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*To*: Steve VanDevender <stevev@efn.org>*Subject*: Re: relativistic acceleration stuff*From*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Date*: Mon, 1 May 1995 13:02:38 -0500 (CDT)*In-Reply-To*: <199504300023.RAA23296@tzadkiel.efn.org>*Reply-To*: Kevin C Houston <hous0042@maroon.tc.umn.edu>*Sender*: Kevin C Houston <hous0042@maroon.tc.umn.edu>

On Sat, 29 Apr 1995, Steve VanDevender wrote: > Here I'm keeping things simple by measuring both mass and energy > in units of mass. What this means, though, is that to accelerate > 1 kg of mass to 0.9996 c, you need to convert a additional 35 kg > of mass completely to energy to apply to the original unit of > reaction mass. That's some power requirement! In the most ideal > case, you have to carry 35 times as much fuel as reaction mass to > get an exhaust velocity of 0.9996 c, _or_ receive that quantity > of energy from an outside source. For your example of wanting to > have 0.011645 kg accelerated to that velocity, the total power > required is 3.701E16 joules; therefore you have to receive beamed > power of 3.701E16 W in order to maintain a rate of 0.011645 kg of > reaction mass accelerated to 0.9996 c every second. This number (3.7 E+16 W) agrees quite well with my results. > > This is complicated further by redshifting; as you reach higher > and higher velocities the effective power from the beam > decreases. In order for the ship to maintain a consistent 10 > m/s^2 acceleration in its frame, the beamer has to continually > increase its power. > I looked at the redshifting problem, and decided that the sensible thing to do is decrease your acceleration as you approach light speed. in the ship's frame, the amount of time spent accelerating less than 10 m/s^2 is only about a week, and if the power starts out slightly higher than needed, then there is no shortfall at midpoint. > If you really could maintain 1 g acceleration in the ship frame, > you run into an even worse problem with getting beamed power -- > during the first year of acceleration, you watch the beam > decrease in power to effectively 0! If you continue the > acceleration using stored fuel on the ship, the beamer is > completely inaccessible to you -- your velocity approaches an > asymptote with slope c that intersects the origin at about t = 1 > yr. You are completely unable to receive anything from the other > side of that asymptote, or in other words power transmitted after > about the first year of beamer (Earth) time; it's very much like > a black hole event horizon opens up 1 lyr behind you. Of course, > in reality maintaining continuous acceleration indefinitely is > pretty much impossible, but you are going to have to expect that > you won't get much beamed power once you get very close to c. > Again, by decreasing your acceleration near the midpoint, you can avoid some of this problem by not getting too close to c. The rest of the problem dissappears (for me) upon noting that the beamer only has to beam about 1.5 years anyway, after that, any light that is beamed to T.C. will get there _after_ we've come to rest. also, the time dialation effect would keep _power_ (not energy) essentially constant wouldn't it? by that I mean that while red-shifted light would be falling on the dipoles, the amount that would fall per second would increase (beacuse our seconds would get longer). Also, because we have to decelerate at some point, power sent from earth after 1 year would still reach us and do us some good, when we are approaching T.C. > Here's a relativity puzzler that you might find enlightening (I > came up with it while I was thinking about these problems, and > while I haven't made a detailed analytical solution, I'm pretty > sure that I know what the answer is): You have two perfect > mirrors facing each other, exactly parallel. You create a > coherent beam of photons that will bounce continually back and > forth between the mirrors. The photons have momentum (and > therefore energy) p; the two identical mirrors each have mass m. > By reflecting off the mirrors, the photons will gradually > accelerate the mirrors, causing them to move apart; what are the > final velocities of the mirrors, and what happens to the beam of > photons? Without going into a detailed analysis, I would say that as the mirrors begin to accelerate, the reflected light would begin falling in energy (wavelength) and that the final speed would depend on the amount of energy in the beam of light and the mass of the mirrors; the final speed should be less than c, unless you have infinite time in which to accelerate. > There are several reasons why I think that a mass-based thruster > may be more advantageous than a photon-based one: > > 1) slowing down. You need a sail three times larger than otherwise > needed, in order to have the incident photons be in the right > direction. > > I don't know that I have the _Flight of the Dragonfly_ book, but > I do have the _Rocheworld_ serial that appeared in Analog, which > I think was the original publication of the book. I just looked > over it for a bit, and it appears to use an idea I hadn't > originally been thinking about, which is to use a sail initially > to accelerate, then to split the sail and use part of it to > reflect and focus energy backwards to the remaining part to > decelerate. I didn't see any hard figures leap out at me other > than the sail sizes, though; those agree with the figures you > imply. > This is one of the monsters in the light sail senario, "Dragonfly" weighs 10,000 tons (inner sail and 3000 tons payload) with a 300 Km dia inner sail, and a 1000 Km outer sail (outer sail alone weighs 72,000 tons) it acelerates at .01g for twenty years, and decelerates at .1 g for 2 years. I meant to say that it'd need three times the _energy_ for deceleration. I now see, that these sail sizes are smaller than the MARS (10000 Km) but the payload size of the MARS is much bigger (2,000,000) tons and we are going to accelerate at 1 Sg, so we wont have any medical problems (and we get that wonderful time dialation) > 2) drag and radiation at relativistic velocity. the best acceleration > is going to be one grav or slightly above, near turnaround point, we > get up to .999999....... of C (you pays your money and you takes your > decimal points) however, this is just where drag starts to play havoc > with your sail. the on coming protons (stripped hydrogens) _appear_ > much more massive than normal (of course). but because of > time-dialation effects, you encounter many more of them in a ship's > "day" than you would otherwise find in interstellar space. near > turnaround, we are going to pass through a volume of space of a > light-year length in around three days. if you have a huge sail, then > the force of the interstellar gale will slow you down fast! > > It's not that the incoming protons are more massive. It's that > they have more energy and momentum. This is kind of a picky > distinction, but at my current level of understanding it is > important. What I mean by saying that their mass has increased, is to say that the momentum and velocity dont add up with the rest mass. p=M*V but the momentum of the oncoming hydrogen coming at near light speed is not consistent with an atomic weight of one AMU it is consistent with a particle of mass of 1 AMU/sqrt(1-v^2/c^2) > > You could also blame the increased incidence of incoming protons > on Lorentz contraction; you get the same result. > Yes I know. > As far as I know, if the beamed power is coming in at radio > frequencies (weren't we talking about a microwave beamer?), the > reflector doesn't have to be one continuous sheet, like it would > be if you were using a visible-light laser. well, even the visible light sail that Foreward uesed, had tiny holes in it that were smaller than the wavelength of laser light ( but when the hero tripled the wavelength of light, (to get around a fensel zone lens that had not gotten enough federal funding), he forgot to mention whether the one-third sized wave lengths would passs through the sail Hee Hee Hee. > > 3)sail size. The sail size needed to accelerate at 1 grav is > horrendous, even for a "small" ship. See Robert L. Foreward's "Flight > of the Dragonfly" (if you haven't already.) > > Again, I don't see why a reflective sail is going to be any > larger than the antenna/collector needed to absorb the energy > rather than reflect it. retro reflection > > What I'm arguing is that if you are relying on beamed power to > reduce your mass load, then you are better off reflecting the > power to get thrust rather than using it to accelerate reaction > mass, at least for the acceleration phase of your trip. Here's > why: > > You don't discard all the reaction mass at once, as in the ideal > equation I derived earlier. In the early phases of the trip, you > are using your reaction mass to accelerate your remaining > reaction mass in addition to the payload mass. So you have to > throw more reaction mass during the early phases of the trip to > maintain constant acceleration. okay, I agree with the esthetics of yur argument, that is it makes sense that the same amount of energy will produce the same amount of acceleration regardless of it's use, but I would like to see a few calculations just to be sure. you've already calculated the amount of energy needed to accelerate the exhaust stream, now convert that energy to photons and calculate the amount of thrust gained by reflecting them. BTW, what kind of acceleration do you gain just by absorbing the photons, and not reflecting them? > > I do realize, however, that one way or another you're going to > have to carry reaction mass to _decelerate_, whether you throw it > out the back with the help of beamed power or not. It turns out > that even if you have the fabled photon rocket, getting to any > high fraction of c requires substantially more fuel than payload; > this fuel/payload ratio _squares_ if you intend to both > accelerate and decelerate under your own power. So a practical > design could very well involve using beamed power to accelerate > and stored reaction mass to declerate. > Actually, we can slow down and scoop up reaction mass needed for the final slow down. I posted a formula by which scoop size needed to get a 10m/s^2 deceleration can be figured.

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