Fuel
To get the liters of fuel I assumed liters of antimatter plus matter.
This results in complete conversion of mass to energy E = mc^2
With c^2 approximately equal to 10^17 this means there are 10^17 joules per kilogram.
The energy needed to accelerate a particle or spaceship to a speed v depends on gamma,
and is equal to (gamma 1) times the rest mass of the particle.
Energy = m0 (gamma1)
The same energy is needed to stop the particle.
So the energy for our trip is double the energy needed to accelerate the spaceship up to speed.
Thus my table uses E = 2 m0(gamma1) to calculate liters of fuel.
If the fuel is anti water plus water it has a density of 1 Kg per liter and the fuel can be easily converted from literrs to kilograms or viceversa.
Scientific Explorations with Paul Doherty 

8 May 2005 