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starship-design: Reactor output losses.
Direct thrust reactor:
For quick calculation of thrust I have a point source of Energy hitting
a square box.
| | Only 1 side of the box gets hit from
the reaction thus 1/6 of the energy
| * | is converted into thrust. Because most
of the energy stikes at a angle
| ______| I guess 2/3 of the energy is turned into
thrust. That is about 11% energy
If a 4500MW reactor is 3 meters. 1/3 of the radius is cube root of the
A smaller reactor 16 MW with a radius 1 meter is my guess for a
practical reactior @ 2050.
Now to find a formula to convert 1.75 MW into thrust...