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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: Re: FTL travel*From*: Timothy van der Linden <Shealiak@XS4ALL.nl>*Date*: Sat, 01 Apr 2000 15:08:02 +0100*In-Reply-To*: <b8.38da630.260fc3fb@aol.com>*Reply-To*: Timothy van der Linden <Shealiak@XS4ALL.nl>*Sender*: owner-starship-design@lists.uoregon.edu

Hi Tom, Some remarks on different letters. >Does the group have any thought, ideas, methods or machines to >solve the problem or know of others attempt or solution to answer >the question "How far is that star?" You mentioned the parallax method. You didn't like the idea to use the starship to deviate from the straight line towards the destination star. I suggest to sent out probes at right angles to the ship's direction of travel. These probes should contain similar equipment as the Hipparcos satellite that Steve mentioned. The probes may be expendible or not. They may even have their own smaller engine system, so that they can accelerate with the main starship. - - - - - - - - >Taking for example the general equation MVof exhaust=MVof payload. >to accelerate a given payload mass of 5 tons twice light speed requires I >accelerate 100 tons of exhaust to 1/10 light speed (at 1 g acceleration). >Using the equation E(kinetic)=1/2 MV we get the energy required. It actually is E(kinetic) = 1/2 m v^2 (Thus v squared.) (The momentum in classical physics indeed is defined as p = m v) - - - - - - - - >Cancellation of the gamma factor is common with the general rocket >equation as (gamma times mass times velocity)of payload=(gamma >times mass times velocity) of propellant. Although the respective >gamma variable are different letters the values calculated are the >same so as gamma subscript payload (gp)=gamma subscript(ge) exhaust >then ge/gp = 1 and makes the cancellation valid. >Gamma is a complex variable with mass, velocity, time and spacial >dimensions. The gamma used most often only depends on the variable called velocity. Furthermore it so simple that even I can remember it. gamma = 1/Sqrt[1-v^2/c^2] The velocity of the exhaust(propellant) is not at all equal to the velocity of the payload. Hence the two gammas ge and gp should be calculated with different v. You'll see that the ge and gp will not be the same, nor cancel each other out. - - - - - - - - >I have no desire to teach the mistaught and misbehaving Somehow I get the idea that this is told *to you* at several times in your life. Otherwise you might have realized that misunderstanding does not have to be a deliberate act (of misbehavior). Timothy

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