Power: fusion
Every ship carries a small fusion reactor rated in gigawatts. Fuel is the easy part. The everyday reaction fuses deuterium, which is found in all water at about 35 g per tonne:
$$\text{D} + \text{D} \rightarrow \begin{cases} \text{T} + \text{p} + 4.0\ \text{MeV} \\ {}^3\text{He} + \text{n} + 3.3\ \text{MeV} \end{cases}$$The tritium and helium-3 it makes are burned in turn. A 10-minute, 5 GW burn then uses about 10 g of deuterium, so the usable energy per kilogram of fuel is:
$$\epsilon_\text{DD} \approx 3.0 \times 10^{14}\ \text{J/kg}, \qquad m_\text{fuel} = \frac{E}{\epsilon}$$Small craft often burn deuterium-tritium instead. It ignites more easily, so the reactor can be much smaller for the same output:
$$\text{D} + \text{T} \rightarrow {}^4\text{He} + \text{n} + 17.6\ \text{MeV}, \qquad \epsilon_\text{DT} \approx 2.9 \times 10^{14}\ \text{J/kg}$$D-T fuel is 40% deuterium and 60% tritium by mass. Tritium has to be bred in refineries and loses about 5% of its strength a year, so a D-T ship is tied to settled space. Explorers run on deuterium alone, because the fuel is everywhere.
The open bubble
A Level 1 bubble stays open to the outside universe. Its wall shimmers like heat haze, and it can steer, stop and be chased. Its power grows with the square of both its radius \(R\) and its speed \(v\):
$$P = P_1 \left(\frac{R}{5\ \text{m}}\right)^2 \left(\frac{v}{\eta \times 1\ \text{AU/day}}\right)^2, \qquad P_1 = 5\ \text{GW}$$The bubble factor \(\eta\) is the same wall-quality figure as for closed bubbles; 1 is standard. With the whole reactor output \(P\) on the drive, the ship moves at
$$v = \eta \times 1\ \tfrac{\text{AU}}{\text{day}} \times \frac{5\ \text{m}}{R} \sqrt{\frac{P}{P_1}}, \qquad R \approx 0.6 \times \text{ship length}$$That's about 1,730 km/s for a car-sized ship, or 0.6% of light speed. Open bubbles get steeply more expensive toward 1% of c (about 1.73 AU per day), so that's treated as a hard ceiling. Anything faster needs a closed bubble; see Interstellar dynamics. The trip itself costs:
$$t_\text{warp} = \frac{d}{v}, \qquad E_\text{warp} = P\,t_\text{warp}$$Plasma thrusters
A bubble drops out at the velocity it went in with, give or take a small trim. Matching orbits, parking at a Lagrange point, or landing takes ordinary thrust. The engines are wildly enhanced ion drives, closer to plasma torches, that throw ionised propellant (usually water) out at 50 to 100 km/s. The rocket equation sets how much propellant a change in velocity \(\Delta v\) costs a ship of dry mass \(m_0\):
$$m_p = m_0\left(e^{\Delta v / v_e} - 1\right)$$At thrust \(F\), propellant flows at \(F / v_e\), and the jet carries a power that the reactor has to supply, with efficiency \(\eta_t\):
$$t_\text{burn} = \frac{m_p\,v_e}{F}, \qquad P_\text{jet} = \tfrac{1}{2} F\,v_e, \qquad E_\text{burn} = \frac{m_p\,v_e^2}{2\,\eta_t}, \qquad \eta_t = 0.8$$If the reactor can't supply that power, the thrust drops to what it can:
$$F_\text{max} = \frac{2\,\eta_t\,P}{v_e}$$Faster exhaust saves propellant but costs power for the same thrust. So pilots run their engines hot and hard for liftoff, then efficient in space.
Gravity wells (not in the calculator)
A bubble can't form or collapse where gravity changes too steeply. For a body of mass \(M\), the minimum distance is:
$$d_\text{min} \approx 20{,}000\ \text{km} \times \left(\frac{M}{M_\oplus}\right)^{1/3} \left(\frac{R}{5\ \text{m}}\right)^{1/3}$$So every trip begins and ends with thrusters: climbing about 20,000 km out of an Earth-sized well costs roughly 9 km/s each way. To include that, add it to the calculator's delta-v.