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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: Fun with Math*From*: wharton@physics.ucla.edu (Ken Wharton)*Date*: Wed, 30 Jul 1997 12:12:29 -0700*Reply-To*: wharton@physics.ucla.edu (Ken Wharton)*Sender*: owner-starship-design

Timothy writes: >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). Okay - but don't read this message if you don't care for the math! The factor you want to optimize is your total change in velocity; you want to slow down as much as possible. My quick derivation of the non- relativistic rocket equation gave me: total delta v = dP/dM ln(Mf/Mo), which seemed to ring a bell. Here dP/dM is the momentum lost per mass that the ship loses. We want to maximize this value. For the beamed particles, let's call the ratio between the rest mass energy and the kinetic energy A, so that: E (rest mass) / E (beam kinetic energy) = A, or M = A E / c^2, where from now on E is the kinetic Energy you put into the beam Plugging M into my good ol' momentum equation, P = {(E^2/c^2) + 2 E M}^0.5 P (beam) = E/c (1+ 2A)^0.5 But don't forget that the beamed power is speeding you up, so this momentum loss is offset by an amount P = E/c (Where E is the same energy you get from the photons and put into the material). 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 In order to maximize this value (which will in turn maximize delta V), we take the derivative with respect to A and set it to zero. It's a quadratic, but I found the correct, exact root to be: Define B = (1/n^2) -1 A = (B^2 +B)^0.5 + B which approximates (very closely for n<0.5) to: A = 2B + 0.5 = (2/n^2) - 1.5 This gives us the optimum "speed" of the beamed particles. The reason that there's an optimum is that all the energy this scheme uses is given by the beamed power, which is in turn accelerating the ship forward. You want to use low energy/mass ratios to counter this added momentum (for a given energy), but you want to use high energy/mass ratios to get the most stopping power (for a given mass). [The big assumption here is that you can't turn any of your sail mass into energy; all the energy comes from the beam, and all the mass comes from the sail] The optimum lies somewhere inbetween. You can then go back and plug this into the equation for delta V, and you'll get the equation I gave earlier. No guarantee I haven't made a mistake, though... >>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. Ken

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