Re: Obliterating the Rocket Equation with a Torusail

IsaacKuo wrote:
> pgarrone@xxxxxxxxxxx wrote:
> I am addressing your objection to using D-He3 fuel
> on the basis of lack of availability. Obviously,
> you can manufacture He3 from Deuterium by your own
> assumptions that D-D fusion is a viable power source.
> An even cheaper way might be get the He3 "for free"
> from the waste products of lithium deuteride fusion.

>>Considered this way,
>>as a fuel deuterium has the highest energy per unit mass

> But not necessarily the highest energy available from
> your available natural resources. If you've got
> lithium around and you've got deuterium around, then
> you can get more energy from lithium + deuterium than
> you can get from deuterium alone.

I get less than half the possible energy per unit mass from
the D + 6Li reaction, as from D + D and taking it all
the way through to 4He.

D + 6Li -> 2 x 4He + 22.4 MeV for 8 atomic masses


And 2 x (D + D) gives 47.7 Mev for 8 atomic masses.

Reactions D + D -> 3He(0.820) + n(2.499)
-> 3H(1.011) + p(3.022)
3He + D -> 4He(3.712) + p(14.641)
3H + D -> 4He(3.561) + n(14.029)
p + n -> D(2.224) (assumed)

Also there is plenty of deuterium,
right here on earth, but lithium-6 is a lot harder
to come by, if millions of tonnes are wanted.

So you are wrong, as usual.

>
> Considering how much easier lithium deuteride fusion
> is than either D-D or D-He3 fusion (we can do it today),
> it's not crazy to suppose future power reactors which
> use lithium deuteride fuel and produce He3 as waste.

I never said anything was crazy.

>
> So basically, your objection to the use of He3 fuel
> based on natural scarcity is not at all a given.

Designing a future mission, decisions have to be made.
Because I picked D-D doesnt mean other choices
cannot be valid for whatever reason.
I merely stated why I choose it.
You seem to dwell on each design choice,
implying that by picking A instead of B,
that I am rejecting B as crazy.
But you are not substantiating your assertion that
fusion-light-bulb is "worthless".

> If you're clever about it, the longer wavelength
> photons get radiated out into space directly. This
> is possible with a solar panel because the Sun only
> subtends a small fraction of the sky.

Its difficult to work out what you mean here.
You are critiquing some aspect of either
my intelligence or my proposed
system, but I dont know how or what.

> But in any case, you're limited to the laws of
> thermodynamics. In order to get any net result
> from radiating a photon, you have to be radiating
> it in some direction where you are NOT also receiving
> radiation of the same (or higher) temperature.

Nothing I've said implies that I rely on radiating from
a lower temperature surface to a higher temperature surface.
You continually state basic laws of physics, imply that
my system breaks them, but dont state how. This is
very boring.

> >With a cyclic heat engine, a fluid is compressed,
> >heated by the heat source, decompressed into kinetic
> >energy, and cooled in a radiator. If the fluid is compressed
> >and heated too much, it will ionize and emit bremsstruhlung
> >radiation, losing energy.
>
> Where do you think that radiation goes? It doesn't
> just magically disappear into nothingness, it's
> turned into photons. Those photons can be reabsorbed
> by the working fluid, if you want to set it up that way.

My own calculations found
that the working plasma would be optically thin, so
it wouldnt absorb radiation easily. The walls
containing the working plasma
might reflect radiation so as to allow it to
be eventually absorbed, but it would be necessary to
keep them cool as well because they are in direct contact
with uniformly super-heated working gas.

You have a hectoring lecturing style that is offensive.

> You can't neglect fluid turbulence and mixing,
> least of all with hydrogen gas. Convection is
> typically the dominant mechanism of transfering
> heat in a gas, even if the "hot" side is "above".

You havent stated why "fluid turbulence and mixing"
cannot be controlled.
In a positive temperature gradient,
the typical heat transfer mechanism through a gas
is radiation. e.g. clear skies at night in the desert.
Your assertion that convection
is "typically" dominant at transferring heat
is incorrect, so you are wrong again.
There is no deliberate forced convection
in this situation.
So you havent made a case that the idea is implausible.

> You reached completely wrong conclusions based on
> completely wrong assumptions.

Empty rhetoric.

> The cost in energy is roughly equal to the amount
> of heat energy being moved--and the result is that
> you STILL have to remove heat from whatever is on
> the "hot" side of the peltier.
>
> Peltiers are just a horrible way to cool things,
> if you can use just about anything else instead.

There is plenty of electric power available,
so the efficiency is irrelevant. The point is
things can be actively cooled if needed.

> Doesn't matter. A magbrake creates its own
> magnetosphere, which ionizes interstellar medium
> with its bow shock.

> A magbrake creates its own magnetosphere, which
> the particle "wind" blows up to many times its
> own diameter.

Nothing says I have to use a magbrake. You haven't
countered the argument about the area needed to make
a significant momentum impact at all.

> >On to the bombtrack. I get equations such as
> >Power = Mass * acceleration * velocity
> >distance = (velocity**2)/(2*acceleration)
> >Accelerating out of the solar system at 1 g,
> >to reach 0.17C, fuel has to be lined up for
> >a distance of 20 times the orbit of Pluto.
> >Nearing the end of the run, the power transfer
> >is 0.51 gigawatt per kg.
> >Given the 85000 tonne payload, this would
> >be 40 petawatts, which is 40 times the power
> >of my engine. Multiply the power and
> >payload by whatever factor you
> >conceed is necessary to slow it down.
>
> The great thing about bombtrack propulsion is
> that it is self-cooling. You simply don't absorb
> energy in the first place. The thrust is produced
> by interaction between charged particles and a
> FIXED magnetic field. To a rough approximation,
> this is a conservative interaction so no energy
> is absorbed by the starship.
>
> Most of the waste energy of the bombs in the
> form of neutral radiation is simply vented into
> space, never even reaching the starship. It's
> the tiny fraction which hits the starship which
> it must contend with. But this could be many
> orders of magnitude less than the amount of
> power going into thrust, so it's a huge win.
>
> The fact that most energy from a bomb is in the
> bomb products cuts out one order of magnitude.
> The distance between the detonation of the bomb
> and the starship cuts out another order of
> magnitude (as the starship goes faster, the
> detonations need to be further forward from
> the ship). The fine geometry of the starship
> cuts out another three orders of magnitude.
>
> So, the waste heat the starship must reject is
> on the order of 1/100,000 as much as the amount
> of power going into thrust. What's more, the
> overwhelming majority of this energy is absorbed
> by the sail lines themselves--already a nearly
> ideal geometry for naturally radiating the
> waste heat.
>
> Amazing power levels are plausible when most of
> your waste energy is simply vented directly
> into space.
>
> Isaac Kuo

Compared to a rocket, the power flux is much higher,
because the fuel must be used much closer to take-off.
It seems implausible that the power transfer is five
orders of magnitude greater than the heat transfer.

You don't address the distance issue at all.
You're proposing this sort of system as being possible with current
technology. To get to 0.3C, as you propose,
you would have to set up vast numbers of
bombs out to a distance of many times the orbit of
Pluto, which is ridiculous.

.

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