starship-design: Reactor output losses.

[Ben Franchuk <bfranchuk@jetnet.ab.ca>]

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
                                 conversion.

If a 4500MW reactor is  3 meters.  1/3 of the radius is cube root of the
power?
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...