On Thu, Oct 10, 2019 at 13:32 Thomas RUX < xxxxxx@comcast.net> wrote:

Afternoon from Roy, WA Cian,

Thank you for the reply and that my first calculation of 240 tons of jump fuel would be required was correct.

Looking at the math:

1. 400 ton hull - 240 tons of jump fuel = 160 Hull tons - 65 tons for a 600 ton hull JD4 Type M drive = 95 hull tons - 20 tons for the bridge = 75 hull tons - 7 tons for a Model/6 Computer = 68 hull tons - 37 tons for a Type M Power Plant 6 = 31 hull tons - 60 tons of power plant fuel = -29 hull tons

Changing the jump fuel from 240 to 160 tons using the 600 ton hull JD drive's 4 parsec:

2. 400 ton hull - 160 tons of jump fuel = 240 Hull tons - 65 tons for a 600 ton hull JD4 Type M drive = 175 hull tons - 20 tons for the bridge = 155 hull tons - 7 tons for a Model/6 Computer = 148 hull tons - 37 tons for a Type M Power Plant 6 = 111 hull tons - 60 tons of power plant fuel = 51 hull tons

Of course this is a possibility that I'm out-to-lunch again.

Tom Rux

On October 10, 2019 at 12:52 PM Cian Witherspoon < xxxxxx@gmail.com> wrote:

The first available drive that provide J4 capability in a 600 ton hull is the model M. In a 400 ton hull, the same model M provides J6. Fuel usage is the 240 tons required for J6 at 400 tons. Just think of it this way, which is implied by the table: each higher letter drive outputs a certain amount of jump power, which is divided by the tonnage of the ship to provide a jump range, and j6 is the maximum survivable.

Now, in modern canon, the actual ability depends on what TL the drive was manufactured at, but that’s just MT onwards.