Re: [Vo]:Rossi: Thermal crosstalk at heatexchanger?
On Oct 14, 2011, at 2:39 AM, peter.heck...@arcor.de wrote: I condensed my thoughts about this into a Youtube video: http://youtu.be/mSyeJa7a3AE Explanation in technical terms: == Primary steam inlet and secondary (warm) water outlet are in direct thermal contact via the brass tube. Because thermal flow behaves like electrical current in ohmic resistive material - it superimposes linear- there must be a temperature gradient. Of course I dont know how thick the brass is, it has without doubt strong coupling to the water. Because the tube is symmetric, there must be a temperature of 65° [(100+30)/2=65] in the middle symmetry point if one end has 100° and the other has 30°. It is then obvious that the sensor must be somewhat warmer than the water, because it is not far from the middle point. Without precise data it can only been estimated, not calculated. Example: Lets assume the distance sensor-water is 3% of the distance sensor- steam. Lets assume there is 3% thermal crosstalk between the 100° hot steam entry and the sensor, which seems reasonable. Then the sensor will report 2 degrees more than the water temperature. Please note: I dont know, if the e-cat works, or if it doesnt work. I would wish by heart, that it might work. But I know, wishful thinking does not work in science and technique. I doubt it works if the resarch is not made much more carefully than the demonstration setups. I guess you missed my analysis of this: http://www.mail-archive.com/vortex-l@eskimo.com/msg52527.html and the analyses of others. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Major Update to Rossi 6 Oct 2011 Experiment Data Review
My review at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf has been very substantially updated. The most important update is inclusion of the following section: SAMPLE SPREADSHEET INCORPORATING POWER ADJUSTMENT FACTOR A sample spreadsheet incorporating flow rates based on water meter readings, and having a delta T, and thus output power, adjustment factor Tadj = 0.25 is located at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011vol1sim.pdf A graph of the important values can be found in Graph 5, appended. A large scale version of Graph 5 can be found at: http://www.mtaonline.net/~hheffner/Graph5.png One key thing to note regarding Graph 5 is that Eout at the end of the run is less than Ein by about a kWh. This reflects energy stored in the heat remaining in the E-cat. Maximum stored energy, 6.727 kWh, 24.2 MJ, occurs right before 15:53, 280 minutes into the run, right before power is turned off, and the “self sustaining running” begins. Storing the 24.2 MJ requires a mean storage Delta T of (2.42x10^7 J)/(2.3x10^4 J/°C) = 1052°C. Assuming the metal started out at 27°C that means an iron temperature of 1079°C. This sets a limit on the period of heat after death boiling that can occur. If the central metal is heated to 1079°C then energy stored for boiling is 979°C * (2.3x10^4 J/°C) = 22.5 MJ. To last through the heat after death period from 280 min. to 476 min. = 196 min., the water boiling power output is limited to an average of 22.5 MJ/(196 min.) = 1148 W. Limiting the mean thermal output of the stored thermal mass to a mean output of 1148 W requires a significant degree of thermal resistance between the thermal mass and the water heat exchanger above the thermal mass. At a midpoint of heat after death, thus a thermal mass delta T of 979°C/2 = 490°C to the boiling water, the thermal resistance required between the thermal mass and the water is (490°C)/(1148 W) = 0.426 °C/W. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: manifold thermal model HERE
On Oct 15, 2011, at 2:05 AM, Robert Lynn wrote: Look in the Temp data Ecat_6_10_11.xls file on nyteknic site at 11:22:01, Could you post a link please? It does disagree with Mat's report, and it is possible there was a transcription error or somesuch. Note that the secondary water flow started at 11:00 On 14 October 2011 20:12, Jed Rothwell jedrothw...@gmail.com wrote: Robert Lynn wrote: It is really perturbing that the initial temperature measurements up to about 11:50 have a 4.3°C temperature difference between the inlet and outlet of the secondary, and even at max power it only rises to about 7.2°C - obviously a huge temperature error, which in itself and alone makes all the secondary loop data totally unreliable. Where do you see that? At 11:52 the log shows a Delta T of negative 0.5°C. That is a bias that they did not bother to adjust. It ranges from 0.5 to 0.7°C. I do not see any signs of huge temperature errors. What are you talking about? Please refer to the specific statements in the log here: http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E- cat+October+6+%28pdf%29 - Jed Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:New Bob Higgins diagram
This is a very nice diagram. However, as I showed in the T2 THERMOCOUPLE LOCATION in: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf the center of the thermocouple comes down precisely on the edge of the fins. On Oct 14, 2011, at 10:34 AM, Alan J Fletcher wrote: Bob Higgins sent me a higher-resolution diagram which I'm hosting at : http://lenr.qumbu.com/111010_pics/Rossi27kwReactorDiagram_lg.png The T2 probe length has been increased to reflect recent discussions. ps If anyone needs image hosting, I've got plenty of storage. Just send me a pic and I'll put it up. At 07:47 AM 10/14/2011, azat avetisyan wrote: Hi Alan Could you please send me Mr Higgins higher-res version of fat- cat's diagram. Please tell him thank you for this wonderful diagram, now I could understand looking on diagram, how e-cat's work. Also, thank you for your website, it's excellent information and analysis source. Thanks Azat You're welcome. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:quantum levitation
On Oct 17, 2011, at 2:19 PM, Esa Ruoho wrote: http://www.youtube.com/watch?v=Ws6AAhTw7RA pretty Very cool! Check out this one too: http://www.youtube.com/watch?v=VyOtIsnG71U Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) * (2.3 cm)) I = 2.9x10^5 s^-1 For 1 kW of 10 keV gammas we have: I = (6.24x10^17 s^-1) * exp(-(0.80 cm^2/gm) * (11.34 gm/cm^3) * (2.3 cm)) I = ~0 s^-1 So, we can see that gammas at 100 keV will be readily detectible, but much below that not so. However, it is also true that 0.2 cm of stainless will absorb the majority of the low energy gamma energy, so we are back essentially where we started, all the heat absorbed by the stainless, and even the catalyst itself, in the low energy range. If the 2 mm of stainless is equivalent to 1 mm of lead, for 1 kW of 100 keV gammas we have: I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) * (0.1 cm)) I = 2x10^16 s^-1 and an attenuation factor of (2x10^16 s^-1)/(6.24x10^16 s^-1) = 32%. Down near 10 keV all the gamma energy is captured in the stainless steel or in the nickel itself. To support this hypothesis a p+Ni reaction set including all possibilities for all the Ni isotopes in the catalyst would have to be found that emitted gammas only in the approximately 50 kEV range or below, but well above 10 keV, and yet emitted these at a kW level. This seems very unlikely. If such were found, however, it would be a monumental discovery. And, it would be easily detectible at close range by NaI detectors, easily demonstrated scientifically. SUBSEQUENT COMMENTS ... can anything said about the inside of the E-cat be believed? There are numerous self-inconsistencies in Rossi's statements, and behaviors. These things may be justifiable in Rossi's mind to protect his secrets. Whether justified or not, such things damage credibility. One thing is for sure: if the E-cat is operated at significant pressure then 2 mm walls would be too thin at high temperatures. Also, there are other limits to surface steam generation I have not discussed, that take precedence at high power densities. One limiting factor is the ability of the catalyst and hydrogen to transfer heat to the walls of the stainless steel container, a process which would likely be mostly very small convection cell driven. Again, we know too little about the internals. Nothing much new about that. A heat transfer limit is reached if a stable vapor film is formed between the walls of the catalyst container and the water. The top of the catalyst container may be exposed to vapor, thereby increasing the thermal resistance, the effective surface area. At high heat transfer rates bubbles can limit transfer rates. It would be an interesting and challenging, though now probably meaningless, experiment to put 4 kW into a small stainless steel container under water and see what happens, see if the element burns out, etc. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
heat. If the energy is in the form of 1 MeV photons or more, then my calculations (caveat: I make lots of mistakes) indicate it is not only readily observable but dangerous with 2 cm or 5 cm of lead.The converse is not inconsistent. There are many theories, including mine which provide logical reasons why the photonic energy from LENR is in the form of EUV or soft x-rays, which will only be observed as heat in normal conditions. Low energy alphas or protons would produce heat and be difficult to detect. You answers to these simple questions would be most appreciated. Dave Simple but somewhat time consuming to answer. I am going to bail out on conversation for a few days. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi 1 MW plant - is there a cooling system?
On Oct 18, 2011, at 10:50 PM, Colin Hercus wrote: Hi Horace, I find your posts quite interesting and you seem to have a rational rather than emotional approach which makes for good reading. I just read your reply to Dave and as it seemed to make the ECat (and my kettle) impossible I thought I'd double check some of your calculations and I think you've made a mistake on the heat flow from the reactor: R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W By my calculations: R = 0.002/(16 * 0.018) = 0.002/.288 = 0.007 °C/W Yes you are right! Another one of my clerical mistakes. The above should be written: R = (0.002 m)/((16 W/(m K)*(1.8x10^-2 m^2)) = 6.94x10^-3 °C/W Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T given as: delta T = (6.94x10^-3 °C/W) * (4390 W) = 30.46 °C Using Fourier's law to check, I get q = (16 W/(m K))*(1.8x10^-2 m^2)*(30.47 K)/(0.002 m) = 4387 W which is well within tolerance. I did not put the review up on my site. I should correct it and put it there. Thanks for the correction! I wish my calculations were checked more often. From engineeringtoolbox.com Fourier's Law express conductive heat transfer as q = k A dT / s (1) where A = heat transfer area (m2, ft2) k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft)) dT = temperature difference across the material (K or oC, oF) s = material thickness (m, ft) So A = 180 CM^2 = 0.018 M^2 K = 16 W/(m K) s = 0.002m Then q = 16*0.018*dT/0.002 = 144 * dT So for 2500W we'd have a temperature difference of 2500/144 = 17 C which is quite reasonable. This is all way out of my area of expertise so I could be messing up units somewhere. Best Regards, Colin Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:quantum levitation
On Oct 18, 2011, at 9:55 PM, David Roberson wrote: Hello Frank, You have an impressive understanding of the flux pinning theory. Can you give me an answer to my question? It appears that energy can be put into the floating disk-magnet combination by pushing or pulling against the disk. Where does the energy show up in the system? Does the disk heat up a small amount as I push or pull on the disk or does the magnet get the energy? This question may be related to the amount of force required to displace the disk. There may be important information revealed as a result of the energy transfer. I eagerly await your answer. Dave Hi Dave, Here is guess for you. The magnetic pressure P = B^2/(2*mu0) is reduced in the volume immediately below and above the puck, except in the thin volumes near the puck of flux transiting the thin vortices in which lines of flux are pinned. The magnetic pressure immediately adjacent to the sides of the puck, and adjacent to the pinning locations is increased. Any movement of the puck relative to a given magnet, provided the movement does not involve a canceling symmetry, such as rotation above a single magnet, or movement on a single magnet track, changes the local B and/or volume in which the B resides, and thus magnetic pressure, and thus energy of the system. Pushing the magnet into place merely involves compressing the B into a higher average pressure, and thus consumes energy. The energy in the B resides in the polarized vacuum. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:quantum levitation
On Oct 18, 2011, at 9:55 PM, David Roberson wrote: Hello Frank, You have an impressive understanding of the flux pinning theory. Can you give me an answer to my question? It appears that energy can be put into the floating disk-magnet combination by pushing or pulling against the disk. Where does the energy show up in the system? Does the disk heat up a small amount as I push or pull on the disk or does the magnet get the energy? This question may be related to the amount of force required to displace the disk. There may be important information revealed as a result of the energy transfer. I eagerly await your answer. Dave Hi Dave, Here is guess for you. The magnetic pressure P = B^2/(2*mu0) is reduced in the volume immediately below and above the puck, except in the thin volumes near the puck of flux transiting the thin vortices in which lines of flux are pinned. The magnetic pressure immediately adjacent to the sides of the puck, and adjacent to the pinning locations is increased. Any movement of the puck relative to a given magnet, provided the movement does not involve a canceling symmetry, such as rotation above a single magnet, or movement on a single magnet track, changes the local B and/or volume in which the B resides, and thus magnetic pressure, and thus energy of the system. Pushing the magnet into place merely involves compressing the B into a higher average pressure, and thus consumes energy. The energy in the B resides in the polarized vacuum. The pinned flux, the flux which travels through the SC, moves relative to the fixing magnet if the SC orientation or position changes. The movement of this close line flux superpositions with, moves relative to, compresses and/or decompresses, the magnetic flux which travels around the SC, resulting in energy changes in the B field there, thus resisting motion. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Is it possible Rossi has already tested his 1 MW prototype behind closed doors?
On Oct 19, 2011, at 5:39 AM, peter.heck...@arcor.de wrote: [snip] Somebody has calculated, at 1MW the steam must go supersonic with this output tube. Then, with 100 kW it must still go some 100 km/h. [snip] I got 803 km/hr, which is less than the speed of sound. You may want to check my calculations! Using the photos here: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3264361.ece The outside width of a standard container is 8 feet, or 2.44 meters From the full photo of the back side: The 8 feet = 129 pixels. The red handle = 16 pixels = (16 px)*(2.44 m)/(129 px) = 30 cm, much larger than I would have thought. In the closeup photo the handle is 94 px, giving (30 cm)/(94 px) = 0.319 cm/px. The cap is 40 px, or 12.8 cm OD. The exit pipe appears to have a 22 px OD, or 7 cm OD. Maybe the pipe is 6.5 cm ID, or 3.25 cm radius, giving an area pi*(3.25 cm)^2 = 33 cm^2. The energy put into the steam depends on the temperature to which it is condensed before being fed back into the E-cat. Assume the condensed water is being fed back at 100°C. The energy to vaporize water at 100°C is 2260 J/g. If 1 MW is heating 100°C water then I estimate the flow has to be 442.5 gm/s, with a volumetric flow of 737.5 liters/sec. This gives a flow velocity of (737500 cm^3/s)/(33 cm^3)= 223 m/s in the pipe, or 803 km/ hr. This is below the speed of sound but over 6 times the recommended speed for the pipe size. If I did the calculations right, then this indicates the device could blow up. If there are emergency steam relief valves on the devices the steam could be released inside the container. Note, if water is fed back at 50°C I get only 675 liter/sec steam flow. Related assessments can be found here: http://www.mail-archive.com/vortex-l@eskimo.com/msg51512.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Update to Rossi 6 Oct 2011 Experiment Data Review
My review at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf has been updated. Improved graph formats were provided. I will be available to discuss this once my finite element analysis is done. Meanwhile, I'll hopefully resume lurk mode. A significant part of the update is inclusion of the following sections: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ACTIVE CONTROL To make any sense of the data with a non-nuclear explanation, it appears the electric heating power is separated into two parts, one part which heats the water directly, and one part which heats an internal metal mass. In addition, it appears there needs to be an active control which affects the thermal conductivity between a large thermal mass and the water, and thus division of the input power into a third part. This control must produce minimum thermal resistance between a hot thermal mass and the water when no power is applied to it. Further, it must be controlled with about 300 mA * 240 V = 7.2 watts of power, because the power from the “frequency generator” must be enough to regulate the thermal output power. When main heater power was cut and when the “frequency generator” power was cut, there was an immediate surge of thermal power out. In both cases, a power cut to the heater(s), and a power cut to the frequency generator, a large thermal pulse resulted immediately upon the power cut. One means of achieving the necessary power control is to use the actuator from a zone valve to make or release contact between large area (e.g. 29 cm by 29 cm) slabs of thermal conductors. This can be accomplished by spring loading the slabs to a closed position and using the actuator from a zone valve (.e.g. Taco Power Head) to press the plates apart. A typical US residential zone valve operates in the appropriate power range, and is activated by about 10 V at 1 A. The power is applied to a resistive material which expands thermally to open a zone valve. In a hot environment such an actuator could expand with less than normal power. An alternative to changing slab separation is to control convective flow of a thermal transfer fluid. In this case when power is applied then flow must be cut off. DYNAMIC FEA SIMULATION A dynamic linear FEA simulation program is being developed to look at potential thermal storage mechanisms. A sample of some run input data is located here: http://www.mtaonline.net/~hheffner/RptR4 Report of the results will be made separately from this review. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Rossi H and Ni consumption
From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms / mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from one E-cat, would be readily detected by a geiger counter at significant range. It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi H and Ni consumption
This is a nonsensical argument. The less hydrogen available for nuclear reactions the *more* the MeV per reaction that is required to make the 1 MW output, thus the less effective any shielding would be, and the *less credible* it is that the MW heat comes from nuclear reactions. On Oct 27, 2011, at 11:14 AM, Axil Axil wrote: There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure. The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E-Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner hheff...@mtaonline.net wrote: From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms / mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container housing the E-cats, and most of the 2x10^18 gammas per second would end up escaping the container. This is an approximate calculation. Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from one E-cat, would be readily detected by a geiger counter at significant range. It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Update to Rossi 6 Oct 2011 Experiment Data Review
My review at: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf has been updated. Improved graph formats were provided. I will be available to discuss this once my finite element analysis is done. Meanwhile, I'll hopefully resume lurk mode. A significant part of the update is inclusion of the following sections: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ACTIVE CONTROL To make any sense of the data with a non-nuclear explanation, it appears the electric heating power is separated into two parts, one part which heats the water directly, and one part which heats an internal metal mass. In addition, it appears there needs to be an active control which affects the thermal conductivity between a large thermal mass and the water, and thus division of the input power into a third part. This control must produce minimum thermal resistance between a hot thermal mass and the water when no power is applied to it. Further, it must be controlled with about 300 mA * 240 V = 7.2 watts of power, because the power from the “frequency generator” must be enough to regulate the thermal output power. When main heater power was cut and when the “frequency generator” power was cut, there was an immediate surge of thermal power out. In both cases, a power cut to the heater(s), and a power cut to the frequency generator, a large thermal pulse resulted immediately upon the power cut. One means of achieving the necessary power control is to use the actuator from a zone valve to make or release contact between large area (e.g. 29 cm by 29 cm) slabs of thermal conductors. This can be accomplished by spring loading the slabs to a closed position and using the actuator from a zone valve (.e.g. Taco Power Head) to press the plates apart. A typical US residential zone valve operates in the appropriate power range, and is activated by about 10 V at 1 A. The power is applied to a resistive material which expands thermally to open a zone valve. In a hot environment such an actuator could expand with less than normal power. An alternative to changing slab separation is to control convective flow of a thermal transfer fluid. In this case when power is applied then flow must be cut off. DYNAMIC FEA SIMULATION A dynamic linear FEA simulation program is being developed to look at potential thermal storage mechanisms. A sample of some run input data is located here: http://www.mtaonline.net/~hheffner/RptR4 Report of the results will be made separately from this review. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Update to Rossi 6 Oct 2011 Experiment Data Review
This post is not archiving for some reason. I have inserted a blank into the URLs as a test. My review at: http://www.mta online.net/~hheffner/Rossi6Oct2011Review.pdf has been updated. Improved graph formats were provided. I will be available to discuss this once my finite element analysis is done. Meanwhile, I'll hopefully resume lurk mode. A significant part of the update is inclusion of the following sections: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ACTIVE CONTROL To make any sense of the data with a non-nuclear explanation, it appears the electric heating power is separated into two parts, one part which heats the water directly, and one part which heats an internal metal mass. In addition, it appears there needs to be an active control which affects the thermal conductivity between a large thermal mass and the water, and thus division of the input power into a third part. This control must produce minimum thermal resistance between a hot thermal mass and the water when no power is applied to it. Further, it must be controlled with about 300 mA * 240 V = 7.2 watts of power, because the power from the “frequency generator” must be enough to regulate the thermal output power. When main heater power was cut and when the “frequency generator” power was cut, there was an immediate surge of thermal power out. In both cases, a power cut to the heater(s), and a power cut to the frequency generator, a large thermal pulse resulted immediately upon the power cut. One means of achieving the necessary power control is to use the actuator from a zone valve to make or release contact between large area (e.g. 29 cm by 29 cm) slabs of thermal conductors. This can be accomplished by spring loading the slabs to a closed position and using the actuator from a zone valve (.e.g. Taco Power Head) to press the plates apart. A typical US residential zone valve operates in the appropriate power range, and is activated by about 10 V at 1 A. The power is applied to a resistive material which expands thermally to open a zone valve. In a hot environment such an actuator could expand with less than normal power. An alternative to changing slab separation is to control convective flow of a thermal transfer fluid. In this case when power is applied then flow must be cut off. DYNAMIC FEA SIMULATION A dynamic linear FEA simulation program is being developed to look at potential thermal storage mechanisms. A sample of some run input data is located here: http://www.mta online.net/~hheffner/RptR4 Report of the results will be made separately from this review. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Best regards, Horace Heffner http://www.mta online.net/~hheffner/
Re: EXTERNAL: [Vo]:Rossi H and Ni consumption
On Oct 27, 2011, at 4:49 AM, Roarty, Francis X wrote: On Thurs Oct 27, 2011 Horace said [snip] It does not seem credible the energy from a Ni-H reaction, at least in the form of one gamma per reaction, provides any explanation for 1 MW of heat, if that thermal power is in fact achieved.[/snip] Horace, Assuming the thermal power is in fact achieved, and the reaction is not Ni-H, what do you feel is the next most credible theory ? Fran A Ni-H or even p-e-p nuclear interaction catalyzed by a Ni nucleus is not ruled out given there is a mechanism to disperse the nuclear energy in small increments and avoid radioactive products. I think the reaction begins with a Ni electron being momentarily delayed in the Ni nucleus in a deflated state interaction with a proton or quark, as defined here: http://www.mta online.net/~hheffner/FusionUpQuark.pdf http://mtaonline.net/~hheffner/DeflateP1.pdf This provides the Ni nucleus with a very large magnetic moment, and magnetic gradient, which permits it to be the target of tunneling of deflated state hydrogen from the lattice. This results in multiple hydrogen nuclei present in the Ni nucleus, and a highly de-energized Ni-H deflated nucleus cluster, with multiple trapped electrons which then radiate energy or transfer it directly to k-shell electrons via near field interactions. Various apparently non-radioactive products are thereby made feasible. Non-radioactive products are the branches nature prefers because they are the least energy products. It is notable that no nuclear reaction may result from a given Ni-H deflated cluster, and yet nuclear heat, in the form of zero point energy, is released and then replenished by the zero point field after the cluster breaks up. See: http://mta online.net/~hheffner/NuclearZPEtapping.pdf Discussion of this could be very academic if there is in fact no excess heat from the Rossi experiments. I am hoping to write a FAQ on deflation fusion, but have not had the time. I will be happy to discuss this at a later time. Best regards, Horace Heffner http://www.mta online.net/~hheffner/
Re: [Vo]:Rossi H and Ni consumption
You are off on a tangent. My point is that Rossi's claims are in conflict with the observed results. I will no longer respond for now. On Oct 27, 2011, at 12:15 PM, Axil Axil wrote: In the Miley presentation that he has recently released, Miley shows transmutation to 39 isotopes over possible contamination levels. The nuclear reactions and transmutation patterns that are going on inside the Rossi reactor are similar to what Miley documents as mentioned in Rossi’s original patent. The presence of a large amount of iron in the Miley results is interesting and similar iron contamination was found in the Rossi ash(10%) when they were analyzed by the swedes. The assumption that the nuclear reactions taking place in the Rossi reaction are exclusively restricted to copper transmutation is mistaken in my opinion. The possibility that the reactions going on are hydrogen only cannot be ignored with the production of copper as only one of many reactions going on. On Thu, Oct 27, 2011 at 3:21 PM, Horace Heffner hheff...@mtaonline.net wrote: This is a nonsensical argument. The less hydrogen available for nuclear reactions the *more* the MeV per reaction that is required to make the 1 MW output, thus the less effective any shielding would be, and the *less credible* it is that the MW heat comes from nuclear reactions. On Oct 27, 2011, at 11:14 AM, Axil Axil wrote: There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure. The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E- Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner hheff...@mtaonline.net wrote: From: http://www.rossilivecat.com/ Quote: - - - - - - - - - - - - - - - - - - - - Andrea Rossi October 25th, 2011 at 4:59 PM Dear Thomas Blakeslee: Grams/Power for a 180 days charge Hydrogen: 18000 g Nickel: 1 g Warm Regards, A.R. - - - - - - - - - - - - - - - - - - - - End quote. At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of H need be provided per 1 atom of Ni. Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. This involves the obviously wrong assumption that all the Ni atoms are transmuted, not a more realistic 3 percent. There is also an outside possibility the H reacts with daughter products, giving the possibility of 10 subsequent daughter reactions per primary Ni+H reaction. Three such reactions is an outside possibility. One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms /mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by single reactions than that is 0.9464 MeV per Ni-H event. The gammas from this would be lethal at short range, even through 2 cm of lead. If it is assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per reaction. If there are an average of 3 daughter reactions per primary reactions that is about 10 Mev per reaction. If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use in protecting the operators. If near 1 MeV gammas are produced the lead shielding is inadequate. One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 gammas per second. using: I = I0 * exp(-mu * rho * L) where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/ cm^3, we have for 5 cm of lead: I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) * (5 cm)) I = 2x10^18 free gammas per second. About half that, or 10^18 gammas/s would be directed toward the interior of the container
Re: [Vo]:Manifold mismeasurement makes models meaningless
I think it is great you are pursuing this Alan. I think the temperature of the thick brass part may play a similar or even larger role than the steel nut. I noted on page 4 of my review: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf - - - - - - - - - - - - - - - - - - - - - - - - - - - This photo by Mats Lewan of NyTeknik of the 6 Oct Rossi Tout thermocouple that it can and probably did extend beyond the steel nut, toward the brass manifold: http://www.mtaonline.net/~hheffner/LewanTcoupleClose.jpg It was thus subject to the air temperature in the volume underneath the insulation and between the brass manifold and steel nut. It is especially notable that the frayed insulation, cut from around the probe tip, was not trimmed. This is very unusual. The frayed electrical insulation may have prevented good thermal contact of the thermocouple with the steel nut, and thus exposed the thermocouple primarily to the air temperature in the vicinity, which would be expected to be higher than that of the steel nut. - - - - - - - - - - - - - - - - - - - - - - - - - - - On Oct 27, 2011, at 1:05 PM, Alan J Fletcher wrote: At 06:51 AM 10/27/2011, Higgins Bob-CBH003 wrote: I examined pictures of the manifold and created a diagram to capture the important features. [I made a small .png version of the diagram that I am trying to include.] I am not sure it is schematically correct yet. A characteristic that I believe is very important in the analysis of the possible temperature contamination is the issue of the fittings used in the manifold. These use pipe threads, and appear to be NPT because of the use of pipe dope. At each junction of pipe threads, there will be a large thermal resistance compared to continuous brass. Analysis of these across-the-thread resistances are going to be hard, particularly with pipe dope and or Teflon tape present as is required to seal NPT. The resistance across the thread boundaries will be high and the net effect will be to significantly decouple the Tout thermocouple from the manifold. These thread boundary effects don't appear to be included in your model. Thanks for the diagram. So far I've just widened my original model to 12 cm ... and get results which are closer to the measured value. http://lenr.qumbu.com/rossi_ecat_oct11_spice.php Update information is copied below : http://lenr.qumbu.com/lenr_spicepics/111027_spice_0001.png The bottom pane shows the new schematic. A is the extreme right, and B and C are the centers of the two steps. The thermocouple is on step C. The center pane shows the temperature across the manifold. A is now at 33.4 C (compared to the secondary water temperature of 30 C). The top pane shows the OFFSET in temperature from A. This new result shows that the result varies dramatically with the geometry -- and since the actual measurements are not known, the results are speculative. - - - - - - - I can easily add in a thread boundary as resistors between the steps. But I don't think I can draw any REAL conclusions from this model ... except to say that the thermocouple should not be ANYWHERE on the manifold! Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Minor progress
line too. http://www.ascovalve.com/Applications/News/NewsASCOMiniatureValve.aspx Interesting that the pulse code modulation PCM variable valves operate at kHz frequency. I suspect I'm looking in the wrong places for the kind of valves desired. If I can get a good realistic computer simulation then it might be fun to build a physical E-cat simulator using that design. OTOH, it would be a much better use of time to resume other experimentation. I slightly changed the format of my web site, but it is all the same old stuff. I thought it felt better starting off with my haiku regarding cold fusion: http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 12:31 AM, John Bresnahan wrote: Dear Mr. (Dr.?) Heffner, I've been eagerly following your posting on the Vortex mailing list, and wish to thank you for the thoughtful analysis you are providing. Regarding the small valve in your model of Rossi's E-Cat device from the October 6th test, could it be that Rossi's frequency generator is used to control and power the valve? Just a thought. Sincerely, John Bresnahan On Nov 7, 2011, at 3:32 AM, Berke Durak wrote: You are proposing a theory where a slug of hot iron releases its stored energy. No. I only used that as an initial example for discussion. I am now looking at, simulating, concrete and other materials, individually or in combination, as noted in my prior post. The e-Cats have enough internal volume to store the reported amount of energy produced in very hot iron, and it is theoretically possible to insulate them using aerogel so that they'll keep their heat for a few hours. Install a controllable heat-exchange mechanism, program the software to emulate a reactor output, et voilà! No computer is required. The heat flow can be controlled simply using the input power profile and a second connection for direct control, e.g. from the frequency generator input. I looked at aerogel, but more mundane insulations are likely good enough. Except that this theory deosn't fly for the 1 MW demo. About 9.5 GJ was produced. There is no credible evidence that 9.5 GJ was produced in my opinion. The 1 MW demo was disgusting scientifically speaking. It was a major step backwards in calorimetry method from the prior test. In any case it is not my subject matter. The 1 MW test was so bad I see no sense in discussing it. I've done some layman calculations, and using aerogel, you could go to 1200 degrees Celsius, and the amount of iron required would be about 250 kg per module. Look at the pictures of the e-Cat. The modules are standing on pieces of metal supported by 5 cm x 5 cm angle sections about 5mm thick. I don't think you can put 500 - 750 kg over 1.5 m on such angle sections. Cement has 3 times the specific heat of iron. Also, the heat output is not substantiated. Neither is the energy input. I will not respond further to comment regarding the 1 MW test as I see it as irrelevant and far less credibly executed and reported than the prior test. However, pure cement thermal output for the 6 Oct test would peak at time T550, 550 minutes after start of the run, which is too late, and the peak too little, without combining the cement with metal slabs or mixtures. -- Berke durak On Nov 6, 2011, at 9:34 PM, Horace Heffner wrote: I continue to plod along on a simulation of prospective E-cat designs to fit the 6 Oct 2011 Rossi test results. I have simulated various combinations of materials for thermal storage and have found that a couple slabs of ordinary Portland cement with a heating resistor sandwiched between them seems to fit the properties of the E-cat fairly well in terms of heat storage dynamics. Call me Horace. I am simply an amateur, not a Dr.. I have appended the ACTIVE CONTROL and DYNAMIC FEA SIMULATION sections from my review, because they provide some clarification. I think the fine temperature control exhibited in response to the very small control current from the frequency generator, in Graph 3; http://www.mtaonline.net/%7Ehheffner/Graph3.png demonstrates the possibility there are two independent, thermally isolated, slabs of material involved. This is also confirmed somewhat by the steep power decline curve at the end of the test. It is also consistent with this assumption that the frequency generator was controlled by a variac. As I noted in my last post, the material used to produce the simulation graphs was *Portland cement*, not iron. (BTW that was a clerical error on my part. The parameters actually shown were those of fire brick. I simply picked iron as an example for my initial calculations in the data review paper because iron has a fairly high specific heat and iron is commonly used in radiators, etc. The following slab commands show some materials I have briefly investigated individually or in combinations: * slab thick spec cond den * . slab description follows slab command ... * slab 20 0.46 80.4 7.874 iron slab 1 0.84 0.01 0.002 aerogel slab 5 2.30 0.64 0.92 HDPE slab 5 1.00 1.31 2.40 ceramic slab 5 0.84 0.166 1.4 asbestos cement slab 20 0.13 35.3 11.34 lead slab 20 0.87 255 2.7 aluminum slab 30 1.05 1.4 2.4 fire brick slab 120 1.55 0.29 1.506 Portland Cement ACTIVE CONTROL To make any sense of the data with a non-nuclear explanation, it appears the electric heating power must be separated into two parts, one part which heats the water directly, and one part which heats an internal mass. In addition, it appears there needs
Re: [Vo]:Minor progress
On Nov 7, 2011, at 5:27 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: I continue to plod along on a simulation of prospective E-cat designs to fit the 6 Oct 2011 Rossi test results. I have simulated various combinations of materials for thermal storage and have found that a couple slabs of ordinary Portland cement with a heating resistor sandwiched between them seems to fit the properties of the E-cat fairly well in terms of heat storage dynamics. I don't get this. What is the point of this simulation? I see no point in debating this with you at all at this point.It it very time consuming, and I prefer to spend the time productively. I still am not at a point where I can even write the paper or the simulation results much less debate results. Nevertheless I'll give you one response and then go back to work on it. It cannot explain the salient facts about the reactor: If you spent an hour or so looking at what I actually provided instead of generating arm waving non quantitative babble then you might gain some understanding. * There is no slab of cement in the reactor. People have looked inside and seen no such thing. There is no record I know of showing anyone having access to the inside of the 30 cm x 30 cm x 30 cm interior box. The slab report shows a probably *insufficient* mass (for fire brick) of 48.4416 kg. See: http://www.mtaonline.net/~hheffner/RptR4 A Slab Report for Portland cement follows: Slab Report i WidthSp.Ht.Cap. Therm.Cond. Den. Description (nwidth) (J/(g K))(W/(m K))(g/cm^3) 1 120 1.550 00.291.506 Portland Cement i WidthThm. Mass Therm.Res.Mass Description (cm) (kJ/°C) (°C/W) (kg) 1 12.0047.11551 2.460125 30.3971 Portland Cement = == === 12.0047.11551 2.460125 30.3971Totals You can see cement only requires 30.4 kg instead of 48 kg, to provide a thermal mass of 47 kJ/°C vs 51 kJ/°C. A significant portion of concrete mass requires replacement with something like aluminum or iron to bring the thermal resistance down to where it needs to be and bring the mass up to where it needs to be. Cement requires even less mass than fire brick because it has 3 times the specific heat of iron, and 50 percent more than fire brick. The device weighed 98 kg. The metal boxes are sheet metal. You think a couple sheet metal boxes, a few pipe fittings and the bolts weigh 98 kg or even 50 kg? To me this makes no sense. Plus the reactor would weigh far more than it does if there was concrete in it. You ignored the numbers I provided. The slab provided actually may provide too little mass. Certainly Portland cement (as opposed to the fire brick actually shown) would have too little mass. It would take a gigantic slab to vaporize 60 L of water, much larger than the reactor. There is no evidence 60 L of water was actually vaporized. This is just an arm waving number without substantiation. No one knows the actual amount of water vaporized in any of the tests due to bad calorimetry. * There is no power going into resistors for 4 hours during the self-sustaining run. If you look at the run data you will see that I used exactly the same power input profile provided by Mats Lewan. The only thing unusual, and wrong, is that I used a starting temperature of 100°C. This has only a small effect on the energy profile, and will be eliminated when I model volume stored, water temperature, etc. * The entire reactor starts at room temperature. There is no way you could hide a very hot object inside it. No insulation is good enough to keep the surface from being quite warm. When people pick up the reactor to prevent weight scale they would feel it is hot. * All of the heat added to the reactor initially is measured and it is far less than the heat that came out. There was no dependable measurement of the heat that came out. So this is not prospective E-cat design. It is not a way to simulate the performance of the E-cat. That depends on exactly what the Pout thermocouple was reading on the heat exchanger. I do not understand what you are getting at here. Yes indeed. Apparently you do not understand. I think this is due to invalid a priori assumptions on your part, especially in regards to thermal dynamics. You think the output temperature curve can only decline after power is turned off. This is clearly not true. - Jed Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 5:25 AM, Stephen A. Lawrence wrote: Quick question, Horace: Are you going for the 470kW which was claimed, or are you working with a reduced number? The 470 value seems to have been predicated, once again, on total vaporization of the input water. If that didn't take place then the generated power may have been substantially lower. I am mainly attempting to reproduce graphs 2 and 5: http://www.mtaonline.net/~hheffner/Graph2.png http://www.mtaonline.net/~hheffner/Graph5.png assuming a much lower energy than the 470kW claimed. To produce Graph 5 I had to assume the calorimetry was off by a factor of 75%, i.e. the calorimeter indicated 4 times the true power. This was a mistakes on my part, as was providing the 0.8°C bias based on the initial thermometer readings. This °C bias, as I noted in my review, added about 37% energy to the output. I should have used the original data, not my biased data. I suspect the displacement in temperature that occurs initially, as shown in Graph 4: http://www.mtaonline.net/%7Ehheffner/Graph4.png may actually be due to the heat exchanger interior and the E-cat interior being cooler than the water temperature at the time of the run start. Perhaps some cold air flow from the E-cat may have made it out of valve as the cold water initially displaced it in the E- cat. In any case, if the actual data is used, then the discrepancy between claimed power out and negative COP power out is not so large. I suspect the Pout thermocouple was not even touching the nut, that the frayed insulation was in part between the thermocouple tip and the nut, thus exposing the thermocouple primarily to the air temperature under the insulation. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 11:24 AM, Stephen A. Lawrence wrote: Ah -- Sorry, Horace, disregard my question. I overlooked the fact that you're ignoring the Oct 28 test, which was the (alleged) 470kW run. (In any case, you obviously are well aware of the heat-of- vaporization issues.) Yes. I am having problems keeping track of things, and have not been able to read much on vortex of late. Sorry I am a bit behind responding, and our messages crossed. Something of possible interest is that the (real) cement slab data is here: http://www.mtaonline.net/~hheffner/Graph6Sb.png This can be compared to the fire brick data here: http://www.mtaonline.net/~hheffner/Graph6S.png The cement data shows the peak energy output much delayed vs the fire brick, around T540 vs T330. However the peak power out delivered is less. A similar phenomenon should occur if controls are used to cut back the power out to make a 6 hour run vs 4 hour run. If the power out were actually from nuclear energy, not thermal energy stored using electric power, then such a large drop in output should not be necessary to accommodate the longer run time. Rossi actually stated on his blog, if I recall correctly, that the power was reduced from 1 MW because the (prospective) customer required 6 hours instead of 4. Obviously the shortcoming from the 1 MW output expectation (objective?) was the fault of the customer! 8^) Obviously, I claim, as I wave my arms wildly. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 3:43 PM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: If you spent an hour or so looking at what I actually provided instead of generating arm waving non quantitative babble then you might gain some understanding. It is not arm waving to point out that THERE IS NO CONCRETE in the reactor. None. You are wasting your time speculating about how this might work, because people have look inside these reactors and they saw no concrete. Are you postulating there is invisible concrete? The device weighed 98 kg. The metal boxes are sheet metal. You think a couple sheet metal boxes, a few pipe fittings and the bolts weigh 98 kg or even 50 kg? To me this makes no sense. Evidently you forgot the thing has lead sheets in it. Look at the photos. If the wrapping was a lead sheet it was very thin. Again, I don't know of anyone being allowed to see the insides of the 30x30x30 interior box. It DOES have lead. It DOES NOT have concrete. Got it? - Jed How about a reference instead of more arm waving? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 4:18 PM, Berke Durak wrote: On Mon, Nov 7, 2011 at 8:12 PM, Colin Hercus colinher...@gmail.com wrote: Or 25kg per module if we just bring the water to 105C and make very little steam But that assumes that the numbers are falsified. In the customer's public report, it says : Water vaporized : 3716 l. So if that figure is false, anything goes and there is nothing left to investigate. You have to put faith in something, otherwise it is pointless to discuss - just call it a scam and move on. -- Berke Durak Anyone who believes that 3716 liter number with so little evidence deserves the E-cat he buys. My work is focused on the 6 Oct. test. I think the 1 MW test too nonsensical and data free to be worthy of a technical discussion. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:My Explantion for Rossi Results
http://img.ound.com/static-data/assets/ 6/3a86f663d804b2877e9dcb0e1f003e699da87b26_m.gif Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
I did try lead in various combinations with other materials. It does not have very good characteristics. I am working to duplicate the output power wave form, given the input power vs time, not just explain the energy balances. I'll have more to say when I finish. Horace On Nov 7, 2011, at 4:05 PM, Colin Hercus wrote: Hi Horace, I was wondering if it's possible to do this with lead rather than another material as long as you have sufficient insulation to reduce the heat flow from the lead to the water. I did a simple simulation and it looked like about 25kg of lead with about 12W/C heat flow would do the trick. I was also thinking that you might get some stratification of the water with cooler water at the bottom and hot near the top. In latter stages of life after death this could be really important to keep the outflow at 100C. For Oct 6th test it also requires water flow to match Mat Lewan's measured rate rather than Rossis 11kg/h, which also leaves the possibility of the flow rate being increased to shut down the reaction. Best Regards, Colin On Tue, Nov 8, 2011 at 4:54 AM, Horace Heffner hheff...@mtaonline.net wrote: On Nov 7, 2011, at 11:24 AM, Stephen A. Lawrence wrote: Ah -- Sorry, Horace, disregard my question. I overlooked the fact that you're ignoring the Oct 28 test, which was the (alleged) 470kW run. (In any case, you obviously are well aware of the heat-of- vaporization issues.) Yes. I am having problems keeping track of things, and have not been able to read much on vortex of late. Sorry I am a bit behind responding, and our messages crossed. Something of possible interest is that the (real) cement slab data is here: http://www.mtaonline.net/~hheffner/Graph6Sb.png This can be compared to the fire brick data here: http://www.mtaonline.net/~hheffner/Graph6S.png The cement data shows the peak energy output much delayed vs the fire brick, around T540 vs T330. However the peak power out delivered is less. A similar phenomenon should occur if controls are used to cut back the power out to make a 6 hour run vs 4 hour run. If the power out were actually from nuclear energy, not thermal energy stored using electric power, then such a large drop in output should not be necessary to accommodate the longer run time. Rossi actually stated on his blog, if I recall correctly, that the power was reduced from 1 MW because the (prospective) customer required 6 hours instead of 4. Obviously the shortcoming from the 1 MW output expectation (objective?) was the fault of the customer! 8^) Obviously, I claim, as I wave my arms wildly. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Andy Findlay's Metal Rod
Andy Findlay mentioned at one time that if you heat the end of a metal rod with a blowtorch it takes time for the heat to reach the other end, and that it continues coming after the heat is removed. The following is a demonstration run showing this using a firebrick rod and the input energy pattern from the 6 Oct E-cat test. The heat ends up eventually evenly distributed (enthalpy works), but it takes quite a while to get there. It is notable that fire brick can withstand 1650°C temperatures without melting. Some ceramics do much better and have similar characteristics otherwise. http://www.mtaonline.net/~hheffner/Graph7Sx.png Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 7:03 PM, Jouni Valkonen wrote: Horace, indeed 2 megaeuros would be good investment to check the validity of Rossi's claim. If it works, then we are hundreds of modules to play around. And if it does not work in means of cold fusion processes then just return the device and get full monetary compensation. Rossi has promised life time warranty and replacing used cells with new ones. (this is the main reason for the high price, that It includes lifetime worth of nickel fuel.) It may come as a surprise, that if you buy a hoax, then it is considered as violation of contract and seller is obligated to refund every monetary losses caused. I do not know Americans, but this is European law. This law applies even for private auctions. It is just completely impossible to sell a 2 megaeuro hoax to buyer who has resources to hire lawyers. —Jouni I think having the money to hire lawyers is meaningless if the money is spent and/or given to charities. This kind of purchase might best be made by people who have badges and carry guns, or have special teams. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 7, 2011, at 7:52 PM, David Roberson wrote: It always works out the way you desire when you cherry pick the data. Throw away data that does not match your needs, keep all that does. This is the common way that some science operates. OK Dave, please supply the input power and flow time stamped data from the time the M-cats are cool initially to power down. I especially would like to have the power supplied via frequency generator. I'll scale it down to a single E-cat and run it through my simulator once it is completed. If you want to be truly honest in your effort, you must explain all of the experimental evidence. And, it is important that you consider the fact that 3 core modules were used in the 1 MW test. The customer acceptance was based upon having 3 cores and thus the results were far more robust. I guess that you are attempting to explain the test in early October by suggesting that there was no active core at all. It might be possible to simulate that since only one core was generating heat. My last conclusion seemed to indicate that the power output was a lot less than the thermocouple data suggested. Try not to be manipulated by Rossi and confused as he seems to enjoy misdirecting us at every possible turn of events. He has done that to me on many occasions. Think of this. One core module supplies 3.4 kilowatts of output in driven mode. In the self sustaining mode it is generally 1/2 that amount. I would not be surprised to see just 1.7 kilowatts under this condition, and that should be easy to simulate with concrete or iron or many other possible materials. But then, with 3 modules of the 1 MW system output power goes up to approximately 10 kilowatts for each ECAT. This will be virtually impossible to simulate in the driven mode where the system puts out 1 MW. You might be able to approximate the self sustaining mode at half that power (500 kW), but it will be much harder. Again, if you expect to convince the majority of us that Rossi is conducting a scam, explain the difficult case. Convincing anyone that Rossi is conducting a scam is not my goal. I *would* very much like to see customers demand at least mediocre calorimetry though. I would like to see credible proof produced, and it could have already been produced at almost no expense, with simply the right experimental plans. I also do not want to see a bad financial deal made of any kind - because it will be used as an example every time serious scientists attempt to obtain research funding. This kind of turn of events could have a serious negative impact on billions of people for years to come. It is important to determine as quickly as possible if nuclear energy is actually being produced and at what power level. That said, consider the fact the less data the more difficult. The most difficult case would be the case where Rossi holds a press conference and states he has been running an E-cat for more than a year to heat a building. Yes, all claim and no data. That's the most difficult case. Really not very far from the MW test though is it? The data from that test, like 9 prior, depends on the assumption that steam was flowing, plus various other assumptions that are unreasonable to make during due diligence. There really isn't much difference between insufficient data and no data. Say, wait a second, Rossi has already specified the the most difficult case. He already stated he has been using an E-cat to heat a building for over a year. Well, I guess I can't run that data through a program, can I? Yep, scientifically irrefutable! Wow, I really had no idea the nuclear energy claim was so well verified. The most difficult experimental evidence can not be explained away! That's iron clad proof! I wonder if dealerships are being offered? 8^) Sorry, the sarcastic nome must have cast a momentary spell on me. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Dave -Original Message- From: Horace Heffner hheff...@mtaonline.net To: vortex-l vortex-l@eskimo.com Sent: Mon, Nov 7, 2011 8:47 pm Subject: Re: [Vo]:Minor progress On Nov 7, 2011, at 4:18 PM, Berke Durak wrote: On Mon, Nov 7, 2011 at 8:12 PM, Colin Hercus colinher...@gmail.com wrote: Or 25kg per module if we just bring the water to 105C and make very little steam But that assumes that the numbers are falsified. In the customer's public report, it says : Water vaporized : 3716 l. So if that figure is false, anything goes and there is nothing left to investigate. You have to put faith in something, otherwise it is pointless to discuss - just call it a scam and move on. -- Berke Durak Anyone who believes that 3716 liter number with so little evidence deserves the E-cat he buys. My work is focused on the 6 Oct. test. I think the 1 MW test too nonsensical
Re: [Vo]:E-Cat / philosophical remarks
On Nov 8, 2011, at 6:41 AM, Jed Rothwell wrote: I wrote earlier that Rossi is in a bind because he has no viable patent. Then just now I wrote that I have urged him to do a proper test, get funding, and then hire experts, the way, Robert Lynn recommends. The problem is, Rossi does not trust outsiders. He cannot even bring himself to give a reactor to the University of Bologna where he has many friends. This is a problem largely of his own making. I understand why he does not trust people. He has had a painful life and he has often been betrayed and unjustly persecuted. For example, one of the charges they sent him to jail for was defrauding the stockholders. He himself was the only stockholder, so this was Kafkaesque. Someone in the Italian justice system had it in for him. I do no see any way for him to escape this conundrum. Rossi says that a public demonstration, controlled by independent engineers, for the benefit of the international media can be beneficial for the dissemination of E-Cat . . . would be completely useless. I expect he sincerely believes this, but it is nonsense. Without question, such a test with be beneficial for the dissemination of the E-cat. But it would destroy his business strategy. He would not think of doing it. Plan B would be to adopt a conventional business strategy like the Lynn and I advocate. I am sure he has never seriously considered doing that. When I and others have suggested this he has brushed us off. As things stand he will never allow a proper test. - Jed Rossi's behavior is absurd, unless he doesn't believe in the technology himself. Then it makes complete sense. If Rossi actually has something useful, and it is not patentable, then he could still make a fortune producing energy and selling it directly to a grid. He could relocate to Mexico and sell power to the west coast of the USA through the existing grid. He could make billions. He could make a fortune with just steam heat by using it to extract oil from Canadian oil sands, though he might have even more trouble with nuclear authorities in Canada than even the USA. In any case, bulk power production would be much easier to beat the red tape on than any kind of small commercial sales. If he produced a just a MW of commercial grid electric power for a few months he would probably have investors flocking to him with money. I would think if he could actually do this he would have done it. If he actually heated a commercial building for more than a year with nickel and hydrogen I would think he would want to show that. If he can produce a COP of 6 or even 3 then it should be easy to drive a sterling generator and turn that COP 6 into COP infinity. I don't see anything happening that is fully consistent with a useful technology being present. There is much happening that is consistent with no useful technology being present. What sane person would invest in E-cats if things are in this status? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 8, 2011, at 5:10 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: Again, I don't know of anyone being allowed to see the insides of the 30x30x30 interior box. 1. Levi and the people at Defkalion say they saw inside. Levi and Defkalion people saw inside the 6 Oct E-cat? I thought Defklion and Rossi had outs. If they saw inside some other device at some other time then that is irrelevant. Levi has been an inside guy from the beginning has he not? I see no difference between him and Rossi in regards to this issue. For that matter Defkalion is or will be selling similar devices, true? What is important, obviously, is access by independent observers. Lewan says you can see more than the photograph shows. There is no sign of concrete. There was undoubtedly no sign of eagles or of diamond rings or elves or many other things either, or for that matter Ni or lead. The phrase see more than the photograph shows can mean anything. BTW, the final large heat pulse before power cut-off could very well be due to water flowing into the 30x30x30 box through a hole. There could be two major slabs, one large and long (and to the left of the wiring input port) with lower thermal conductivity, and one short one with higher thermal conductivity material (to the right of the wiring input port). The water access port would provide access of the water to the larger left slab. Access of water to the smaller higher thermal conductivity slab would be the result of removal of the signal generator signal. Just to be clear, not that it is very important or relevant, I did not use the term concrete to mean ordinary concrete made with sand and rocks. Ordinary concrete has poor thermodynamic properties compared to Portland cement. If you see me use the term concrete please assume it is one of my many typos. I actually mean cement. Cement delays the heat pulse too long.Ceramics or fire brick delay the post power cut off heat pulse to a time closer to the observed data. 2. In previous tests observers dumped out the water from the vessel after the run and measured the volume. There is no space unaccounted for in the vessel. There is no place to put concrete. These are meaningless words. I specified *inside* the 30x30x30 cm inner box. What happens outside that box is obviously immaterial. Why would you bring such a red herring into the discussion? 3. The previous cylindrical reactors were easy to see inside of. There was no concrete in them. It makes no sense to claim that the previous reactors were real and this one is fake. This is nonsense, and yet another red herring. You are digging pretty deep to respond! 8^) The calorimetry for those devices was entirely different. They were not designed by Rossi to demonstrate heat after death. The obvious flaw in the demonstration of those devices is the output was never observed - it was simply sent down a drain. Furthermore, you claim that output power is not measured accurately but this is incorrect. This analysis shows that the temperature of the cooling loop thermocouples was correct to within 0.1°C: http://lenr-canr.org/RossiData/Houkes%20Oct%206%20Calculation%20of% 20influence%20of%20Tin%20on%20Tout.xlsx Take a look at this photo again: http://www.mtaonline.net/~hheffner/LewanTcoupleClose.jpg There is a good possibility the thermocouple did not even touch the metal of the steel nut. Why would anyone with any experience at all leave the mess of ragged insulation around the thermocouple? It looks to me the thermocouple was probably exposed primarily to the air temperature under the insulation. At any rate, any one with nominal experience should know to place the thermocouple down the rubber hose a bit to avoid thermal wicking in the metal. No one has challenged this analysis. Besides, even if this is incorrect and half of the input power is being stored while the electric power is turned on, What do you mean half the input power is being stored? It is all being stored (except for leakage through the insulation) until heat shows up at the heat exchanger. the overall output profile is still correct, and output greatly exceeds input. In other words, in the storage scenario, you lower the output curve to half of the input, while power is on, and then measure the area of stored energy, and compare it output energy during the time power is on, and afterwards. The area of the latter greatly exceeds the former. All you are saying here is the output energy is larger than the input energy. We can not know that without good thermocouple readings. This is not inferable from a measurement. This is a rehash of old well trodden material. Storage cannot explain these results. Sure it can, if the thermocouple readings are not reliable. Most importantly, simple
Re: [Vo]:Minor progress
On Nov 8, 2011, at 10:02 AM, Jed Rothwell wrote: David Roberson dlrober...@aol.com wrote: Jed, I have reason to believe that the output thermocouples are reading incorrectly. Then I suggest you address the paper uploaded by Houkes, and show where it is in error. - Jed Why is this material not in pdf format like other material on LENR- CANR.org? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:E-Cat / philosophical remarks
On Nov 8, 2011, at 10:10 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: Rossi's behavior is absurd, unless he doesn't believe in the technology himself. Then it makes complete sense. His behavior is irrational and absurd. However, such behavior is common among inventors and discoverers, and it has been throughout history. There are many famous examples such as John Harrison. There are many in the present day and among cold fusion researchers, such as Patterson. I do not think it makes complete sense that Rossi does not believe in the technology himself. This is a different statement from the one I made. I implied Rossi's behavior makes complete sense if he does not believe in the technology himself. I did not say it makes complete sense that Rossi does not believe in the technology. There is a difference. The question though should be which premise is more consistent with Ross not believing in the technology? If he did not believe in it, he would gleefully promote it and he would put on more impressive demonstrations. Fake but impressive. He would gladly accept money from investors since the only point of doing this would be to fleece people. That is not what he is doing. He is, in fact, beating off investors with a stick. He is turning down money. I know several people who offered him large sums. He refused them all. He did not even answer some of them. This is not characteristic of a fraud who does not believe in his own work. Well, that depends on what the terms of the offers was doesn't it? Whether performance clauses were discussed, for example. Also, from whom the offers were made. It is characteristic of a lone inventor who does not want to give up control. Patterson was the same way. I know people who offered him funding, which he turned down. As I said, he was determined to have 100% market share. And yet he is considering a stock offering? If Rossi actually has something useful, and it is not patentable, then he could still make a fortune producing energy and selling it directly to a grid. He could relocate to Mexico and sell power to the west coast of the USA through the existing grid. He could make billions. I do not think the power companies would allow this. You think Mexico would not cooperate with this on a shared profit basis? A chance to make billions? I think someone at some level and above would support it. Mexico is moving into the solar business now I believe. Also, by the time he set up and was able to do this, the secret of this technology would be out and he would be reverse engineered by every major industrial manufacturing company on earth. - Jed How long could it take to have a bunch of E-cats, say 6 M-cats, made and shipped to Mexico? After that it is just a matter of driving an appropriate generator. The ones used for solar thermal should do nicely. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:E-Cat / philosophical remarks
I wrote: This is a different statement from the one I made. I implied Rossi's behavior makes complete sense if he does not believe in the technology himself. I did not say it makes complete sense that Rossi does not believe in the technology. There is a difference. The question though should be which premise is more consistent with Ross not believing in the technology? I guess I need more sleep. This should read: This is a different statement from the one I made. I implied Rossi's behavior makes complete sense if he does not believe in the technology himself. I did not say it makes complete sense that Rossi does not believe in the technology. There is a difference. The question though should be which premise is more consistent with Rossi's behavior, he believes his own claims, or not? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:E-Cat / philosophical remarks
On Nov 8, 2011, at 10:54 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: The question though should be which premise is more consistent with Rossi's behavior, he believes his own claims, or not? The premise that best fits his behavior is the same one that fits Harrison, Patterson, William Shockley, and many other people with a personality similar to Rossi's. They are intensely possessive. It is difficult to believe that Harrison, Patterson, or Shockley would put on about a dozen demonstrations of their technology, repeatedly botch the scientific aspects of the demonstrations, and refuse to acknowledge or fix the problems. When Patterson could no longer reproduce his results he freely admitted it. These were scientifically trustworthy people. How about Rossi's self contradiction record with regard to the E-cat? Does that put him in the same camp with Harrison, Patterson, and Shockley, or in a different class? At what point does the balance of probability tip? How many failed failed demonstrations and absurd refusals to correct does it take to seriously question whether there is anything at all to the technology. So far Rossi has left a critical degree of freedom without observations in each experiment. His behavior is completely consistent in this regard. Can this be considered random? At what point does a Baysian model provide a sufficient confidence level the behavior is not random? Which premise is most consistent with Rossi's actions? He believes in the technology and has nothing to hide from a black box evaluation of energy out? He doesn't believe a rigorous test will confirm his claims? The answer seems fairly obvious to me which premise is most consistent. However, others can look at the same set of facts and draw opposite conclusions. This makes for an interesting world I guess. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 8, 2011, at 11:15 AM, Jed Rothwell wrote: Horace Heffner hheff...@mtaonline.net wrote: Levi and Defkalion people saw inside the 6 Oct E-cat? So they say. Just to be clear, they say they saw inside the 30x30x30 cm inside box in the 6 Oct E-cat demo? Do you have a reference on this? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
Well I got some sleep and am catching up on this thread. I am very disappointed. The confusion here is incredible. It also appears no one has read my paper at all: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf especially the sections T2 THERMOCOUPLE LOCATION and VOLUME CALCULATIONS, wherein I analyze the photos, Photo 1 and Photo 2 in my paper, which for some reason everyone confuses as showing the inside of the 30x30x30 cm inside box that supposedly houses one to three 1 cm thick reactors (or 3 cm thick reactors if you please, Rossi made both statements), and to which I referred when I said no one saw inside it at the demo. I was *not* referring to the roughly 50x60x35 cm *exterior* box. The posters on this for some reason seem to confuse the two boxes. Jed calls the 30x30x30 cm inside box the reactor, though it clearly is much more than the reactor. It is a reactor housing that supposedly keeps the reactor dry and protected, and to which 1 /4 inch and 1 inch water sealed conduit pipes connect which carry water, main power, and the frequency generator power from the outside to the stuff inside the box. The material I have analyzed fits inside the 30x30x30 cm box. The 50x60x35 cm exterior box to which others refer is irrelevant, except when water levels and temperatures are simulated. It is disappointing that people would think I have not even seen the photos I so carefully analyzed and described in my paper. This reinforces the feeling I have had that this is all a waste of time. Here are the important facts: 1. No one at the 6 Oct demo saw inside the 30x30x30 cm box. It was not opened. 2. Mats Lewan did not see any features of the box aside from what was shown in the various photos. He did not see any exterior structures that might be important, such as holes, vents, fins underneath, etc. The only features visible were the bolted flanges and the pipe feed throughs. 3. The small interior 30x30x30 box was bolted to the bottom of the exterior box. It is thus unlikely a set of fins like those on top are present on the bottom of the 30x30x30 cm box. 4. No one would have been able to observe cement, ceramic tiles, fire brick, iron slabs, lead slabs, Ni containers, valves, wiring, hidden water access ports, etc., because the inside box was not opened. 5. The inside and outside boxes, and the contents of the inside box, together weigh 98 kg. Clearly the inside and outside boxes, pipes and bolts that are visible do not weigh anything like 98 kg. The boxes are made of sheet metal. Therefore the density of the 30x30x30 cm box and its interior contents is very high. I am attempting to construct my simulation within these constraints. I think Bob Higgen's diagram at: http://lenr.qumbu.com/rossi_ecat_oct11_a.php is inaccurate. The reactor is enclosed inside the 30x30x30 cm interior box. The fins are not as big as shown. There is only one set of fins, on top. The thermocouple is much longer than shown and likely rests against the edge of the inside box, and probably on the flanges of the inside box, which are not shown. The gaps between the inside box and the edges of the outside box are too large in proportion. The 50x60x35 cm exterior box dimensions include the flanges to which the top panel is bolted. This only leaves a few centimeters gap (5 cm on the ends, 3 cm on the sides, excluding the flanges) between the inside box and the outside box. See the sections of my paper referenced above. I should note here that I am working on an update of those sections based on an improved photo analysis. Here are my best numbers so far: Width of E-cat inside box: 30.3 cm Interior width of E-cat outside box, flange to flange: 49.6 cm Interior width of E-cat outside box, side to side : 40.6 cm Interior length of E-cat outside box: = 46.3 cm Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
I wrote: It is a reactor housing that supposedly keeps the reactor dry and protected, and to which 1 /4 inch and 1 inch water sealed conduit pipes connect which carry water, main power, and the frequency generator power from the outside to the stuff inside the box. That should read: It is a reactor housing that supposedly keeps the reactor dry and protected, and to which 1 1/4 inch and 1 inch water sealed conduit pipes connect which carry water, main power, and the frequency generator power from the outside to the stuff inside the box. I wrote: 1. No one at the 6 Oct demo saw inside the 30x30x30 cm box. It was not opened. That should read: 1. None of the observes at the 6 Oct demo saw inside the 30x30x30 cm box. It was not opened. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:What Happened with DGT
On Nov 8, 2011, at 3:19 PM, Daniel Rocha wrote: That's the rest mass of the electron... So, any idea? 2011/11/8 Terry Blanton hohlr...@gmail.com From Jeane Manning: http://changingpower.net/articles/physicists-insights-on-greece-and- rossi%E2%80%99s-e-cat/#more-716 Her source says: It is known that the type of asserted reaction, namely Ni -Cu transition, must release gamma photons in the 511 Kev range but this was never actually measured. 0.511 MeV . . . now where have I seen that before? ;-) T This is baloney. One of the early tests involved use a coincidence counter, a pair of gamma counters with coincidence circuitry, which picks up the gamma pairs from positron annihilations. None were observed above background. It was used up close to the reactor. Here is a photo in which the pair of opposed coincidence counters can be seen: http://www.ccemt.org/Energy%20Alternatives/cold_fusion/files/ rossi_cold_fusion_aparatus_scintillator_300.jpg posted on this blog: http://www.cce-mt.org/Energy%20Alternatives/cold_fusion/cold_fusion.html regarding a February 2011 test. Part of the second counter can be seen protruding below the surface on which the E-cat rests. Celani has observed some single (not positron annihilation) counts : I brought my own gamma detector, a battery-operated 1.25″ NaI(Tl) with an energy range=25keV-2000keV. I measured some increase of counts near the reactor (about 50-100%) during operation, in an erratic (unstable) way, with respect to background. See: http://blog.newenergytimes.com/2011/01/18/rossi-and-focardi-lenr- device-celani-report/ http://tinyurl.com/4djya8 Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 8, 2011, at 9:52 PM, Aussie Guy E-Cat wrote: I have spent some time on working out what is what in the Exposed E- Cat photos. What can be seen is boiler scale on the reactor heat radiation fins, external conduits and assembly bolts which seems to indicate water and steam occur in the outer box as the Higgins drawing suggests and not inside the reactor core as you suggest. Nonsense! That water and steam are present in the outside box has never been in doubt by anyone that I know of. What I suggested is the possibility ports can be opened to the inside box to permit timed and limited water exposure to selected slabs of material, and the resulting steam emissions. The source and destination of the water/steam is of course the outside box, and then the top vent. You assertion that you can determine whether or not this occurs from the photos is the nonsense. The steam outlet from the outer box is via a fitting on the top and not from the reactor core as you suggest. You must think I am and idiot to say such a thing about me. Did you not read my estimates of the location of the port in my photo analysis? http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Do you think I am unaware of the T fitting in the top of the outer box through which the thermocouple also is fitted, the location of which I determined by photo analysis? This would suggest the water input is to the outer box (inlet fitting on the bottom lower front left and not from the side as the Higgins drawings suggests) Well of course there is a water inlet on the outside box, on the left front. and there is no water inside the smaller finned reactor core. This you have no way of knowing. See attached photo. From what I can see there are 3 conduits connections into the reactor core to supply H, heater power and RF energy. There are actually four: 1 water, 1 gas, 2 for frequency generator input. Based on my measurements of the photos and assuming a symmetrical reactor core design, there is room for the fins on the bottom of the reactor core as Higgins suggests. Of course there is room. The problem is the fins were not observed there by Mats Lewan who had extensive access at the demo being discussed. AG On 11/9/2011 4:53 PM, Horace Heffner wrote: Well I got some sleep and am catching up on this thread. I am very disappointed. The confusion here is incredible. It also appears no one has read my paper at all: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf especially the sections T2 THERMOCOUPLE LOCATION and VOLUME CALCULATIONS, wherein I analyze the photos, Photo 1 and Photo 2 in my paper, which for some reason everyone confuses as showing the inside of the 30x30x30 cm inside box that supposedly houses one to three 1 cm thick reactors (or 3 cm thick reactors if you please, Rossi made both statements), and to which I referred when I said no one saw inside it at the demo. I was *not* referring to the roughly 50x60x35 cm *exterior* box. The posters on this for some reason seem to confuse the two boxes. Jed calls the 30x30x30 cm inside box the reactor, though it clearly is much more than the reactor. It is a reactor housing that supposedly keeps the reactor dry and protected, and to which 1 /4 inch and 1 inch water sealed conduit pipes connect which carry water, main power, and the frequency generator power from the outside to the stuff inside the box. The material I have analyzed fits inside the 30x30x30 cm box. The 50x60x35 cm exterior box to which others refer is irrelevant, except when water levels and temperatures are simulated. It is disappointing that people would think I have not even seen the photos I so carefully analyzed and described in my paper. This reinforces the feeling I have had that this is all a waste of time. Here are the important facts: 1. No one at the 6 Oct demo saw inside the 30x30x30 cm box. It was not opened. 2. Mats Lewan did not see any features of the box aside from what was shown in the various photos. He did not see any exterior structures that might be important, such as holes, vents, fins underneath, etc. The only features visible were the bolted flanges and the pipe feed throughs. 3. The small interior 30x30x30 box was bolted to the bottom of the exterior box. It is thus unlikely a set of fins like those on top are present on the bottom of the 30x30x30 cm box. 4. No one would have been able to observe cement, ceramic tiles, fire brick, iron slabs, lead slabs, Ni containers, valves, wiring, hidden water access ports, etc., because the inside box was not opened. 5. The inside and outside boxes, and the contents of the inside box, together weigh 98 kg. Clearly the inside and outside boxes, pipes and bolts that are visible do not weigh anything like 98 kg. The boxes are made of sheet metal. Therefore the density
[Vo]:Krivit names some Rossi customer names
http://blog.newenergytimes.com/2011/11/09/poor-journalism-by-wired-u- k-on-rossi-story/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
On Nov 8, 2011, at 10:35 PM, Aussie Guy E-Cat wrote: Mate I'm not a physicists or an antagonists. Just a very practical old power systems engineer. You have come up with a exotic theory of scam that requires you to prove it. Not true. It is not I who is making the claims. I merely intend to show some of the arguments put forth here that the data provided indicate Rossi's clamis have to be real are false. If the data can be reproduced with a device which produces no nuclear energy, whether that device actually exists or not, then it should be pretty obvious the data does not support Rossi's claims. I am advocating for better testing procedures. The actual existence or not of my simulated device is irrelevant. The important point is the quality of the data. I made suggestions in my report for specific ways to improve the quality of the data. I am not alone in this. Many other people have suggested numerous similar things over recent months. Rossi's behavior is potentially seriously damaging the future of LENR research and the future of billions of people. I think it is important to speak out about this. If I say I doubt your theory, that is my right and you have no right to say Nonsense cause you have absolutely no proof of what you suggest is even remotely true. I have the right. In fact exercised it. 8^) Your statement made no sense at all. You wrote: ... water steam occur in the outer box as the Higgins drawing suggests and not inside the reactor core as you suggest. The observation that ... water steam occur in the outer box... does not preclude in any way that water and steam can occur in the inner box under limited control. You made an erroneous inference, a logic error. It makes no sense. You also grossly underestimate my understanding of the structure of the E-cat in question. As a point of interest do you accept the significant and long term reports of excess heat generation in Ni-H LENR cells? If you knew anything of my history, or looked at my web site, you would know I am an LENR advocate and experimenter, and that I accept that some experimental reports of light water excess heat are likely correct. I have done some experimenting myself and put forth some amateur theories: http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf http://www.mtaonline.net/~hheffner/dfRpt http://www.mtaonline.net/~hheffner/DeflationFusion.pdf The question in my mind is not whether LENR exists, but rather whether any evidence exits at all that supports Rossi's claims of commercially viable nuclear energy production. These are two very different things. If not why? If yes then why do you doubt Rossi? I see Rossi as potentially the biggest threat to the field that has ever come along. I also think I made fairly clear in my data review my position with regard to the 6 October 2011 test I have been addressing of late: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf I think it was the best of the tests so far, but still obviously inconclusive. AG On 11/9/2011 5:39 PM, Horace Heffner wrote: On Nov 8, 2011, at 9:52 PM, Aussie Guy E-Cat wrote: I have spent some time on working out what is what in the Exposed E-Cat photos. What can be seen is boiler scale on the reactor heat radiation fins, external conduits and assembly bolts which seems to indicate water and steam occur in the outer box as the Higgins drawing suggests and not inside the reactor core as you suggest. Nonsense! That water and steam are present in the outside box has never been in doubt by anyone that I know of. What I suggested is the possibility ports can be opened to the inside box to permit timed and limited water exposure to selected slabs of material, and the resulting steam emissions. The source and destination of the water/steam is of course the outside box, and then the top vent. You assertion that you can determine whether or not this occurs from the photos is the nonsense. The steam outlet from the outer box is via a fitting on the top and not from the reactor core as you suggest. You must think I am and idiot to say such a thing about me. Did you not read my estimates of the location of the port in my photo analysis? http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf Do you think I am unaware of the T fitting in the top of the outer box through which the thermocouple also is fitted, the location of which I determined by photo analysis? This would suggest the water input is to the outer box (inlet fitting on the bottom lower front left and not from the side as the Higgins drawings suggests) Well of course there is a water inlet on the outside box, on the left front. and there is no water inside the smaller finned reactor core. This you have no way of knowing. See attached photo. From what I can see there are 3 conduits connections
[Vo]:Back to lurk mode
I'm going back to lurk mode to try to get something done ... or maybe just go on vacation. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Minor progress
First let me correct an earlier statement in this thread. In regards to the pipe conduits to the interior box from the front of the outer box I said: There are actually four: 1 water, 1 gas, 2 for frequency generator input. That was meant to say: There are actually four: 1 gas, 1 main power, and 2 for frequency generator input. I think it is especially odd that the two frequency generator conduits, one above the interior box flanges, one below, are 1 1/4 inch pipe, while the conduit for the main power is only 1 pipe. It seems reasonable to speculate as to what might require, and be located inside, the large pipes. On Nov 9, 2011, at 10:35 AM, Jouni Valkonen wrote: 2011/11/9 Horace Heffner hheff...@mtaonline.net: The material I have analyzed fits inside the 30x30x30 cm box. The 50x60x35 cm exterior box to which others refer is irrelevant, except when water levels and temperatures are simulated. I am responding to this post only because words I did not issue have been put in my mouth. If you think that there is a 30×30×30 cm³ black box Black is your wording, not mine, in relation to color. Those dimensions came from Mats Lewan's report which I reference in my paper: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf I also determined from the photos that the actual dimension is closer to 30.3 cm. Any reference to a black box I might have made in my writing was not literal, but I don't recall referring to the interior box as black. The color might be called rusty dirty scale deposited on aluminum. (it was not mine impression, but my impression is based on indirect conclusion made that I do not remember anyone saying seen such a large black box inside), If you had read my paper you would have seen a photograph appended of the 30x30x30 cm interior box, with sealed pipe fittings going into it from the front of the larger box. and you think that Rossi is an evil criminal and fraudster, I did not at any time say that. Those are your words, not mine. It is you who repeatedly jumps to the fraud conclusion, not me. Fraud or self delusion are of course possibilities I recognize, as do many others, especially given Rossi's inability numerous times to provide anything other than highly flawed calorimetry data, or refusal to admit the importance of such mundane scientific concepts as controls, etc. The lives of billions of people are affected by Rossi's actions now, regardless the outcome. Why will he never make the tiny incremental effort required to properly demonstrate he produces nuclear heat? If he does not give a damn about the rest of the world, only his marketing strategy, then that indeed does not speak highly of his morality, does it? His bizarre behavior raises logical questions. Has he no faith in himself to produce his claimed results? Has his discovery gone the way of Patterson's beads? Are his results now merely amplified artifacts, or insufficient to be commercially viable? Is he unable to run for multiple days, much less multiple months as claimed? Only Rossi himself is responsible for creating these doubts. What I *would* be happy to do is show the possibility that a logical construction can produce the observed results. Given the 37% extra output heat that I mistakenly built into my spread sheet by biasing the temperature, it does not take an unfeasible error in the Tout reading to accommodate a good match of result by simulation. Given it is not even known for sure the Tout thermocouple was in direct contact with metal, this is not a far reach. However, if I could show even a possible fraud based mechanism exists which simulates the results with the given inputs, that would be sufficient to demonstrate the calorimetry requires improving. It should be sufficient to quell at least some of the ridiculous non-quantitative arm waving true believer arguments made here, but probably won't. You do see the difference between calling Rossi an evil criminal fraudster and showing a logical mechanism exists which reproduces the experiment outputs given only the experiment inputs, don't you? The purpose for the latter is to provide some motivation or justification for a customer demand for appropriate due diligence. The former would serve no purpose. Many people in the blogosphere have said or implied the E-cat is a fraud, so the former would be useless, in addition to being unsubstantiated arm waving. then why do you cannot understand, that it is also trivial to fit internal chemical power source to 30×30×30 cm³ black box? If you had read my paper, especially the section CHARACTERISTICS OF THE CENTRAL MASS you would have understood. There is a logical explanation for using slabs of material to retain and stabilize heat. Thin layers of insulation can be placed between the iron and the catalyst, and the catalyst and the water
Re: [Vo]:comment on New Energy Times' editorial about MeV/He-4
not effect heavy element transmutation LENR with the closest atoms, the lattice heavy elements. Fusion with a hydrogen atom that is typically even further away than the nearby lattice heavy elements is then also precluded. CF is known to happen below the surface, within the lattice. Whether it also happens on the surface due to collective surface oscillations as suggested by Windom and Larsen is immaterial. An explanation of CF needs to cover all observations, not just a select few. The distance between lattice sites, i.e. the distance from the potential well an absorbed hydrogen nucleus occupies (a lattice site) and the adjacent potential well another hydrogen atom can occupy, is less than the distance between a lattice site and the adjacent lattice atoms. Windom and Larsen estimate slow neutrons to be absorbed in less than a nanometer, 10^-9 meter, about 10 angstroms. That is about 10 hydrogen atoms, or 3 Pd atoms in width. If neutrons can make it 3.5 Å [ note typo fixed here, said 0.5 Å previously ] into a nearby hydrogen nucleus they can make it 1.79 Å into Pd or another lattice element just as well. There are no other nuclei in the way, so cross sections are not even an issue. Heavier atoms are not all that much bigger than light ones because atomic radius does not grow much with atomic number, e.g. radii in angstroms: Pd 1.79, Au 1.79, Ni 1.62, Li 2.05, K 2.77, Al 1.82, Cu 1.57, Pb 1.81. If fusion is occurring at a rate sufficient to account for excess heat then NA should occur at a huge rate also, one that could not possibly be missed. Heavy LENR is known to occur, has been observed, and thus requires just as much explanation as other CF results. The lack of high energy radiation signatures for both CF and heavy transmutation LENR, both of which are known to occur both very close to and below the surface, requires an explanation. The unusual branching ratios observed require an explanation. The presence of ultra-slow neutrons in the lattice provides no explanation for these things. Gammas from NA should be readily observed from heavy element transmutation if it is due to neutrons. The presence of hypothesized high mass electrons on a cathode surface, near surface hydrogen fusion reactions, were suggested to absorb fusion gammas in less than a nanometer. This explanation can not account for gamma absorption near heavy elements. NA gammas should be readily detectable. I think the presence of a free electron in the nucleus at the time of fusion is the logical explanation of all these things, how the Coulomb barrier is breached, why high energy particles and gammas are not seen from hydrogen fusion reactions, why the branching ratios are so skewed, and why almost no signature, including heat, is seen corresponding to nuclear mass changes from heavy lattice element transmutation. How this is proposed to happen is described in Cold Fusion Nuclear Reactions at: http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Doing the Bosenova
On Feb 1, 2010, at 7:21 AM, Jones Beene wrote: ZPE is looking more and more like an “energy sink” instead of an “energy source” … … but do not fear vorticians – perhaps it can be both. Particle physicists know well the vacuum acts as both a mass/energy source and sink - even at GeV levels. But if it only operates an energy sink, It doesn't. Borrowed mass/energy is part of every weak interaction, because the W-, W+, and Z have huge masses. These are borrowed on short term loan. However, the fact the universe is here indicates that in the right circumstances the vacuum can put mass/energy on a long term loan. I think nature is making long term loans all the time, as noted in: http://www.mtaonline.net/~hheffner/GravityPairs.pdf and pp 21-24 of: http://www.mtaonline.net/~hheffner/GravityPairs.pdf If the latter is correct, then we can possibly do this in a controlled fashion. then even so, as anyone who has followed this emerging viewpoint is aware, an energy sink can be almost as valuable for alternative energy, if not more so - than a true energy source … since we already have an adequate source. It is hidden away in the 300 degree K “ambient” blackbody field that we are so accustomed to living in (“swimming in”) – that we seldom look at it as a source. And ironically, it is not a source without the sink. There is a pun in there somewhere (sink or swim?) but I will leave it for the others punsters to take a shot at. Jones Analogically speaking, since the capacity of the drain matches that of the faucet, the question at hand is how to plug the drain. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Roald Hoffmann poem
http://www.roaldhoffmann.com/pn/modules/Downloads/docs/ An_Unusual_State_of_Matter.pdf http://tinyurl.com/ykmwoxe Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Doing the Bosenova
On Feb 4, 2010, at 7:30 PM, mix...@bigpond.com wrote: Note that in the neighborhood of Rb there is a slight kink in the curve. This may allow reactions like 2 Rb - A + B + energy, where A is lighter than Rb and B is heavier than Rb. Perhaps Horace can run his program and see if there are any exothermic reactions possible? Here are the results: Energetically Feasible Stable Bose Condensate Pairs X, Y Resulting from Reactions of the Form: Rb + Rb -- X + Y + energy --- Equations follow for Rubidium, Rb, element 37 --- 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV 85Rb37 + 87Rb37 -- 86Kr36 + 86Sr38 + 1.024 MeV 85Rb37 + 87Rb37 -- 86Sr38 + 86Kr36 + 1.024 MeV 85Rb37 + 87Rb37 -- 88Sr38 + 84Kr36 + 3.588 MeV 87Rb37 + 87Rb37 -- 88Sr38 + 86Kr36 + 1.992 MeV Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Doing the Bosenova
On Feb 4, 2010, at 7:30 PM, mix...@bigpond.com wrote: Note that in the neighborhood of Rb there is a slight kink in the curve. This may allow reactions like 2 Rb - A + B + energy, where A is lighter than Rb and B is heavier than Rb. Perhaps Horace can run his program and see if there are any exothermic reactions possible? Here are the results if you give the reactions 2 MeV margin for catalytic electrons: Energetically Feasible Stable Bose Condensate Pairs X, Y Resulting from Reactions of the Form: Rb + Rb -- X + Y + energy --- Equations follow for Rubidium, Rb, element 37 --- 85Rb37 + 85Rb37 -- 86Kr36 + 84Sr38 + -0.425 MeV [-0.425 MeV] (R2_Rb:1) 85Rb37 + 85Rb37 -- 85Rb37 + 85Rb37 + 00.000 MeV [00.000 MeV] (R2_Rb:2) 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV [2.620 MeV] (R2_Rb:3) 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV [00.527 MeV] (R2_Rb:4) 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV [4.177 MeV] (R2_Rb:5) 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV [1.342 MeV](R2_Rb:6) 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV [2.193 MeV] (R2_Rb:7) 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV [1.145 MeV] (R2_Rb:8) 85Rb37 + 85Rb37 -- 94Zr40 + 76Se34 + -1.816 MeV [-1.816 MeV] (R2_Rb:9) 85Rb37 + 87Rb37 -- 86Kr36 + 86Sr38 + 1.024 MeV [1.024 MeV] (R2_Rb:10) 85Rb37 + 87Rb37 -- 87Rb37 + 85Rb37 + 00.000 MeV [00.000 MeV] (R2_Rb:11) 85Rb37 + 87Rb37 -- 86Sr38 + 86Kr36 + 1.024 MeV [1.024 MeV] (R2_Rb:12) 85Rb37 + 87Rb37 -- 88Sr38 + 84Kr36 + 3.588 MeV [3.588 MeV] (R2_Rb:13) 85Rb37 + 87Rb37 -- 90Zr40 + 82Se34 + -0.404 MeV [-0.404 MeV] (R2_Rb:14) 85Rb37 + 87Rb37 -- 92Zr40 + 80Se34 + -0.551 MeV [-0.551 MeV] (R2_Rb:15) 87Rb37 + 87Rb37 -- 87Rb37 + 87Rb37 + 00.000 MeV [00.000 MeV] (R2_Rb:16) 87Rb37 + 87Rb37 -- 88Sr38 + 86Kr36 + 1.992 MeV [1.992 MeV] (R2_Rb:17) Total number of reaction equations: 17 Maximum number of D fused with X: 0 Adjustment factor to compound nucleus radius: 1 Energy threshold for including reaction, in eV: -200 Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Doing the Bosenova
On Feb 5, 2010, at 6:57 AM, Jones Beene wrote: As we mentioned in previous postings, any nuclear reaction with Rb is extremely unlikely, if we assume it is related in any way to a thermonuclear reaction. I think this is true. OTOH, the fact that a gas, Kr, would be produced from a Rb Bose condensate wavefunction collapse, it is very tempting to think such a thing is possible. The Bosenova was created using 85Rb: http://www.nist.gov/public_affairs/bosenova.htm This gives the following potential reactions to stable products: 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV It is notable that one of the potential products is a gas, krypton, which might escape detection in the experiment if produced. The nucleus 85Rb has an even number of neutrons, 48, plus 37 protons and electrons. Provided the electrons and protons pair spins, the net spin of the 85 Rb atom is zero. At one time I suggested the possibility that an (extrenal source provided) energetic particle could collapse the wave function of a Bose condensate to a point: http://mtaonline.net/~hheffner/BoseHyp.pdf This would mean that both the nuclei and electrons would condense to (approximately) a point. Such a collapse would create a highly negative energy entity, having possibly on the order of many GeV negative energy. However, as the electron wavefunctions expand, the negative energy would be restored from the vacuum, and the nuclei would have the energy to react, producing nearly zero net energy reactions. The reaction that would be triggered first, from paired rubidium nuclei, would be: 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV Thus producing a large proportion of krypton gas. The 2.620 MeV is otherwise irrelevant, because it is essentially consumed by the electron negative energy. The explosion would be produced with nominal energy. This is admittedly far fetched, for various reasons, one of the most obvious ones being this: an amount of strontium corresponding to the krypton created would be left behind. Surely this strontium would have been noticed, if present in such a large proportion. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Doing the Bosenova
I wrote: The reaction that would be triggered first, from paired rubidium nuclei, would be: 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV That was one of my typical clerical errors. The reaction triggered first as condensed electrons inflated would be: 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV as can clearly be seen from the list of the only energetically feasible reactions producing stable nuclei: 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV It is notable that, unlike LENR heavy element transmutation, where experimental data indicates otherwise, large number Bose condensate triggered reactions, if they exist, may not be limited to fairly stable nuclei. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Doing the Bosenova
On Feb 5, 2010, at 11:13 AM, Jones Beene wrote: -Original Message- From: Horace Heffner JB: As we mentioned in previous postings, any nuclear reaction with Rb is extremely unlikely, if we assume it is related in any way to a thermonuclear reaction. HH: I think this is true. OTOH, the fact that a gas, Kr, would be produced from a Rb Bose condensate wavefunction collapse, it is very tempting to think such a thing is possible. JB: Well, I'm not sure why 'any gas' would be preferential, It is not that gasses per say are preferential at all. It is a matter of energy. Only those reactions that yield net energy occur (at least according to conventional theory). When I gave the following reactions as the only Energetically Feasible Stable Bose Condensate Pairs X, Y Resulting from Reactions of the Form: Rb + Rb -- X + Y + energy, this means I checked all such possible reactions, and these were the only 85Rb + 85Rb reactions yielding positive energy: 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV When I backed off the energy by 2 MeV I got more: 85Rb37 + 85Rb37 -- 86Kr36 + 84Sr38 + -0.425 MeV 85Rb37 + 85Rb37 -- 85Rb37 + 85Rb37 + 00.000 MeV 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV 85Rb37 + 85Rb37 -- 94Zr40 + 76Se34 + -1.816 MeV Now here is a much more interesting variation. When I allowed short half-life product nuclei and required net energy from each reaction I still obtained the first set of reactions. But, when I permitted radioactive products and an energy deficit of up to 2 MeV, this is the much larger list I obtained: 85Rb37 + 85Rb37 -- 85Kr36 * + 85Sr38 * + -1.752 MeV 85Rb37 + 85Rb37 -- 86Kr36 + 84Sr38 + -0.425 MeV 85Rb37 + 85Rb37 -- 85Rb37 + 85Rb37 + 00.000 MeV 85Rb37 + 85Rb37 -- 86Rb37 * + 84Rb37 * + -1.838 MeV 85Rb37 + 85Rb37 -- 87Rb37 + 83Rb37 * + -0.662 MeV 85Rb37 + 85Rb37 -- 85Sr38 * + 85Kr36 * + -1.752 MeV 85Rb37 + 85Rb37 -- 86Sr38 + 84Kr36 + 2.620 MeV 85Rb37 + 85Rb37 -- 87Sr38 + 83Kr36 + 00.527 MeV 85Rb37 + 85Rb37 -- 88Sr38 + 82Kr36 + 4.177 MeV 85Rb37 + 85Rb37 -- 89Sr38 * + 81Kr36 * + -0.431 MeV 85Rb37 + 85Rb37 -- 90Sr38 * + 80Kr36 + -0.501 MeV 85Rb37 + 85Rb37 -- 89Y39 + 81Br35 + 1.342 MeV 85Rb37 + 85Rb37 -- 90Y39 * + 80Br35 * + -1.958 MeV 85Rb37 + 85Rb37 -- 91Y39 * + 79Br35 + -1.921 MeV 85Rb37 + 85Rb37 -- 90Zr40 + 80Se34 + 2.193 MeV 85Rb37 + 85Rb37 -- 91Zr40 + 79Se34 * + -0.527 MeV 85Rb37 + 85Rb37 -- 92Zr40 + 78Se34 + 1.145 MeV 85Rb37 + 85Rb37 -- 94Zr40 + 76Se34 + -1.816 MeV The radioactive products above are flagged with a asterisk. I think this speaks as to one aspect of why heavy element LENR tends to create stable products. The unstable products have larger masses, thus leaving no energy (or less energy) for the reaction to be pulled off as the deflated electrons gradually reduce their binding energy and escape the nucleus, thus permitting the most energetic reactions to occur first, thus tending to prevent the reactions which create radioactive nuclei. In the case of Rb + Rb it is somewhat coincidental that *no* reaction is energetically feasible that creates a radioactive product. It is also true that no deflated electron energy deficits were involved in the calculations, but this case still demonstrates one aspect of how the creation of radioactive products is suppressed in heavy element LENR. There is also the question as to why fission would be expected, and not typical small particle decays, e.g. beta, proton, or alpha decay, etc. One reason is that fission occurs when Z^2/A 47. In this case (2*37)^2/47^2 = 32.2, so no fission should be expecte, conventionally speaking. However, It seems to me that expanding wavefuction electrons in the nucleus, post BEC collapse, likely exert a powerful influence on fission via nucleus kinetic interaction, vacuum energy supplied Schrodinger pressure, and negative nuclear halo production, all of which which expand the nucleus and tear it apart. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:comment on Violante data as covered by Steve Krivit
Two things to consider: (1) reversing the current *does* dissolve the Pd surface, and (2) previous work has shown that helium production takes place near but below the surface (order of microns), while tritium production tends to take place on or very close to the surface (within a few atomic widths). This has been a classic problem with CF, converting the process into a bulk effect instead of a surface effect for all practical purposes. On Feb 7, 2010, at 2:58 AM, Michel Jullian wrote: 2010/2/2 Abd ul-Rahman Lomax a...@loma xdesi gn.com: ... A single SRI experiment has been published that made strong efforts to recover all the helium, and it came up with, as I recall, about 25 MeV. That experiment was discussed in the paper submitted by Hagelstein, McKubre et al to the DOE in 2004: http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf They flushed helium out by simply desorbing and reabsorbing deuterium several times, by varying the cell current, which they reversed in the end to get all the D out. It seems to me that if they actually managed to extract all the helium this way, which their resulting Q value suggests (104±10 % of 23.8 MeV), the reaction can't possibly happen in the bulk. Not even subsurface. It has to happen exactly on the surface, with some (about half) of the produced helium nuclei going slightly subsurface. If the reaction itself was subsurface, surely about half of the produced helium couldn't be recovered without more radical means such as the one you suggested below. ... 2. Recovery of *all* the helium -- except perhaps for minor and unavoidable leakage, which should, of course, be kept as small as possible. What occurs to me is to dissolve the cathode. This seems a good idea. I forget the best acid to use, but I do know that palladium can be dissolved. As I recall, Aqua Regia is the best for Pd. Michel Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Underground neutron counting
the background is shown. It is close to zero in the region of channels 1600, but in clear excess (34.8 neutrons) in the region below this. The Fig. states that channel 1400 lies at 760 keV thermalised neutrons. There was no observable effect of D2 pressure (1.1 to 101 kPa), nor of the addition of LiD. No excess neutrons were found when Ti or Pd metal was crushed under D2O, to emulate the Russian work (Klyuev et al), which is thus not confirmed. 021993|051993 # # Tomas P, Blagus S, Bogovac M, Hodko D, Krcmar M, Miljanic D, Pravdic V, Rendic D, Vajic M, Vukovic M; Fizika (Zagreb) 21 (1989) 209--214 Deuterium nuclear fusion in metals at room temperature. ** Experimental, Pt, electrolysis, neutrons, surface analysis, res- Starts with an interesting historical introduction on cosmic ray mesons and discussions of 1947 and thereafter. This team tried to reproduce the FPH electrolysis experiment. X-ray fluorescence after long electrolysis showed Pt deposition of the Pd. A (6)Li-glass scintillation (NE 912) counter was used to used to detect neutrons. The experiment took place in an underground lab, and no neutrons above the low background were seen. The authors promise results from tritium analysis of both the electrolyte and palladium, as well as from proton measurements, to be done. 051989|061989 # Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:CF in Physics Today
On Feb 6, 2010, at 3:51 PM, Jones Beene wrote: ... IOW muon fusion is ongoing but rare. I think cosmic ray triggering may be very important to triggering cold fusion burst events. Also to surface volcano creation frequently observed. Small but important distinction. Therefore, I think it is safe to say that MCF *always* occurs in palladium deuteride CF as a matter of course, in fact it would be difficult to restrain it from occurring, but the number of fusion events is so low over any given time span that it cannot explain the excess heating ... or rather it can explain only a small fraction of the excess heat - probably far less than one percent. It could, however, serve to explain a small diurnal variation. As for a larger diurnal variation, or as a real triggering event, that would be where the difficulty lies. Cosmic rays are isotropic. At the surface their effect is not isotropic due to a slight east-west bias due perturbation of cosmic rays by the earth's magnetic field, however diurnal *flux* variation is small. I think it is neutrino flux that varies daily due to the sun being the primary local source, and the earth (or in the case of an eclipse the moon) absorbing some of the neutrinos. Some component of the solar wind might be important? Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:comment on Violante data as covered by Steve Krivit
On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote: 2010/2/7 Horace Heffner hheff...@mtaonline.net: Two things to consider: (1) reversing the current *does* dissolve the Pd surface, True, but extremely slowly I believe. A Pd anode is known to dissolve relatively fast in acidic electrolytes such as D2SO4, but I don't think that's what they used. It is doubtful whether they reverted the current long enough to dissolve more than a few atomic layers. I think the experimenters were competent. They knew what they were doing. Using a Faraday constant of 96,485 C/mol, and (conservatively) a valence of 4, n for moles produced, I for current = .2 A, t for time = 1 s, we get: n = I * t / (96,485 C/mol * 4) n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or 3.12x10^17 atoms per second. We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/cm^3 = 7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3, and the atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per second is (3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per atom) = 4.45x10^-6 cm/s, or 445 angstroms per second. The number of layers of atoms removed is (4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s. If this is correct (highly suspect! 8^), then at a current density of 200 mA/cm^2 we have a thickness of 183 atoms removed per second, or 445 angstroms per second. and (2) previous work has shown that helium production takes place near but below the surface (order of microns), while tritium production tends to take place on or very close to the surface (within a few atomic widths). I guess you mean they are *found* there, couldn't they be both produced on the surface, only with more kinetic energy in the helium nuclei (alphas) than in the tritium nuclei for some reason, so that the helium is implanted more deeply? I find the idea of two different nuclear reaction sites producing different products a bit unlikely. No, most of the 4He reactions occur sub-surface. What do you think produces a volcano? A surface reaction? The typical 4He produced by CF does not have MeV kinetic energy, and is not surface produced. If it were there would be massive alpha counts. There is not sufficient kinetic energy to push alphas that deep into the Pd. This has been a classic problem with CF, converting the process into a bulk effect instead of a surface effect for all practical purposes. Maybe it's just not possible, because you can't make large D fluxes collide head-on Head on collisions, i.e. kinetics, can not possibly account for cold fusion. in the bulk, this can only happen at a significant scale on the surface (desorbing vs incident fluxes). In the bulk, it seems to me the deuterons just push and follow each other down the lattice's concentration gradients, and never really collide hard. Also, if Bose Einstein Condensates are involved, they requires cold bosons for their formation. Head-on collisions may be a plausible mechanism for deuteron kinetic energy removal. This would only be the case if the collisions were almost all totally inelastic. The only way that can happen is if they are fusions. Michel On Feb 7, 2010, at 2:58 AM, Michel Jullian wrote: 2010/2/2 Abd ul-Rahman Lomax a...@loma xdesi gn.com: ... A single SRI experiment has been published that made strong efforts to recover all the helium, and it came up with, as I recall, about 25 MeV. That experiment was discussed in the paper submitted by Hagelstein, McKubre et al to the DOE in 2004: http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf They flushed helium out by simply desorbing and reabsorbing deuterium several times, by varying the cell current, which they reversed in the end to get all the D out. It seems to me that if they actually managed to extract all the helium this way, which their resulting Q value suggests (104±10 % of 23.8 MeV), the reaction can't possibly happen in the bulk. Not even subsurface. It has to happen exactly on the surface, with some (about half) of the produced helium nuclei going slightly subsurface. If the reaction itself was subsurface, surely about half of the produced helium couldn't be recovered without more radical means such as the one you suggested below. ... 2. Recovery of *all* the helium -- except perhaps for minor and unavoidable leakage, which should, of course, be kept as small as possible. What occurs to me is to dissolve the cathode. This seems a good idea. I forget the best acid to use, but I do know that palladium can be dissolved. As I recall, Aqua Regia is the best for Pd. Michel Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Super-radiance 2.0 Was: comment on Violante data
On Feb 8, 2010, at 11:38 AM, Jones Beene wrote: -Original Message- From: Stephen A. Lawrence I'm not going to pretend I can follow the reasoning here. Sorry... Well, you can lead a horse to water, but you can't make it drink. If it helps to slake your thirst for nano-insight on this subject, here is the same story from a different slant - the breakdown of Planck's law at the nanoscale: http://www.physorg.com/news168101848.html http://www.nanowerk.com/news/newsid=11917.php For the first time, however, MIT researchers have achieved this feat, and determined that the heat transfer can be 1,000 times greater than [Planck's] law predicts. This statement strikes me as rather nonsensical. Why would anyone expect near field effects, virtual photon exchange, to operate in the same manner as far field effects, photon exchange? Note: no one suggests a violation of CoE, and therefore greater emission on the nano-structured surface (superradiance) will be compensated elsewhere. That can't be too difficult to grasp, once you get past the false belief that Planck's law is really a Law, instead of a general observation that proved correct within the limitations of its relevant time frame. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Liquid Glass
Jones, Thanks for posting that reference! Cool! Actual desktop USB interface computer laser cutters. And they sell used ones on occasion too. That stuff reminds me of the liquid sodium silicate I used to play with as a kid. It was sold under the name Eisenglass I think. It came as a viscous liquid in quart cans. It was used to paint eggs (still in the shells) in order to preserve them longer I think. This lost importance when refrigerators became common. I added chemicals like copper sulfate to the Eisenglass to grow a chemical garden in a glass jar. It formed neat plant-like tentacles. I don't know where I got the recipe for that. I think it might have been Sci. Am. or Pop. Sci. I am curious as to why you think circuits have to be etched? To use silicon for a solar cell I think it has to be doped, so as to create a PN boundary. It is the potential drop across the PN boundary that actually drives a solar cell. The sun merely creates the ions in the gap so they can be accelerated across it. I do wonder if it might be possible to use a zinc or zinc plated substrate (zinc is a hole conductor) coated with silicon that is chemically doped as an N (electron) conductor. If so, the remaining things necessary to create a solar cell are a transparent conductive overcoating, and possibly the printing of a very conductive metallic collector circuit on top. On Feb 9, 2010, at 6:51 AM, Jones Beene wrote: Ron, You have to wonder - with liquid glass and a commercial laser engraver (etcher) which is similar to an ink jet printer - http://www.epiloglaser.com/product_line.htm and some imagination and metal-coated film, if one could not etch the circuits with the printer, then coat this film with the glass, and thereby make a large and fairly efficient homemade nano-solar thin film photocell array... Jones -Original Message- From: Ron Wormus This sounds very cool. http://www.physorg.com/news184310039.html Ron Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:comment on Violante data as covered by Steve Krivit
On Feb 9, 2010, at 2:09 AM, Michel Jullian wrote: Hi Horace, sorry for the late response, my comments below. 2010/2/7 Horace Heffner hheff...@mtaonline.net: On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote: 2010/2/7 Horace Heffner hheff...@mtaonline.net: Two things to consider: (1) reversing the current *does* dissolve the Pd surface, True, but extremely slowly I believe. A Pd anode is known to dissolve relatively fast in acidic electrolytes such as D2SO4, but I don't think that's what they used. It is doubtful whether they reverted the current long enough to dissolve more than a few atomic layers. I think the experimenters were competent. They knew what they were doing. Using a Faraday constant of 96,485 C/mol, and (conservatively) a valence of 4, n for moles produced, I for current = .2 A, t for time = 1 s, we get: n = I * t / (96,485 C/mol * 4) n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or 3.12x10^17 atoms per second. We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/ cm^3 = 7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3, and the atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per second is (3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per atom) = 4.45x10^-6 cm/s, or 445 angstroms per second. The number of layers of atoms removed is (4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s. If this is correct (highly suspect! 8^), then at a current density of 200 mA/cm^2 we have a thickness of 183 atoms removed per second, or 445 angstroms per second. This would be correct if palladium, when driven as an anode, did dissolve in an alkaline electrolyte (they classically used LiOD in that M4 experiment, according to their original report at http://newenergytimes.com/v2/archives/1998epri/TR-107843-V1.PDF , thanks to Steve Krivit for the link), which it doesn't, see the Pd/H2O Pourbaix diagram at http://www.platinummetalsreview.com/jmpgm/data/datasheet.do? record=532database=cesdatabase which shows that such corrosion only occurs in an acidic electrolyte (pH 3). It has been pointed out to me privately that hydrogen charge transport has to be accounted for as well, i.e. that hydrogen evolution reduces the effective corrosion current. However, since the reversed current cleaning process was carried out in part to degass the Pd, I expect the hydrogen contribution to the positive surface charge of the Pd anode would be extremely diminished in the latter part of this cleaning process. Well, this is indeed an interesting electrochemical problem. My experience is that nothing, including platinum, totally avoids anodic corrosion if there is a current present. Passification works in part by eliminating the current at the potential at which the passivation is occurring, or less, i.e. by building an insulating layer. I do not think passification of *highly loaded* Pd is possible. The evolving hydrogen would prevent accumulated oxidation of the Pd surface. I have done various passivation experiments (not with Pd though) and my experience has been that passification takes considerable time, even for metals that are not loaded with hydrogen, and once it does occur, the current is highly reduced. Further, if a constant current source is used then the voltage rises to the point where the passified surface is breached. Beyond that, and this is a fairly irrelevant point I know, I think Pd corrodes as an anode in the presence of current in neutral Ph salt electrolytes. The EPRI article states: They accomplished loading with a combination of initial low cathode current densities of ~20-50 mA/ cm2, followed by current ramps up to ~1.0 A/cm2. Current reversals to deload or “strip” the cathodes of D and clean the surface by temporarily making it an anode resulted in high loadings. It seems to me the Pd would be dissolving during the deloading process when the current is reversed. Also, apparently my estimate of 200 mA/cm^2 was too low - it was probably 1 A/cm^2. It would be interesting to actually do an experiment with Pd wire, loading and then reversing the current repeatedly for a long period and then weighing the wire. It seems to me the experimenters would not have gone thorough this procedure if the current reversal did not actually clean the electrode surface, i.e. expose a pure Pd surface. A fully passified surface would not be effective at loading hydrogen as a cathode because it would not even be conductive. If pure Pd was exposed to the electrolyte as an anode it seems to me certain that Pd was being dissolved in the process. One thing I take to be self evident to anyone who has practical experience with electrochemistry experiments. If you have current through an anode then *some* of that anode is going into solution, and that includes platinum. I
Re: [Vo]:Britz versus Huizenga's totals
On Feb 9, 2010, at 12:27 PM, Jed Rothwell wrote: To be fair to Britz, he has also at times defended the reputations of researchers, if not their results. He has distanced himself from the extreme tactics of his fellow skeptics. Also to be fair to Britz, ever since the early days of the true believers and skeptics on sci.physics.fusion, Dieter Britz has maintained that he is neutral on the subject, keeping an open mind. He is neither one of the faithful nor an atheist, but rather an agnostic. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:speaking of printing solar cells ...
http://fastflip.googlelabs.com/view?q=view% 3Apopularsource=news#ark6CLCHHYH9NM http://tinyurl.com/yfztyg9 Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Quantum computer simulates hydrogen molecule
http://www.wired.com/wiredscience/2010/01/quantum-computer-hydrogen- simulation/ http://tinyurl.com/yff5q8r This is an amazing accomplishment. Makes me wish I were a young guy just starting his education. The next 50 years of physics should be extraordinary. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Test message with tabular data 4
I just realized Microsoft Outlook users might not have seen my last email correctly, so here it is again in rich text. On Feb 12, 2010, at 7:07 AM, Jed Rothwell wrote: I just tried this with three messages which were probably too big. I think 40 KB is the limit. Microsoft Word put this table out in HTML with enormous overhead. This version is prepared with an HTML editor. Britz database stats [snip] It came through nicely on my Mac. However, HTML in general, and graphics in particular, are not good in the archives. Graphics and some HTML or even rich text also do not show up properly at the online archives, www.mail-archive.com. Happily, ascii graphics and blank line white space, which many Microsoft users don't get properly via plain text, show up nicely in the archives. See: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - http://www.mail-archive.com/vortex-l%40eskimo.com/msg37685.html [Vo]:Test message with tabular data 4 Jed Rothwell Fri, 12 Feb 2010 08:08:37 -0800 I just tried this with three messages which were probably too big. I think 40 KB is the limit. Microsoft Word put this table out in HTML with enormous overhead. This version is prepared with an HTML editor. Britz database stats Res+Res-Res0NoYearTotalPositiveNegativeUndecidedEvaluationPositive + Undecided198920546832254681990248757641561161991130462918376419926522131 119 331993663110817391994422033162319952919361251996482410773119973219247231 998 331923922199923180141920001510014112001171120411200218920792003721042200 4640 02420056222042006640115200755000520086200422009001007388238128253516 - Jed - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - And see: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - http://www.mail-archive.com/vortex-l%40eskimo.com/msg37686.html Re: [Vo]:Test message with tabular data 4 Jed Rothwell Fri, 12 Feb 2010 08:11:50 -0800 Ah. That worked. Let me try HTML from on-line article: The following table shows just how much more people consume in a serving today compared to people in the 50s: *Serving Sizes Then and Now**Food or beverage**1950s**Expanded 2003 portion*French fries 2.4 ounces up to 7.1 ouncesFountain soda 7.0 ounces 12 to 64 ouncesHamburger patty 1.6 ounces up to 8.0 ouncesHamburger sandwich 3.9 ounces 4.4 to 12.6 ouncesMuffin 3.0 ounces 6.5 ouncesPasta serving 1.5 cups 3.0 cupsChocolate bar 1 ounce 2.6 to 8 ounces - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Test message with tabular data 4
On Feb 12, 2010, at 9:25 AM, Jed Rothwell wrote: Horace Heffner wrote: It came through nicely on my Mac. However, HTML in general, and graphics in particular, are not good in the archives. Graphics and some HTML or even rich text also do not show up properly at the online archives, www.mail-archive.com. Ouch. Happily, ascii graphics and blank line white space, which many Microsoft users don't get properly via plain text, show up nicely in the archives. That's sorta hard to make. It there a program to convert tables into ascii graphics and blank spaces? It sounds like something I would have written in 1978. - Jed If you leave out the graphics and just print a table as rich text, that might work. If it does not come out in columns, due to a variable font, you might be able to fix that by selecting the text and converting it to a fixed font. Converting to a small pitch (as I did below) is sometimes useful for getting everything to show up without line wrap for some people, and some lists, but the pitch in www.mail-archive.com is fixed, as is the line width, so everyone sees line wrap there is you exceed the maximum. I'm not sure what the maximum line width is for www.mail-archive.com but the scale following will help tell that. If it is a spreadsheet important to you and that you might refer to at a later date you might just covert it to a pdf, put it on your web site, and provide the URL instead. Even better is to convert to modifiable HTML spreadsheet format, so people can plug in their own data to affect the totals, do projections, etc. Following is 10 pitch Courier, at least as I sent it anyway. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - . 1.2.3.4.5.6.7.8. 9 123456789012345678901234567890123456789012345678901234567890123456789012 345678901234567890 http://www.mail-archive.com/vortex-l%40eskimo.com/msg37685.html [Vo]:Test message with tabular data 4 Jed Rothwell Fri, 12 Feb 2010 08:08:37 -0800 I just tried this with three messages which were probably too big. I think 40 KB is the limit. Microsoft Word put this table out in HTML with enormous overhead. This version is prepared with an HTML editor. Britz database stats Res+Res-Res0NoYearTotalPositiveNegativeUndecidedEvaluationPositive + Undecided198920546832254681990248757641561161991130462918376419926522131 119 331993663110817391994422033162319952919361251996482410773119973219247231 998 331923922199923180141920001510014112001171120411200218920792003721042200 4640 02420056222042006640115200755000520086200422009001007388238128253516 - Jed - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - And see: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - http://www.mail-archive.com/vortex-l%40eskimo.com/msg37686.html Re: [Vo]:Test message with tabular data 4 Jed Rothwell Fri, 12 Feb 2010 08:11:50 -0800 Ah. That worked. Let me try HTML from on-line article: The following table shows just how much more people consume in a serving today compared to people in the 50s: *Serving Sizes Then and Now**Food or beverage**1950s**Expanded 2003 portion*French fries 2.4 ounces up to 7.1 ouncesFountain soda 7.0 ounces 12 to 64 ouncesHamburger patty 1.6 ounces up to 8.0 ouncesHamburger sandwich 3.9 ounces 4.4 to 12.6 ouncesMuffin 3.0 ounces 6.5 ouncesPasta serving 1.5 cups 3.0 cupsChocolate bar 1 ounce 2.6 to 8 ounces - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Lead as a possible nuclear catalytic LENR agent
. Numerous lead alloys are commercially available, including lead-calcium alloys used commonly in battery electrodes. See: http://www.keytometals.com/Article10.htm Key is utilization of a hydrogen permeable high lead content lattice. Lead alloys have low melting points, which might be useful for helium removal. Of further interest is that the Pd + 2D* compound nuclei spontaneously alpha decay with practical half-lives, even if their source is not LENR. The lead isotopes and natural abundances are: El.Abundance 204Pb 1.4% (1.4x10^17 y half-life) 206Pb 24.1% 207Pd 22.1% 208Pd 52.4% The primary lead nuclear catalytic reactions are: 208Pb82 + 2D* -- 212Po84 -- 4He2 + 208Pb82 (3x10^-7 s half life) 207Pb82 + 2D* -- 211Po84 -- 4He2 + 207Pb82 (56 ms half life) 206Pb82 + 2D* -- 210Po84 -- 4He2 + 206Pb82 (148.37 d half life) with some much less common (and desirable) additional reactions: 204Pb82 + 2D* -- 208Po84 -- 4He2 + 204Pb82 (2.898 y half life) 204Pb82 + 2D* + e- -- 208Po84 + e- -- 208Bi83 (2.898 y half life) The half lives given, though they apply to the compound nuclei, are only upper limits in these cases, because they only apply to nuclei without tightly bound free electrons in them. Such electrons generate much shorter half-lives and precipitate heavy fragment fissions. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Lead as a possible nuclear catalytic LENR agent
On Feb 14, 2010, at 11:18 AM, Horace Heffner wrote: The lead isotopes and natural abundances are: El.Abundance 204Pb 1.4% (1.4x10^17 y half-life) 206Pb 24.1% 207Pd 22.1% 208Pd 52.4% Note: pd above is a typo, and should be Pb. Unfortunately, of the above isotopes of lead, only 207Pb has a nuclear moment, which is useful in increasing tunneling probability. More interesting from the perspective of magnetic moment is 209Bi, which has a 100% abundance, and a large nuclear magnetic moment in comparison to 208Pb. The nuclear catalysis LENR reaction is: 209Bi83 + 2 D* -- 213At85 -- 4He2 + 209Bi83 (125 ns) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Bismuth as a nuclear catalytic LENR agent
Bismuth-209 appears to be an excellent CF nuclear catalysis agent when used with deuterium. It has 100% abundance. The nuclear catalysis reaction is: 209Bi83 + 2 D* -- 213At85 -- 209Bi83 + 4He2 + 23.847 MeV [-8.560 MeV] (125 ns) The major potential drawbacks are the presence of energetically feasible fission reaction channels not deflated electron confined: 209Bi83 + D* -- 198Pt78 + 13C6 + 21.660 MeV [5.599 MeV] 209Bi83 + 2 D* -- 198Pt78 + 15N7 + 37.819 MeV [5.412 MeV] bismuth has a typically low melting point, even in many alloys, and bismuth does not sustain a viable CF lattice by itself, i.e. must be imbedded in a useful lattice. It has a lattice constant of 4.75 angstroms, as opposed to Pd at 3.89 Å, and iron at 2.87 Å. Interesting coincidence that the average of iron and bismuth lattice constants is within about 2% of that of Pd. A bismuth-iron alloy might provide a feasible CF lattice at high loading temperatures. Bismuth has a spin of 9/2, a large value of mu = 4.5444 mu_N, and gyromagnetic ratio of 43.75 x 10^6 rad s^-1 T^-1. It has a nuclear magnetic resonance frequency of 6.963 MHz in a 1 T field. Nuclear catalysis is carried out best in as large a magnetic field as possible, using as large a B field gradient as possible. Other considerations are documented here: http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf http://www.mtaonline.net/~hheffner/dfRpt Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Plasma fusion budget
On Feb 17, 2010, at 7:11 PM, Jed Rothwell wrote: See: http://aries.ucsd.edu/FPA/OFESbudget.shtml MFE = Magnetic confinement (Tokamak) ICF = Inertial confinement fusion This was linked to from the plasma fusion lobby: http://fusionpower.org - Jed The numbers are in 2000 dollars. Adjusted to 2010 dollars the totals are: MFE adjusted to 2010 dollars: 21,141 billion dollars ICF adjusted to 2010 dollars: 14.985 billion dollars Total 2010 budget: 36.126 billion dollars The 2010 budget for both is $426 M + $458 M = $884 M. Just 1% of the hot fusion budget could go a long ways toward advancing low energy nuclear research. It would be feasible to develop an LENR lab, and staff it for research and for providing mass spec, radiation counting, and other services to researchers. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Short poem about cold fusion
Cold fusion metal unseen binding mystery silent winter's fire. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Extraordinary Error -- no electric field exists inside a conducting liquid in an insulated box with two external charged metal plates, re work by SPAWAR on cold fusion since 2002 -- also hot
On Feb 22, 2010, at 6:32 PM, Rich Murray wrote: Extraordinary Error -- no electric field exists inside a conducting liquid in an insulated box with two external charged metal plates, re work by SPAWAR on cold fusion since 2002 -- also hot spots from H and O microbubbles: Rich Murray 2010.02.22 http://rmforall.blogspot.com/2010_02_01_archive.htm Monday, February 22, 2010 http://groups.yahoo.com/group/astrodeep/message/42 _ I hesitated to respond to this because we beat the electric field subject to death here on vortex years ago and I don't have time for idle discussion, and don't want to be involved in a sci.physics.fusion style rancorous debate. I'll just state the obvious once and hopefully leave it at that. If you were correct regarding your assertion then there would be no observable effect from the experiment with imposed field vs the control. There *is* a significant effect on surface morphology, as can be seen in Fig. 3 of the old SPAWAR (Szpak et al) article here: http://lenr-canr.org/acrobat/SzpakSprecursors.pdf Any effect at all on surface morphology will likely affect LENR. The important thing related to this discussion, however, is that the above article negates your premise entirely. There *is* an effect within the cell due to the imposed externally imposed HV field. Experiment only trumps theory. Bockris' text on electrochemistry states that ion motion in the electrolyte between the interface layers is not primarily due to electric field, because most all the potential drop occurs across the two molecule thick interface layers. Therefore ion motion in the electrolyte proper is mostly due to random walk and concentration gradients. It is notable that even if ion redistribution fully negates a field within the electrolyte, that negation occurs via an ion redistribution, and thus an artificial concentration gradient is obtained. A change in surface electron density and distribution in the electrode is thereby obtained as well, and that is significant to cold fusion. I had more to say regarding the Szpak cell in particular in 2004: http://www.mtaonline.net/~hheffner/Szpak.pdf and much more to say here later regarding surface electron distribution: http://www.mtaonline.net/~hheffner/DeflationFusion.pdf and its prospective effect on cold fusion nuclear reactions: http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf http://www.mtaonline.net/~hheffner/dfRpt Last year I suggested here another variation of the Szpack cell design, the edge-on grid co-deposition method, which has a number of benefits for research purposes: http://www.mtaonline.net/~hheffner/EdgeOnGrid.pdf and which preserves the original edge-on orientation of the cathode surface with respect to the field. As far as any assertion the SPAWAR CR-39 pits are due to chemical etching or surface damage, if you read the many SPAWAR publications you will see things have moved far beyond that issue, by use of barriers between the CR-39 and electrolyte, and via various control experiments. Regarding bubbles and hot spots, theses were also discussed here previously. Here are posts in a couple threads of possible related interest: http://www.mail-archive.com/vortex-l@eskimo.com/msg35917.html http://www.mail-archive.com/vortex-l@eskimo.com/msg35608.html Note the thread links at the bottom of these pages. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Extraordinary Error -- no electric field exists inside a conducting liquid in an insulated box with two external charged metal plates, re work by SPAWAR on cold fusion since 2002 -- also hot
On Feb 23, 2010, at 12:05 PM, Stephen A. Lawrence wrote: On 02/23/2010 03:32 PM, Horace Heffner wrote: It is notable that even if ion redistribution fully negates a field within the electrolyte, that negation occurs via an ion redistribution, and thus an artificial concentration gradient is obtained. A change in surface electron density and distribution in the electrode is thereby obtained as well, and that is significant to cold fusion. Thanks, Horace! I didn't recall the earlier discussion and I totally missed that point. (This has been my day to overlook things, it seems.) Gee, *every* day is one of those overlooking things days for me! Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Extraordinary Error -- no electric field exists inside a conducting liquid in an insulated box with two external charged metal plates, re work by SPAWAR on cold fusion since 2002 -- also ho
On Feb 23, 2010, at 2:09 PM, Michel Jullian wrote: 2010/2/23, Horace Heffner hheff...@mtaonline.net: ... Therefore ion motion in the electrolyte proper is mostly due to random walk and concentration gradients. ... The ion motion is due to a force, what kind of force do you think, the concentration gradient force? It's of course an electric force, There are three ways a flux of ions can come about (Bockris p. 288): diffusion, migration (also called electromigration and conduction) and hydrodynamic flow. entirely due to an electric field. Ultimately, that's what it is. Same thing for electrons in a metal. Michel Consider Frick's first law of steady state diffusion, which states the flow vector J_i for species i is proportional to the concentration vector (d c_i)/( d x) in typical cell conditions, i.e., one dimensionally speaking: J_i = - D (d c_i)/( d x) where D is called the diffusion coefficient. There are other adjustments to be made, but this version of Frick's law is adequate for this discussion. Notice it works for species independently, and can work in a zero electrostatic field environment with respect to subsets of species. The electrostatic gradients measured in the center of very large cells can not possibly account for the species flows required to support the current. Just because diffusion of charged species is happening doesn't necessarily even mean there exists an electrostatic potential gradient. In an electrolyte having more than two charged species (e.g. Na+, H3O+, and OH-), just because a large concentration imbalance develops between two species at the electrodes, (e.g. H3o+ at one end and OH- at the other) doesn't mean a correspondingly large electrostatic gradient necessarily develops. A very small simultaneous migration of *all* the charged species not involved in the electrolytic reaction (e.g. Na+) happens very quickly to neutralize any large potential gradient in steady state conditions, with the very slow diffusion velocities involved having very little effect due to the motion in unison. It then only remains for the electrode created species to reach concentration gradients that handle the individual specie currents. This was a key point in my discussion of how the hot spots formed on the back side of the cathode mesh in the SPAWAR experiments. Hydrogen ion (H3O+) conduction (and OH- in the opposed direction) can occur right through the mesh, right through a zero electrostatic field surface, by diffusion alone. However, there is necessarily a line somewhere on both sides of the periphery of each mesh wire where zero potential exists, and thus no interface layer. This is an ideal location for recombination to occur right on the mesh wire, because the H3O+ and OH- species are migrating in opposed directions and can attach directly to the metal surface there. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:I'm at the Rocket Conference and I will speak tomorrow
On Feb 23, 2010, at 4:09 PM, fznidar...@aol.com wrote: I am here. http://www.ias-spes.org/SPESIF2010/Registration/ SPESIF2010_Registration.php I went to supper with Prof. George Miley. We had a good talk, however, he does not fully accept what I am saying. He will be speaking at 8 AM. I will see him. There was a cancellation and I now have the opportunity to speak at the gravitational anomaly conference. I will present my paper The Control of the Natural Forces I'll be on stage at 10:45 at Johns Hopkins. Martin Tajamer from the ESA will be there at my lecture. I look forward to meeting him. I have met many people including Len Danczyk of Energetics Technologies. This is a good opportunity for me and I will do my best to pull it off. They will not let me take pictures. I'll try to get a few anyway. Frank Znidarsic Congratulations Frank! I noticed it says Papers approved for SPESIF are reviewed by the technical staff, Chairs and Co-Chairs and other Committee Members needed for a proper peer review and are published by the American Institute of Physics (AIP) in an AIP Conference Proceedings here: http://ias-spes.org/SPESIF.html Sounds like you might get published in the AIP Conference Proceedings. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:for Extraordinary Error, with the blog, click on the Go to Homepage: while for #42, use http://groups.yahoo.com/group/AstroDeep/messages and then go to #42: Rich Murray, Santa Fe, New Mex
On Feb 23, 2010, at 9:48 PM, Rich Murray wrote: for Extraordinary Error, with the blog, click on the Go to Homepage: while for #42, use http://groups.yahoo.com/group/ AstroDeep/messages and then go to #42: Rich Murray, Santa Fe, New Mexico 2010.02.23 Sometimes yahoogroups is slow to make posts available, although it was sent to me immediately -- I hadn't noticed any previous problems with blogspot. Rich I post to vortex-l only. I don't wish to post at yahoo.com or read stuff at yahoo.com. I'm happy here. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Extraordinary Error -- no electric field exists inside a conducting liquid in an insulated box with two external charged metal plates, re work by SPAWAR on cold fusion since 2002 -- also ho
On Feb 23, 2010, at 4:24 PM, Horace Heffner wrote: Consider Frick's first law of steady state diffusion, which states the flow vector J_i for species i is proportional to the concentration vector (d c_i)/( d x) in typical cell conditions, i.e., one dimensionally speaking: J_i = - D (d c_i)/( d x) where D is called the diffusion coefficient. I accidentally left out a word above: concentration vector above should say concentration gradient vector. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Extraordinary Error -- no electric field exists inside a conducting liquid in an insulated box with two external charged metal plates, re work by SPAWAR on cold fusion since 2002 -- also ho
On Feb 24, 2010, at 9:45 AM, Michel Jullian wrote: Hi Horace, Another typo: Frick instead of Fick. That's a funny one! Must have been a Freudian slip. 8^) All these macroscopic phenomena you discuss regarding the motion of ions in an electrolyte boil down, at the atomic scale, to the electric force, don't you agree? Of course I don't agree. Neither does Bockris. Did you check the reference? You are *assuming* that only a force affects random walk. The gradient induced flow is not due to an actual force, even though it acts just like a separate force on each species. If you draw a plane perpendicularly through a concentration gradient, one side has a higher concentration than the other. Random crossings therefore occur more often going from the strong to the weak side of the gradient. This results in a net flow that reduces the gradient. In any case, in a dense conductor, whether liquid or solid or even a dense gas such as atmospheric air, if you have a _steady_ current of charged particles, then there exists a net DC electric field provoking it, and in the absence of a magnetic field each charged particle does a random walk whose average is the electric field line. Things don't work this way in an electrolyte, especially if you have a long inter-plate distance and a long equilibrium time. Proof: the average velocity (drift velocity) of each charged particle is equal to its mobility times the local electric field, see e.g. http://en.wikipedia.org/wiki/Electron_mobility for the case of electrons, or look up drift velocity in the Feynman Lectures on Physics. The electric field between the anode and cathode interfaces of an electrolytic cell may be very small (it's indeed immensely larger in the interface regions), but it explains entirely the steady cell current. Michel I am already familiar with these concepts. These are *not* complete electrochemistry concepts. The proof is invalid because the conditions differ. Michel, an electrolyte is *not* like a conductor, for the reasons I already described. I repeat, The electrostatic gradients measured in the center of very large cells do not come close to accounting for the species flows required to support the steady state current that develops. I think there are some experiments you can do to demonstrate the effect of diffusion gradients. One is to send a short pulse between planar electrodes chosen such that the anode is partially dissolved (with Faradaic efficiency of about 1) into the solution by the pulse. This creates a planar concentration gradient near the anode and results in a delayed current trace. I would do some homework for you and type up some quotes from Bockris, and draw some pictures, but I'm short on time right now. Again I say, Bockris' Modern Electrochemistry, p. 288 ff is a good place to start. It suffices to say ion concentration gradients in an electrolyte produce ion flows, even bi-directional opposed ion flows, even *without* the presence of an electrostatic field. Diffusion based ion flow is not a trivial concept or second order concept. It is an important concept for things like flow battery design and modeling the mechanics of inter-interface currents. 2010/2/24 Horace Heffner hheff...@mtaonline.net: On Feb 23, 2010, at 4:24 PM, Horace Heffner wrote: Consider Frick's first law of steady state diffusion, which states the flow vector J_i for species i is proportional to the concentration vector (d c_i)/( d x) in typical cell conditions, i.e., one dimensionally speaking: J_i = - D (d c_i)/( d x) where D is called the diffusion coefficient. I accidentally left out a word above: concentration vector above should say concentration gradient vector. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:SRI Experiment HH
On Feb 23, 2010, at 9:51 AM, Steven Krivit wrote: http://newenergytimes.com/v2/news/2010/SRI-Expt-HH/SRI-Experiment- HH.shtml Vorts, I have deliberately not provided any explanation, analysis or interpretation. Instead, I'd like to hear your thoughts first. In particular, I'd like to hear your thoughts on the graph drawn by McKubre as compared to the graph I drew. I'd also like to hear your thoughts and analyses on the green and blue lines. Steve I haven't taken the time to look into this in detail, but my first impression is that, unless there is a typo, it makes no sense at all to attempt to draw the 23.82 MeV line through Fig. 1, or to draw any conclusions from the graph as to energy per helium atom produced. Perhaps I'm misreading the x axis labeling Excess Power/Current (mW / A), or the intended meaning of the x axis values. To be sensible the x axis should simply be excess energy, i.e. the integral of mW over time. It looks like voltage was roughly uniform, so the (input) mW/A should roughly be a constant, given power P = I * (V - v0). So, basically, the x axis is a constant times excess power. It should be a constant times excess energy to make any sense, or to plot the green line on it. Alternatively, at a constant power the helium could be measured over equally spaced intervals, and then the green line should be horizontal, i.e. fixed amount of helium produced per interval of time corresponding to the mean excess power for the interval. Maybe if someone took the time to look deeper into this they could make some sense of it. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
On Feb 24, 2010, at 1:39 PM, Harry Veeder wrote: Has Naudin just made his transformer more efficient...or is it really a generator? http://www.youtube.com/watch?v=_xS6Fknxv18 Harry The site says Wooww, the power at the OUTPUT is greatly increased without significant change at the DC input , yet there is no effort made to measure input power, only current. It would make more sense to get the I and V traces for the input coil. It pretty obvious how the thing works. The torus field, which remains inside the torus, deflects the permanent magnet field away from the torus, and thus the permanent magnet's field oscillates, cutting back and forth across the secondary coil windings and generating power there. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
A variation. Pulse in primaries P1 or P2 cuts permanent B through secondaries S1 or S2. inline: sketch.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
A silly variation, but closer in nature. inline: 100_2469.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
On Feb 24, 2010, at 5:58 PM, OrionWorks - Steven Vincent Johnson wrote: From: Horace: ... The site says Wooww, the power at the OUTPUT is greatly increased without significant change at the DC input , yet there is no effort made to measure input power, only current. It would make more sense to get the I and V traces for the input coil. It pretty obvious how the thing works. The torus field, which remains inside the torus, deflects the permanent magnet field away from the torus, and thus the permanent magnet's field oscillates, cutting back and forth across the secondary coil windings and generating power there. Can you clarify something for me, Horace. The conjecture that the field oscillates, as you state, cutting back and forth across the secondary coil windings... is intriguing, particularly since you seem to be saying the field is dynamically oscillating even though there are no moving parts. In layman's terms - what does that mean, particularly energy-wise. My prosaic thinking patterns keep wanting to envision MOVING magnets passing across coils of wire that in turn generate electricity. But nothing seems to be physically moving in this configuration. I'm confused! /:-\ Transformer parts don't move, but they still get energy transferred from a primary to a secondary. In your opinion, does this demonstration allegedly show real OU configuration, or is this device demonstrating something else that's not really OU. As I said earlier, there is no measurement of input energy. So who knows? I think the odds of free energy from this are *very* slim. The responsible thing to do is to measure it, but that spoils all the fun I assume. Where's the energy coming from that allegedly powers the LEDs? It is coming from the signal generator that drives his primary. Speaking of energy, I wonder if the magnet inserted into the torus will eventually lose its magnetism if the device continues to power the LEDs? Sure, all magnets do. His will faster because there is no keeper. The little drawings I sent avoids the keeper problem by providing multiple paths for the permanent field. The primaries simply divert the permanent magnet B field. It is interesting that, in my drawings, the primary current increasing eliminates the secondary coil magnetic field rather than increasing it. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
inline: 100_2470.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ 100_2470.JPG
Re: [Vo]:Naudin's improved generator
On Feb 24, 2010, at 5:58 PM, OrionWorks - Steven Vincent Johnson wrote: From: Horace: ... The site says Wooww, the power at the OUTPUT is greatly increased without significant change at the DC input , yet there is no effort made to measure input power, only current. The above should say RMS current and RMS voltage, which is not necessarily the same thing as power. It would make more sense to get the I and V traces for the input coil. It pretty obvious how the thing works. The torus field, which remains inside the torus, deflects the permanent magnet field away from the torus, and thus the permanent magnet's field oscillates, cutting back and forth across the secondary coil windings and generating power there. Can you clarify something for me, Horace. The conjecture that the field oscillates, as you state, cutting back and forth across the secondary coil windings... is intriguing, particularly since you seem to be saying the field is dynamically oscillating even though there are no moving parts. In layman's terms - what does that mean, particularly energy-wise. My prosaic thinking patterns keep wanting to envision MOVING magnets passing across coils of wire that in turn generate electricity. But nothing seems to be physically moving in this configuration. I'm confused! /:-\ I haven't been following any of this so I should have kept quiet. Sorry if I duplicate what has been said. Also, I should have answered this more thoroughly, sorry. Transformer parts don't move, but they still get energy transferred from a primary to a secondary. They can be viewed as creating magnetic field line loops that cut through the secondary coils and then retreat, cutting the coil again. These field lines can be visualized as moving through the center of the transformer core - even though it has a low mu, in order to form the flux loop that goes through the core. Their density in the hole of the core is low so they have to move faster when traversing the hole in the core. It appears the primary core in the video is small compared to the magnets. This means there is magnetic flux that extends out beyond the core and circles back to the south end of the permanent magnet stack, i.e. that does not go through the core. When the current is high in the primary coil, then only one return leg through the primary torus core is available, thus even more flux is diverted out into the space around the primary. To the degree the primary current plus permanent B field saturates the core then even more flux is diverted out into the nearby space. The nearby space is occupied by the primary. As the primary current oscillates, the B field that projects into the secondary coil grows large to the side of the primary where the primary flux opposes it, and diminishes where the primary flux reinforces it, but then increases on that side if saturation occurs. I just posted a drawing, Fig. 3, in a separate email that shows how the ejected flux cuts through the secondary coil. The alternating current in the primary ejects one side of the flux and then the other, cutting the secondary coils in the process. It would be interesting to know how much power is being drawn by the LEDs. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Edge-on grid method document updated
The Edge-on Grid Method document, describing a variation of the SPAWAR cell, has been edited and a drawing included: http://www.mtaonline.net/~hheffner/EdgeOnGrid.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
On Feb 24, 2010, at 7:11 PM, Harry Veeder wrote: - Original Message From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Wed, February 24, 2010 9:30:23 PM Subject: Re: [Vo]:Naudin's improved generator On Feb 24, 2010, at 1:39 PM, Harry Veeder wrote: Has Naudin just made his transformer more efficient...or is it really a generator? http://www.youtube.com/watch?v=_xS6Fknxv18 Harry The site says Wooww, the power at the OUTPUT is greatly increased without significant change at the DC input , yet there is no effort made to measure input power, only current. It would make more sense to get the I and V traces for the input coil. Are we watching the same video? ;-) Did you not see my correction?? Near the beginning of the video this green blurb appears briefly: voltage and current are measured at the input of the controller... The digital meters display the input voltage and the input current and both numbers remain constant. Harry __ Looking for the perfect gift? Give the gift of Flickr! http://www.flickr.com/gift/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
On Feb 24, 2010, at 7:04 PM, Horace Heffner wrote: On Feb 24, 2010, at 5:58 PM, OrionWorks - Steven Vincent Johnson wrote: From: Horace: ... The site says Wooww, the power at the OUTPUT is greatly increased without significant change at the DC input , yet there is no effort made to measure input power, only current. The above should say RMS current and RMS voltage, which is not necessarily the same thing as power. It would make more sense to get the I and V traces for the input coil. It pretty obvious how the thing works. The torus field, which remains inside the torus, deflects the permanent magnet field away from the torus, and thus the permanent magnet's field oscillates, cutting back and forth across the secondary coil windings and generating power there. Can you clarify something for me, Horace. The conjecture that the field oscillates, as you state, cutting back and forth across the secondary coil windings... is intriguing, particularly since you seem to be saying the field is dynamically oscillating even though there are no moving parts. In layman's terms - what does that mean, particularly energy-wise. My prosaic thinking patterns keep wanting to envision MOVING magnets passing across coils of wire that in turn generate electricity. But nothing seems to be physically moving in this configuration. I'm confused! /:-\ I haven't been following any of this so I should have kept quiet. Sorry if I duplicate what has been said. Also, I should have answered this more thoroughly, sorry. Transformer parts don't move, but they still get energy transferred from a primary to a secondary. They can be viewed as creating magnetic field line loops that cut through the secondary coils and then retreat, cutting the coil again. These field lines can be visualized as moving through the center of the transformer core - even though it has a low mu, in order to form the flux loop that goes through the core. Their density in the hole of the core is low so they have to move faster when traversing the hole in the core. It appears the primary core in the video is small compared to the magnets. This means there is magnetic flux that extends out beyond the core and circles back to the south end of the permanent magnet stack, i.e. that does not go through the core. When the current is high in the primary coil, then only one return leg through the primary torus core is available, thus even more flux is diverted out into the space around the primary. To the degree the primary current plus permanent B field saturates the core then even more flux is diverted out into the nearby space. The nearby space is occupied by the primary. . . The last sentence above should read: The nearby space is occupied by the secondary. . . As the primary current oscillates, the B field that projects into the secondary coil grows large to the side of the primary where the primary flux opposes it, and diminishes where the primary flux reinforces it, but then increases on that side if saturation occurs. I just posted a drawing, Fig. 3, in a separate email that shows how the ejected flux cuts through the secondary coil. The alternating current in the primary ejects one side of the flux and then the other, cutting the secondary coils in the process. It would be interesting to know how much power is being drawn by the LEDs. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:SRI Experiment HH
On Feb 25, 2010, at 5:40 AM, Jed Rothwell wrote: Horace Heffner wrote: I haven't taken the time to look into this in detail, but my first impression . . . With all due respect, it is a bad idea to discuss these things without looking into them in detail, and a person's first impressions are likely to be wrong. If I were afraid of being wrong it would destroy my creativity, I would learn little, and I would post nothing at all. That of course might be a good thing from your perspective, but not mine. 8^)In this particular case, for example, I would have no excuse for procrastinating on doing my tax return! is that, unless there is a typo, it makes no sense at all to attempt to draw the 23.82 MeV line through Fig. 1 . . . That is an expectation value. Here you have missed the point entirely. There is no such expected value of energy per helium atom as a function of excess heat power. There is an expected value of energy per helium atom as a function of excess heat *energy*. That shows how much helium there would be if the ratio of helium to heat was 23.82 MeV per reaction, and if every atom of helium were recovered. Apparently it does not. It shows a ratio of helium to excess power, not excess heat. Obviously, not every atom is -- or can be -- recovered. As the text points out a lot of the helium is stuck in the cathode and can only be recovered after the experiment. Perhaps I'm misreading the x axis labeling Excess Power/Current (mW / A), or the intended meaning of the x axis values. To be sensible the x axis should simply be excess energy, i.e. the integral of mW over time. Those are instantaneous power readings taken at different times, arranged in ascending order. . This makes the graph seem nonsensical. . The helium does not stay in this cell; it is open, like the Miles cell, and the helium is collected from the effluent gas. This is not a time graph of the run, and that is not the integrated energy. . Then the green line makes no sense at all without a further explanation. . In other words, at one point when the cell was producing about 70 mW the helium reading came out 2.4 +/- 0.8, and another time when power was ~100 mW, a helium reading came out 2.8 +/- 1.2. These numbers do not relate to The points at the bottom are either experimental error or caused by helium being trapped in the cathode. It is difficult to say which. Quoting the paper, p. 2 and 3: Figure 1 presents the results of concurrent excess power and helium measurements performed during open cell electrolysis using two different Pd and Pd-alloy cathodes. In three instances where excess power was measured at statistically significant levels, 4He also was found to be conveyed out of the cell in the electrolysis gases (D2 + O2). This makes total sense. The solid line in Figure 1 plots the regression fit of these data to a line passing through the origin; the dashed line is that expected for 4He generation according to the reaction: d + d -- 4He + 23.82 MeV (lattice) [1] This is the part that needs clarification. There is no clear link established between helium concentration and power produced. It is clear from the slopes of these two lines that the observed 4He constitutes only 76 ± 30% of the 4He predicted by equation [1]. The helium concentration is not predicted by equation 1. Equation 1 only establishes a relationship between helium atoms created and excess *energy* produced. It has nothing to do with power. A more significant problem in Figure 1 is that three further 4He samples, taken at times of non-zero excess power (open diamonds), exhibited helium concentrations only at the level of the analytical uncertainty, as did numerous samples taken in the apparent absence of excess power production. Clearly if 4He is produced in association with excess power, it is not released to the gas phase immediately, or completely. http://lenr-canr.org/acrobat/McKubreMCHtheemergen.pdf That seems pretty clear to me. I do not understand why people here are confused by it. Maybe if someone took the time to look deeper into this they could make some sense of it. I didn't have to look very deeply. And you didn't make any sense. Look folks: An author may not present data the way you would choose to present it. I often find that a graph shows something other than what I assumed; i.e., it shows power rather than integrated energy. Oops. I usually have to read a paper several times to figure out what's what. So let's not jump to conclusions about these things, or assume that X or Y doesn't make sense. You need to cut the authors some slack. It is tough writing papers and explaining things. Someone once complained to Oliver Heaviside that his papers were very difficult to read. He responded, That may well be -- but they were much more difficult to write
Re: [Vo]:SRI Case Expt 4He
On Feb 24, 2010, at 7:37 PM, Steven Krivit wrote: http://newenergytimes.com/v2/news/2010/SRI-Case-Repl/SRI-Case.shtml Has anybody ever noticed that 2 out of 3 runs that show 4He growth in the SRI Case replication show a peak and then a decrease in 4He? This is a helium leak-tight chamber. Where does the 4He go? s The helium flows out of the cell with the evolved H2 and O2 gasses. If helium production slows and cell gas production remains constant then the concentration of helium in the gas produced diminishes. Modeling this over time is not so simple in the real case, as it is a multi-compartment flow model, and one in which one of the flow compartments (the Pd) performs in an unpredictable manner. To answer your question more directly, the helium continually flows out of the cell with the effluent, which is sampled periodically. If helium production stops, then the helium concentration necessarily must eventually drop to zero because the cell gas is continually produced and water is periodically added to the electrolyte to continue operation. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:SRI Experiment HH
On Feb 25, 2010, at 7:26 AM, Jed Rothwell wrote: Horace Heffner wrote: is that, unless there is a typo, it makes no sense at all to attempt to draw the 23.82 MeV line through Fig. 1 . . . That is an expectation value. Here you have missed the point entirely. There is no such expected value of energy per helium atom as a function of excess heat power. Obviously I meant that. Please do not nitpick. That shows how much helium there would be if the ratio of helium to heat was 23.82 MeV per reaction, and if every atom of helium were recovered. Apparently it does not. It shows a ratio of helium to excess power, not excess heat. I meant that was the power (or I guess the average power) during the time it takes to collect the sample of effluent gas. They let the collection cylinder fill up many times, to purge atmospheric helium. If this were Arata he would list the energy and time unit, with some unit that is hard to translate back into power, such as kilojoules per hour. This is technically correct because of course helium is proportional to energy not power, but I find it confusing. 60 minutes time 60 seconds and so on . . . As I recall we have the Mesopotamians to thank for that. Why we can't have time in base-10 I do not know. They tried it after the French Revolution but people didn't buy it. But I digress. Those are instantaneous power readings taken at different times, arranged in ascending order. . This makes the graph seem nonsensical. It doesn't seem nonsensical to me. Maybe those are average power readings during the time they collected the sample. Excess power does not fluctuate quickly with a Fleischmann Pons bulk palladium cell, so it could be both. Quoting the paper, p. 2 and 3: Figure 1 presents the results of concurrent excess power and helium measurements performed during open cell electrolysis using two different Pd and Pd-alloy cathodes. In three instances where excess power was measured at statistically significant levels, 4He also was found to be conveyed out of the cell in the electrolysis gases (D2 + O2). This makes total sense. Good. Next time read the paper before commenting. Never!! 8^) Well, maybe sometimes. Jed, that is only one sentence that makes sense without further explanation, not the whole paper or even just the graph. I think this issue was well worth discussing, and I feel totally justified in discussing it at even a superficial level since the question had been put the list. It seemed to me reasonable to comment on the obvious elephant in the room because it appeared there was a present tendency to ignore it. This is the part that needs clarification. There is no clear link established between helium concentration and power produced. Well, it isn't clear, because helium production is so complicated, but I think it is a pretty strong case. Again I think you miss my point, or I didn't make it clear. I agree there is a good case for helium production. There is even some support for sporadic proportional heat to helium production. The point was in regard to the sensibility of the graph axes and the green line. The complexity of helium production and even measurement is a side issue. I would say the whole paper is an attempt at clarification. A pretty good one at that, but you can't expect much detail from only 9 pages. - Jed So true. It seems to me most scientific papers leave out critical details or explanations. I think writers are too close to their own thoughts, assumptions, and expectations, and don't even realize what has been left out or what help the reader might need for easy comprehension. When I read my stuff after it has aged I'm amazed at the critical things I left unsaid, how far what I actually said was from the meaning I intended to convey, and how many ways my remarks could be easily misinterpreted. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:SRI Case Expt 4He
On Feb 24, 2010, at 7:37 PM, Steven Krivit wrote: http://newenergytimes.com/v2/news/2010/SRI-Case-Repl/SRI-Case.shtml Has anybody ever noticed that 2 out of 3 runs that show 4He growth in the SRI Case replication show a peak and then a decrease in 4He? This is a helium leak-tight chamber. Where does the 4He go? s Oooops! I didn't notice the subject was on the Case experiment. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
On Feb 25, 2010, at 11:01 PM, Harry Veeder wrote: - Original Message From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Thu, February 25, 2010 12:25:58 AM Subject: Re: [Vo]:Naudin's improved generator On Feb 24, 2010, at 7:11 PM, Harry Veeder wrote: Are we watching the same video? ;-) Did you not see my correction?? the part about rms voltage and current? harry That's right. It's the same old issue we've seen over and over again, and discussed here ad nauseam. The I*V is not a measure of power when there are phase angles, transient demand, varying frequencies, or square waves involved. It takes a fast integrating power meter to measure input power. This applies to battery DC input to a device with these kinds of power demands as well. Another issue is there is no apparent measurement of power output. As we have seen before, driving LEDs with transients can cause the perception of an amount of light that requiring more power than actually used. It appears the power produced is a small proportion of the power applied. Lastly, as we all know, if there is a claim of significant overunity, then the loop has to be closed for the claim to be credible. I see no reason to think the device is not a transformer that works by displacing a high mu material field. This is not a new idea. There are commercially produced power supply transformers that work on this principle. They are not overunity. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Naudin's improved generator
On Feb 26, 2010, at 11:15 AM, Harry Veeder wrote: The controller is powered by DC. He measures voltage and current going into into the controller. The controller converts this to electrical pulses which feed the toroidal coil, so don't these measurements give you an upper bound on the input power? No. THey give a lower bound. Another issue is there is no apparent measurement of power output. As we have seen before, driving LEDs with transients can cause the perception of an amount of light that requiring more power than actually used. It appears the power produced is a small proportion of the power applied. I sent him an email asking for the wattage of the LEDs. Lastly, as we all know, if there is a claim of significant overunity, then the loop has to be closed for the claim to be credible. I see no reason to think the device is not a transformer that works by displacing a high mu material field. This is not a new idea. There are commercially produced power supply transformers that work on this principle. They are not overunity. Probably not... Definitely not. Their performance was measured in the high 90's percent range if I recall. They were similar to the Fig. 1 drawing I sent, except they had 8 (or more) legs instead of two. This kept the permanent magnet flux more constant and permitted an 8 (or more) phase output which was rectified to make DC. They were used in electronics power supplies. That's all I remember. I wouldn't know how to find them without doing a patent search, which is how I originally found them in the first place, I think. It was years ago. Harry Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Younger Dryas event extended to Andes
Wow, that's a big event. http://www.sciencedirect.com/science? _ob=ArticleURL_udi=B6V93-4XHVGW7-1_user=10_coverDate=03%2F15% 2F2010_rdoc=1_fmt=high_orig=search_sort=d_docanchor=view=c_acct=C 50221_version=1_urlVersion=0_userid=10md5=62d1dd4fb0e08c37ecf1a2 bf25b0f66d http://tinyurl.com/yaeodb4 doi:10.1016/j.geomorph.2009.10.007 W.C. Mahaney et al., W.C.,Oct,2009,Geomorphology,Evidence from the northwestern Venezuelan Andes for extraterrestrial impact: The black mat enigma ABSTRACT: A carbon-rich black layer encrusted on a sandy pebbly bed of outwash in the northern Venezuelan Andes, previously considered the result of an alpine grass fire, is now recognized as a ‘black mat’ candidate correlative with Clovis Age sites in North America, falling within the range of ‘black mat’ dated sites (~ 12.9 ka cal BP). As such, the bed at site MUM7B, which dates to 11.8 ka 14C years BP (raw dates) and appears to be contemporaneous with the Younger Dryas (YD) cooling event, marks a possibly much more extensive occurrence than previously identified. No fossils (megafauna) or tool assemblages were observed at this newly identified candidate site (3800 a.m.s.l.), as in the case of the North American sites. Here, evidence is presented for an extraterrestrial impact event at ~ 12.9 ka. The impact-related Andean bed, located ~ 20 cm above 13.7–13.3 ka cal BP alluvial and glaciolacustrine deposits, falls within the sediment characteristics and age range of ‘black mat’ dated sites (~ 12.9 ka cal BP) in North America. Site sediment characteristics include: carbon, glassy spherules, magnetic microspherules, carbon mat ‘welded’ onto coarse granular material, occasional presence of platinum group metals (Rh and Ru), planar deformation features (pdfs) in fine silt-size fragmental grains of quartz, as well as orthoclase, and monazite (with an abundance of Rare Earth Elements—REEs). If the candidate site is ‘black mat’, correlative with the ‘black mat’ sites of North America, such an extensive occurrence may support the hypothesized airburst/impact over the Laurentide Glacier, which led to a reversal of Allerød warming and the onset of YD cooling and readvance of glaciers. While this finding does not confirm such, it merits further investigation, which includes the reconnaissance for additional sites in South America. Furthermore, if confirmed, such an extensive occurrence may corroborate an impact origin. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:A link between Alzheimer's and herpes zoster?
Cold Sore Virus Linked To Alzheimer's Disease: New Treatment, Or Even Vaccine Possible http://www.sciencedaily.com/releases/2008/12/081207134109.htm http://tinyurl.com/5ujyxx It is not a giant leap to consider the possibility that herpes zoster might be linked to Alzheimers. Like Alzheimer's, and unlike cold sores, which are caused by the herpes simplex virus, Shingles occurs late in life. http://www.mayoclinic.com/health/shingles-vaccine/AN01738 http://tinyurl.com/create.php Since Singles vaccine is recommended by the Mayo Clinic for everyone over 60 anyway, it might be a very good investment, one earning unexpected dividends. Shingles is pretty horrific as it is: http://www.skinsite.com/info_herpes_zoster.htm http://tinyurl.com/4aeeg The somewhat obvious potential link I suggest between Alzheimer's and herpes zoster (varicella zoster), and possibly the Shingles vaccine, a vaccine against herpes zoster, could be studied by survey. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Alzheimer's and herpes zoster should be studied.
hypothesis that genital Herpes may also play a role, perhaps an increasing role with time since the sexual revolution in the 1960's which was brought about by birth control and cultural changes. Genital Herpes can infect anyone who has sex, even if only once. An estimated 25% of adults from varying backgrounds, income levels and ethnic groups have Herpes type 2 causing genital Herpes. Herpes type 2 is often so mild that an estimated two thirds of those infected don’t even realize they have it. Herpes type 2 rarely causes complications and more rarely spreads to other parts of the body. It is estimated that herpes simplex 1 now accounts for as many as 30% of all genital herpes cases in the U.S. and 2-5% of the recurring outbreaks are associated with the herpes type 1 virus. See: http://www.herpesonline.org/articles/herpes_virus.html http://tinyurl.com/yddtj3f Herpes prevalence, in particular HSV-2 prevalence, varies with geographic location. Information on age- and sex-specific prevalence of herpes simplex virus (HSV) types 2 and 1 infections is essential to optimize genital herpes control strategies, which increase in importance because accumulating data indicate that HSV-2 infection may increase acquisition and transmission of human immunodeficiency virus. This review summarizes data from peer- reviewed publications of type-specific HSV seroepidemiologic surveys. HSV-2 prevalence is, in general, highest in Africa and the Americas, lower in western and southern Europe than in northern Europe and North America, and lowest in Asia. HSV-2 and -1 prevalence, overall and by age, varies markedly by country, region within country, and population subgroup. Age-specific HSV-2 prevalence is usually higher in women than men and in populations with higher risk sexual behavior. HSV-2 prevalence has increased in the United States but national data from other countries are unavailable. HSV-1 infection is acquired during childhood and adolescence and is markedly more widespread than HSV-2 infection. See: http://www.ncbi.nlm.nih.gov/pubmed/12353183 http://tinyurl.com/ye4me24 A study of Alzheimer's vs HSV-1 and HSV-2 virus prevalence across age groups and geographical location may be useful. Even an amateur created web based survey might be useful to explore some of the relationships, in order to justify the expense of a professional survey. Such a survey might even be conducted as a high school science project. Such a survey might include the following questions: 1. Geographical location? (providing a multiple choice of regions) 2. Age? (providing a multiple choice of age groups) 3. Have you had cold sores or Herpes of any kind? 4. Have you had Shingles? 5. Have you had chicken pox? 6. Have you had a chicken pox or MMRV vaccination? 7. Have you had a Shingles vaccination? 8. Have you been diagnosed with Alzheimer's? 9. Have you been diagnosed with dementia? It would be important to obtain a large proportion of Alzheimer's respondents. Perhaps the Alzheimer's foundation would be helpful in this regard. See: http://www.alzfdn.org/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Alzheimer's and herpes zoster should be studied.
On Mar 5, 2010, at 6:02 AM, Taylor J. Smith wrote: Hi Horace, Thanks for the info. I take minocycline every day to keep the Lyme spirochetes in my brain at bay. So I should probably protect myself from herpes as much as possible. Jack Smith Thanks for the feedback. I expect to get a Shingles vaccination ASAP myself. I have corrected the post as best I can and put it on my web site here: http://www.mtaonline.net/~hheffner/AlzheimersShingles.pdf http://tinyurl.com/y9okwpe I quoted extensively because web articles tend to go away after some time. The quotes were posted in blue type because they are so extensive and I wanted to make clear my content vs authoritative content. I hope this post will be of use to someone. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Alzheimer's and herpes zoster should be studied.
On Mar 6, 2010, at 9:43 PM, Harry Veeder wrote: Nice study. Hope it holds up. What would be nice to discover is that a Shingles vaccination could reduce or delay the onset of Alzheimer's for a substantial percent of the population, say 15 percent or more. [snip] one data point: I know someone who had shingles about 14 years ago, but he has not developed Alzheimer's. Possibly a new method of detecting Alzheimer's: Eye test that spots Alzheimer's 20 years before symptoms: Middle- aged could be screened at routine optician's visit http://www.dailymail.co.uk/news/article-1243181/Simple-eye-test- Alzheimers-catch-disease-crucial-early-stage.html http://tinyurl.com/ycgesjs Harry There is a well known genetic predisposition factor in Alzheimer's. On Mar 5, 2010, at 1:38 AM, Horace Heffner wrote: The relationship between Alzheimer's and herpes zoster should be studied. A solid link between herpes simplex virus-1 (HSV1 and Alzheimer's has been found. See: Cold Sore Virus Linked To Alzheimer's Disease: New Treatment, Or Even Vaccine Possible: http://www.sciencedaily.com/releases/2008/12/081207134109.htm http://tinyurl.com/5ujyxx In regards to HSV1, the above reference explicitly states: The team had discovered much earlier that the virus is present in brains of many elderly people and that in those people with a specific genetic factor, there is a high risk of developing Alzheimer's disease. For this reason there should be a large percentage of the population that does not develop Alzheimer's regardless their history - those who do not have the genetic predisposition. I suggested Shingles may play a direct role, or possibly a role in the onset of Alzheimer's, and a survey may be useful in indicating whether such a link exists. The viral DNA found in the plaque is that of HSV1. That indicates an HSV1 vaccine, once developed, may be effective in preventing the disease. Meanwhile, if Singles does indeed play a role in the onset of the disease, merely obtaining a Shingles vaccination, which is recommended by authorities for those over 60 anyway, may prevent the onset of Alzheimer's in a significant percentage of the population. Thanks for posting that new Alzheimer's screening test information. I hope that becomes commonly available soon. It sounds like it should work for more than just Alzheimer's. It should determine the existence of other forms of dementia as well. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
