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*To*: Steve VanDevender <stevev@efn.org>*Subject*: Re: (Re:)^4 starship-design: The Size of the Problem*From*: big Spamma Jammer <ll5800@loop.com>*Date*: Tue, 20 Aug 1996 06:35:31 -0500*Organization*: Spam patrol*References*: <960820015349_263881579@emout10.mail.aol.com> <199608200637.XAA06993@tzadkiel.efn.org>

Steve VanDevender wrote: > > While I haven't had time to do a more complete write-up, I thought I > would also mention an interesting corollary to Rex's analysis of the > energy requirements of beaming power to accelerate a relativistic > spacecraft. Not only is a large amount of power required, but the > beaming equipment must be capable of (typically) output at a rate that > can be over two orders of magnitude larger than is needed to accelerate > the spacecraft at the start of the trip. Or, put another way: The beaming station uses only a tiny fraction of it's maximum possible power at the start of the journey > > I'm going to present some of the math without proof or demonstration at > this time, but I'm sure it will be interesting fodder for discussion > (either because Timothy or Rex will find any mistakes I might have made > or because it shows another facet of difficulty to the problem of > beaming power). > > I've recently been working on the physics of light signals between a > "stationary" object and an object undergoing relativistic acceleration > relative to it. Consider an object undergoing uniform accleration > relative to itself; its frame position at its proper time t1 is: > > [ t x ] = [ 1/a * sinh(a * t1) > 1/a * cosh(a * t1) ] > > At time t1 = 0, its position is [ 0 1/a ] (note again that for > simplicity I am using geometrized units where c = 1 and acceleration has > units 1/s (acceleration is fraction of c per unit time)). Consider an > object beaming power to the object to accelerate it that also starts at > that position, so that at time t=0 it coincides with the accelerated > object at its proper time t1=0. If energy (light) from the beamer is > emitted at time t, then the time t1 at which the accelerated object > receives the energy is: > > t1 = -(ln(1 - a*t))/a > > Note that this is an asymptotic relationship -- as the frame time t of > the beamer approaches 1/a, the object proper time t1 approaches > infinity. This consequently means that the beamer must send energy for > any possible trip within a time 1/a, no matter how far the acclerated > object goes, and that the rate at which power is sent increases > asymptotically to infinity as t approaches 1/a. The relative rate of > time passage between the beamer and the accelerated object at frame time > t has the relationship > > dt1 / dt = 1/(1 - a*t) > > In the case where a = 9.8 m/s^2 (or in geometrized units, 3.267e-8 c/s), > the asymptote is reached within about year of beamer time (3.06e8 s). > The good news is that to boost an object at 1 g up to its turnaround > point and then provide deceleration power to its destination, you beam > power for no more than two years, no matter how far away you send the > object. The bad news is that at the turnaround point you are beaming > some large multiple of the power needed to keep the object accelerating > at 1 g at the beginning and end of the trip, because of the relative > rate of time lapse between the beamer and the accelerated object. > > In fact, given the relationship between t1 and t, we can characterize > just what this multiple is based on halfway trip time of the object. > Solving t1 = -(ln(1 - a * t))/a for t, we get: > > t = (1 - e^(-a*t1))/a > > Substituting into 1/(1 - a*t), we get: > > dt1 / dt = e^(a*t1) Okay Steve, could you put that into real numbers for me. How much bigger is the turnaround power vs. the beginning/ending power? > > In other words, the maximum power output at turnaround is exponentially > related to trip time for the object. > > Perhaps the worse news is that because the relative rate of time lapse > is asymptotic, even the schemes proposed for exponentially > self-reproducing power generation equipment ultimately run up against > the asymptotic limit; the asymptotic relationship always reaches a point > where it is growing faster than the exponential function. Only if they both start at the same time. If you start the exponential function first, and then wait until it is capable of producing the proper amounts of power, before starting the asymptotic relationship, (ie you complete your beaming station before the ship ever leaves orbit.) Then I don't see the problem. Of course, you could always have a coast phase somewhere up near .9 C Coast phases at hefty fractions of C I don't mind so much. The math and physics always gets weird near the turn-around point anyways. > > Hopefully this makes sense to at least some of you. I hope to have time > to cover more of the background soon, as I'm sure this is confusing > without it. Timothy knows I've been working on analysis of accelerated > objects in .... -- Kevin "Tex" Houston http://umn.edu/~hous0042/index.html

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