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*To*: starship-design@lists.uoregon.edu*Subject*: Re: starship-design: Fun with Math*From*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Date*: Thu, 31 Jul 1997 15:53:38 +0100*Reply-To*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Sender*: owner-starship-design

Ken, I've one question left: >So now your total delta P (for 100% efficiency) is > >P = E/c {(1+2A)^0.5 - 1} > >With a light -> particle efficiency, n (n<1) this becomes > >P = E/c {n(1+2A)^0.5 - 1} > >You lose stopping power by a factor of n, but the "pushing" power from the >beam is just as large! Solving for P/M, we plug in the earlier formula >for M, the E's cancel, and > >dP/dM = c {n(1+2A)^0.5 -1}/A Is dP/dM equal to P/M ? >>>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 > >Unfortunately, I already subtracted the 1; Mo = 2.24*Mf, Msail = 1.24*Mf. Oh wait a minute... Max Speed = 0.134c ln(Mo/Mf) 0.3c=0.134c ln(Mo/Mf) 2.24=ln(Mo/Mf) 9.38=Mo/Mf Mo=9.38Mf We both forgot the Logarithm! (did you lure me into that?) This really makes the mass-ratios dramatic for low efficiencies... (This also clarifies my doubts about the apparent low ratios.) Timothy

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