Ship's characteristics:

  • Powerplant : fusion
  • Size : 35 m (length) x ~ 17.5 m (width) x ~ 10m (height)
  • Fuel : offboard
  • Desired Performance : deep interplanetary (kuiper belt)

Power per reaction:

  • Deuterium ($^2H$) + Tritium ($^3H$) = Helium ($^4He$) + 3.5 MeV (source)
  • Helium to Carbon ($^42He + ^42He \rightarrow ^{12}_6C$) = 7.275 MeV (net; by way of Beryllium) (source)
  • Helium + Carbon to Oxygen = 7.162 MeV

Events per volume of material:

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Thinking that "100 years from now" technology could give you the best available $10^{-21}$ reactions per cubic meter, per second.

Let's say most of the ship (35 meters long, 17.5 meters wide, 10 meters high) is engine, and about nearly all of that is reaction vessel (which is a big assumption compared to today's technology).

Reaction vessel is $35 \times 17.5 \times 10$ = 6,125 cubic meters Power generated is $3.5 MeV \times 10^{-21} \times 6,125 $ = $2.14 \times 10^{-16}$ MeV

That's not gonna work.

Assuming "100 years from now" technology can also give you better reaction rates. Since we're going bold, let's say 25% of $\dot{m}$ is reacting.

So, each gram of 50-50 Deuterium/Tritium mix (assuming hydrogen fuel) (avg. molar mass 2.5) contains ${1 \over {2.5}} \times 6.02 \times 10^{23}$ reagents, 25% of which react, and produce 3.5 MeV on each reaction (and assuming the energy imparted the fast neutrons is unrecoverable). 2.107 $\times 10^{23}$ MeV $\approx$ 3.3 $\times 10^{10}$ Joules of energy (33 gigajoules) per second. Or 33 GW. Scaling up to kilograms of reactants would give 33 TW.

Let's say the deductions of power to run accessories is negligible (but maybe it isn't). What's the $\dot{m}$ and $ve$ out the back? $3.3 \times 10^{13} = {1 \over 2} \times 1 \times v^2 \rightarrow 6.6 \times 10^{13} = v^2 \rightarrow ve = $ 8,124,038 meters per second (2.7% c)

$\dot{m}$ is 1 kilogram per second. The rocket equation (excluding nozzle effect) is $F = \dot{m} v_e$

Thrust, therefore, is 8,124,038 Newtons (8.1 MN) per kilogram of fuel mix. That's in the same region as the 764 kN produced by space shuttle main engines.

What's the peak amount of fuel that can flow through the ship?

Say the fuel is stored in liquid form, and the fuel line can be no bigger than the ships' width (17.5 meters). And let's say it's circular. $A = \pi r^2 \approx 240$ square meters. The volume is equal to that area times the rate in which fuel is being brought aboard by the pumps. The density of liquid hydrogen is 70 kg per cubic meter.

Let's say 1 $m \over s$ for now, giving $\dot{m}_{fuel} = 70 \times 240 = $ 16,800 kilograms per second.

Applying that to the thrust per engine : 16,800 $\times$ 8,124,038 Newtons $\approx$ 136 giganewtons.

You can go higher or lower on fuel flow rate. And select a different fuel. Or re-adjust down the huge "future technology" boost to reaction rate.


The mass-energy efficiency of this set-up is the $E_{extracted} \over {\dot{m} c^2}$. You only get 33 TJ per kilogram. That's a mass-energy efficiency of $3.6 \times 10^{-7}$ or 0.000036%

This is important for evaluation long-range performance (like interplanetary travel). To accelerate your load to a cruise velocity of 0.01c requires a mass-energy of $(0.01c)^2$, times the mass of your ship (or 0.0001 m in this scenario).

How much fuel mass your motor requires to confer this energy is taken by dividing the mass-energy requirement (0.0001 m) by your mass-energy efficiency (3.6 \times 10^{-7}) getting 277 kilograms of fuel mass required for every kg of payload (and does not include deceleration).

That's clearly not going to work. So, let's try a lower top speed: 0.001c ~ 2.7 kilograms per kg of payload.

Considering both acceleration and deceleration, then, for every kg of payload, you'll need 2.7 / (2.7 + 1) = 72% of your mass to be fuel. And another 72% of the residual to be fuel for deceleration, giving 72% + (72% x 0.28) = 92% of your payload mass will need to be fuel.

Maybe an even slower top speed. Let's try 0.0001c ~ 0.027 fuel/pay = ... most of your cargo capacity can be payload, instead of fuel.

What's that do to performance? 0.0001c is a cruise velocity of ~ 30,000 meters per second / 100,000 kilometers per hour. To travel a distance of 1 AU at this cruise speed (approx. 8 light-minutes) takes about 55 days.

Thrust Performance

To push a 150 ton load (150,000 kg) with 4 ships each providing 8.1 MN of thrust up to a cruising velocity of 0.001c, you could get accelerations up to 216 $m \over {s^2}$ (or ~21 gees). Not sure you wouldn't want to limit to only a few gees. At that acceleration, you'd reach cruise velocity of 300,000 $m \over s$ in 23.1 minutes.

If 150 tons is not inclusive of fuel, you'll need 1,725 tons of fuel. Total weight = 1,875 tons. Acceleration in this case would be 17.28 $m \over {s^2}$, and it would take 4 hours, 49 minutes (approx) to reach cruising speed.

Acceleration will actually vary over the course of the burn (as the weight being pushed starts to drop), being lower than the average value at the beginning of the maneuver and higher than the average at the end.