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*To*: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, bmansur@oc.edu, DotarSojat@aol.com*Subject*: Re: Explorer Power Gain Problem*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Fri, 15 Mar 1996 12:59:05 +0100

>> The formula is right, but the number is only valid when the velocity of the >> starship is 5 m/s. The problem with the document is that the meaning of the >> number is not told. > >Now I remember. I calculated the power the drive system would need for a one >second boost, and assumed it would be constant for the flight. I.E. if >it takes X watts to push the ship at a ship G for a secound. Multiply that >by the number of seconds of boost to get to desired speed, and .... > >Given that the engine should need to output the same power, to accelerate the >same mass, at the same rate, at diferent speeds. It seems like it should >work. Ah that makes some sence, you indeed can use this trick to calculate the power needed for a self-fueled ship. But keep in mind that the mass of the ship may decrease significantly due to the "burning" of the fuel. Timothy

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