Re-imagining the Book-2 drives

[infojunky]

FREE TRADER ENGINEERING (M1/J1/P1)

• LBB2 = 15 dT, MCr 22
• HG:TL9-12 = 14 dT, MCr 22
• HG:TL13-14 = 12 dT, MCr 22
• HG:TL15 = 10 dT, MCr 22

Stolen... And dropped into the Blackhole of Quality.

My back burner project surfaced again this week (Obviously, my family is visiting from Japan so I should be cleaning in prep for having a full house). As such the various flavors of Power Plant and EPs have been floating around my head. As Power is the real difference between a military ship and a civilian one.

 

[Straybow]

Wet blanket alert

• Because of the liabilities of the Model A maneuver drive, the Model A power plant is designed to function also as an auxiliary propulsion unit. By passing additional hydrogen fuel through the plant, it acts as a fusion thruster to produce another 1000 metric tons thrust to compensate for the maneuver drive's loss of thrust.

Nah, not possible. A 500 MW power plant, consuming 5 tons H2/day, can't produce 1000 ton·gees of thrust. At 5 tons/day the mass flow rate would only be 58 grams (yes, grams) of H2/second. Based on conservation of momentum, exhaust velocity would have to be:

1,000,000 kg · v/s(ship) = 0.058 kg/s · v(exhaust)

1,000,000 kg · 10 m/s² = 0.058 kg/s · v(exhaust)

v(exhaust) = (1,000,000 kg · 10 m/s²) / (0.058 kg/s)

v(exhaust) = 172 Mm/s = 0.574c

Well, that is physically possible. Before I ran the number I thought it might require Ve > c.

But total energy (neglecting relativity for back-of-envelope calc) required is:

½ · 0.058 kg/s · v² = 862 TW

which is more than 1,000,000 times more power than the 500 MW power plant.

 

[aramis]

Nah, not possible. A 500 MW power plant, consuming 5 tons H2/day, can't produce 1000 ton·gees of thrust. At 5 tons/day the mass flow rate would only be 58 grams (yes, grams) of H2/second. Based on conservation of momentum, exhaust velocity would have to be:

1,000,000 kg · v/s(ship) = 0.058 kg/s · v(exhaust)

1,000,000 kg · 10 m/s² = 0.058 kg/s · v(exhaust)

v(exhaust) = (1,000,000 kg · 10 m/s²) / (0.058 kg/s)

v(exhaust) = 172 Mm/s = 0.574c

Well, that is physically possible. Before I ran the number I thought it might require Ve > c.

But total energy (neglecting relativity for back-of-envelope calc) required is:

½ · 0.058 kg/s · v² = 862 TW

which is more than 1,000,000 times more power than the 500 MW power plant.

All of which points towards an N-space deformation drive. Rather than expanding/contracting space, it creates a gravity gradient using much lower energies which you fall through...

 

[kilemall]

Huh.

I was figuring the jump was a massive muon-catalyzed fusion event, which created the huge pulse of power, and then all that 10% per jump mass got ejected at once, which is fine, ship gets powered through the 'whateveryouwannacallit' and we're done.