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*To*: KellySt@aol.com, hous0042@maroon.tc.umn.edu, stevev@efn.org, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl*Subject*: Re: Engineering Newsletter*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Wed, 15 Nov 1995 15:22:36 +0100

The first part of this letter has been retarded a few days because of a bad working mail-demon... I added the newer part after the first two older messages. Subject : Photon energy ReplyTo : Timothy ReplyFrom: Kevin >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 < Some text left out> >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 I'm happy to hear this. Unfortunately your conclusion was based on wrong calculations. I will explain here: You used: g * M /Ve * c but that formula is non relativistic. The right formula would be: g * M /(Ve * c * gamma) This gives the mass exhaust per second. That mass does not need to be translated to 'out of the tanks'-mass. After that you use a strange and wrong method to calculate the needed energy. The right formula for kinetic energy would be: E=m c^2 (gamma-1) (You don't need to create the rest energy, that part is already onboard the ship in the form of mass) So re-doing your first calculation: Take a=13.34 m/s^2 (this is the minimum to get even with the received photons) gamma=35.36 for v=0.9996c (gamma=1/SQRT(1-v^2/c^2) m/t=M a/(gamma Vexh) = 2.5E9 13.34/(35.36 0.9996 c) = 3.15 Kg needed E=m c^2 (gamma-1)=3.15 c^2 (35.36-1)= 9.73E18 So 1E19-9.73E18=2.7E17 Watt left over, that will give an acceleration of about 0.38 m/s^2 (This all assumes 100% efficiency) So I was not completely right saying it was impossible, but very efficient is it certainly not. I'm very happy that you tried to do the calculations, it makes it easier to find the difficulties. From my physics background I'm used to do more in formulas and less in numbers. Next time I will try to remember and use some numbers also. Subject : Solar array ReplyTo : Kevin ReplyFrom: Timothy >> 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! There is still a factor 1E5 involved. That is still much more, so a diet is not necessary and not possible if we want a relativistic ship. Meaning our ship could be 1E5 times heavier. >> 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.) You mean 1.000.000 disks with radius of 25 metres! 1000 disks or 1 million disks that makes quite a difference. >> 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. I agree. But launching 1 million disks? Two other problems I could think of: - Any black spot on the solar array will probably melt away. - Each disk would be about 6 kilometres apart if they are in the same orbit. ------------------------------------------------------------------------------ NEWER PART ------------------------------------------------------------------------------ ReplyTo : Keving ReplyFrom: Timothy Subject : The bathtub is flowing over (EUREKA) >I HAVE SOLVED THE PROBLEM OF SLOWING DOWN!!!! <smug grin> First convince me... then celebrate... >First, there is _no_ way to reduce the momentum imparted by the photons. >However, we _can_ change the _direction_ of the thrust. NO and yes, you cannot change the initial size and direction of the photon thrust, but you can (of course) choose the direction of the thrust when transmitting a photon. In better words: You should see the reflection of a photon as two steps, independant of each other. The first step is receiving the photon where momenum to your mirror is added in the same direction as the photon CAME FROM. The second step is transmitting(reflecting) the photon, hereby is the momentum of the photon added in the opposite direction the photon GOES TO. So if you shoot a photon from the negative x direction to a mirror which is has 45 degree angle(on the x=y line) it first gets the momentum p in the x direction and second the momentum p in the y direction. This makes a total momentum of Sqrt(p^2+p^2)=Sqrt(2)*p in the xy direction. ^ | y | / | ______|/ | / +--- x / >Total energy needed Et = Er + Ke >Et = 2.12 E+17 + 1.46 E+19 = 1.48 E+19 > >Energy Available Ea = Eb * .8 (conversion losses) >Ea = 3.24 E+19 * .8 = 1.66 E+19 I think I know what you mean: (better formulas, see the previous message) Pp=10 m/s^2 Mr = Pe*Ms/(gamma Ve *c) = 20*2.5E9/(35.4 0.9996 3E8) = 4.72 Kg/sec Et = Me c^2(gamma-1) = 4.72 9E16 (35.4-1)=1.46E19 So, indeed much less than 3.24E19 But now how do you capture the photons? You are talking about mirrors (sail) all the time but not about capture. (I included a GIF-image of how I think you would do that) I still don't know why you used an angle of 76.6, doing the calculation with an angle of 85 degrees you need even less energy. My guess is that you forgot the capturing of the photons and therefore forgot to add momentum somewhere. You tried to trick the photons and thereby violated the preservation of momentum: If you receive an amount of photons, all their momentum is transferred to you. Once more, whatever ingenious construction you can think of, to receive a certain amount of photons and use their energy, you ALWAYS get ALL their momentum in the same direction as they went to. ReplyTo : Kelly ReplyFrom: Timothy Subject : Drawings >I was thinking of having the shielding in a fixed U shaped shielding trough >that runs around the inside of the outer hull (the open end of the U points >inward). If the shielding is made of steel instead of lead, the hab >centrafuge tracks can run around on the top edges of the U. (of course the >steel would be at least 3 feet thick!) That seems to be almost the same as what I wrote. OK. >>> (Tip: jpeg images would be about 4 times smaller, so would >>> mean a less bytes with almost the same image quality) > >Now if Rick would get a JPEG viewer. ;) I can't help with that, I don't know much about Mac-computers and their software. ReplyTo : Kelly ReplyFrom: Timothy Subject : Multi generation ship >Of course instead of a 5 year (ship time) trip, your talking about centuries >to thousands of years, and a crew 10-1000 times larger, and a proportianatly >larger ship, etc... You might make an energy savings, but it wouldn't be as >dramatic as you might expect. Hmm, yes you're right when using these numbers. But remember, my approach was VERY rough. In fact each human needs about 1.3E7 joules per day (=4.7E9 Joules per year). In the previous calculation I assumed Earth's biosphere balance. Which seems to be quite inefficient if it was designed only to keep people alive. Assuming Earth's biosphere and 100 square metres per person (which may be less) I calculated that every person used 4.4E14 Joule per year. So this means an efficiency of 0.001 % (My guess is that we could do far better that that.) >Building things takes skills, tools, and materials, as well as power. Of >course we didn't give the ship tremendous power reserves. The power needed to build the ship can probably be neglected with respect to the energy needed for the 1000 year trip. Materials should be mined at those asteroids. After the ship is build these mining-plant could be sold for much money if necessary. Tools could be made there or on Earth where ever it is cheapest. About the skills, I'm not sure. I guess working in space has to be common before we can build any interstellar ship.

**Follow-Ups**:**Re: Engineering Newsletter***From:*Kevin C Houston <hous0042@maroon.tc.umn.edu>

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