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*To*: Timothy van der Linden <T.L.G.vanderLinden@student.utwente.nl>*Subject*: Re: Engineering Newsletter*From*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Date*: Mon, 13 Nov 1995 04:38:45 -0600 (CST)*Cc*: KellySt@aol.com, stevev@efn.org, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl*In-Reply-To*: <199511102257.AA24464@student.utwente.nl>*Reply-To*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Sender*: Kevin C Houston <hous0042@maroon.tc.umn.edu>

On Fri, 10 Nov 1995, Timothy van der Linden wrote: > ReplyTo : Kelly > ReplyFrom: Timothy > > >We seem to have talked oiurselves out of all the engine concepts? We can't > >launch fuel over interstellar distences. The MARS system seems to need > >impossibly large collector arrays, and eiather system would cook the ship. > > > >bummer. > > Cheer up Kelly, we will find a way, we can always make a multi-generation > ship, I guess. <shudder> Subject : Photon energy ReplyTo : Timothy ReplyFrom: Kevin > > OK, I agree with you about the aiming and timing 'problem' I wrote you > about. But I think I found another problem, so read on. > > >> What matters is the energy:momentum ratio of the beamed power. > >> For the photons in the beam that ratio is: > >> > >> E:p = c:1 where c=3E8 > >> > >> For the Asimov moving at 0.7 c that ratio is: > >> > >> E:p = 0.4c:1 > >> > >> The beam needs to have about 1/0.4=2.5 times less momentum than the Asimov, > >> but that is still a lot, so I can't agree with you that the momentum of the > >> beam is neglectable. > > WARNING! > I made a mistake here, I forgot that the momentum of the Asimov increased > while it was receiving photons. So the best thing to do is forget the above. > That's good, because i didn't understand it anyway. > >no, I am talking about the momentum of the photons as opposed to the > >momentum of the ions the "Asimov" will eject as exhaust. Those will be > >Hydrogen ions or maybe Xenon moving at .9996 or (.99996, depending on how > >much energy you can invest) C at these speeds, a small mass flow is > >sufficient to slow us down (or speed us up depending on which phase of > >the mission we are in) at a constant 1 G. > > This was clear to me, but rethinking this made me realize that your method > can't work. Because adding momentum to the Asimov will only make it move > faster. Transforming it to reverse momentum would surely break one of the > basic physic laws. > Huh? We won't be "subtracting" momentum, We'll just be adding it in the direction opposite to our direction of travel... thus we slow down and no physical laws get broken in the process. That's why a linear accelerator is better than a sail, you can aim your exhaust out of either end. > In formulas: > > - You shoot some photons at the Asimov, that will give an energy Up=p*c > where p is the momentum of the photon. > - To accelerate a mass M to speed v to decelerate the Asimov, we need an > amount of energy Uk=M*c^2*(gamma-1). > - We use all the energy of the received photons for that acceleration, > so Up=Uk --> p*c=M*c^2(gamma-1) --> p=M*c*(gamma-1) > - Thus to accellerate a mass M to speed v we need to receive photons which > will add us a momentum of p=M*c*(gamma-1). > - But of course we can subtract some momentum that we created by shooting > that mass M away. That momentum equals to p=gamma*M*v. > - Now the problem arises because by receiving photons we gain more momentum > that we loose by shooting that mass M away. > In physics: gamma*M*v is always less than M*c*(gamma-1) > All these formulas have me confused, let me see if I can work this out using some real (made-up) numbers. assume E=1 E+19 Watts, mass of "Asimov" = 2.5 E+09 Kg, wavelength of beam = 21 cm. I think we both agree that during the accel phase, the momentum from the photons helps us accelerate. But you are saying that the momentum of the photons is more than the momentum of the exhaust stream during the decelleration phase. hmm. let's see... p=E/c therefore p=3.34 E+10 Kg m/s the "Asimov" masses at 2.5 E+09 so if p=m v then v=p/m and v, (the amount of velocity change) = 13.34 m/s so every second that the beam is on us, we get "pushed" toward T.C. at 13.34 m/s faster than we were before. now let's use that energy we absorbed (minus 20% for conversion losses, now E= 8 E+18) to accelerate some amount of material to allow us to slow down. to find out how much material (per second) we must eject, let's use the rocket equation first. the apparent mass Ma (as seen by the crew) of the exhaust is g * M /Ve * c where Ve is .9996 and g is 23.34 (10 m/s^2 for us, and 13.34 m/s^2 to counteract the photonic thrust) so Ma= 194.71 Kg/sec moving at .9996 of c. note that this is only 3.89 Kg out of the tanks so thus the rest mass of the exhaust is 3.89 Kg/sec. Now let's see how much energy it takes to accelerate 3.89 Kg to .9996 of c the energy will equal the kinetic energy of accelerating the mass to the required speed, plus the energy of the "mass increase" E=Ke + Re Ke=1/2 m v^2 = 1/2 *3.89* (.9996*c)^2 = 1.74 E+17 Re= m c^2 = (194.71 - 3.89) * c^2 = 1.71 E+19 total energy required = 1.73 E+19 But we only have 1 E+19 (8 E+18 after conversion losses) what if we slowed down at 5m/s^2 the rocket equation: g is 18.34 (10 m/s^2 for us, and 13.34 m/s^2 to counteract the photonic thrust) so Ma= 153.00 Kg/sec moving at .9996 of c. note that this is only 3.06 Kg out of the tanks so thus the rest mass of the exhaust is 3.06 Kg/sec. Now let's see how much energy it takes to accelerate 3.89 Kg to .9996 of c the energy will equal the kinetic energy of accelerating the mass to the required speed, plus the energy of the "mass increase" E=Ke + Re Ke=1/2 m v^2 = 1/2 *3.06* (.9996*c)^2 = 1.34 E+17 Re= m c^2 = (153.00 - 3.89) * c^2 = 1.34 E+19 Hardly any decrease at all. Okay, Okay, I see your point (finally). so we can speed up, but we can't slow down even using beamed power. unless we use a retro reflecting ring sail, and that seems like such a waste > > Subject : Solar array > ReplyTo : Kevin > ReplyFrom: Timothy > > >increasing the number of transmitters will: > >reduce the amount of photon thrust that each transmitter is subjected to. > >reduce the amount of solar array that each trnsmitter must have. > >reduce the amount of beam jitter (by averaging the errors, they are reduced) > >reduce the heat load of each transmitter. (the non-Sol side of the solar > > panels will make an excellent heat radiator) > >increase the total cost of the mission (hey, you don't get nothing for free) > > But it won't reduce the total solar array, which is really big: > > Total Solar Power : 4E26 Watts Not really all that much more than the power a 2.5 E+09 Kg ship needs to get up to lightspeed in a decent amount of time. I think the ship needs to go on a diet! > Area of a globe with radius 1E9 metres : 7.9E17 square metres > -> Solar power per square metre : 5E8 Watt > > Mean amount of power needed by the Asimov : 1E18 Watt > -> Size of solar array : 1E18/5E8 = 2E9 square metres > = disc with radius 2.5E4 metres. > or 1000 disks with radius of 25 metres (Okay, Okay, a little more than 25 meters, but you get the point.) > Remember 1E9 metres is quite near Sol. You wouldn't like to be there in your > space suit, because 5E8 Joules would be added to your body temperature every > second. No one would be there, the transmitters would be built near earth, and launched into close solar orbit. if one or another fails, you launch another one rather than try and fix it. Kevin

**References**:**Re: Engineering Newsletter***From:*T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)

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