Hello Cian,

>On July 18, 2019 at 11:15 PM Cian Witherspoon <xxxxxx@gmail.com> wrote:

>It might be a typo, depending on when the design was written up in regards to the design sequence being created.

>So far it looks like they set aside 10% of the hull for fuel usage, or they didn’t do the numbers right and did

>consumption as roughly 10kl/day. I don’t have any of the specs or the MT design stuff available, but see if they

>accidentally swapped the numbers on fusion plant size and fuel usage.

Thank you for providing plausible scenarios of why I'm not able to get the right numbers for the Guardian.

In "A MT Starship Design Example" Mr. Fugate's method followed the suggestion to calculate the fuel load of 30-day of power plant operation per MT Referee's Manual page 60. The process used was Fuel Consumption x Power Plant Volume x 24 x 30. To get done with the Design Evaluation and filling out the UCP specification sheet I've used this method substituting 120 for the 30 days in the design example.

1. Continuous Operation

"Divide the power plant fuel tank volume in liters (do not include starship jump drive fuel) by the fuel usage per hour to determine the hours of continuous operation. Divide the number of hours by 24 to arrive at the number of full days of continuous operation."

I feel that the continuous operation instructions should have left the fuel tankage in kiloliters instead of liters and use kl when I calculate the continuous operation duration.

My calculations return a fuel usage for a TL 12 Fusion Power Plant Volume 3 kl x Fuel Consumption 0.003 = 0.009 kl/hr and 0.216 kl a day. For 120 days the fuel load would be 25.92 kl.

Days of Continuous Operation = Fuel tankage 25.92 kl/Fuel Consumption 0.009 kl/24 = 2880/24 = 120 days.

Eight-hour Work Day Operation = Fuel tankage 25.92 kl/Fuel Consumption 0.009 kl/24 = 2880/8 = 360 days.

I used the wrong fuel consumption rate in my first go round, but even using the correct fuel consumption will not bring 120 kl of fuel to be operate for 120 days.

Days of Continuous Operation = Fuel tankage 120 kl/Fuel Consumption 0.009 kl/24 = 13,333.3333/24 = 555.5556 days.

Thank you again for your reply and for catching me in another math error.

Tom Rux