Morning Phil for Roy WA,

Thank you for the reply and the explanation which I can understand.

I probably miss understood the comment posted somewhere in the 'Jump Drive Question' thread relating to a TL 14 and TL 15 JD.

Tom Rux
On October 17, 2019 at 9:37 PM "Phil Pugliese (via tml list)" <xxxxxx@simplelists.com> wrote:

 
That depends on what you mean by "better" & also whether or not you using CT LBB2 (standard mass produced commercial drives primarily for civilian use) or CT HG LBB (custom drives produced primarily for military/paramilitary use) 

I would consider the J6 (TL15) to be better cuz' it can do in one J6 jump what the J5 (TL14) takes two to accomplish.
(Using CT HG, it really doesn't matter what TL the gear is produced at except that higher TL's give a greater # of selection options)

The fact that the J5 drive is 'smaller' (takes up less hull space' doesn't necessarily make it 'better', IMO.

It's just smaller. 

While a argument could be made that a J5 'drive *could* or *should* get better 'numbers' if manufactured at TL15 (vs TL14) that doesn't seem to be the case.

For a detailed explanation of that I refer you to Prof Shulashgili's seminal work, "The Transition from TL14 to TL15 & it's Financial Effects Upon the Efficiency of Starfaring", University of Rhylanor, 1050.

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On Thursday, October 17, 2019, 07:37:05 AM MST, Thomas RUX <xxxxxx@comcast.net> wrote:


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

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