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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*: Tue, 21 Nov 1995 21:31:53 +0100

ReplyFrom: Timothy ReplyTo : Kevin H. Subject : Sail >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 Yes agreed, somehow I missed that. I derived the formula using relativistics from the start. It can be dangerous to use non-relativistic formulas and later subtitute relativistic ones! >However, I will in the future use: >KE=M * C^2 * (gamma - 1) >where >gamma=1/SQRT(1-Ve^2) Great, but I think Steve does not like them... :) >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 OK, let me rewrite this and tell me if you agree: incoming outgoing ray ray \ / \ / \ / a (\/ ------------------- mirror (sail?) || || \/ resulting momentum The ray reflects at an angle a, this gives the mirror a momentum of 2*p*SIN(a) where p the momentum of the incoming and outgoing photon (this assumes the mirror does not move). The (kinetic) energy gain of the mirror in this proces is 2*p*c*SIN(a) the other part of the energy of the photon is still in the outgoing ray. >Question: If a photon has a waveLENGTH of 21 cm, what's it's waveWIDTH? > >Hint: It ain't zero. waveWIDTH ? I've never heard of this, please explain this new phenomenon if relevant. >> 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. Then where is the conical section sail used for if it does not reflect? >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. It was a 2 color (1 bit) GIF image created with CorelDraw. Maybe that 1 bit thing is why it did not work. What image viewer do you use?. Anyway, I will put it on the web, use your web browser with the next URL: http://rugth10.th.rug.nl/~linden/ray.gif >> 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. I still think you forgot something, but also I still don't know how and where you receive the momentum. I think I don't understand the explaination in the first letter you wrote about this. >> 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. Ah, do I get it right if I think you mean that the sail absorbs the photons? And that you think that if the photons are absorbed at an angle the forward momentum is less than if the photons are absorbed perpendicular? If these last guesses are wrong please answer the lines above the last quotes. ReplyTo : Kevin ReplyFrom: Tim Subject : Question on diodes (schottsky's) >As you no doubt know, conversion from microwaves to elec power is >achieved with diodes. I only have a fague idea of how this works. A Skottky diode is just a fast diode, right? First of all do we really need direct current or can we make an alternating current work also? If you really create a direct current, how do you complete the current loop so that the antenna does not become positive or negative loaded. >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. That's why I didn't get it the first time I guess. >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? When and why is the circuit completed? I know some things about electronics but how to use antennas to make a direct current from a wave. If someone is able to explain in an other way, I would be very happy.

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

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