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*To*: starship-design@lists.uoregon.edu*Subject*: Re: starship-design: Deceleration scheme*From*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Date*: Wed, 30 Jul 1997 18:52:45 +0100*Reply-To*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Sender*: owner-starship-design

Ken, You did some calculations: >For any efficiency percentage, n (n<1), there is an optimum ratio >between the kinetic energy you should impart to the beam and the rest >energy of the beam. For n<0.5 the following formula is an excellent >approximation: It is? Please show your derivation... I've doubts about your mass-ratios (they are too optimistic). >E (rest mass) / E(kinetic energy) = [2/(n^2)] - 1.5 >So: the equation for the maximum speed that this method will allow you >to decelerate from is the following: > >Max speed = c ln(original Mass/final Mass) > x {[(4-2n^2)^0.5]-1}/[(2/n^2)-1.5] Once again, I'd like an explanation/derivation >This is a strange enough formula that I'll plug in a few numbers. For >50% efficiency, this becomes: > >Max Speed = 0.134c ln(Mo/Mf) > >So if you want to decelerate from .3c you need the sail to be 1.24 as >massive as the rest of the ship. (I didn't do this relativistically, so >any number higher than .3c is probably suspect) Shouldn't that be: you need the sail to be 0.24 as massive as the rest of the ship? (Ie. a zero instead of an one) Mo=Mship+Msail and Mship=Mf Mo=1.24*Mf --> Mf+Msail=1.24*Mf -> Msail=0.24*Mf Timothy

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