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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*: Tue, 21 Nov 1995 06:40:07 -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*: <199511151422.AA06167@student.utwente.nl>*Reply-To*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Sender*: Kevin C Houston <hous0042@maroon.tc.umn.edu>

To: Timothy From: Kevin H. Subject: Various On Wed, 15 Nov 1995, Timothy van der Linden wrote: > Subject : Photon energy > ReplyTo : Timothy > ReplyFrom: Kevin > > >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. But thats what I used, granted I did it in two steps, but multipling the result of g * M / (Ve *C) by SQRT(1-Ve^2/C^2) is the same as your formula > > 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) > No, I'm not creating rest energy, I'm creating the relativistic energy. my energy formula (which may well be wrong) was broken into two parts: 1) a non-relativistic KE formula for getting the rest mass up to Ve KE=1/2 m (Ve*C)^2 2) a way to account for apparent mass increase (sorry steve, I know that grates on you, but it don't make sense to me any other way) by using E=mc^2 and may I point out that my answers were within 5% of the "correct" answers. not bad for a first guess. However, I will in the future use: KE=M * C^2 * (gamma - 1) where gamma=1/SQRT(1-Ve^2) Note, in my formulas Ve is expressed as a fraction of C anyway, so I don't have to divide by C^2 > Subject : The bathtub is flowing over (EUREKA) > > > >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 > / > No, the momentum of absorbing the photon: cos (45) * p and the momentum of transmiting a photon: cos (45) * p adding them together would give you 2 cos(45)*p and since cos (45) = SQRT(2)/2, the total is: (drum roll please) Sqrt(2)*p Which of course is the same result as your formula, but for different reasons. and would not be the same for angles other than 45 Question: If a photon has a waveLENGTH of 21 cm, what's it's waveWIDTH? Hint: It ain't zero. > > > 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) No, _you_ keep talking mirrors all the time. I never once talked about reflecting the photons anywhere. BTW, I've never gotten one gif image to work. I can save 'em, and I can get them to my home machine, but when they get there, I get a "Not a vaild GIF file" when I try to display. And that includes a file that Kelly FTP'd me, so I know it's not my mail program. Maybe I'm just a lone IBM'er in a MAC group. any other IBM'ers here? Did you get the GIF's to work? let me know so's I can get working copies please. > 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. Every time I worked out the formulas, (and I just did it with your equation for KE) I ended up with not enough energy if I used an angle of less than 76.6 degrees (photon thrust exceeds engine thrust) and too much left over energy if I used an angle much greater than 76.6 (engine thrust exceeds photon thrust) > 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. Yes, agreed. I received all of the photons Monmentum, but only cos(76.6) of it is in the direction of ships travel. > 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. No, not true. Tilted surfaces receive all the momentum, but at an angle which is normal to backside of the surface. To : All From: Kevin Subject: Revised numbers using Timothy's Kinetic Energy formula: Velocity of exhaust: .99996 C (yes, this has changed also) gamma of exhaust 111.8045 Energy Beamed from Earth 2.08 E+19 Watts Energy after conversion 1.66 E+19 Watts (80% eff) angle of antenna: 76.6 cosine of angle : 0.360702 at this angle and energy, the photons impart a 10 m/s^2 foreward thrust so the ship's engine must impart a 20 m/s^2 backward's thrust. to accomplish this we need an exhaust rest mass of G*M/(Ve*C*gamma) which equals 1.49 Kg/sec out of the tanks getting this up to .99996 C will require M*C^2*(gamma-1) or 1.49E+19 Watts leaving 1.77 E+18 left over. more than enough energy to keep the crew alive. I'm still working on the heat balance, more when I finish To: All From: Kevin Re: Question on diodes (schottsky's) As you no doubt know, conversion from microwaves to elec power is achieved with diodes. I've been assuming that a diode on the antenna but not connected to a circuit is just sitting there, and the antenna will reflect. when the circuit is completed, the diode will convert the microwaves to elec with some heat left over, and the antenna wil absorb. is this correct? Kevin P.S. to Timothy: I can't speak for Steve, but yes, I do actually write these at 2:08 am or whatever time appears on the header. I often stay up late or get up early to write these. and of course, the time you see is the time I sent it, not the time I started, so you can subtract about two days for writing time ;) PPS to All: I heard a couple of good jokes: 1) How do a Mathamatician, a Physist, and an Engineer _prove_ all odd numbers are prime? The Mathamatician says: 3 is prime, 5 is prime, 7 is prime. That's three examples, the rest follows by induction. The Physicist says: 3 is prime, 5 is prime, 7 is prime, 9 is... experimental error, 11 is prime, 13 is prime .... The Engineer says: 3 is prime, 5 is prime, 7 is prime, 9 is close to being prime, and as we get into higher and higher numbers, the difference between the prime and non-prime portions is so small that it can be ignored. And the Technician says: 3 is prime, 5 is prime, 7 is prime, 9 is... Hey Joe, hand me hammer. second joke: The Mathamatician, Physicist, and Engineer were all at a conference (on prime numbers no less) and that night, a fire broke out in each room. The Engineer woke up, saw the fire, grabbed a bucket, and ran into the bathroom. he estimated the size of the fire, rate of spread and available oxygen, and drew that much water plus 10% to cover his errors. He ran back into the bedroom, poured 95% of the water onto the fire, checked that it was out, then poured the rest just to make sure, and went back to bed. The Physicist woke up, saw the fire, grabbed a bucket, and ran into the bathroom. he wrote down a few equations detailing the size of the fire, the rate of spread and the available oxygen, accounted for the ambient temperature and humidity of the air, as well as the heat of combustion of the materials that were burning, calculated the precise amount of water to put out the fire, drew that amount and ran back into the bedroom, poured the water on the fire, checked the results against the predicted results and went back to bed. The Mathamatician woke up, saw the fire, grabbed a bucket, and ran into the bathroom. he turned on the faucet, saw the water coming out. wrote down a few equations, proved a solution existed. and went back to bed.

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

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