## Expense Epstein Drive## Over and over again, the SF series Expense is praised for its realistic Epstein Drive spaceship engine. Here is the physics behind the Epstein Drive.
SpaceX uses methane and oxygen for its new Raptor engines. Methane gives off 55.6 MJ/kg when burned with oxygen. Because, after all, a rocket engine has to take the oxygen with it: 2 O are needed for the C, which then makes one CO2 molecule. For the 4 H again 2 O are needed, that makes 2 H2O molecules. C has 12, H has 1, so CH4 has atomic weight 16. Oxygen too. With an ideal combustion mixture of 1 part methane, 4 parts oxygen by weight, you then get 11.12 MJ out of 1 kg. E = mv²/2. With the formula you can then calculate back the theoretical maximum exit velocity: Square root of ( 2 * 11.12 MJ ) = 4716 m/sec. If 100% of the combustion of methane with oxygen were converted into thrust, then the jet would come out of the engine at 4716 m/sec. SpaceX would like to achieve 3750 m/sec. Since energy increases with the square of velocity, this is 63% efficiency.
Let's take the simplest type of nuclear fusion for this. 4 hydrogen atoms fuse to form a He atom. Hydrogen has an atomic weight of 1.008, helium has an atomic weight of 4.002602. It is from this weight difference, 4 * 1.008 = 4.032, on which we live here, that is our sunshine. Let's take 1 kg per second again, as in the first example. In nuclear fusion, the 1 kg of hydrogen becomes 992.7 g of helium and 7.3 g of energy. Einstein's e=mc² is responsible for the conversion of matter into energy. 0.0073 kg * 299,792,458² = 655 TJ. T stands for Tera which means 10 to the power of 12. Of course, there is a mass defect in the combustion of methane and oxygen as well, but it hides so many places after the decimal point that it is meaningless in the calculation. Here, however, the mass defect is already 0.73%. Also here the ejection velocity is already so high that the Newton formula is no longer correct and one would have to go over to relativistic Einstein formulas. Is however for a first overview much too complex. It must be multiplied then still with an estimated efficiency. Per kg helium there are 660.1 TJ. This would result in an output speed of 36,335 km/sec. But since there are only 992.7 g helium this gives a thrust of 36 MN. The Raptor engine has 63% efficiency. Let's rather assume 100% for the Epstein Drive, because with only 63% efficiency a lot of waste heat would have to be handled somehow. This 7.3 g of energy is equivalent to 156,549 TNT. A little more than 10 Hiroshima bombs per second. If only 10% of that ends up as waste heat in the engine, it could cause overheating.
In Expense, if you are not in a particular hurry, simply accelerate half the distance at 0.3 G and decelerate the other half at 0.3 G. With an ideal engine with 100% efficiency, I don't want to do less efficiency here because of the waste heat problem, 1 kg hydrogen helium fusion would bring 36 MN thrust per second. With that, you can accelerate a 12,000 ton spaceship at 3 m/sec². To put this 12,000 t in perspective. The Mars battleship of the Donnager class has 250,000 t. So with 12,000 t you are somewhere between Rocinante and Donnager. So now let's take a little trip to Uranus and back. With 1 million second acceleration the first 1.5 billion km would be put back, then 1 million second braking, there we are with 3 billion km. The same again for the way back. 4 million seconds every second consuming a kg of hydrogen is unfortunately also 4,000 t.
250,000 t at 30 m/sec² is 7,500 MN of thrust. That would be 208 kg of hydrogen per second. The fusion energy corresponds to 33 Mt TNT per second. Why are you fooling around with railguns and such? If the jet of the engine is halfway focused, then it makes sense to blow the enemy away with 33 Mt TNT per second. |