1995.09.15 / I Johnston / Re: Practical utilization of the Griggs device Originally-From: ianj@castle.ed.ac.uk (I Johnston) Newsgroups: sci.physics.fusion Subject: Re: Practical utilization of the Griggs device Date: 15 Sep 1995 12:41:48 GMT Organization: Edinburgh University Horace Heffner (hheffner@matsu.ak.net) wrote: : One is as a boiler replacement for a hot water baseboard (HWBB) heating : system. If the Griggs device were to be driven by a gas (or oil if : location demands) powered engine, it could economically replace both : boiler and pump in a HWBB system. A really good sound abatement system : would be necessary, but such things exist and are in use by the military : for some ground vehicals. Location of the system outside the building is : also an option already often used in Alaska. It would also be nearly : trivial to add a generator to the shaft to supply electricty to drive : controls, starting battery charging, etc. A combined heat and electrical : power unit would be very useful in remote Alaska locations, all you would : need is separate clutching of the Griggs device and generator to do this. Sorry, it's been done. MicroCHP (Combined Heat and Power) systems use an IC engine and a generator, and use the excess heat (calm down Jed, I mean the stuff in the cooling water) for heating. Very efficient. The mechanical conversion efficiency of an IC engine is pretty lousy anyway - so why convert to mechanical then back to heat? Ian cudkeys: cuddy15 cudenianj cudfnI cudlnJohnston cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Jim Carr / Re: French nuclear test agenda Originally-From: jac@ds8.scri.fsu.edu (Jim Carr) Newsgroups: sci.physics.fusion Subject: Re: French nuclear test agenda Date: 14 Sep 1995 17:11:56 -0400 Organization: Supercomputer Computations Research Institute In article <42rp76$sjr@gamera.zeta.org.au> rrydge@zeta.org.au (Richard Rydge) writes: > >Behind France's resumption of nuclear testing lies a secret agenda. Nothing secret about it. They have been quite upfront about why they want to do this series of tests. They want to update their weapons and still comply with the CTBT. What I don't understand is why, if it is so safe, they don't do the tests in the south of France or under an Alp. ;-) >France is rushing to perfect simulation techniques so that it can >create the next generation of nuclear weapons outside of the >Comprehensive Test Ban Treaty. > >Other nations would then have to follow suit, and the nuclear arms >race would resume with a vengeance. This one (the US) already did that. It is rather surprising that the French did not take good enough data on earlier tests to analyze their codes, so it must be more than that. I would guess they are testing the codes by testing a specific design developed with those codes. >It's kick-starting an arms race that could engulf us all. I have news for you. The arms race engulfed us a long time ago. There is actually much less total megatonage around now than in the early 60s, but the many tens of metric tons of Pu sitting in storage could change that if anyone felt the need. -- James A. Carr | What a long strange trip it's http://www.scri.fsu.edu/~jac/ | been. Supercomputer Computations Res. Inst. | Jerry Garcia Florida State, Tallahassee FL 32306 | 1942-1995 cudkeys: cuddy14 cudenjac cudfnJim cudlnCarr cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Jim Carr / Re: French nuclear test agenda Originally-From: jac@ds8.scri.fsu.edu (Jim Carr) Newsgroups: sci.physics.fusion Subject: Re: French nuclear test agenda Date: 14 Sep 1995 17:30:40 -0400 Organization: Supercomputer Computations Research Institute In article <4343nj$h6j@anemone.saclay.cea.fr> Mario Pain writes: > > But the nuclear tests by France have also another goal: to make credible >the french nuclear deterrent, by showing that France is still a sovereign >country and is ready to take soverign decisions and carry them out whatever >the rest of the world may think of it. Does the big bad Deutsch Mark scare the poor little old franc? Chunnel tourists from Britain ordering boiled beef in your restautrants? EuroDisney still open? Nothing like a sovereign decision to make you feel better. If France *really* wanted to demonstrate credibility, they would set off the tests right there in France. To prove you are ready to use nuclear weapons, you have to prove that you are ready and willing to commit national suicide. -- James A. Carr | What a long strange trip it's http://www.scri.fsu.edu/~jac/ | been. Supercomputer Computations Res. Inst. | Jerry Garcia Florida State, Tallahassee FL 32306 | 1942-1995 cudkeys: cuddy14 cudenjac cudfnJim cudlnCarr cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.15 / Mitchell Jones / Re: The Protoneutron Theory of "Cold Fusion" Originally-From: 21cenlogic@i-link.net (Mitchell Jones) Newsgroups: sci.physics.fusion Subject: Re: The Protoneutron Theory of "Cold Fusion" Date: Fri, 15 Sep 1995 08:15:14 -0500 Organization: 21st Century Logic In article , hheffner@matsu.ak.net (Horace Heffner) wrote: > In article <21cenlogic-1109951052420001@austin-1-9.i-link.net>, > 21cenlogic@i-link.net (Mitchell Jones) wrote: > > > In article , > > hheffner@matsu.ak.net (Horace Heffner) wrote: > > > > > In article <21cenlogic-0509951113520001@austin-2-12.i-link.net>, > > > 21cenlogic@i-link.net (Mitchell Jones) wrote: > > > > > > [snip] > > > > > > > ***{The point here is to accurately visualize what is going on in the > > > > palladium cathode. To do that, we must accept the fact that the hydrogen > > > > ions (protons and/or deuterons) retain their positive charges until > > > > *after* they enter the cathode. Think about it: if, when a hydrogen ion > > > > appeared at the surface of the cathode, it were to be handed an electron, > > > > it would immediately become electrically neutral, and would cease to be > > > > attracted to the cathode. In that case, it would bubble away as a gas and > > > > loading would be impossible. Here, instead, is what must happen: (1) When > > > > an H+ reaches the surface of the cathode, it retains its charge, and > > > > enters the lattice structure. (2) The presence of H+ ions within the > outer > ^^^^^^ > It just can't do this without detection of a voltage reduction and current > reduction at the cathode, which doesn't happen. What does happen is an > inflow (current) of electrons to balance the charges in the conductor. > This was understood in Faraday's time. ***{Yes, and I never denied it. However, I get the distinct impression from your series of comments on this topic that you are not clear about when the principle of no net charge within a conductor applies and when it does not apply. Therefore, let me set up a concrete example. Suppose we mount a silver sphere on a glass rod, and then touch it with an instrument bearing a large negative charge. At the moment of contact, the charge will begin to flow into the ball, and, until the charges have equalized at every point, the flow will continue. Clearly, this period of disequilibrium will be very brief, but it will exist, and during that period, there will *not* be equality of charges at every point within the conductor. When inequality of charges exists, a voltage gradient exists, and electron current flows from regions of lower charge to regions of higher charge. The point: to say that inequality of charges exists within a conductor amounts to the same thing as saying that a voltage gradient exists, and that a current flow is taking place. Thus the fact that the positive charges (the H+ ions, or protons) are moving in the opposite direction to the electrons indicates that a gradient exists and that, therefore, the surface of the cathode is more positive than the interior. I have said, several different ways, that I do not claim that the entire positive charge of the anode migrates to the cathode, but you seem hell bent on ignoring those statements. I therefore can only conclude that you deny that there is a voltage gradient in the cathode, however tiny, and that you claim as a consequence that H+ ions are handed their electrons when they reach the surface of the cathode! If true, this would mean that loading was impossible, since upon receiving their electrons, each and every H+ ion would immediately be transformed into a neutral hydrogen atom, its radius would expand by more than 4 orders of magnitude, and it would become too large to enter the lattice! Is this what you are saying? If so, then I have to tell you in advance that what you are saying is mind-bogglingly ridiculous. Faraday's principle of equal charges at every point in a conductor is a statement about equilibrium conditions. It applies exactly only to a static charge in a conductor after enough time has elapsed to permit equilibrium to be achieved. Of course, because of the high speed of current flow, it applies imperfectly even to most disequilibrium situations. All this really means, however, is that in a low resistance conductor a small gradient will produce a large flow, and, thus, a small flow will tend to be accompanied by a gradient that is too small to be measured with ordinary instruments. *This means it is an error to assume that, in ordinary conductive situations, the voltage gradient has been eliminated completely.* (If it were not an error, then Ohm's law would be invalid.) If, therefore, you uncritically transfer the principle that voltage gradients are zero inside conductors from the situations to which it exactly applies (i.e., isolated conductors bearing static charges) to situations where it only applies approximately (i.e., circuit elements in which current is flowing) you will render yourself unable to analyze enormous numbers of common problems in electrical engineering. To repeat: under ordinary circumstances, the principle that charges are uniform inside a conductor applies exactly *only to static charges at equilibrium.* When you assume that it applies *exactly* to circuit elements in which there are ongoing current flows, you are guaranteed to eventually get yourself into deep trouble--as, in fact, you have done here. The correct interpretation, in such cases, is always that the voltage gradient is present, even if it is too small for your instruments to pick up. You know it has to be there, because the current is flowing. (Sometimes, of course, electrons flow against a voltage gradient (e.g., when driven by a collapsing magnetic field) or when there is no gradient (e.g. inside a superconductor). Such cases, however, are irrelevant to the present discussion.) Bottom line: your persistence in arguing this point has convinced me that you are making a serious error regarding the way you visualize current flow. --Mitchell Jones}*** > > > > > layers of the lattice gives those layers a positive charge, while the > > > > inner layers retain their negative charges. (3) When enough H+ ions have > > > > packed into the outer layers of the cathode to neutralize the charge of > > > > the cathode, the voltage gradient between the anode and the surface of the > > > > cathode disappears, and the H+ ions in solution are no longer attracted to > > > > the surface of the cathode. ***{You passed over this seemingly false statement here, probably because I had already stated elsewhere that the gradient did not have to be eliminated entirely, merely slightly reduced, in order for the H+ ions to move into the cathode. This original statement was made in the sense of an illustration, but it was poorly worded and misleading. Therefore, for the benefit of those who may have missed the subsequent qualifications: the entire anode charge, under ordinary circumstances, does not migrate to the cathode. However, a voltage gradient does have to exist across the cathode (from the electron source to the entry point of H+ ions), in order for entry to take place. --Mitchell Jones}*** Thus the + charge in the outer layer of the > > > > cathode does not build up to "millions of volts," but only to the level > > > > necessary to eliminate the voltage gradient from anode to the surface of > > > > the cathode. In effect, the anode charge migrates into the outer surface > > > > of the palladium electrode, and the cathode charge retreats into the inner > > > > portion of the palladium electrode. (4)) Once the anode's charge has > > > > migrated to the surface of the cathode via this mechanism, loading of the > > > > palladium electrode does *not* cease, because H+ ions on the inner surface > > > > of the positive region are constantly having electrons passed to them from > > > > the negatively charged central region. Each time this happens, the > > > > affected H+ ion becomes a neutral H atom. However, it can't bubble out of > > > > the solution, because it is trapped inside the palladium electrode. (5) > ^^^^^^^ > > This just can not happen, can it? Where is the ionization potential > coming from to dislodge this trapped atom? If there is none the atom is > trapped. There can be no H diffusion into the electrode because all the > surface sites would be occupied and thus blocked, which is contrary to > fact. ***{Apparently you did not read, or have forgotten, that I stated explicitly that neutral hydrogen atoms do *not* diffuse in the cathode. They cannot because, as just noted, they become trapped in the unit cells of the lattice. Instead, it is the H+ ions (i.e., protons) which move into the lattice. They move in because (a) the radius of such an ion is some four orders of magnitude less than that of a neutral hydrogen atom, and (b) despite your denials, a small voltage gradient exists in the cathode. Since it is the H+ ions that are moving into the lattice, they do not "diffuse." To speak of diffusion implies that a barrier to their movement exists which they must overcome. That is nonsense. As I have noted previously, they move as freely in the lattice as flies in an amphitheater. So how do neutral hydrogen atoms get loaded into the interior of the lattice? The answer: because that is where they form. H+ ions move into the lattice toward the source of the - charge, and electrons move through the lattice toward the source of the + charge. When an electron and an H+ ion meet in an empty unit cell, they form a neutral hydrogen atom, which is subsequently trapped there. --Mitchell Jones}*** Massive volumes of H diffuse into the lattice, more than one for > every site. If what you are saying were true, diffusion would be only a > surface effect. ***{See above. --MJ}*** > > > > > > Each time a new, neutralized hydrogen atom is added to the pool of those > > > > trapped inside, the positive charge of the surface region drops slightly, > > > > and a new H+ ion enters the surface region from the outside, again > > > > eliminating the voltage gradient between the anode and the surface of the > > > > cathode. (6) Then, another H+ ion inside the palladium electrode is handed > > > > yet another electron and neutralized. (7) Then, another H+ ion enters the > > > > surface. And on and on it goes, until saturation loading is approached, > > > > and, hopefully, "cold fusion" kicks in. Bottom line: it isn't necessary > > > > that the cathode achieve a charge of "millions of volts;" but it *is* > > > > necessary that it build up a positively charged region on its outer > > > > surface, in order for the loading process to take place at all. --Mitchell > > > > Jones}*** > > > > > > > > > > > > In article , > > > > hheffner@matsu.ak.net (Horace Heffner) wrote: > > > > > > > > > > Requoted here with minor typo's corrected to aid in voltage calculation to > > > follow: > > > > > > > > I would like to take this opportunity to clear up some mistaken > thoughts I > > > > > previously posted here. First, some data from the Handbook of Chemistry > > > > > and Physics (HCP) and a Sargent-Welch periodic table. All sizes in A > > > > > (1e-10 m). > > > > > > > > > > Radius of Pd atom: (Pd) 1.79, (Pd covalent) 1.28, (Pd+2) .80, (Pd+4) .65 > > > > > Radius of H atom: (H) .79, (H-1) 1.54, H (covalent) .32 > > > > ***{Note: the radius of the innermost Bohr orbit is .53 Å, and that of the > > H+ ion (a proton) is .0000137 Å. --Mitchell Jones}*** > > > > > > > Bond lengths: (H-H) .746, (Pd-Pd) 2.751 > > > > > Density of Pd: 12.0 g/cm^3 > > > > > At. Wt. of Pd: 106.4 > > > > > > > > > > Now, the Pd lattice is cubic, face centered. > > > > ***{It is misleading to refer to this lattice as cubic. The smallest unit > ^^^^^^ > > Please note above, I referred to Pd as "cubic, face centered", which it > is, as opposed to "cubic" or "cubic, body centered". > > > > cell consists of two four-sided pyramids (e.g., like the Egyptian > > pyramids) joined base-to-base. The faces of each pyramid are equilateral > > triangles with a side length of about 2.67 Å. The cubic unit cell to which > > you refer is three times as large, and is not a satisfactory conceptual > > unit for present purposes. The reason: when loading takes place, it is the > > double-pyramid unit cells that are loaded--i.e., that have H atoms placed > > in their centers. --Mitchell Jones}*** > > > > You could imagine building > > > > > this lattice by taking 1 atom thick sheets of spheres arranged in square > > > > > patterns and laying them one on top of each other, but the top > layer atoms > > > > > resting naturally in the spaces between the bottom layer atoms. > The bonds > > > > > formed are between atoms in seperate layers only. If we call the > distance > > > > > between alternate sheets S, then the bond length D will be 1/2 the > length > > > > > of the diagonal of the cube S^3, or .866 S. I seem to recall the jist of > > > > > this being posted before. > > > > > > > > > > For simplicity sake let's assume all atoms are spheres. Now, applying > > > > > common sense, suppose we considered an imaginary Pd made up of only > > > > > alternate sheets. It would have half the density, or 6.0 g/cm^3. Also, > > > > > each atom would occupy a cube (S^3) of volume V. Applying Avogadro's > > > > > number we get 6.79E22 atoms/cm^3 for Pd. Our half density sheets would > > > > > therefore have 3.40E22 atoms/cm^3, so V = 2.94E-23 cm^3, and S = 3.09E-8 > > > > > cm = 3.09 A. This gives D = 2.67 A, which corresponds within about 3 > > > > > percent with the published bond length of 2.751. So this is good, common > > > > > sense and published values agree. > > > > > > > > > > Now this means we have a radius R for Pd in the lattice because R > > =(0.5)D = > > > > > 1.38 A. This radius corresponds best with a covalent radius. > > > > > > > > > > Now, the largest sphere that can be placed between the Pd spheres is > > > > > diameter d = S - R - R = S - D = .134 S. Since S = 3.09 A, d = (.134 * > > > > > 3.09) A = .414 A. This means the H atom will have a radius r = .5 > d = .207 > > > > > A. > > > > ***{Using the calculated bond length of 2.67 Å rather than the published > > one, the size of the entry hole in one of the triangular pyramid faces is > > large enough to permit the passage of a sphere with a radius of .206 Å. > > This means entry is permitted only to H+ (protons) or D+ (deuterons). A > > neutral hydrogen atom is too large to pass through the face and into the > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > > And thereby you have disproved your own theory! The surface of the > elctrode must become blocked. No diffusion is possible. ***{See above. --MJ}*** > > > > unit cell. Once inside the cell, of course, there is room in the center > > for a sphere of radius .55 Å, which will accomodate one neutral H atom > > with its electron orbiting at the innermost Bohr radius of .53 Å. To get > > there, however, its parts must enter as separate charged particles--i.e., > > as a proton and an electron---and combine after they are inside. At this > > point, it is appropriate to ask ourselves what sort of loading ratios > > result from such a process. For example, if there is one H atom inside > > each unit cell of the lattice, what is the loading ratio? Well, for an > > infinite lattice the count of unit cells is simply three times the count > > of the number of Pd atoms in the lattice. (Since each Pd atom in a layer > > is the top of a vertical unit cell in the two layers below it, the number > > of vertical unit cells equals the number of Pd atoms in the lattice. The > > same is true of unit cells oriented right to left and front to back. Hence > > there are 3 times as many unit cells as Pd atoms in the lattice.) If we > > stuff a single, neutral H atom into each unit cell, therefore, we would > > have a ratio of 3 H to every Pd! It therefore follows that, for the > > > > I just don't understand this. Maybe you could explain further. I suspect > we might be talking about two different geometries, in which case it is > highly probable my conception is wrong. I did the geometry above off the > top of my head, so that adds credence to what *you* are saying. I guess > I'll just have to make a trip to the library to find something simple with > pictures! ***{Actually, I think your geometry is correct. The problem is that it is difficult to see the relationship between the number of atoms in the lattice and the number of unit cells. Others have had the same problem you are having, and I have responded to their comments in other posts. See, for example, my response to Robert Eachus in this same thread. --Mitchell Jones}*** > > However, let me explain something else I am thinking so it will be easier > to clear this up. Each cell has one and only one top vertex. Each atom is > the top of only one cell. So it seems that this represents a 1-1 > relationship of cells to atoms. ***{A full two-thirds of the cells do not have a top vertex because they are laying on their sides. You ignore the existence of unit cells that are oriented left-to-right and front-to-back. Vertically oriented unit cells are only 1/3 of the the total. --Mitchell Jones}*** A loading above 100% implies there are at > least two H stuffed into at least 1 cell, by the pidgeonhole pricipl. ***{Not true: if you count the unit-cells oriented left-to-right and front-to-back, you will discover that there are 3 unit cells per atom of palladium, not 1 cell per atom. At that point, you will see that a loading ratio of 1 hydrogen per palladium means that 1/3 of the cells are loaded. --Mitchell Jones}*** > Extending the logic, unless there is sufficient lattice expansion or > cracking to take care of it, every H atom over 100% loading must exist in > a cell with other H atoms. ***{Nope. --MJ}*** > > > > loading ratios that are actually seen (not much above .8), more than two > > thirds of the unit cells are unoccupied. --Mitchell Jones}*** > > > > This appears to be too small to be covalent (i.e. .32), so we are led > > > > > to think the bond must be ionic. This agrees with other chemistry if I > > > > > recall correctly, and is the basis for Marshall Dudley's Hypothesis. > > > > > Loading above 1-1 leaves no room for H orbitals. > > > > > > > > > > Anyone, please comment on flaws in this line of reasoning. > > > > > > > > > > > [snip] > > > > > > Now, to check your model of the electrode surface, let's assume a single > > > H+ ion blocks each entry hole to the lattice (i.e. the ion layer is only > > > one atom thick.) > > > > ***{I never said they would block entry. They are, in fact, too small to > > block entry. They do, however, carry their positive charges with them into > > the lattice until such time as they meet up with outgoing electrons. Since > > protons are more than 1837 times as massive as electrons, their drift > > velocities in the lattice are vastly reduced, and they tend to not > > penetrate very deeply before being neutralized. Thus the positive charge > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > > And therefore block the surface to additional diffusion. ***{Nope. To repeat: the radius of the H+ ion is four orders of magnitude less than the radius of a hydrogen atom. H+ ions can move through occupied unit cells with ease. --Mitchell Jones}*** > > > > that migrates to the cathode will be concentrated in the outer regions of > > the lattice. --Mitchell Jones}*** > > > > From the above calculations we have surface atoms in a > > > square array, with the sides of each square rougly S = 3.09E-8 cm. > > > Therefore each Pd atom presents an area of roughly S^2 = 9.55E-16 cm^2. > > > So we have roughly 1/(S^2) = 1.05E15 ions in a 1 cm^2 area. For > > > simplicity, let's assume we area talking about a 1 cm^2 surface area > > > cathode. This cathode would then have a charge Q=(1.06E-19C/ion)*(1.05E15 > > > ions) =1.11E-4 Coulombs. > > > > > > Now, what we are talking about is essentially a capacitor with area = 1 > > > cm^2 and a plate separation of S. This let's us use the capacitance > > > formula: > > > > > > C = e_k * A / S = (8.85E-12 F/m) (.01 m)^2 / (3.09E-10 m) > > > C = (8.85E-12 * 1E-4 / 3.09E-10 ) F > > > C = 2.86E-6 F (about 3 uF) > > > > > > Now volts = Q/C = (1.11E-4)/(2.86E-6) = 38 volts. > > > > > > This is still a pretty big barrier. This implies that to just balance an > > > electrode voltage of 2 volts, thus terminating electrolysis, that only > > > about one site in 19 could be occupied by a + ion. > > > > ***{The fact that a minority of the sites are occupied by H+ ions at a > > given instant is not a problem. The absolutely crucial point here is not > > that the surface of the cathode acquires the full positive charge of the > > anode, but rather that H+ ions *must* penetrate into the lattice before > > being neutralized in order for loading to take place. The reason: > > electrons that are given over to H+ ions before they enter the unit cells > > This is not true if "given over" does not include stripping the electron > from the conduction band. All that is required is that an electron charge > be paired, i.e., brought to close proximity, not that a bond forms or that > the electron must leave the conduction band. ***{At this point, we are leaving the realm of physics and entering never-never land. You are postulating a known falsehood: that electrons do not enter orbits around hydrogen nuclei, given the conditions that are known to lead to that outcome. Why are you impelled to such extremes? Because, I think, you are imbued with the false notion that "no charge differences within a conductor" applies *exactly* (rather than approximately) to a conductor in which a current is flowing. By virtue of that false premise, you are led to suppose that each and every H+ has its charge neutralized by an electron *before* it enters the lattice. But if that were to happen, you note, then the resulting neutral hydrogen atoms would be too large to enter. Since you *know* that lattices somehow become loaded, you are forced to the absurd conclusion that the bound state would not be formed! Your premise is false, Horace: a small voltage gradient exists across the cathode, and that gradient pulls H+ ions into it. Forget the silly rule you learned in school about "no voltage gradients inside a conductor." It was a rule of thumb, not a conceptually exact statement of fact. --Mitchell Jones}*** > > > > of the lattice result in gas production, not loading. Once a particular > > unit cell is occupied by an H+ ion, the next step is to hand it an > > electron, converting it into a neutral H atom which, because the radius of > > the innermost Bohr orbit is .53 Å, is now trapped inside the unit cell. > > This is the way the lattice is loaded, *because it has to be.* To repeat: > ^^^^^^^^^^^^^^^^^^^^^^^ > > This is not true. For example, I gave an alternative theory in my Straw > Horse Hypothesis now posted. ***{See above. --MJ}*** > > > > electrons that are given over to H+ ions before they enter the unit cells > > of the lattice result in gas production, not loading. --Mitchell Jones}*** > > > > > > > > I have a different mental model I would like to share. I don't know if it > > > is correct, but I'll spell it out, as if it's a fact, so it can be > > > dissected. > > > > ***{I am not sure why you are presenting an alternative model here. None > > of the statements that you made above revealed any reason for rejecting > > the model that I have proposed. If your difficulty lies solely in the > > My difficulty lies in the fact that your model does not take into account > the bipolar nature of water, nor the fact that electrodes are perfectly > capable of conducting the most minute currents and voltages, and that > unbalanced charge inside a conductor, especially as a macro phenomena, > does not agree with experimental facts, or even Faraday's theory and > experiments. You ask me to give up too much personal experience without > any apparent reason. ***{It isn't "personal experience," Horace. It is a silly rule of thumb that they teach in undergraduate courses on electrical engineering. Basically, all it means is that if you connect the probes of an off-the-shelf voltmeter to two points on a conductor in which current is flowing, the instrument will register a difference of 0 volts. But that is merely a statement about the accuracy of the instrument, not a conceptually exact statement of the reality of the situation. Think about it: if a current is flowing, and the conductor has resistance, then by Ohm's law there must be a voltage drop between the two probes! (If you compute such voltage drops, you will discover that they are, in fact, too small to register on off-the-shelf instruments.) --Mitchell Jones}*** > > > > implication, admittedly conveyed by my early description, that the full > > anode charge migrates to the surface of the cathode, then let me emphasize > > again that this is not necessary. What is necessary--absolutely > > necessary--is that H+ ions enter the lattice cells before they receive > > Not true again, for reasons above. > > > electrons. This fact implies that there is going to be some degree of > > migration of the anode charge into the outer portion of the cathode, but > > does not necessitate that the full anode charge do so. --Mitchell > > Jones}*** > > > > > > Water molecules are bipolar, electrostatically speaking. So, if a + ion is > > > in solution, the adjacent water molecules align their - sides toward the + > > > ion. Now we have a sphere with all + sides of the water molecules on the > > > surface. The water molecules adjacent to the sphere align their - sides > > > toward that layer around the sphere, and so on. There is in effect a hugh > > > ionically bonded +1 charge molecule (clump) that migrates through the > > > electrolyte. When the clump reaches the cathode, the elctrostatic force > > > of the cathode, if sufficient voltage, can separate the + ion from the > > > clump. > > > > > > It seems like, if this is true, H atoms preceeding a clump to the cathode, > > > and at the surface of the cathode, would have a very high probability of > > > being forced into a site by a clump. Also, some would H atoms would > > > escape, combining with other H atoms to form H2 and bubble out. > > > > > > What do you think? > > > > ***{To repeat: I do not see a need for an alternative model. A voltage > > gradient is all that is required to attract an H+ ion into the unit cells > > of the cathode. It doesn't have to be "forced" in: the openings in the > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > Even though pressure increase at the surface of an electrode is an effect, > I think you are right about this. The H+ atom will induce a - charge on > the cathode in it's vivinity, thus further attracting itself to the > electrode, generating a demand for more current, and electrostatically, > but losely bonding itself to the lattice. By losely I mean both the H+ ion > and the induced electron are free to move through the lattice, it's just > that motion generated phonons would slow them down. ***{You are getting more and more tangled up here because you continue to analyze the situation based on a false premise. To repeat: a small voltage gradient exists, under ordinary circumstraces, across conductors that are carrying current. --Mitchell Jones}*** > > > > triangular faces of the unit cells are large enough for it to move in > > *very* easily--as easily as a fly into an amphitheater, in fact. Remember: > > an H+ is simply a proton--i.e., the nucleus of a hydrogen atom. The > > standard formula, curve-fitted to the experimental data, for computing the > > radius of a nucleus is simply r = (1.37E-13)A^(1/3), where r is the radius > > in cm and A is the mass number. For H+ the mass number (number of protons > > plus number of neutrons) is 1, so r = 1.37E-13 cm = .0000137 A. Once > > inside the cathode, such an H+ ion will drift toward the electron source > > until it acquires an orbital electron. At that point, it will become an > > electrically neutral H atom, with an effective radius as per the innermost > > Bohr orbit (.53 Å or slightly larger). This means, roughly speaking, that > > its radius will increase by four orders of magnitude when it captures an > > electron, and it will become trapped in the unit cell which it is > > Trapped, thus preventing futher diffusion, means what you say is at odds > with the facts. ***{No, it is merely at odds with your false premise. --Mitchell Jones}*** > > > occupying at that time. --Mitchell Jones}*** > > > > > [snip] > > > > > > ***{At this point, I would like to toss out what I call the protoneutron > > theory of "cold fusion." If this theory is correct, then "cold fusion" > > ain't fusion! > > > > This is about all I can deal with at this time. I have various private > email discussions going, and no time for real world stuff like sleep. I am > beginning to wonder if getting this internet accout was such a good idea! ***{In fact, it was a great idea, and I'll bet you are loving every minute of it! The internet, after all, is the true education system of a free society. The net is a place where real learning takes place, as opposed to the fake learning imposed in government controlled "education" institutions. And real learning, shock of shocks, is actually fun! Result: parents may have to beat their kids to force them to go to the government's fake schools, but they don't have to beat them to force them to surf the internet! Instead, they have to pry their little fingers away from their keyboards and force them out the door, so they can waste their days listening to statist propaganda instead of learning! --Mitchell Jones}*** > > > You have my admiration and respect for publishing your theory I and look > foreward to delving into it. > > > > Needless to say, I present the protoneutron theory as a hypothesis subject > > to verification or falsification. If it proves to be indefensible, then I > > will abandon it. On that, you have my absolute guarantee. --Mitchell > > Jones}*** > > > > =========================================================== > > -- > Horace Heffner 907-746-0820 > PO Box 325 Palmer, AK 99645 =========================================================== cudkeys: cuddy15 cuden21cenlogic cudfnMitchell cudlnJones cudmo9 cudqt3 cudszL cudyr1995 ------------------------------ 1995.09.14 / Jim Bowery / Re: Idea Futures Originally-From: jabowery@netcom.com (Jim Bowery) Newsgroups: sci.physics.fusion,sci.physics,sci.energy Subject: Re: Idea Futures Date: Thu, 14 Sep 1995 17:15:19 GMT Organization: NETCOM On-line Communication Services (408 261-4700 guest) [ Discussing an http://if.arc.ab.ca/IF.shtml claim at 5% odds: ] Richard Schroeppel (rcs@cs.arizona.edu) wrote: : yarvin-norman@CS.YALE.EDU (Norman Yarvin) writes : Of course 5% interest for 10 years is a lot less than my bank pays me. : My bank is likely to actually pay the money, too. :-) : The IF concept is that in a real-money game, the stakeholder would invest : the stake in some low-risk way. You would, in effect, be gambling with : 1995 dollars, and the winner would receive the stake + interest. : This till leaves the annual ROI at only interest+.5%, which would not be : enough to attract a savvy investor who can do better. The IFE has serious fundamental problems but it is better than net-news word games. It is ONE step toward a real money system that incentivizes long-term accuracy in forecasting and lets people "put up or shut up" without going through all the hassles on high risk investment imposed by the SEC. If governments can grab money out of the hands of poor people to subsidize political corruption and rich education districts using KENO gambling games, it seems only reasonable that technical people, particularly those involved in issues invaded by technosocialism, in a period of political chaos, should be able to sort out the signal-providers from the noise-makers in a way that has lasting effect and real behavior-modification potential. -- The promotion of politics exterminates apolitical genes in the population. The promotion of frontiers gives apolitical genes a route to survival. Change the tools and you change the rules. cudkeys: cuddy14 cudenjabowery cudfnJim cudlnBowery cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Robert Eachus / Re: Potential-Power transformation Originally-From: eachus@spectre.mitre.org (Robert I. Eachus) Newsgroups: sci.physics.fusion Subject: Re: Potential-Power transformation Date: 14 Sep 1995 17:42:41 GMT Organization: The Mitre Corp., Bedford, MA. In article <4378cl$qvn@watnews1.watson.ibm.com> prasad@watson.ibm.com (prasad) writes: > Look what I found in sci.energy... Here's at last the Chernetsky data > I had been asking around for! The newspaper clipping I'd got in India > in late '89 had mentioned something about virtual photons and ZPE, and > that had had me intrigued. Well, no accelerators, no vaccuum energy.. > > alex (alex@frolov.spb.su) wrote: > > Present article classify extraction of excess power by means of > > electric discharge ( arc ) as special case for application of electric > > potential field for creation of power. Sorry to disillusion people, but the earliest reference I ever found to using a carbon arc to produce electric power was from 1897. I actually replicated the experiments to see if by using DC arcs we could provide customers with more light for less money. (This was at Macbeth Arc Lamp Company, around 1965.) Turned out we could get 60% conversion efficiency, which tain't bad McGee! (Whoops! That is converting energy from the burning carbon to useful light, net of input eletrical power. If we were look for excess power generation, it was about 30%) But since it meant the carbons had to be changed more often, and copper coated graphite rods are not the cheapest form of carbon, it turned out not to make economic sense. The same process works in all arcs--forming the arc puts energy into a constrictive magnetic field which confines the plasma--just like the central current in a tokamak. This energy can be recovered almost reversibly. In fact in pulsed Xenon arcs, the real problem is to keep it from overwhelming the power supply. (Which is why I started studying the Project Sherwood results, we literally had pwer supplies burn up from the reverse EMF if a pinch extinguished the arc.) With carbon arcs, the carbon burning adds to the plasma pressure, and you can literally extract more energy from the expansion of the plasma than was put into confining it. If you want to try this trick at home, it is really quite simple: ========|||| <-- Extraction electrode |||| |||| <-- Carbon arc plasma ========|||| <-- Positive Electrode \||/ =========|| <-- Negative Electrode (Usually the bottom electrode is fed up from the bottom, but that is a detail.) Put 10 VDC between the bottom pair of electrodes and extract power at 30 VDC between the outer electrodes to run a motor generator pair. (For other than an experiment, you need a flywheel on the motor generator pair, a servo for feeding the electrodes and striking the initial arc, etc. Also note that if you try to extract more power than is possible, the arc will destabilize at the extraction point.) To get really fancy, use a hollow center electrode and blow air in. -- Robert I. Eachus with Standard_Disclaimer; use Standard_Disclaimer; function Message (Text: in Clever_Ideas) return Better_Ideas is... cudkeys: cuddy14 cudeneachus cudfnRobert cudlnEachus cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Horace Heffner / Practical utilization of the Griggs device Originally-From: hheffner@matsu.ak.net (Horace Heffner) Newsgroups: sci.physics.fusion Subject: Practical utilization of the Griggs device Date: Thu, 14 Sep 1995 10:37:26 -0900 Organization: none The Griggs device, fusion or not, seems to be a very good method of converting mechanical motion into heat. It seems like this would be useful provided the driving energy were not from electricity. I would like to suggest a couple practical ways to use this capability as is. One is as a boiler replacement for a hot water baseboard (HWBB) heating system. If the Griggs device were to be driven by a gas (or oil if location demands) powered engine, it could economically replace both boiler and pump in a HWBB system. A really good sound abatement system would be necessary, but such things exist and are in use by the military for some ground vehicals. Location of the system outside the building is also an option already often used in Alaska. It would also be nearly trivial to add a generator to the shaft to supply electricty to drive controls, starting battery charging, etc. A combined heat and electrical power unit would be very useful in remote Alaska locations, all you would need is separate clutching of the Griggs device and generator to do this. I would think the military would also have a use for a portable unit of this nature. The efficiency of the unit would be improved further by heat exchanging between the exhaust gas and air intake, and using a water cooled ingine heat exchanghed with the HWBB system, as well as heat exchanging the HWBB system with the exhaust. Such a system should be near 100 percent efficient. Which side of the 100 percent is of course debatable. Another potential use is as a heat generator on a windmill. Solar is not much of an option here in Alaska in winter, but when the wind blows (and *does* it blow!) the heat bill goes up. One problem with windmills is a lack of a means of energy storage. However, a new phenominon in Alaska could take care of that - heat exchange wells. Wells drilled to provide thermal mass for heat pumps are becoming more commonplace. Such a well would provide a perfect storage medium for heat generated by a windmill. These may or may not be new ideas, but these ideas are very relevent to life in Alaska. It is possibly interresting to know that the Alaska Science and Technology Foundation (with which I am not associated), 550 Wast 7th Avenue, Suite 360, Anchorage, AK 99501-3555, 907-272-4333 provides some very large grants (average size $120,000, some more than double that) for research in energy related fields, plus advisory assistance in tapping capital resources for subsequent stages of development, as well as bridging grants. Unfortunately, nuclear energy stuff is specifically excluded. But like the patent office, if you don't mention CF (and don't generate radioactive stuff), you might be able to get somewhere. Perhaps Griggs and others should consider moving to Alaska. Regards, Horace -- Horace Heffner 907-746-0820 PO Box 325 Palmer, AK 99645 cudkeys: cuddy14 cudenhheffner cudfnHorace cudlnHeffner cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Horace Heffner / Re: Potential-Power transformation Originally-From: hheffner@matsu.ak.net (Horace Heffner) Newsgroups: sci.physics.fusion Subject: Re: Potential-Power transformation Date: Thu, 14 Sep 1995 11:33:26 -0900 Organization: none I wouldn't get too excited by all this. Check out the rest of the threads by "alex" in sci.energy. (Frolov changes the thread name on just about every post, makes it pretty hard to follow unless you look for the name he uses, "alex".) When I suggested he use calorimitry to prove his concept, Frolov got in a huff. I further went to the trouble of providing a nifty simple circuit so he could prove his device, but he had stopped posting altogether. That should tell you something. I would also like to mention that I and others expressly supported Frolov's right to post this stuff, technobabble or not. In article <4378cl$qvn@watnews1.watson.ibm.com>, prasad@watson.ibm.com (prasad) wrote: > Look what I found in sci.energy... Here's at last the Chernetsky data > I had been asking around for! The newspaper clipping I'd got in India > in late '89 had mentioned something about virtual photons and ZPE, and > that had had me intrigued. Well, no accelerators, no vaccuum energy.. > > (Looks like Don Kelly's still in the business, too.) > > ------ > > In article <42bap3$mte@sundog.tiac.net>, conover@max.tiac.net (Harry H Conover) writes: > |> Here's what happens when someone puts pen to paper without even the > |> most elementary understanding of physics or engineering. Enthusiasm is > |> not an effective substitute for knowledge. > |> > |> Sorry to be this blunt, but the net already has too much of this > |> meaningless techno-babble. > |> > |> Harry C. > |> > |> ------------------------------------------------------------------------ > |> > |> > |> alex (alex@frolov.spb.su) wrote: > |> : POTENTIAL-POWER TRANSFORMATION > |> : Theory and experiments > |> > |> : by Alexander V. Frolov > |> : P.O.Box 37, St.-Petersburg, 193024, Russia > |> : email alex@frolov.spb.su > |> > |> : INTRODUCTION [clip] -- Horace Heffner 907-746-0820 PO Box 325 Palmer, AK 99645 cudkeys: cuddy14 cudenhheffner cudfnHorace cudlnHeffner cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Kennel / Re: The Protoneutron Theory of "Cold Fusion" Originally-From: mbk@jt3ws1.etd.ornl.gov (Kennel) Newsgroups: sci.physics.fusion Subject: Re: The Protoneutron Theory of "Cold Fusion" Date: 14 Sep 1995 19:34:21 GMT Organization: Oak Ridge National Lab, Oak Ridge, TN Mitchell Jones (21cenlogic@i-link.net) wrote: > OK, you guys, I've been waiting for two days to see if somebody was going > to earn three ataboys by noticing that the protoneutron theory explains > the Griggs result! Time's up! The ataboy window is closed! > Here's the way Griggs' machine generates excess heat: > (1) The 3500 rpm rotation of the "cylinder full of holes" produces massive > turbulence in the water, and generates millions of tiny cavitation voids. > (2) Since "there ain't nothin in them voids," they collapse as soon as > they move to a less turbulent point in the flow. > (3) When a void collapses, the walls crash together at the center, > producing momentary giant overpressures. (This is a phenomenon well known > to plumbers. When, for example, a long column of water running in an empty > pipe slams into a closed valve, the giant overpressure that results is > known as a "surge," and a blown valve is the frequent result.) > (4) In the moment of final collapse, there is a tiny sphere at the center > of the former void, where the situation is sufficiently crowded at the > microscopic level so that electrons in the lowermost (K) shells of > hydrogen atoms do not have room to orbit. While the duration of this > moment is short in human terms, it is an eternity on the time scale of a > hydrogen atom, whose electron revolves around its nucleus at a frequency > of 10 quadrillion revolutions per second. This is well above Fermi degeneracy pressure. > (5) For an instant, therefore, such spheres are packed with protoneutrons. Not even a hydrogen bomb, in its instant of fusion, has enough density and pressure to convert atoms into neutronium. > --Mitchell Jones > =========================================================== cudkeys: cuddy14 cudenmbk cudlnKennel cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.14 / Horace Heffner / Re: For lunatic eyes only: the Straw Horse Hypothesis Originally-From: hheffner@matsu.ak.net (Horace Heffner) Newsgroups: sci.physics.fusion Subject: Re: For lunatic eyes only: the Straw Horse Hypothesis Date: Thu, 14 Sep 1995 19:35:08 -0900 Organization: none I have recieved a number of private emails about this thread and other posts of mine. Hopefully people are not concerned about replying to a thread of this title, especially after some of the bizarre and/or pornographic thread titles of late. First, I would like to apologize to anyone who might have been offended by my brand of humor. Since it was primarily self depricating humor, I though it might go some ways towrd breaking the ice in this group between the pro's and the am's. A little laughter is good for the soul I hear. If I offended anyone, it is my hope I offended both groups equally. Below I will clarify what I said: [snip] > Now, looking at this from a particle point of view, as the electron > approaches the end of it¹s free proton partial orbital, the possibility > for tunneling exists. This is because there is an energy barrier, which > if suddenly leaped, would result in a one way jump of the electron to the > conduction band. An electron so doing would carry to the metal lattice > the net momentum and energy gained by it¹s acceleration toward the free > proton initially, which has not been paid back, plus emit a photon to > correspond to the original transaction that freed it. > > The process described would generate most of the free energy of CF. > There is no ash. The conditions described could also generate fusion > through mechanisms as described in other posts, but the size of the > tunneling effect should be much larger than any fusion effects due to a > higher probability of occurrence and the longer average tunneling distance > of an electron. > To clarify the hypothesis, I am not talking about *any* nuclear reactions or electron captures. The tunneling referred to is *electron* tunneling through the potential barrier at the end of it's partial orbital. An assumption is there are lots of infr-red photons in the lattice space due to heat. These photons convert photon mass to electron energy when they become massless (at rest) and eject band electrons from their orbitals. Similarly, when the ejected electron completes a partial orbital, it re-inters a band orbital and gives up an infra-red photon, thus converting energy to the mass of the photon released. Further, since the ejected free electron is acting in a more particle like role, there is reason to believe it (some) could accelerate to the close proximity of the free proton, in some cases gaining KeV energies before completing an eliptical course around the nucleus. Before returning to a band, this energy would be lost due to work overcoming the coulomb force of the proton. In addition, a free electron approching, circling, or departing a free proton would be experiencing significant acceleration, thus would be expected to give off one or more photons, converting some of it's energy to the mass of these photons. As the electron continues on toward the band termination of it's partial orbital it is climbing an energy hill. However, if it can reach the band, there is a sudden energy cliff at the end, the electron falls back into a band and gives up a photon, converting energy to mass and releasing a photon of at least the mass of the photon that started the sequence to begin with. Now, since there is an energy barrier preceeding entry to the conduction band, the possibility exists for the electron to tunnel through it. If the electron can do so, there are two possibilities for the creation of extra mass/energy: (1) any photon released while in the partial orbital would have been free, and (2) if the electron can tunnel the last bit of it's way there, it will arrive with more momentum (and energy) than if it did not tunnel, because it does not have to do the work to overcome the coulomb force of the proton. Though this hypothesis contradicts the first law of thermodynamics, it seems to make more sense to me than looking a *nuclear* effects which invariably create neutrons detectable by beta decay and transmutations, or gamma radiation, neither of which have been detected in sufficient quantities to account for the heat generated. The reason it makes more sense to me is that what is hypothesized is happening in a "mass to energy to mass exchange" regiem (photon/electron) that is very active in the lattice at thermal lattice energies. There is a lot of infra-red photon activity in the lattice, activity which increases with heat. The only assumption that needs to be made is that tunneling, in some cases, is a borrowing transaction that does not have to be paid back in full. An alternative would be that, after emitting a photon, the electron in the suborbital has lost momentum so can not return to a conduction band, so spirals into the proton, emitting photons as it goes, but can not combine with it because of the extra energy needed to create a neutron. Thus you end up with a protoneutron, as described in Mitchell Jones' Protoneutron Theory of CF in another thread. This neutral charge protoneutron would be unable to stabilize and would float around until it could find a nucleus or another protoneutron to combine with, either of which it could do without penetrating a coulomb barrier. If, at thermal energies, protoneutrons were unable to combine with anything but protoneutrons, the potential and likelyhood also exists for creating nnnn particles as described in another recent post, which result in T3 and H4, but no neutrons. Either way, through tunneling or protoneutron creation, the possibility of sub-orbitals gives rise to a potential explanation for excess heat. Regards, Horace -- Horace Heffner 907-746-0820 PO Box 325 Palmer, AK 99645 cudkeys: cuddy14 cudenhheffner cudfnHorace cudlnHeffner cudmo9 cudqt3 cudszL cudyr1995 ------------------------------ 1995.09.14 / Bill Page / deutron bombardment of Ti foils - a negative result Originally-From: wspage@ncs.dnd.ca (Bill Page) Newsgroups: sci.physics.fusion Subject: deutron bombardment of Ti foils - a negative result Date: 14 Sep 1995 13:40:24 GMT Organization: Daneliuk & Page In the most recent issue of Wayne Green's "Cold Fusion" newsletter (issue no. 13) M. Srinivasan reports on current cold fusion research at BARC. Among the several notable items of interest was a report by Anand, Basu, Ramakrishna, and Srinivasan of the Neutron Physics Division at BARC entitled "15. A search for anomalous high-energy charged particle emission during bombardment of 140-keV deuterons on Ti foils (Takahashi experiment)". Quote: In a recent experiment carried out by Akito Takahashi of Osaka University, when pulsed deuteron beams of 243 KeV were bombarded on Aluminum-coated Ti and Pd foils of 3-22 micro-m thickness, some unusual high-energy charged particles were detected in the region higher than the well-known proton peak for the conventional D-D reaction. The charged particle spectra were measured using a Si-surface barrier dectector and some of the high-energy charged particles were identified as helium. These helium peaks and other unusual counts in the high-energy region are difficult to explain and might have something to do with the multibody fusion reactions proposed by Takahashi. In order to confirm the emission of such high-energy charged particles, a similar experiment was carried out at BARC. About 2 - 20 micro-A steady beams of 140-keV deuterons were bombarded on 10-25 micro-m thick Ti-foils using a deuteron accelerator. In the first set of measurements, the Ti foils were vacuum-depostied with 1 micro-m thick Ag only on the back surface (detector side), whereas in the second set the Ti foils were also coated with 0.1 micro-m thick Ag on the front surface (beam side). A surface barrier detector was used to measure the spectra of charged particles emitted from the back side of the Ti-foils. The Ti-foils were thick enough to stop the 140-keV deuterons. A prominent peak in the energy region of around 0.7 - 1 MeV was observed, which has been ascribed to the normal proton peak from the expected D-D reaction. We did not observe the presence of any other high-energy charged particles in any of the experimental runs, which were conducted for a period of 2-3 hours. During the period of deuteron bombardment on Ti foils, the neutrons produced due to normal D-D reactions were also counted with the help of a BF3-counter embedded in polypropylene. The rate of neutron production was found to be of the same order as that of proton production. These experiments, performed with steady deuteron beams as compared to Takahashi's pulsed deuteron beams, could not confirm the emission of anomalous high-energy charged particles. End Quote. Now, Takahashi's protocol is not identical to the Kasagi et al. protocol (Kasagi used Ti deuteride as the target) but there are certainly similarities in the observations. And the protocol at BARC is different again, but I think the BARC results may suggest that the specific surface preparation of the Ti foil (and the Ti deuteride targets) may be a very important factor. Cheers, Bill Page. cudkeys: cuddy14 cudenwspage cudfnBill cudlnPage cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ 1995.09.15 / Bruce Simpson / Re: French nuclear test agenda Originally-From: bruce@faxmail.co.nz (Bruce Simpson) Newsgroups: sci.physics.fusion Subject: Re: French nuclear test agenda Date: Fri, 15 Sep 1995 21:10:32 GMT Organization: FaxMail Technologies mbk@caffeine.engr.utk.edu (Matthew Kennel) wrote: >Bruce Simpson (bruce@faxmail.co.nz) wrote: >: {Nasty stuff about how agressive and non-democratic france is >: and how it's building its arsenal} >OK, all that may be true. But it seems that every single point you >make applies to China at ten times the force. Not quite. China is testing on land and in their own back yard. If something goes wrong, containment and decontamination of a land-based facility is *much* easier than an oceanic one. It is much easier to remove contaminated soil and dispose of it than it is to isolate and dispose of trillions of litres of seawater! Sea currents can move huge amounts of water very long distances with no regard for territorial limits. Likewise, many of the South Pacific nations rely heavilly on fishing as a food source and key component of their economy. China does not come into the South Pacific and endanger the lives and environment of other countries with its testing as does France. I do not agree with *any* nuclear testing but at least China has the grace and courtesy to do it in their own back yard. >In Australia, are you more worried about France or China? Well I'm from New Zealand actually and if history is anything to go by, France poses a *much* greater threat to our nation's safety and sovereignty than China. China has not sent its agents to untertake covert missions of terrorism on New Zealand soil - France has! >Are you so afraid of offending China's 'sensibilities' that nobody >really dares to confront them? {PRC's algorithm: do whatever it >wants, screw anyone. Act incredibly insulted when anybody dares complain.} Our government has objected to China in the strongest possible terms, we have done the same to the French. There is no hipocrasy involved. >Besides, it's not even clear that France has multistage fusion weapons, >though China definitely does. I can't see that this is much of an issue. >Testing a few low-yield bombs really doesn't make a real difference: it's >the attitudes, intentions combined with the capabilities of the regime in >power that does. It is *very* easy to take this perspective when these potentially dangerous tests aren't being performed in *your* area of the world. The French argue that the risks are *very* low - not zero, but very low. I would respond by saying: 1. If the risks are that low then perform your tests on continental France. Do you really think that the French people and neigbouring nations would tollerate this? I think not. 2. *Any* risk is too much. Given the sheer scale of the problem which could occur in a worst-case scenario, the number of people it could affect and the total inability of France or anyone else to "put things right" in an oceanic environment I think risk is something we are not prepared to accept. Note also that it is not France which is taking the risk - in the event of a problem, France can simply walk away and ignore the problem. The legacy of the tests will remain with the people of the South Pacific region for a significant time to come. >Witness USA and the Soviet Union. As *soon* as the Communist Party and >USSR was dissolved, the major threat disappeared even though the >hardware stayed the same. But Russia is not testing its nuclear weapons in Cuba and the US is not testing theirs in Alaska. The issue of sovereignty is always raised by the French - Mururoa is *their* territory and therefore they are entitled to do what they choose there. Might I query their attitude to NZ undertaking nuclear tests beneath the New Zealand embassy in Paris using the same rational? France can argue all they like but there is no escaping the fact that their actions have been universally condemned by the rest of the world. To continue in the light of such universal criticism is plain and simple arrogance. *----[Fixed-price software development over the net ]----* | bsimpson@iprolink.co.nz or bruce@faxmail.co.nz | *--[C/C++, Win, OS/2, POSIX, device-drivers, fax, comms]-* cudkeys: cuddy15 cudenbruce cudfnBruce cudlnSimpson cudmo9 cudqt3 cudszM cudyr1995 ------------------------------ processed by cud.pl ver. 0.5 Sun Sep 17 04:37:04 EDT 1995 ------------------------------