[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

*To*: Steve VanDevender <stevev@efn.org>*Subject*: Okay, I give up ( well, not exactly ;)*From*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Date*: Sun, 26 Nov 1995 12:40:51 -0600 (CST)*Cc*: Timothy van der Linden <T.L.G.vanderLinden@student.utwente.nl>, KellySt@aol.com, rddesign@wolfenet.com, RUSSESS@cellpro.cellpro.com, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, David Levine <David@InterWorld.com>*In-Reply-To*: <199511260653.WAA09366@tzadkiel.efn.org>

On Sat, 25 Nov 1995, Steve VanDevender wrote: > Look, I'm not doing this to dump on your parade, but it's a very > simple check that I'm doing on your proposed solution to verify > that it satisfies the laws of physics. i know you aren't doing this to be mean, I'm just frustrated by the situation. I checked my spreadsheet again, and found a cos where a sin should be, and vice versa. > > We are looking at a system consisting of some quantity of photons > with total momenergy [ p p 0 0 ] and a spacecraft with momenergy > [ m 0 0 0 ], as seen in the frame where the spacecraft is > initially at rest. No matter what, whether or not these two > components interact, then the final total momenergy of all the > resulting components must sum to [ m+p p 0 0 ]; in other words, > the system has net forward momentum p always and forever. This > is fundamental and inescapable. If you start out with a big box > containing the photons and the spacecraft, then that box always > has the same forward momentum for as long as all the original > components remain in the box and nothing new is put in the box. > Even worse for your argument, you are specifying that the photons > are entirely absorbed by the spacecraft, meaning that the only > component that is left after the interaction is the spacecraft, > which must then have all the forward momentum. > Yes, I should have known this. in Chem Eng, we use the concept of a control volume all the time. > I know that my argument barely even uses math and must seem too > simple to refute all the work you've put into this, but your > solution fails to conserve momentum and is therefore physically > invalid. > Simple arguments can convince simple minds (like mine) > > it does have the same momentum, just that some of that momentum has been > > made to cancel out. equal but opposite momentums cancel out > > I think what your analysis is failing to account for, at some > level, is that momentum is a vector quantity. Your think you > have found a way to turn around part of the momentum and pit it > against itself. Unfortunately that can't satisfy conservation of > momentum. Not just the magnitude of the total system momentum, > but the direction, must be conserved. You can't "dump momentum > in the perpendicular direction.". Any forward momentum that you > take away from the beam _must_ be transferred to the ship, > whether the beam is reflected or absorbed, no matter the angle of > the reflecting or absorbing surface; it cannot be transferred to > momentum in any other direction. The structural loading caused > by the beam does not absorb momentum continuously, because > momentum implies motion -- unless you unhook the reflector or cut > it into pieces, the reflector cannot accept momentum separately > from the ship because it cannot move relative to the ship; a tiny > amount of momentum goes into stretching the ship's structure when > the beam is first turned on, and accounts for no more absorption > after that. The only way to cancel out a quantity of momentum is > to accept it from another source; you cannot take a single source > of momentum and use one part of it to cancel another part of it. Agreed that the way i proposed won't work, why can't a single source be decomposed and used to cancel itself? It can be done with light, a single beam can be split, half the beam can have it's phase delayed by 1/2 wavelength, and then recombined with the other half of the beam, the net result is _nothing_ no beam. they destructively interfere. Since any particle or energy can be considered a wave, (debroglie wave for momentum) why can't this same trick be used? > > "I canna change the laws of physics, Captain." > -- Chief Engineer Scott "any science, sufficiently advanced, is indistinguishable from magic" -- Arthur C Clarke Okay, If I can't raise the bridge, perhaps I can lower the river. in studying the equations again, the big problem seem to be the exhaust velocity. (I think we knew all along that Ve= .99996 C was impossible) if i turn down the Exhaust velocity, then the exhaust invarient mass must increase. The original reason to increase the exhaust velocity was to conserve RM. but if we can use a maser sail for the first half of the trip, we might be able to accept a higher RM rate for the decell portion of the trip. so here's what I need: a simple easy to use, relativisticlly correct formula that tells me how much energy I need to accelerate a given exhaust mass to a given speed. I'll be using Me=G*(Ms+Mf) * gamma/(Ve *C) where Ve is exhaust Velocity expressed as a fraction of C. Me is Exhaust Mass. Ms is ship's Mass. Mf is Reaction Mass. G is ship's acceleration. gamma is SQRT(1 - Ve^2) I know Tim likes gamma= 1/sqrt(1-Ve^2), but then he divides by it instead of multipling. all mass is invarient or rest mass. Ve is with respect to the ship, and I'm thinking it'll be in the .80-.9? perecent of C range. every time I try to calculate this, I get flamed for using the wrong formula. Timothy gave me one, but then said that you (Steve) had problems with it. so I thought I'd head off any potential problems by asking for the right formula ahead of time. Kevin

**Follow-Ups**:**Okay, I give up ( well, not exactly ;)***From:*Steve VanDevender <stevev@efn.org>

**References**:**Re: Engineering Newsletter***From:*Steve VanDevender <stevev@efn.org>

- Prev by Date:
**Engineering Newsletter** - Next by Date:
**Re: Okay, I give up ( well, not exactly ;)** - Prev by thread:
**Re: Engineering Newsletter** - Next by thread:
**Okay, I give up ( well, not exactly ;)** - Index(es):