Morning from the Pacific Northwest,

The discussion on the Jump Drive Question thread has me wondering about my understanding of how TL and the percentage of the hull required for the jump drive, maneuver drive, and power plant work.

I build a hull at TL 15 with a J6 drive is available and requires 7%, a M6 drives requires 17%, and a power plant that is 1% x Hull tonnage x Pn of the hull

Instead of the TL 15 J6 drive I install a J5 drive which on the Drive Tech Table is TL 14 and on the Drive Potential Table requires 6% of the hull.

How can the J5 drive perform better at TL 15?

Tom Rux