Morning from WA,

I'm stepping through design evaluation section of MT Referee's Manual pages 84-89 for Guardian using the specifications on MT COACC page 93.

The Hull inoperative/destroyed damage points of 90/225 are for a 100 UCP hull. My calculations for a 120 UCP hull has the inoperative damage points as 1,620/15 = 108 and the destroyed damage points as 1,620/6 = 270.

I've run into a small issue with Power specification duration calculations.

The Guardian's duration specification is 120 days of continuous operation and 360 days using the power plant for eight hours a day with a fuel load of 120 kl.

MT Referee's Manual page 85 the instructions are to

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."

2. Work Days

"Divide the number of hours by 8 to determine the number of eight-hour work days."

3. Annotation

"The two numbers, 24-hour continuous days and 8-hour work days, are separated by a slash."

1. Continuous Operation

Guardian power plant fuel tankage = 120 kl or 120,000 liters

A Fusion 12 power plant's fuel consumption is found on MT Referee's Manual page 62 as 0.003 kl/hr unmodified by the use of fuel purification plants since none are installed.

Fuel Tankage =120,000 liters/Fuel Consumption 0.003 kl per hr./24 = 40,000,000/24 = 1,666,666.7 days. This does not equal 120 days of continuous operation shown on the Guardian's specifications. Leaving the fuel tankage as kiloliters results in 1,666.7 days of operation, which is still not 120 days.

120 days = Fuel tank/0.003/24 = 120 x 0.003 x 24 = 8.64 kl of power plant tankage which does is not 120 kl as indicated by the Guardian fuel tankage.

The Guardian's TL 12 fusion power plant has an output of 6 Mw. The power plant has an output of 2 Mw per kiloliter of plant which means that the plant is 6/2 = 3 kiloliters. Fuel Consumption = Power Plant volume 3 kl x Fuel Consumption 0.003 kl/hr. = 0.009 kl/hr. To operate for 24 hours the plant consumes 24 x 0.009 = 0.216 kl. To operate for 120 days the Guardian's fuel is 120 x 0.009 = 25.92 kl.

I was hoping that all my approaches would work out with at least two matches and now I'm confused. Can someone help me out please?

Tom Rux