LENR FAQ

We’ve added a LENR FAQ page to the LENR Knowledge Base which gives some frequently asked questions about LENR, along with brief answers and links to further information. I’ve also added a link to it in the navigation bar of E-Cat World.

Here’s a link to the FAQ: http://kb.e-catworld.com/index.php?title=LENR_FAQ

You might find this to be a useful page for people new to the topic, or if you want to share a quick link with someone.

Many thanks to Sanjeev for creating this document! Anyone who would like to contribute to the Knowledge Base as an editor please let me know ([email protected]), and I will set you up with a user account.

  • Obvious

    The Li6 was higher, and Li7 was lower in test results. This does suggest that the Li7 lost neutrons, but that is still conjecture, since isotope fractionation cannot be ruled out at present. However, isotope fractionation by the device is almost as surprising as moving the neutrons around without external radiation.

    • Mats002

      Obvious, what is your comment on Svein Arild Utne above, about Li6? Is it likely that LiAlH4 have ‘wrong’ isotopes to make the effect?

      • Obvious

        If it is old stuff, maybe. I don’t think it is much of a problem with new supply, unless it comes from some old stockpile somewhere. One would have to test your lithium to be sure. A lot of lithium got gobbled up for lithium ion batteries in the past 10 years, which should have sort of flushed the old stuff from the supply chain. It also goes into ceramics, albeit without as much refining. Companies like FMC might know their product isotope distribution, maybe.
        I think it is feasible that the reproducibility problem in the main F&P era could be related.

    • lars

      yes sorry, you are right, so can Li7 loosing neutrons and becoming Li6, lost its neutrons to Ni62 that has increased a lot while Ni58 has decreased a lot?

      • Obvious

        Several people have done the math, and there is a neutron shortage using this simple pathway, and even a neutron shortage when all the H is turned as well. But the assumptions used in these attempts to figure out the neutron math might not be totally accurate.

  • Svein Arild Utne

    Lithium-6

    Lithium-6 is valuable as the source material for the production of tritium
    (hydrogen-3) and as an absorber of neutrons in nuclear fusion
    reactions. Natural lithium contains about 7.5 percent lithium-6, with
    the rest being lithium-7. Large amounts of lithium-6 have been separated
    out for placing into hydrogen bombs. The separation of lithium-6 has by now ceased in the large thermonuclear
    powers, but stockpiles of it remain in these countries. Lithium-6 is
    one of only three isotopes with a spin of 1 and has the smallest nonzero
    nuclear electric quadrupole moment of any stable nucleus.

  • http://www.animpossibleinvention.com/ Mats Lewan

    My understanding is that LENR is a wider concept than ‘cold fusion’ or other similar processes supposedly generating excess heat, rather referring to all kinds of nuclear reactions not involving high energy particles or radiation that are normally observed in relation to known nuclear reactions. But I believe we should let the scientists with long experience in the field set up the definition. Another concept not yet mentioned in the FAQ is CMNS, Condensed Matter Nuclear Science.

    • SG

      Just completed your book Mats and about all I can say is … wow. I encourage all to obtain and read Mats’ book if you want to gain incredible insight into the past few years and the remarkable events that have shaped the current cold fusion arena.

      • http://www.animpossibleinvention.com/ Mats Lewan

        😉

  • mecatfish

    I have a little trouble equating LENR with COP. At what COP do we consider it being overunity? What exactly does COP of 1 mean? Is COP = 1 mean it isn’t any more efficient than the heater wire used to heat the reaction chamber? Anyway I would like to see some more discussion on COP and how it relates to LENR. I looked up COP on wiki and it was a very large and confusing article and was written so as not to be specific to LENR.

    • Ken

      My understanding is that the way LENR enthusiasts use COP is different than the actual term when referring to heaters. A COP of 2 in an LENR device means that it produces 2x the energy put into the system, COP 10 is 10x, etc.

      • builditnow

        A COP of 1 = a typical heater.
        A COP of 2 = a heater + LENR output equal to the heater, so twice the heat of COP=1
        A COP of 10 = 10 times the heat of COP = 1, LENR putting generating 9 out of 10.
        (some expert can correct this if it’s wrong, I’m just an old EE).

        The complicating factors are things like, the “grade” of the energy.
        Electricity is “high grade” in that you can use electricity in say an arc to produce extremely high temperatures, possibly millions of degrees or to charge up lasers for a burst of extreme temperatures for very brief moments.

        A heat pump however can only produce a modest temperature difference, enough for warm air in a house.
        Heat from burning gas is limited to something like 2000c (rough guess here).

        So, Rossi’s warm cat that once operated at about 100C could be considered a low grade heat, not really useful for generating electricity. However, Rossi’s Hot Cat operating at 1400C is a much higher grade energy, capable of being converted into electricity and doing many useful things.

        So, there has been confusion on COP, for instance, about comparing reactor COP with the COP of a heat pump, the discussion seems to have gone away however, possibly because it is not valid to compare the COP of a heat pump with the COP of a Hot Cat because the usefulness of the Hot Cat output is much greater. Also, the COP of a heat pump is not producing any energy, it’s just pumping heat and actually loosing energy in the process and, it uses high grade energy like electricity to produce low grade heat.

        • Obvious

          There are Coefficient Of Performance (COP) and Coefficient Of Efficiency (COE). The latter is generally substituted for the former in LENR.
          Compare to a lever, where one pushes 1/3 as hard to lift something without the lever. The COP would be 3: one third of the pushing force makes the thing lift. 3 times the weight can be lifted with the same force.
          The COE would be 1, since the same energy amount in total is being expended as without the lever to lift the object to the maximum reach of the lever.
          At least that is how I understand it.

          • US_Citizen71

            COE is what most mean when they refer to COP in regards to LENR. We probably need a new term to avoid confusion. Something like CEI Coefficient of Energy to Input.

            • Obvious

              Energy Ratio Relating Output/Received….kidding….

              • Zack Iszard

                What is the minimum number of metrics needed to fully describe the hypothetical performance of a standalone LENR reactor system? I don’t want to posit any myself, I merely ask the best starting question.

                • Obvious

                  That’s a tricky question. If heat is the product, then comparing heat in to heat out is the main goal. But both heat measurements need to be verified by more than one reading, and by more than one type of reading to gain validity to the measurements. Consider each heat measurement as resting on a tripod of proofs. If a leg is missing, the measurement might not stand up to scrutiny. Electrical heat in is fairly straightforward: Resistance, Watts, Amps…
                  Heat out is where all sorts of measurements get problematic. If the heat out derived from electrical heat only (calibration) is done repeatedly, minor adjustments made, and the effects noted and quantified, when excess heat occurs it will be much clearer. A single experiment proves next to nothing, including for calibration. It is a lot more work than it seems at first, but in the long run it will be less work, because one won’t be chasing ghosts due to uncertainties in the calibration that reduce the certainty of the output results, should an excess heat effect occur.

                • Mats002

                  With risk for bias Zacks question which is not my intention I wonder what is the minimum effect difference that is hard to deny in such a measurement?

                • Obvious

                  That is defined by the total cumulative inaccuracy and imprecision determined by the series of calibrations. A result lying outside the 3 sigma range of experimentally derived “errors” (they aren’t errors, really, they are uncertainties) in the calibration set would be significant, IMO. A set of anomalous results, with their 3 sigma range of values all outside the 3 sigma control range would be ideal. But first you must define the 3 sigma range of experimental control values. This is impossible with one or two control/calibration tests. 10 control runs would be statistically weak, but probably OK for a first pass. With electrical heat and a very consolidated and consistently designed reactor setup, the control data should exhibit a tight range of values. Any variation in the reactor bodies (wire wraps, alumina thickness, coating thickness, wire metallurgical properties, thermocouple hysteresis, etc.), electrical profile, ambient temperatures, thermocouple position, etc., will contribute to the uncertainties that must be defined in order to detect anomalous behavior compared to the control. The better the control, the tighter the 3 sigma distribution curve, and the easier it is too see when something is outside the defined calibration uncertainties.

                • Mats002

                  But if the difference is huge, several 100% as in COP 2+, what are the requirements to achive 3 sigma in that case?

                • Obvious

                  If the difference is huge, AND is outside of all possible uncertainties (determined), the you probably have something. 100% more than the average of a bunch of noise that spreads 50% either way around the average is not as compelling. 100% itself means nothing if the noise is huge. A COP of 1.2 would be huge if the 99% of the control data was in the one sigma range (near perfect control repeatability).

                • Mats002

                  So the recipie after seeing excess heat would be repeat the reference (without fuel) many times to see the ‘spread’ of the curves? Then I guess rerun with fuel many times but here I guess the ‘spread’ can be wider and that would not affect the sigma?

                • Obvious

                  Yes. 3 sigma by definition contain 99.7% of all the results. Anything outside this in a control may actually be an error, or the data isn’t sufficient to define the curve properly yet. These outlier results can usually be discarded for general comparisons, but the circumstances around them should be investigated. Once the curve is very tight, there probably always will some outside, since 0.3 % by definition should be there… The practical limit of uncertainty for the experiment is found once the area of the curve cannot be tightened/reduced any more by refining the control procedures and equipment.

                • Mats002

                  Thanks Obviuos, I hope this is the day that makes history, we’ll see.

                • Obvious

                  I hope so. Although, I would say it might be beginning of making history, since apparent success enough times is needed to weed out the uncertainties. A “COP” of 12, on the other hand would encourage me to pop the champagne anyways. That would be hard to explain by any measure if the experiment is reasonably and professionally done (and it seems to be so far, within the current reasonable limits of the equipment and funding level, etc.).

        • mecatfish

          So if I have a space heater with nichrome wire as the element and a ecat that has a cop of 2 and it has the same length of nichrome wire as the element, Then I can say that it costs half as much in electricity bill to operate the ecat heater?

  • LENR4you

    a LENR-Simpleshow movie would be perfect to share. http://simpleshow.com/

    • Mats002

      Yes, I have seen some of those before, they are nice and informative! Who pay for the job?

      • LENR4you

        As the inventors of the simpleshow, we are in a perfect position to help you explain your issue in a friendly and open manner in just three minutes, no matter how complicated your topic is. Register now for the simpleshow initiative!

        With this opportunity, we are giving selected non-profit organisations a chance to put the spotlight on their issue in an effective and striking manner with a short simpleshow. Our programme includes direct support in the design, creation and production of the clip and is partial funded by simpleshow.

        All orders will be realised. The finished explainer video can be used on the Internet, in Social Web and at events free of further license fees.

        The price for a supported simpleshow is € 5.000 plus 7% VAT.

        http://simpleshow.com/initiative/

      • Owen Geiger

        I found the Fiverr.com site interesting. You can find video and animation artists who do small projects for $5. Choose someone with high customer satisfaction. Might be worth a try. https://www.fiverr.com/