Re: starship-design: Fun with Math

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

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