1) Let's assume we have a fixed tonnage budget for propulsion, i.e. engines and fuel, and we're interested in the trade-off between performance and fuel efficiency.

Let x be the proportion of fuel in our propulsion tonnage.

(1-x) is therefore our engine proportion. Keeping speed constant, this is proportionate to the inverse of our power multiplier.

Specific fuel consumption is proportionate to power multiplier ^2.5

Our range is therefore proportionate to x(1-x)^2.5

I find it more useful to map fuel use to speed at constant range than to range at constant speed: (x(1-x)^2.5)^0.4

Standardising for 1.0 as the highest possible speed, achieved at x=2/7, we get

(x(1-x)^2.5)^0.4*1.4/(2/7)^0.4Google calculator (just put the last function into a Google search field) gives us a nice graph with value pairs if we mouse over it.

This is ready for use with no further thought, I'm not aware of any hidden pitfalls.

*

2) Now let's look at the effect of engine tonnage on cost and fuel efficiency. Speed is kept constant. We use a standard size for every single engine.

We simplify things a little and think of our ship as consisting of engine and payload, x being the proportion allocated to engines and (1-x) therefore being the payload.

Power is kept constant, so power multiplier is proportionate to 1/x.

Fuel consumption is therefore proportionate to (1/x)^2.5

Fuel efficiency is therefore proportionate to (1/x)^2.5/(1-x)

Standardising for 1 at the most efficient setup at x=0.714286, we get

((1/x)^2.5/(1-x))/8.11686With power multiplier above 1.0, the engine cost is the same no matter the size. Cost efficiency of the propulsion plant is therefore

1/(1-x), the theoretical optimum being an infinitesimally small engine with infinite power multiplier.

With power multiplier below 1.0, cost of a single engine scales quadratically with power multiplier, so engine cost is proportionate to 1/x. Cost efficiency of the propulsion plant is therefore proportionate to (1/x)/(1-x). Standardised for 1 at our most cost-efficient setup, achieved at x=0,5, we get

1/(4x-4x^2)*

Putting

((1/x)^2.5)/(1-x))/8.11686 , 1/(1-x), 1/(4x-4x^2) into the Google search field gives you these graphs (fuel efficiency, cost efficency > 1.0, cost efficiency <1-0). For a useful overview we may want to zoom to have about 0.15 to 0.75 for x and 0.5 to 6 for f(x).

While I find this useful, this needs a little more caution than 1).

We count everything that isn't engine as payload. That includes bridge, engineering spaces, armour (can't do away with the first layer), and all sorts of other things we may not think of as mission tonnage. So a seemingly very fuel-efficient design with 70% engines may not carry a lot of stuff we actually care about, dramatically lowering practical efficiency.

What we count as overhead and what as mission tonnage isn't even fixed: for a general purpose warship armour may be mission tonnage, if a given speed in nebulae is a non-negotiable requirement some or all may be overhead.

Likewise, our ship will need to carry some fuel at the cost of payload, so our ship with the compact high-power engines may not be quite as cost-efficient as the graph implies. And if we'd need >40% of engine weight in fuel, corresponding to the 2/7 of propulsion tonnage as in 1), we probably dropped the ball.

Comments about correctness, clarity, usefulness or considerations I neglected are warmly appreciated.

**EDIT**: Assuming 5%/10% of ship size as overhead to be subtracted from the effective payload, one gets

((1/x)^2.5)/(0.95-x)/9.7130565 , 1/(0.95-x)*0.95, 1/x/(0.95-x)/4.43213

or

((1/x)^2.5)/(0.9-x)/11.7365 , 1/(0.9-x)*0.9, 1/x/(0.9-x)/4.93827

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