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*To*: "starship-design@lists.uoregon.edu" <starship-design@lists.uoregon.edu>*Subject*: starship-design: Reactor output losses.*From*: Ben Franchuk <bfranchuk@jetnet.ab.ca>*Date*: Fri, 13 Nov 1998 16:51:32 +0000*Reply-To*: Ben Franchuk <bfranchuk@jetnet.ab.ca>*Sender*: owner-starship-design@lists.uoregon.edu

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...

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