. Reapers also can do 30Ly in a 2km ship but that maybe do to a different velocity curve than normal.
Does that help or do you need some other info?
Hey
@UberJJK can you tell me what you'd put for a "normal" 1000m Dreadnought, I want to see how much money Yog has to play with to make core costs non-linear.
Ok, so, I can start with these. The question is "how much power is needed to achieve FTL of a given speed?", right? I am not mistaken here, am I? So, first several assumptions:
1) "amount of eezo and power increase exponentially with the mass affected and the top speed" - I'll interpret it like this: a given current (total charged particle flux) will generate a given K (ME coefficient) in any sufficient amount of eezo. Ie if you pump 1 A of current into 1 kg of eezo, the effect will be the same as if you pumped it into 10 kg of eezo. Secondly, I'll say that eezo has "saturation stage" ie you can't pump more than a given amount of current into a given amount of eezo (it'll either overheat, or won't produce stronger field or will discharge immediately, etc). Thus, I'll say that
the total amount of power is proportional to the amount of eezo in the core. This also means that, basically, costs are linearly dependant on the total power.
2) For the sake of the first approximation, I won't account for potentially different discharge rates (ie that a larger core generating the same effect as a smaller core will discharge less frequently) - I'll do this a little bit later
3) I'll assume a fighter to be around 5 meters long and having the cruising speed of 5 light years per day - it's not like they need more, and you put this as a basic low limit of speed, and it's a reasonable one
4) I'll assume all ships to be proportional to each other, so M~L^3
5) I'll completely divest the total sublight top speed from the power of the eezo core. Ie I'll say that the engines and the eezo core reactor are two separate systems. And it makes sense, as in the engines you need high energy particle generation (high momentum), and in the core you need high currents, but low energy particles.
6) I'll use the numbers from
my paper for C and K. Meaning that the average travelling speed V=C_int/20=C/sqrt(K)/20. Meaning that K=(20*V/C)^2
The formula I am going to be fitting with the data we have is going to be like this
W(M,K)=C1*exp(C2*(abs(K)-1)+C3'*M)
W(L,V)=C1*exp(C2*((20*V/C)^2-1)+C3*L^3)
This gives me three variables to play with. That's quite a lot, really. Substituting cost for power, I have the following data points:
Cost, millions
|
Speed, LY/day=365.25*C
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Size, meters
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6*10^4
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15
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135
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10
|
5
|
5
|
28*10^6
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15
|
1000
|
This is actually quite neat as I have three variables (C1, C2, C3) and three data points. They give me the following resulting formula:
Cost(L,V) [millions]=3.377*exp(0.0434*((V[LY/day])2-1.87*10-8)+6.16*(L[m]/1000)3)
Now, let's stress test this thing, ok?
First, let's check how civilian (5 light years per day) vessels compare to military ones (15 LY/day). And the result is they are
5884 times cheaper. Which is pretty damn close to your stated goal of 5000 times cost reduction.
What is the cost of a 2 kilometer long military vessel? Prohibitively expensive standing at an astronomic
1.5*1026 millions of credits. Meaning that no economy can handle those, and, at those scales, you either take severe speed reduction, or you optimize mass a lot or you figure out better technologies.
What is a cost of a Reaper, 2 km long and 30 LY per day fast?
7.8*1038 millions of credits. Again, I sorta kinda see why they make so few of those per cycle, even accounting for completely different logistics and economics and superior technology.
Now, please keep in mind that this is a first approximation preliminary analysis. With comments and feedback it's going to be revised a lot. I certainly am going to include discharge times (probably reversely proportional to the size of the core) at least. Other factors too
(point me to them).