pp 294&295 contradiction on drive efficiency meaning, and strange formula
If I'm not mistaken, there's a contradiction between the notes at the bottom of table X on p 294 and table Y on page 295, and a rather crucial one to actually choosing drives.
p294 says
Efficiencies round down (thus Early Jump-1 at
90% becomes Jump-0).
p 295 says
Example2. At TL12, maximum Jump for the drive is (Table W) Jump-3. The builders acquire a Generic Jump Drive-D TL13
with the same tonnage, Efficency=90%, Fuel Requirement= 1.1. The Drive can produce Jump-4 while costing only MCr12.5.
The ship, assuming a Power Plant with equal potential, requires 4 x 10% x 200 x 1.1 tons = 88 tons of fuel for a Jump-4.
[Hm. Firefox points out that there's a typo there, in fact: "Efficency" instead of "Efficiency". So an erratum in any case.]
If we applied the "efficiencies round down" rule to example 2, it would mean you could NOT get Jump-4 out of that Generic drive. This is a rather drastic rule, but if it is intended to be applied, it should be spelled out in big red letters (on a black background no doubt), and Example 2 should be amended.
If this is indeed the way it's supposed to work, though, it would be nice to have a more detailed formula somewhere that allows you to choose drives.
For example, if I want Jump-4 for a larger ship, say 400T (hull D), the tables say I could use either Jump drive H or J. H is the minimum. But say I want to save money and buy a generic drive. By the round down rule, Generic drive H gives 90% efficiency and drops me to Jump-3. But surely drive J, though the Drive Potential table also rounds its performance to Jump-4, is actually stronger than drive H. Presumably, or at least potentially, a Generic drive J would still perform at Jump-4.
As a referee, I would absolutely say so as a judgement call. A formula would help.
Now here's where we get to another very confusing thing, possibly an erratum: in two places there's a formula referenced (bottom of pp. 322 and 346, directly in reference to Maneuver drives and Power Plants, but apparently referring to Jump drives as well) that says
Drive Potential for a specific drive is the EP (Energy Points) per Hull Ton; ignore fractions and round down.
The Starship Drive Potential table pre-calculates values for most common combinations of drive and hull.
The Drive Potential table is table Z1 on p 296. But that's not the formula it's using. It is using the formula Potential=EP*2/ton. I'm guessing this came from some weird idea of adding the drive EP to the power plant EP and dividing that by the hull tonnage, but it doesn't make that clear anywhere, and it doesn't really make sense, since that would mean drive potential could be boosted by a big power plant to something higher than its rating.
So anyway, I think this should be laid out much more clearly, right up front in the discussion of Starship Design on p 278. It should state what EP are (which is not mentioned on the tables in question and was
very hard to find), and give the formula
Drive Potential = (Drive EP)*2/(Hull Tonnage). Then when you get to Tech Level Stage Effects you can refer back to that formula for a proper calculation. In my example above, I get that a Generic Drive J, with 90% efficiency, gives (900EP*2/400Tons)*0.9 = Jump-4.05; success!