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# Nanotube Deuteron Aiming

For the "Geodesic Fusion Reactor"
 (+7) [vote for, against]

This Idea is an attempt to address the aiming problem that was described in another Idea (linked).

A carbon nanotube is basically a sheet of graphene (one single layer of graphite) that is rolled up into, obviously, a tube shape. We want a range of tube-diameters, including the very smallest we can make, for this Idea.

On the outside of each tube we apply a layer or two of electric insulation, like boron nitride --since it normally occurs in sheet form, it logically can also exist in tube form.

We now take a sort of "perfect series" of these different-sized nanotubes and connect them end to end, like a pipeline. By "perfect series" I'm talking about the number of atoms in the circumference of the nanotube. Suppose the smallest-diameter nanotube has six carbon atoms (the classic "benzene ring") in its circumference. The next-size nanotube would have seven atoms in its circumference. The next size after that would have eight, and so on.

Anyway, we put the smallest of the series at one end of our tiny pipeline, and the largest of the series at the other end. It may be necessary for the length of each tube to be longer, as its diameter gets smaller. That's not really saying much; the longest smallest nanotube might never need to be more than a couple of centimeters in length.

Since graphene is an electrical conductor, we can put an overall electric charge on our tiny pipeline. We want this to be a positive electric charge.

Now consider a particle accelerator. One common type was found in "cathode ray tube", the heart of the old-fashioned TV set, before all the manufacturers switched to making flat-screens. A CRT contains "electron guns", so obviously the particles it accelerates are electrons.

The point to remember here is, unlike huge facilities like the Large Hadron Collider, which is nine kilometers in diameter, particle accelerators don't have to be very large to do useful work. Small CRTs were often very portable.

Different particle accelerators can accelerate other particles than electrons, of course, and, as proved by something known as the Farnsworth Fusor (linked), they can be powerful enough to cause nuclear fusion reactions, yet still be small enough to fit on a table-top. We want some of THAT type of particle accelerator, to accelerate "deuterons" the nuclei of deuterium-hydrogen atoms.

After you accelerate a deuteron, now you need to aim it. That's what our nano-pipeline is for. Our ordinary particle accelerator has as its "target" the big end of our nano-pipeline. We want that "big end" to be big enough that we can guarantee that all deuterons shot at it, by the accelerator, will enter the pipeline, instead of hit the substance of the pipe, or miss it entirely.

Deuterons are positively charged, so when they enter the nano-pipe, they will be repelled such that they will travel down the center of the pipe. Obviously we want to accelerate them enough so that they always enter the pipe despite this repulsion.

As the deuteron travels down the nano-pipeline, it also encounters smaller and smaller-diameter sections of nanotube. By the time it exits the small end of our nano-pipeline, it will be about as well-aimed as it is humanly possible to aim it.

Will this be enough, so that four such nano-pipes, their exits all very close to each other, can aim four deuterons precisely enough so that they can reliably fuse to form Beryllium-8? To be determined, of course!

 — Vernon, Jul 13 2012

Geodesic Fusion Reactor Geodesic_20Fusion_20Reactor
As mentioned in the main text. [Vernon, Jul 13 2012]

Farnsworth Fusor http://en.wikipedia.org/wiki/Fusor
As mentioned in the main text. [Vernon, Jul 13 2012]

Making smaller nanotubes Oxygenated_20Benzene_20Nanotubes
Perhaps we can have benzene-circumference nanotubes, after all. [Vernon, Jun 22 2015]

Okay... bun pending, but how are you going to build it ?
 — FlyingToaster, Jul 13 2012

[FlyingToaster], you are forgetting this is the Half-Bakery. For full details like that, you need a Full Bakery.
 — Vernon, Jul 13 2012

Bun for mentioning the Farnsworth Fusor.
 — DIYMatt, Jul 13 2012

 First the minor points. Not all nanotubes are conductive (atom arangement matters). Also, the smallest diameter is considerably larger than a benzene ring. The structure of nanotubes is such that the smallest possible is six benzene rings in circumfrence, and the step size is going to be a ring at a time. Finally, you can't really "coat" carbon nanotubes with an insulator.

Now the major point. This has no advantage over a simply maintained electrical field, since that can be tuned to the level of precision needed without mechanical guidance.
 — MechE, Jul 13 2012

 [MechE], in an optical fiber that has "cladding", a photon can enter at an angle and bounce off the cladding as it makes its way along the fiber. I was assuming that a particle in a hollow tube might do the equivalent, and wanted to reduce the bouncing (and flatten bounce angle) as the particle progressed.

 I'm sorry to hear that carbon nanotubes can't be as small as I imagined, but that doesn't mean some other type of nanotube can't be that small (like, if it was made of metal, and built an atom at a time).

 I'm almost certain that our existing aiming techniques are not accurate enough to ensure that two atomic nuclei can always collide; in particle accelerators that collide particles, two groups of particles are aimed at each other, and only a few collide as the groups pass through each other. A collision-fusion reactor needs better aiming accuracy than that, especially if we want to make Beryllium-8 from four deuterons.

I do think nanotubes can help with more-accurate aiming, although how much so remains to be seen, since nuclei are vastly smaller than atomic structures, after all.
 — Vernon, Jul 13 2012

I'm not sure if a charge on the tube will keep a particle in the centre. For magnetism, the field inside a solenoid is (if I remember correctly) uniform. If the same is true of the electric field in a nanotube, it won't centralize the deuteron.
 — MaxwellBuchanan, Jul 13 2012

 //since nuclei are vastly smaller than atomic structures, after all.//

 I think that's a nunderstatement. Nuclei are, what, 10^-15m, and a carbon nanotube is, what 10^-9m? So this is the equivalent of using a ten kilometre-wide pipe to guide a marble. Given the six-log difference in scale, you might as well go another three logs and use a micron-scale "rifle" whose properties you can tweak as far as you like.

 As regards precision of aim, etc, it's possible to control the position of large objects with well- sub-nanometre accuracy.

Still, this did introduce me to the "Farnsworth fusor", for which thanks.
 — MaxwellBuchanan, Jul 13 2012

 // Ensure to get a Corgi registered one //

It's been GasSafe for a looong time now ... please, do try to stay current.
 — 8th of 7, Jul 13 2012

Atomic nuclei are cheap. At a certain point, spend less on ever higher levels of aiming accuracy and just fire off more atomic nuclei.
 — AusCan531, Jul 14 2012

 [lurch], an appropriate repulsive electric charge was proposed, to keep the deuterons away from the nanotube walls. The remark about a uniform electric field may need some attention. If the charge is AT the wall, then it should diminish with distance away from the wall, both inside and outside the tube. However, I do see that this can be related to Gravitation and a spherical shell (no net force upon anything anywhere inside the shell). So, yes, a magnetic field will probably be required, instead of an electric field, to guide the deuterons down the nanotubes.

[MaxwellBuchanan], we will be making nanotubes suitable for this purpose soon enough. No need to hurry and use today's much-larger accelerator tubes.
 — Vernon, Jul 14 2012

 // If the charge is AT the wall, then it should diminish with distance away from the wall//

 Yes, but (as you know from your comments on gravity inside a sphere), the further you are from one wall, the greater the influence of the opposite wall. I am not sure, but I think this gives a uniform field across the inside of the tube.

There's also quantum uncertainty to consider. You're trying to constrain something very small to a path which is also very small. Perhaps this can be done, and you will just have a very, very large uncertainty in either the speed of the particle or its position along the axis of the tube; but I'm not optimistic that you can squeeze something down to such a narrow linear path.
 — MaxwellBuchanan, Jul 14 2012

Further reading about fusors says that " Nonetheless, grid collisions remain the primary energy loss mechanism for Farnsworth-Hirsch fusors", because the field within the spherical electrode is uniform, allowing the ions to contact the electrode. Intuition suggests the same thing would happen for a tubular electrode, but I'm still not sure.
 — MaxwellBuchanan, Jul 14 2012

 I was wondering also about the alleged "island of stability", whereby some superheavy nuclei are predicted to be very stable.

If this island exists, does this mean that one could get energy by fusing relatively large (eg, iron) nuclei? Would the greater inertia and cross-section of large nuclei make them easier to fuse (if the resulting superheavy nucleus were stable)?
 — MaxwellBuchanan, Jul 14 2012

 If the Island of stability exists, It will still be higher energy than iron, as it's a local, not global minimum. This should make fusion of iron very difficult. It should, however be possible to fuse very large (Radium on up?) unstable nuclei to reach the island.

Regardless the relative increase in nucleus diameter is probably not significant enough to make a major difference in ease of impaction.
 — MechE, Jul 14 2012

 [MaxwellBuchanan], In a spherical shell, the inverse-square law of force-diminishing with distance is balanced by the area of the shell. An off-center object may have a greater attraction for the nearer part of the shell than for an equal area on the other side, but there is actually a larger area of the farther part of the shell, so it balances that attraction out.

 In a tube, the math will certainly work differently, so I can't be sure the effect will be the same. Regarding fusion and the island of stability, forget it. All atoms heavier than iron are on the wrong side of the "curve of binding energy"; it takes energy to make them (which is why we get energy out when uranium or plutonium fissions).

I daydream about building these tubes from neutronium, because THAT stuff could certainly be small enough. If it could stably exist in such small quantities/shapes. And then there is the problem that neutronium is electrically neutral stuff, and we still need to apply some force to keep the deuterons away from the tube walls.
 — Vernon, Jul 14 2012

 // the problem that neutronium is electrically neutral stuff //

 To begin with, it's a fluid - or at least, it behaves like one. Then, it only exists in staggeringly high gravitational fields. It's non-conductive, but do you actually realise what its values for relative permeability and permittivity are ? Perhaps you would like to go and look that up.

And since you're relying on the gravity field to keep your neutronium stable - possibly as a stable vortex rather than a physical tube - how exactly are you going to stop the deuterons collapsing into degenerate matter too ? Wrapping them in aluminium foil probably isn't going to work - it may be OK for keeping out the CIA's secret Brane Ray, but gravity is rather more subtle and insidious.
 — 8th of 7, Jul 14 2012

 Hey, I did say the notion of using neutronium was a daydream. I'm well aware that the currently known facts about the stuff make it unsuitable even for a HalfBakery Idea like this one.

I think I'll try writing a computer program to calculate crude sections of a short charged cylinder, to see if the force is unbalanced when a point-charge is off-center. I'm suspecting it will be.
 — Vernon, Jul 15 2012

Try looking at the math behind a mass-spec tool called a Cylindrical Mirror Analyser.
 — 8th of 7, Jul 15 2012

 Here's my QBASIC program, folks, and the results. There is one part of it where I am doubtful about doing it right. Feel free to correct me.

 DIM a AS DOUBLE, b AS DOUBLE, c AS DOUBLE, d AS DOUBLE, e AS DOUBLE DIM f AS DOUBLE, g AS DOUBLE, h AS DOUBLE, i AS DOUBLE, j AS DOUBLE DIM k AS DOUBLE, m AS DOUBLE, n AS DOUBLE, o AS DOUBLE, p AS DOUBLE DIM q AS DOUBLE, r AS DOUBLE, s AS DOUBLE, t AS DOUBLE, u AS DOUBLE DIM v AS DOUBLE, w AS DOUBLE, x AS DOUBLE, y AS DOUBLE, z AS DOUBLE 'not using "l" because looks like a "1"

 'F = (q1)(q2)/r^2, formula for force between charged objects. I think there is normally some sort conversion constant in there, which can be ignored here because we are merely comparing different gatherings of the same type of force. 'pi = 3.14159265358979323846... 'circle circumference = (pi)(diameter) 'diameter = 2 * radius

 For a point (H) located halfway between the center of the circle and the circumference of the circle, the distance from that point to some other point on the circle can be found by first imagining two right triangles, and the line (L) which contains both the circle's central point and the off-center point.

 The first right triangle includes part of (L), and all of a radius-line (R) of the circle, from the central point to a given point (G) on the circle. The sine (S) of the angle between (L) and (R) is the distance between (G) and (L). The cosine of that angle can be subtracted from (H); the absolute value of that difference (D) is needed.

 The second right triangle now has two known sides, (D) and (S). Its third side is the hypotenuse, which can be computed from the Pythagorean formula. That third side is also the desired distance from (G) to (H), which can be fed into the charge/force equation.

 That force has to be decomposed into vectors, to allow computing the total force along line (L), between (H) and the point on the circle where (L) intersects it. Fortunately, the relevant vectors are directly related to the previously computed values of (S) and (D) --we only need the (D) vector.

 p = 3.141592653589793# 'pi q = p / 180 'for converting degrees to radians r = 1000 'radius of chosen circle d = 2 * r 'diameter c = p * d 'circumference i = 180 / 10000 'ten thousand increments over 180 degrees 'the other half of the circle is just like this half a = 0 'accumulator for increments h = 0 'holding variable for vector sum

 FOR x = 1 TO 10000 a = a + i b = a * q 'convert to radians s = SIN(b) * 1000 'incorporate radius of circle t = COS(b) * 1000 IF (t > 0) THEN u = ABS(t - 500) '(G) is on the same side of the circle as (H)

 Note for (t > 500), the force is between (H) and the "near wall" of the circle; all other values of (t) relate to force between (H) and the rest of the circle.

 ELSE u = ABS(t) + 500 '(G) is on the other side of the circle from (H) END IF v = s ^ 2 'begin Pythagorean calculation w = u ^ 2 y = v + w 'not taking square root; next equation would square it! f = 1 / y 'assume two equal charges separated by square root of y g = f * (u / (u + s)) 'decompose assuming direct proportions of vectors '(I have my doubts about this being the right method) IF (t > 500) THEN h = h + g ELSE h = h - g END IF NEXT x PRINT h

 Some results are: 2.708079157232185D-04 'for x = 1 to 1,000 ("i" was also modified) 2.72805918747158D-03 'for x = 1 to 10,000 2.730058987800122D-02 'for x = 1 to 100,000 ("i" was also modified)

Remember this is for half a circle; double it for the full effect. It appears that the near side interacts more forcefully than the far side. Not much more forcefully! Still, the tendency, if we are talking about a repulsive force, would be for the deuteron to stay in the center of the circle. That tube will have to be SERIOUSLY charged up to be effective!
 — Vernon, Jul 15 2012

 Folks, the code I posted contains something of an oversight. I set the radius of that circle to 1000, thinking about how finely I wanted to divide it up. However, with double-precision floating-point math, that was kind of silly.

 So I've re-run the program using a radius of 10, and adjusted the halfway mark, also. The new result, for 10,000 point on the circle, is: 27.28059187471586

A nanotube, with its extremely small radius, should actually have a quite-large force centering the deuterons.
 — Vernon, Jul 17 2012

If you stick a metal lid on it, you short out the electrodes. But maybe I miss your point.
 — MaxwellBuchanan, Jul 17 2012

 [lurch], I won't say you're wrong, partly because I respect those heavyweight names you mentioned (Gauss, et al). I still have a question. If we talk about electrons on a charged can, and say the electrons repel each other to stay at the surface of the can, then ... if there is no electric field in the body of the can, exactly what is doing the pushing that makes electrons stay diametrically opposite each other? Imagine only two electrons used to "charge" that can, then four, then six, ....

 It seems like we are only considering forces between pairs of electrons nearby each other, and ignoring forces between more-separated pairs, where, logically, those forces should be crossing the interior volume of the can.

I will completely agree that inside a SPHERICAL container, forces will cancel out, because the inverse-square law relates strongly to spherical surfaces. I'm just not so sure about a cylinder.
 — Vernon, Jul 18 2012

//hair stands on end// because the arse end of the charge is at the bottom of the van der Graaff. Likewise your tinsel would repel if you touched the pipe to the business end of the VDG.
 — FlyingToaster, Jul 18 2012

well, no: I actually do know... I just have no clue what you're talking about in general 'cuz I've been saving trying to follow the annos on this post for a rainy day.
 — FlyingToaster, Jul 18 2012

 [lurch], regarding a circle, and a "chord", and a cylinder vs a sphere: The little program I wrote clearly indicates that a circle can exhibit an unbalanced force upon an object inside the circle. Let us assume that the off-center object is half-way between the center and the circumference, like I did in my program. Let us assume the line through the center and the off-center object is horizontal. The relevant chord is a perpendicular/vertical line passing through the off-center object.

 When the circle is compared to a sphere, one way to imagine it is as a sort of connected series of smaller circles. The chord-line should be rotated to create a disk-chord inside the sphere. Let us also take our original horizontal line and turn it into a horizontal plane bisecting the sphere; our off-center object is on that plane. The intersection of this plane with the disk-chord is a perpendicular/horizontal line (I'll call it PH) through our off-center object.

 Now, as we compare those smaller circles making up the sphere, to the chord and our original circle, we can see that the location of the PH line is no longer halfway between the center and the circumference; it is closer to the circumference. With respect to forces between this circle's circumference and that point, it would still be unbalanced.

 However, where the PH line intersects the sphere's interior surface is the place where we are beginning to describe circular sections of the sphere that are entirely "on one side" of the off-center object. It is these circles that get to counter-balance the previous unbalanced forces, leaving no net force upon the off-center object, inside the sphere.

Now for a cylinder. Our chord-disk becomes a rectangle stretching through the whole length of the cylinder. Even an infinite cylinder. And every circular slice we consider, of the cylinder, still computes an unbalanced force upon the off-center object. So, what am I missing?
 — Vernon, Jul 18 2012

I just want to say that this thread completely satisfies my visions of a few bakers here, wearing lab coats and playing with VandeGraff Generators.
 — RayfordSteele, Jul 18 2012

I just want to say that this thread completely satisfies my visions of a few bakers here, wearing lab coats and playing with van de Graaff Generators.
 — RayfordSteele, Jul 18 2012

[lurch], I'm still wondering about your response to my last post here. Where is the flaw in my figurings, such that a balance of forces can exist inside a cylinder, upon an off-center object?
 — Vernon, Oct 30 2013

[lurch], your July 17, 2012 annotation here contains this statement: "The electrical field inside of a conducting cylinder is zero." --and I don't disagree with SOME of that. But we don't have JUST a conducting cylinder here! We have a positively charged cylinder with a positive charge inside. And positive charges ALWAYS interact with each other, one way or another. In a sphere, the interaction is balanced. In a cylinder ... well, that's what we are discussing.
 — Vernon, Nov 04 2013

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