Playing around with the hyperspace fuel equation

The hyperspace fuel equation is:

fuel = k \times ( \frac{dm}{o} )^p

Where:

  • fuel is the amount of fuel consumed in tons
  • k is based on the rating of the FSD:
    • A: 0.012
    • B: 0.010
    • C: 0.008
    • D: 0.010
    • E: 0.011
  • d is the distance in light years
  • m is the mass of the ship in tons
  • o is the optimized mass rating of the FSD
  • p is based on the class of the FSD:
    • 2: 2.00
    • 3: 2.15
    • 4: 2.30
    • 5: 2.45
    • 6: 2.60
    • 7: 2.75
    • 8: 2.90

Solving for d:

\begin{align} k \times ( \frac{dm}{o} )^p & = fuel \\ (\frac{dm}{o})^p & = \frac{fuel}{k} \\ \frac{dm}{o} & = \sqrt[p]{\frac{fuel}{k}} \\ dm & = \sqrt[p]{\frac{fuel}{k}} \times o \\ d & = \frac{\sqrt[p]{\frac{fuel}{k}} \times o}{m} \end{align}

If we set fuel to be equal to the maximum fuel per jump of the FSD, then we can determine the “distance factor” for the FSD since k, p, and o are all known based on the FSD. For example, given a 5A FSD with a maximum fuel per jump of 5 tons and an optimized mass of 1,175.4 tons, the distance factor would be:

\begin{align} \sqrt[2.45]{\frac{5}{0.012}} \times 1175.4 & = \\ \sqrt[2.45]{416.\overline{6}} \times 1175.4 & = \\ 11.730 \times 1175.4 & \approx \\ & \approx 13787.442 \end{align}

Divide that value by the current mass in tons of the ship and you’ll get the max distance that the ship can jump in light years.