Re: starship-design: Deceleration scheme

[TLG.van.der.Linden@tip.nl (Timothy van der Linden)]

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