Drive Efficiency

[RogerD]

I'm trying to understand the efficiency rules. For power plants (and related) it is straightforward enough - an Experimental Power Plant-B would produce 200x50%=100EP.

I'm pretty sure the rule as written would be that all drives use this same math and that the Power Plant must supply EP at least equal to the modified EP input.

However, this seems backwards for drives who take power as input. Shouldn't we divide instead of multiply to get the required input? So an experimental Jump Drive-B would require a power plant to supply 200/50% = 400EP. You would need a standard Power Plant-D or better to supply these power needs...

 

[AnotherDilbert]

You need the same potential power plant, not the same EP.


But you certainly don't need a 400 EP power plant to supply a 100 EP drive, you need an (about) 100 EP power plant.

 

[kilemall]

You need the same potential power plant, not the same EP.


But you certainly don't need a 400 EP power plant to supply a 100 EP drive, you need an (about) 100 EP power plant.

I would think both, an experimental drive would demand more power and the power plant would be inefficiently larger to supply the demand.

 

[AnotherDilbert]

Even T5 has a limit to the level of complication: It doesn't track power separately.

The only requirement is equal or higher potential.

 

[RogerD]

I guess I see what you mean about the rules deal with potential, not EP. In the design I was working on (the first ship with a jump drive for a civilization), it struck me funny that I could put a lower power plant in when it is driving a less efficient jump and maneuver drive. Given the rules, I guess this is OK though. It's not like the experimental drive is not being penalized

Example:
Standard TL9 Hull-A, 100 tons
Experimental TL9 Maneuver Drive-B, Potential=1, 9 tons, MCr 180, EP=100
Experimental TL9 Jump Drive-B, Potential=1, 45 tons, MCr 450, EP=100
Improved TL9 Power Plant-A, Potential=1, 4 tons, MCr 4, EP=110

By both the potential and the EP, the Power Plant A should be able to drive the two experimental drives.

 

[AnotherDilbert]

By both the potential and the EP, the Power Plant A should be able to drive the two experimental drives.

Yes, the drive letter is irrelevant, it's just the potential that matters in T5.