First, I made a mistake.
I wrote:
 > In Traveller: The New Era TL 15 fusion consumes 100L per 6MW-years
 > or 6/1400 = 0.43%

This should be 6/140 or 4.3%. 1L hydrogen = 1.4MW-years fusion power.
Still not very good in terms of power plant performance.

Richard Aiken wrote:
 > So . . . when I give my Fireflyesque multi-function reaction engines
 > enough on-board fuel endurance to handle 4 round trips to jump point
 > and back from a typical planet for ~5% of ship volume, I'm in the
 > right ballpark, physics-wise?

Hmm.
Using the first row of the travel time table in MT Ref's Companion (p.21):
.28 days at 1G, .2 at 2G, .16 at 3G, .14 at 4G, .12 at 5G, .11 at 6G
equals 6.72, 9.6, 12, 13.44, 14.4 and 15.84 G-hours respectively.

4 round trips = 8x above G-hour figures.

Reaction drives that only need 5% of ship volume in fuel to get
multi-G-hour performance are unlikely, thinking about the standard
rocket equation. [delta-v = exhaust velocity x ln(initial mass/final
mass), where ln = log to base e]

I get required delta-v on the order of 1930-4500km/sec assuming 4 round
trips over the 1-6G acceleration range.

5% volume implies (initial mass/final mass) gets close to one. If the
fuel is liquid hydrogen, the mass is about 0.4%. So ln(1/0.996) is
~0.004.... The required exhaust velocity exceeds the speed of light.

You're going to need lots more fuel.

I decided to push things to the limit with my previous post. What would
it take to get the performance level required?

Direct conversion of most of the fuel's mass-energy to kinetic energy.
Exhaust (waste energy) is a plume of neutrinos to avoid too many
unpleasant interactions with nearby matter.

180G-hours thrust? 0.4% of the volume of the hull is fuel. 3 parts per
10,000 mass units is fuel.

Rob O'Connor