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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 19:05:38 -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*: <199511212031.AA00782@student.utwente.nl>

To: Timothy From: Kevin > > >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... :) arrgghh! will you two make up your mind! either figure it out, consult a prof. or a book or something. and when you do get it right, please tell me and I'll use that method. <sheesh!> > > > >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. > Okay, I see, it does give the same result for angles other than 45 degrees. but it's cos not sin. and now your drawing shows just what I'm saying... if the incoming ray (in your drawing) is from Sol, then the resulting momentum would not be in the direction of T.C. > >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. > Not new at all, tbhe width of a EM wave is the same as it's length. That's why an absorbed photon gives a momentum normal to the absorbing surface. > >> 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? > It reflects during the accel portion of the trip. During decell phase, the conical section is an antenna. and it absorbs. > >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. > > <snip> 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? Yes, That seems right. Let me try again. y (T.C.) ^ | | +---> x Note: the "real" thrust should be normal to the | surface, ascii graphics prevents this | (abs) means absorbed V -y (Sol) Exhaust /\ real +y-component || +y-component thrust | || | real \ | / / |~~| \ \ | /thrust \|/ / | | \ \|/ -x <-----/^(abs) / | C| \ (abs)^\------> +x component / | /a) | o| \ | \ component | | r| | | | e| | incoming incoming photon photon Also note, I've omitted cabling across the T.C. side and Soll side of the sail. This cabling mechanically transmitts the force from one side of the sail to the other. > > I only have a fague idea of how this works. A Skottky diode is just a fast ^^^^^ I'm not trying to flame you Tim, I think you meant "vague" and since you're not sure either, Let's wait to hear from someone who is. Kevin

**Follow-Ups**:**Re: Engineering Newsletter***From:*Steve VanDevender <stevev@efn.org>

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

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