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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: We need to get on the same (pellet) track first*From*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Date*: Sun, 10 Aug 1997 15:54:42 +0100*Reply-To*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Sender*: owner-starship-design

Hello Isaac, In order not to loose track of the situation, I left out the rest of you last letter. I will come back to the rest of that letter as soon as we agree about the following. >>>3. The fuel requirements for a given cruise velocity go up roughly >>> linearly, as opposed to exponentially for a fusion drive. > >>With "fusion drive" I assume you mean a design that takes all its fuel with it. >>You have to specify what you mean with fuel: mass or energy >>The energy (not mass) requirements for a selffueled design do not increase >>exponentially, but with a 3th power. > >Actually, the energy _does_ increase exponentially. The amount of >propellant required increases exponentially, so the amount of energy >required also does (because the amount of energy used goes up >linearly with the amount of propellant used). This is well established >in the rocket equation. That is only so if you keep the exhaust velocity constant. You can also keep the mass ratio constant and increase the exhaust velocity. You'll see that by doing that, the cruise velocity increases linearly with exhaust velocity. An increase of exhaust velocity will give you a second power increase in energy. (My 3th power results may be the result of using relativistic calculations which have higher orders) (If you want to see for yourself you can check out http://www1.tip.nl/users/t596675/sd/calc/calc.html) Here are some numbers that are the result of the calculations on that page. The numbers are for a ship of 1 kg (The numbers scale up linearly with the payload mass). Vcruise Total ship mass Energy_needed Exhaust_Velocity 0.1c 4.935 6.99 E14J 0.0629c 0.2c 4.975 2.863E15J 0.1263c 0.3c 5.047 6.710E15J 0.1912c 0.4c 5.158 1.267E16J 0.2582c 0.5c 5.322 2.156E16J 0.3285c 0.6c 5.568 3.49 E16J 0.403 c 0.7c 5.954 5.61 E16J 0.486 c I should note that anything above 0.3c doesn't apply to pure fusion designs, but only for designs that use a more energy rich fuel. >>I'm not so sure that a pellet track uses linearly more fuel if it increases >>its cruising velocity. My guess is that it will be 3th power too. >>Can you show/explain me why you think it will increase linearly? > >Because you do not have to carry the propellant with you. The >amount of thrust you get is still roughly proportional to the >amount of propellant used, although you are right that it will >get more difficult with speed. Due to the way the drive works, >this increase shouldn't be much (mostly because it's not very >efficient at low speeds--the amount of momentum added is limited >by the magnetic fields). You say that I'm right that it will get more difficult with speed. But then you say that is neglectable due to your design. This discussion is not about design inefficiencies. It is about the elementary physics that are involved. Once we agree about that, we can discuss the flaws of specific designs. The trust or force you get is not proportional to the amount of mass used: To accelerate a fast moving mass needs quadratically more energy than accelerating a slow moving mass. If you mean "quadratic" instead of "roughly", then I can agree. But the use of "roughly" in this context would be quite odd. p = Momentum E = Energy Mpellet = Pellet's mass Vrelative = Relative velocity of the pellets dV = Velocity increase of a pellet The momentum gained by a pellet : p=Mpellet*dV Energy needed for that velocity increase: E=0.5*Mpellet*(Vrelative+dV)^2 So for a constant momentum (constant dV), you need quadratically more energy since Vrelative will increase while the ship accelerates. >However, assuming a perfectly ideal accelerator track scheme, >the increase will be with the square of the velocity. In the >ideal situation, the same amount of energy is imparted to each >incoming pellet, so the amount of thrust you get from a pellet >is inversely proportional to its relative velocity. This will >blow up track mass requirements as the square of the desired >cruise velocity. No, in an ideal track you will add as little energy as possible to each unit of mass. That will give you the most momentum per unit of energy. An ideal(=minimum energy requirement) track therefore has unlimited amounts of mass available. p=m*v E=0.5*m*v^2 --> p=SQRT(2*E*m) So the lower the velocity increase of a single pellet, the more momentum per Energy you get. Example: To accelerate 100 kg with 0.1414 m/s I need one Joule 100 kg with 0.1414 m/s has a momentum of 14.14 Ns To accelerate 1000 kg with 0.0447 m/s I also need one Joule 1000 kg with 0.0447 m/s has a momentum of 44.7 Ns Timothy

**Follow-Ups**:**Re: starship-design: We need to get on the same (pellet) track first***From:*kuo@bit.csc.lsu.edu (Isaac Kuo)

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