by
Wladimir Guglinski
retired, author of the Quantum Ring Theory
.
.
Abstract
Dr. Wilfried Nörtershäuser of the Helmhotz Center for Heavy Ion Research at the University in Mainz says on the 2009 experiment which had detected a neutron halo in 4Be11 with distance 7fm from the cluster:
“By studing neutron halos, scientists hope to gain further understanding of the forces within the atomic nucleus that bind atoms together, taking into account the fact that the degree of displacement of halo neutrons from the atomic nuclear core is incompatible with the concepts of classical nuclear physics”[ 2 ]
In the case of 4Be11, the halo neutron and the nuclear core are separated by the distance of 7fm, and so such isotope represents the experimental proof that the cohesion of nucleons within the light isotopes cannot be promoted by the strong nuclear force.
Such experimental discovery published in 2009 had been predicted years ago, because according to the new nuclear model proposed in Quantum Ring Theory, published in 2006, the cohesion of the nucleons within the light nuclei is not caused by the strong nuclear force.
Here in this paper the new nuclear model is submitted to a scrutinity so that to verify whether from its structure it’s possible to explain the stability of the light nuclei and to reproduce the nuclear properties as nuclear spins, electric quadrupole moments, and magnetic moments. Nuclear magnetic moments are calculated from two different and independent methods. In the second, named “method of equilibrium between nucleons”, it’s presented the Lagrangian for nuclei with Z < 8. The results obtained from them agree each other, and are corroborated by nuclear spins and electric quadrupole moments suplied by nuclear tables.
In this Part One are presented calculations on magnetic moments for the isotopes of lithium, beryllium, and boron. In the next paper Part Two will be exhibited calculations for carbon, nitrogen, and oxygen. In the paper Part Three the author will exhibit calculations for electric quadrupole moments.
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Dear Brian:
We are manufacturing our plants in the USA and I am mainly helping in this the US Partner. This will be my job in the next 6 months, together with the development of the Hot Cats.
Warm Regards,
A.R.
Mr. Rossi
I hope that you are well. I was wondering if you could tell us a little more about the short-to-medium terms goals of your work and what you’ll be working on for the next 6 months to a year.
Thank you for taking the time to answer my question
Brian
Joe wrote in May 13th, 2013 at 3:16 AM
Wladimir,
1. If the Least Action Principle does not apply to excited nuclei, how do you determine the placement and orientation of nucleons along n(o)?
RESPONSE:
Joe,
I dont know all the laws that connect the flux n(o).
It needs investigations, based on experiments.
I only know that, when energy is supplied for a non excited nucleus, it can change its status as follows:
1- by changing the spin of a nucleon regarding the flux n(o), as happens in the case of excited 6C12.
2- by changing the position of the nucleons, as happens when 5B10 is excited (see Fig. 25 page 35.
.
2. Do not spin and magnetic moment always share the same sign, since it is the spin that generates the magnetic moment? Why is the magnetic moment of D-2 not positive in Fig.4? Why does the flux invert the magnetic moment from positive to negative? And how does it do this while maintaining the exact same value for the magnetic moment (0.857)?
RESPONSE:
No, Joe,
spin and magnetic moment always share the same sign in CLASSICAL Nuclear Physics.
Faraday had discovered the laws of electromagnetism for the macros-phenomena.
In QRT I had discovered the laws of electromagnetism for the micro-phenomena, and I had discovered that the electromagnetic phenomena depends on the flux n(o).
regards
wlad
Dear Stefano:
Yes, you got it.
Warm Regards,
A.R.
Dear Dr. Rossi
Thank you for your effort in keeping alive the comunicatikn with your readers. In practice I understand that the system overall consumes gas or grid power for the 35% of the time but that energy is returned in any case to the costumer. And in that time the system has a cop of about 1. In other words the real gain is in the 65% of the time with a large production of energy without consume. Is it correct.
Dear Hank Mills:
Obviously I cannot answer you, but let me confirm one thing: I meant exactly what I wrote.
Reading with the necessary attention and without any bias what I wrote, the description of the operation of the apparatus with Activator and E-Cat is very simple. I made what I wrote read by a 13 years old middle school girl, and she understood perfectly. So should do anybody. One thing: remember that the Activator anyway gives to the Customer the heat he needs with a COP 1.02 when it is turned on, and this heat produced by the activator is totally paid by the very Activator, whose heat is not lost, is entirely utilized by the Customer. Therefore the energy produced by the E-Cat has no input energy to pay for, because the energy that activates the cat, after activating he cat, goes to the Customer who anyway had to pay for it to use it. As a consequence of this fact, the E-Cat energy output is not gravated from any energy consume.
Warm Regards,
A.R.
Warm Regards,
A.R.
Dear Andrea,
I think one issue of confusion is the issue of what being “on” and “off” means for the activator and the cat. Some think that you mean when the activator is on the cat is off and producing zero thermal power. My understanding is that the activator turns on when the cat is producing too much heat (the nuclear reactions becoming too intense). During this period, the cat may be off, but being off does not mean it is not producing power in the form of heat via nuclear reactions. It just means that it is being stabilized via the resistance element of the activator, and, most likely, the radio frequency stimulation to the cat has been shut off. The cat is producing power during this time, but is simply revving down so to speak, like taking your pedal off the accelerator to slow down a speeding car to a more appropriate speed. So, can you confirm that this is the proper interpretation of what the cat being “off” means?
Thank you.
Hank
Dear daniel De Caluwe’:
1- please read what I have already explained
2- see 1
Warm Regards,
A.R.
Dear Dr. Rossi,
1. Refering to your recent answer to Dr. Joseph Fine, and the fact that the Activator has a lower energy-production (about 1/10th of the Tiger in the example you gave in your answer to Joseph Fine) than the main part (Tiger) itself, do you agree that 1 system/E-cat (= 1 Tiger + 1 Activator) alone produces 35% of the time (when only the Activator is active) a lower heat output (in your example 1/10th) than the other 65% of the time (when the main part (Tiger) of the E-cat is active)? So, the thermal energy production of 1 element (= 1 Tiger + 1 Activator) fluctuates?
2. And to make the (thermal) energy production of a bigger system (with many individual E-cats (many Tigers and as many Activators, with each Tiger his own seperate Activator)) constant, and not fluctuating (between 1/10th when the Activator is active and 1 when the Tiger is active), do you solve that problem via the control system, making the different Activators work at different times, so that you get a smooth and constant (thermal) power output?
Kind Regards,
Dear Andrea Rossi and Dr. Joseph Fine,
I to swoon at the thought of the possibilities of the Activator – Reactor eCat coupled with direct thermal-to-electric conversion.
In order to have a constant power output, would not an energy storage device (such as a battery) be needed to power the Activator when it is on, coupled with a battery charger using the thermal-to-electric converter as a source? Otherwise, would not the available electrical output power drop due to the Activator load?
If we use a point design of:
eCat output thermal power = 10 kW
Thermal-to-electric conversion efficiency = 30%
Assume Activator power load of 1 kW
then the available electric output would be changing between 2 and 3kW depending whether the Activator was on.
With a 35% activator duty cycle, the battery charger would need to be about 400W and would decrease the electric output from 3 kW to about 2.6 kW on the average. Thoughts?
Wladimir,
1. If the Least Action Principle does not apply to excited nuclei, how do you determine the placement and orientation of nucleons along n(o)?
2. Do not spin and magnetic moment always share the same sign, since it is the spin that generates the magnetic moment? Why is the magnetic moment of D-2 not positive in Fig.4? Why does the flux invert the magnetic moment from positive to negative? And how does it do this while maintaining the exact same value for the magnetic moment (0.857)?
All the best,
Joe
Dear John,
You are right, this technology can help to resolve the CO2 issue. As for how the system activator-Ecat works I already have answered to your questions: please go through the many, many, many answers I already gave regarding this issue. I have nothing to add to what I have already said.
Warm Regards,
A.R.
Andrea — With atmospheric CO2 levels now reaching 400 ppm, your 100 – 200 COP hot-cats are urgently needed World wide.
You indicated that your basic hot-cat activator operates at 1kW when it is turned on 35% of the cycle. When the 1kW activator is turned off and the hot-cat is turned on for the remaining 65% of the cycle how much energy does that hot-cat produce? Does the 1 kW activator trigger a 100 kW hot-cat?
Dear Dr Joseph Fine:
Please don’t go too far: just, for now, let’s limit to what I wrote about the Activator/E-Cat cycle. Please read carefully what I wrote. More than that is not possible to get, so far. Our basic module is made by an apparatus in which we have 2 components: an activator, which consumes abour 900 Wh/h and produces about 910 Wh/h of heat. This heat activates the E-Cat and then goes to the utilization by the Customer, so that its cost is paid back by itself. This activator stays in function for the 35% of the operational time of the syspem of the apparatus. The E-Cat, activated by the heat of the Activator, works for about the 65% of the operational time, producing about 1 kWh/h without consuming any Wh/h from the grid. Combining these modules we can make E-Cats of 1 kW , 10 kW, 100 kW, 1 MW , respectively, of power.
Warm Regards,
A.R.
Dear Andrea Rossi,
The initiator/Mouse is at low (or zero power) when switched off. It/(the Mouse) consumes 1 kW only 35% of the time (and produces essentially 1.02 kW of heat during this time). Instead of using a (Joe)COP of 2.86(P2/P1) = 286, assume the actual COP is only 100-125. This is admittedly much higher than the presumed COP of 10-12, but lets continue along this path.
If the “ETA”, or thermal-to-electric conversion efficiency at 350-400 C, is between 30% and 40% (40% is on the high side), and COP is between 100-125, (let alone 200-250) then each 100 kW Hot Cat would be able to drive from .3*100 to .4*125 or 30 – 50 other Hot-Cats. So, each 1 KW input to a first-stage Mouse can produce not 100 KW-th but, by generating 30+ kW-Elec, may be able to drive 30 other second-stage 100 kW Hot Cat devices.
Even if it is possible to produce 3 MW-thermal with one (1st stage) 1 kW Mouse, it may be much simpler to have 30 1-kW Mice to drive 30 separate 100 kW devices.
Without getting too exuberant, three stages of multiplication by a factor of 20, instead of 30, is already 20^3 = 8000. That is, even if a single 1st stage kW Mouse can only drive 20-100 kW ‘Cats’, three such stages, using a single 1 kW input (gas or electric), ultimately might produce 8000*100 kW or 800 MW of heat.
And if you can produce 800 MW of heat, or even much less, why do you need a 1 kW input? (Other than for start-ups.)
0) Is that the basic concept of what you are trying to accomplish?
1) When a 100 kW HotCat or Tiger-Cat is not in the self-sustained mode (SSM), what mode or state is it in? In other words, how much thermal power is produced during the 35% of time the 100 kW Cat is not in SSM? Does it have an electrical input to bring it back under control? Or do you simply remove its input Hydrogen (Idrogen) supply?
Multi-stage regards,
Joseph Fine
PS I apologize for this comment, as I may be engaging in irrational exuberance.
Joe wrote in May 12th, 2013 at 4:31 AM
Wladimir,
Nuclear scientists should always examine new alternative models because science is always progressing. This includes QRT. (It is odd that you condemn the possibility of a 5th force, since QRT introduces its own new force – REPULSIVE gravity.)
RESPONSE:
Joe,
there is a big difference between the proposal of mine on the repulsive gravity and the proposal of the nuclear theorists proposing the 5th force.
Look:
1- I had developed my theory along 20 years.
Step by step I was discovering the models and the laws that must rule their working.
Along the years I had faced many troubles, and had undertaken many efforts to eliminate the inconsistencies.
The proposal of the repulsive gravity had been a consequence of such effort of mine:
After long meditations, I finally had concluded that my nuclear model cannot work without a repulsive gravity.
So, that was a theoretical conclusion, required in order to get a coherent working of my nuclear model.
I even did not know that there are pear-shaped nuclei at that time.
2- Nowadays the theorists are thinking about a 5th force because their theories do not fit to the recent experimental findings.
For instance, the pear shape of high nuclei do not fit to what we expect from the classical principles of the Standard Nuclear Physics.
So, the theorists today are not thinking about the existence of a 5th force from the theoretical viewpoint,
They are thinking about to propose the 5th force from an ad hoc viewpoint, with the aim to ADAPT their nuclear theory to the results of experiments.
An ad hoc solution is not desirable.
Because even if they succeed to adapt the pear shape to the results of the experiments by keeping the fundamental principles of the Standard Nuclear Physics, it is possible such sort of solution will hide the true cause of the existence of the pear-shaped nuclei.
In QRT the nuclei with pair number of complete hexagonal floors have elipsoidal shape (for example, the 92U238)
But nuclei with odd number of hexagonal floors have tendency to be pear-shaped.
And if the nucleus has odd number of hexagonal floors and also incomplete hexagonal floors, the tendency to be pear-shaped is greater.
regards
wlad
Joe wrote in May 12th, 2013 at 4:31 AM
Wladimir,
1. I had asked you what causes the sign of the spin of a nucleon to change abruptly and work against the Least Action Principle. You asked me where this occurs. I now answer that it occurs in Fig.26 (page 37) in the excited 6C12. Can you explain why one deuteron suddenly changes its spin?
RESPONSE:
Joe, the 6C12 is excited.
The Least Action Principle cannot be applied, since the 6C12 receives additional energy, which breaks the conditions existent for the application of the Least Action Principle.
The principle is not applicable for excited nuclei.
.
2. The deuteron D-2 in Fig.4 (page 6) seems contradictory. It has a positive spin, yet a negative magnetic moment. How is this understandable?
RESPONSE:
Joe, dont you see the difference between D-1 and D-2?
You have to pay attention to the direction of the flux n(o) crossing the both D-1 and D-2.
Look:
D-1: it has i=-1, and the flux n(o) enters through the pole “+” (blue) of D-1
D-2: it has i=+1, but the flux n(o) enters through the pole “-” (red) of D-2
regards
wlad
PS: sorry the delay, I didnt see your comment because of the publication of the new article Electrical Catalyst.
Dear Dr Joseph Fine:
1- The Activator provides heat to the Customer when it is turned ON, while it also activates the E-Cat. When it is turned off his production is very low.
2- yes
3- I would say: since the Activator pays for itself, being its COP ~1.02 , the E-Cat has a COP difficult to evaluate: we say 100-200, but, as a matter of fact, at the denominator there is zero.
4- this depends on the model of the apparatus. In the basic Hot Cat it is about 1 kW
5- no
Warm Regards,
A.R.
Dear Pekka Janhunen:
It’s simple: if the COP of the activator is >1, whatever the number, it gives heat that repays for itself, besides activating the E-Cat.
The economy of the system is: the Activator consumes 1, yields 1.x as heat to the Customer AND activates the E-Cat: the E-Cat consumes nothing from the grid , gives its energy to the Customer.
Warm Regards,
A.R.
Dear Andrea Rossi,
That the activator with COP 1.1 would pay for itself is still something that I do not understand because it still appears to need 1.0 worth of external energy to run. Do you perhaps mean that the activator is run with its usual cooling turned off and stabilised by some endothermic reaction in the E-cat charge? In a closed box with no cooling, any process with COP>1 will go to thermal runaway, I guess. But if an endothermic reaction provides a “thermostat”, then such process might be stable, until the endothermic reaction stops (at which point the E-cat starts). But a chemical endothermic reaction would be too weak to cause this kind of behaviour: maximal chemical reaction (4 eV per atom) of 2 grams of Ni-H charge is only 25 kJ or 0.007 kWh. This sounds far-fetched so I think I must probably have some other misunderstanding.
regards, /pekka
Andrea Rossi, Pekka, Steve Karels, Italo and readers”
In the Cat and Mouse configuration, another way of looking at COP is the output power produced by the E-Cat (P2 = 100 KW) divided by the input power provided to the Mouse ( P1 = ?). The “Mouse” has a COP of 1 or 1.02, so almost all the thermal output of the Mouse provides the thermal input to the Cat. And, this Mouse activity is only “ON” 35% of the time.
1) When the Mouse is “OFF”, does it provide direct heat output to the customer?
2) When the Mouse is “ON”, does it provide direct heat output to the customer as well as heat input to initiate the Cat?
3) If COP = P2/P1 and P1 uses external power(or gas?) for 35% of the time, my estimated COP is: (Joe)COP = 2.86*(P2/P1). Here, 1/(0.35) = 2.86. So, if P1 = 10 kW, and P2 = 100 kW, the (Joe)COP = 28.6 or if P1 = 25 KW, then (Joe)COP = 4*2.86 or about 11.4. This agrees with Steve’s value. Probably for the same reason.
4) What is the power output of the Mouse?
5) Is there a relationship between the power output of the Mouse/Initiator and the percentage of time that the Cat can be in self sustain mode (SSM). For example, if the Mouse power were doubled, would the Cat be able to be in SSM for 75% of the time, etc.?
Best regards,
Joseph Fine
Dear John:
Done.
Warm Regards,
A.R.
Dear Pekka Janhunen:
Please read the answer I gave to Steven N. Karels few minutes ago. The source of the activator can be electric power, gas, or any heat source.
And again congratulations for your very smart ESTCube-1.
Warm Regards,
A.R.
Dear Steven N. Karels:
a- when the activator is on the E-Cat is off, when the E-Cat is on the Activator is off. When the Activator is turned off the temperature rises, becauce the E-Cat is activated, when the Activator is turned on the temperature lowers, because the E-Cat goes off. If we consider
100 hours of operation, for about 35 hours the Activator is on and the E-Cat is off, while for about 65 hours the E-Cat is on and the Activator is off. We reached a good stability for this reason: the Activator gives to the E-Cat enough energy to give good performance, but not enough to escape from control, like a Mouse which make a Cat nervous, but not too much; then, the Activator stops, the CaT goes on, until he returns to sleep; at this point the Mouse- activator is turned on, but the temperature goes down because the E-Cat is off; at this point the Cat becomes again nervous, and immediately the Mouse- activator is turned off, while again the temperature raises, and so on. The invention of this cycle, regulated by a complex software, allows to reach high temperatures in good stability. The important thing is that also the Activator has a charge, so that it reaches a COP more than 1, paying for itself: for this reason the energy that the Activator consumed is paid for by itself and does not affect the COP of the E-Cat. You know how I invented this system? I was in North Carolina and observed in a garden of the hotel a sleeping cat: a squirrel passed fast close to the cat, and the cat made some move to reach the squirrell, but the squirrel disappeared and the cat returned to sleep…that’t learning from Nature!
b- No
c- So far we are making R&D on the 100 kW Tigers, I can’t answer
d- see 3
e- see 3
Warm Regards,
A.R.
Dear Andrea Rossi,
Many thanks for the continued information and clarifications of the Activator and Reactor. A very interesting implementation. Also, thanks to Pekka Janhunen for the closed form solution to average COP. A few additional questions:
a. In the Activator – Reactor system, essentially you have a Reactor without electrical heaters but it receives its control heat from the Activator. So at thermal steady-state operation, the Reactor is in SSM 65% of the time and in the other 35%, heat is being applied by the Activator? Is this essentially correct?
b. Conceptually, if you were to construct a 1MW industrial device using 10 of the 100 kW Tiger units, would you be limited in output power control to 10%?
c. If the 100 kW Reactor just begins its SSM operation mode and the user demand significantly decreases, can the 100 kW unit be immediately throttled back or must one wait for the cessation of SSM before an output reduction occurs?
d. In the 100kW design, is the Activator an internal part of the Reactor body or an external unit (for example, it wraps around the Reactor)?
e. What are the dimensions of the 100 kW unit? – I would like to run a heat transfer analysis, please.
Wladimir,
Nuclear scientists should always examine new alternative models because science is always progressing. This includes QRT. (It is odd that you condemn the possibility of a 5th force, since QRT introduces its own new force – REPULSIVE gravity.)
1. I had asked you what causes the sign of the spin of a nucleon to change abruptly and work against the Least Action Principle. You asked me where this occurs. I now answer that it occurs in Fig.26 (page 37) in the excited 6C12. Can you explain why one deuteron suddenly changes its spin?
2. The deuteron D-2 in Fig.4 (page 6) seems contradictory. It has a positive spin, yet a negative magnetic moment. How is this understandable?
All the best,
Joe
Dear Andrea Rossi,
If the activator would run with thermal energy, then I would understand that its consumption does not matter because it could then be driven by thermal energy produced by the E-cat (produced in an earlier cycle and stored or produced by another anti-phased reactor in the same plant).
However if the activator uses resistive heating as earlier indicated, then I am confused, because electric energy is about 3 times more valuable than thermal energy.
So I’m wondering if the activator uses thermal or electrical energy? Maybe it is only started up by resistors at the first time and on subsequent cycles uses thermal input from the E-cats?
regards, /pekka
PS. By the way many thanks to Argon and AR for the kind words concerning ESTCube-1, which is working well.
Dear Dr. Rossi,
You should patent the cascade “mouse-cat” or “Rat-tiger” setup (if you have not already done so) separately. Leaving out the “cold fusion” bit because this setup may be applicable to other more conventional reactions, that may help your patent being granted faster.
Dear Italo R.:
1- the cycle depends, is regulated by a central control system
2- no, when the Activator is on ( and the Cat off) all the energy consumed by the Activator is given as a heat production with a COP slightly superior to 1 ( 1.02): for this reason the energy consumed from the Activator has not to be accouinted for as a consume of the E-Cat: all the energy consumed by the activator is recovered and used by the Customer.
Warm Regards,
A.R.
Dear Dr. Rossi, may I ask something about the last stunning development?
1 – How long in minutes is the cycle: mouse (35% of total) + cat (65% of total)?
2 – During the first 35% when the cat is off (no ssm also), the heat produced by the whole apparatus is tending to zero?
Dear Ron Stringer:
The Activator pays for itself, having a COP around 1.02, so its consumption of energy has not to be accounted for in the balance of the E-Cat, because it produces enough heat to pay for its own consumption with a 100% efficiency. The COP of the E-Cat is therefore net of the Activator’s consumpton. The COP of the E-Cat, as a matter of fact, has a zero at the denominator, but we rate it between 100 and 200 considering some parasitic consumption of energy, also bacause “infinity” makes no sense in Physics.
Warm Regards,
A.R.
Dear Readers of the Journal Of Nuclear Physics:
today has been published the paper “Electrical Catalyst” by Tadej Bajda ( Slovenia).
JoNP
Dear Andrea Rossi,
In the mousecat case with no time overlap between activator mode and E-cat mode, the total COP is
COP = (t1*P1 + t2*P2)/(t1*P1/C1 + t2*P2/C2)
where t1=0.35 is the duration of the activator, t2=0.65 is the duration of the E-cat, P1 is the output power of the activator, P2 is the output power of the E-cat, C1 is the COP of the activator and C2 is the COP of the E-cat. For example if P1=25 kW, P2=100 kW, C1=1.1 and C2=100, the total COP is 8.6.
Near these parameters, approximation COP=2*C1*(P2/P1) can also be used. With the above numbers the approximation gives COP=8.8.
The difference between C1=1.0 and C1=1.1 is only 10% in the overall COP. Essentially, the total COP is twice the output power ratio of the E-cat and its activator, and also multiplied by C1. Twice, because 0.65/0.35 is roughly 2.
regards, /pekka
Plagiarism in the journal Nature – AGAIN ???????
From: wladimirguglinski@hotmail.com
To: nature@nature.com
CC: epja@itkp.uni-bonn.de; helayel@cbpf.br; jyeston@aaas.org; prc@aps.org; apr-edoffice@aip.org; cjp@fzu.cz; ver@cisp-publishing.com; pnj@bauuinstitute.com; johna_6@yahoo.com; chupp@umich.edu
Subject: Plagiarism in the Journal Nature: AGAIN ???????
Date: Sat, 11 May 2013 12:27:24 -0300
Dear Dr Karen Howell
Senior Editor , Journal Nature
In 2012 the journal Nature had published the paper How Atomic Nuclei Cluster, where there is a plagiarism of an argument proposed in the page 137 of my book Quantum Ring Theory, publihed in 2006.
Now it seems that the journal Nature had published a plagiarism again, in the paper of May-2013 entitled Studies of pear-shaped nuclei using accelerated radioactive beams.
http://www.nature.com/nature/journal/v497/n7448/full/nature12073.html
Tim Chupp, one of the authors of the paper, gave an interview for the Brazillian blog Inovaçao Tecnológica, where he says:
”The pear shape is special. It means that neutrons and protons which compose the nucleus take positions a litle different along an internal axis“
Well, the existence of the internal axis of the nuclei is proposed in my book.
According to the new nuclear model proposed in my book, the nucleons gyrate about the z-axis (see the nucleus 46Pd in the page 13 of the article Stability of Light Nuclei, published by Andrea Rossi’s Journal of Nuclear Physics):
http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf
According to my theory, deuterons form hexagonal floors about the z-axis.
In my email to the editor of the European Physical Journal I told him that each day the journals of Physics will continue to publish plagiarisms of my book, because as many experiments are suggesting that my theory is correct, then it is obvious that new experimental findings (which are coming to light now and will be coming in the future) will oblige the theorists to addopt the arguments of mine proposed in my book.
In 2012 the journal European Physica Journal had published a plagiarism of the idea of mine on the space filled by particles and antiparticles.
http://www.zpenergy.com/modules.php?name=News&file=article&sid=3464&mode=&order=0&thold=0
Now the journal Nature had published the plagiarism of my argument that there is a central axis within the nuclei.
So, new plagiarism are coming, of course…
Regards
Wladimir Guglinski
Dear Giovanni Guerrini,
No, just follow what I wrote. I cannot give more particulars now.
Warm Regards,
A.R.
Dear Ing. Benedetto Schiavone:
1- while the activator is on the E-Cat is off
2- the E-Cat, when operates, is always in ssm
3- with this configuration, since the Activator has a COP that allows to produce energy equal or more ( even if slightly) to the energy consumed, the resulting COP is the one of the E-Cat, otherwise you can make the average, but at this point the concept of COP changes foundamentally.
Warm Regards,
A.R.
Dear Giovanni Guerrini,
The activator works better than the resistances inside the E-Cat, it is true.
Warm Regards,
A.R.
Dear Joe
A new experiment (May-2013, journal Nature) had detected that some heavy nuclei are pear-shaped:
http://www.nature.com/nature/journal/v497/n7448/full/nature12073.html
”This contrasts with the more prevalent rugby-ball shape of nuclei with reflection-symmetric, quadrupole deformations. The elusive octupole deformed nuclei are of importance for nuclear structure theory, and also in searches for physics beyond the standard model”
One of the authors is Tim Chupp, who gave an interview for the Brazillian blog Inovaçao Tecnologica:
http://www.inovacaotecnologica.com.br/noticias/noticia.php?artigo=nucleo-atomo-formato-pera&id=010130130509
In the interview he says:
”O formato de pera é especial. Ele significa que os nêutrons e os prótons que compõem o núcleo estão em posições ligeiramente diferentes ao longo de um eixo interno”
Translation:
”The pear shape is special. It means that neutrons and protons which compose the nucleus take positions a litle different along an internal axis“
Joe,
I think I dont need to remember you about the z-axis in the nuclear model of QRT. Therefore the nuclear physicists are taking a way which corroborates the existence of the z-axis in QRT.
In the interview Tim Chupp also told that the nuclear theorists are thinking about the existence of a 5th fundamental force in nature:
” Os núcleos em forma de pera seriam assimétricos porque os prótons estariam sendo empurrados para longe do centro do núcleo por alguma força nuclear ainda desconhecida –
Podendo estudar e comparar as duas peras, os físicos esperam não apenas descobrir uma nova força fundamental da natureza, como também explicar…
”
Translation:
The pear-shaped nuclei should be asymmetric because the protons would be pushed far away the center of the nucleus by some yet unknown nuclear force-
From the study and comparison of the two pearls, the physicists hope not only to discover a new fundamental force of nature, but also to explain…
Then we have to consider some points, dear Joe:
1- The speculation on the existence of a new fundamental 5th force of nature do not fit to the Heisenberg’s scientific criterium. So, the theorists are thinking about to reject the Heisenber’s criterium.
But instead of to propose a speculation on the existence of a 5th force…
… why do not think about a speculation not so drastic, as the existence of the flux n(o)?
As the theorists had now concluded that it is impossible to keep the Heisenberg criterium, and speculations cannot be avoided, then why do not consider the flux n(o) ?
2- If a 5th fundamental force should really exist, it would have to manifest itself in other phenomena. Why do the 5th force manifests itself only in the shape of some heavy nuclei?
Why the 5th force do not manifest itself in other heavy nuclei?
Why the 5th force do not manifest itself in other phenomena?
Joe,
dont you think that such proposal of a 5th force is very most speculative than the proposal of the flux n(o) proposed in my new nuclear model ?
Regards
Wladimir Guglinski
Bounding the Activator Power Size and Upper COP Limit
Determining the Activator power output requirements is essential in estimating the average COP of the 100kW Tiger eCat. This analysis attempts to estimate the needed size for the Activator and a resulting bound on the Tiger COP.
We know from previous postings that the start-up time for eCats is in the 1 – 4 hour range, so I will assume an Activator time of 1000 seconds although it could be up to 10,000 seconds. It is more likely the time needed to bring the eCat from room temperature to eCat ignition temperature is on the shorter side.
We assume the Activator must bring the Tiger from room temperature to over 350C to initiate the reaction.
We know the mass is 200 kg. We assume a specific heat capacity of 0.466 J / (kg * K) – we assume the mass is steel.
Power = 200 kg * 350K * 0.466 J / (kg * K) / 1000 sec = 32.62 W
Activator COP is stated at 1.1 so the needed output is ~ 30W.
Multiply by 10 to account for ambient heat transfer losses, unknown system losses, etc., we assume a working Activator power output of 300W.
Feeding that back into the Average COP computation:
Tiger: COP of 100 – 200, output power = 100 kW, 65% of time, therefore, input power when Activator if off = 0.5 – 1.0 kW
Activator: COP of 1.1, assume output power 300W to 1KW (Rossi – mentioned smaller 1KW eCats), 35% of the time, input power of 0.3kW to 1kW
Average input power: 0.4 – 1.0 kW
Average COP: 100 kW / (0.4 to 1.0 kW) = 100 to 250
This does not include any mechanical (pumps, etc) power consumption but this should bound the upper COP side.
Dear Andrea,
I have an other question:
I don’t ask what rat gives to tiger,but I suppose that it reorganizes the E-cat for the next reaction better than could do a resistance.
Sorry,but we are curious like cats !
Thank you
Regards G G
Gent.mo Ing Rossi,
prima di tutto i miei più sinceri complimenti per i progressi ottenuti; un paio di anni fa l’E-Cat sembrava avesse già caratteristiche assolutamente incredibili; oggi con i nuovi Hot-Cat, doppio sistema Activator/CAT e i “big” E-Cat/Tiger ci troviamo davanti ad una tecnologia ancora più incredibile! Non oso immaginare quali altri sviluppi si potranno ottenere nei prossimi mesi e anni :-).
Le scrivo chiedendoLe un aiutino per comprendere meglio il funzionamento Activator/E-Cat che tanto ci sta entusiasmando: ha detto che “Activator and E-Cat never go at the same time” e che “The activator works for the 35% of the operational time of the system, while the E-Cat works for the 65% of the time”.
1) Nel 35% del tempo Lei intende che l’Activator è in SSM o che in tale periodo sta assorbendo energia (resistenze accese)?
2) Nel 65% del tempo quando dice che l’E-Cat lavora, intende che è in SSM mode o che semplicmente produce calore?
3) Ci può dire, in base alle prime stime,immagino prudenti, quale potrebbe essere il COP dell’intero sistema Activator/E-Cat?
La ringrazio e Buon Lavoro
Ing. Benedetto Schiavone
Dear Dott Rossi,
summarizing:
1)activator and E-cat never go at the same time
2)the only source of external energy is the resistance of the Activator
3)the activator runs for the 35% of the time and the E-cat for the 65%
4) the COP of the first is 1.02-1.1 and of the second is 100-200.
Could we think that (?):
the activator have a low cop to have a high stability for driving to restart the E-cat when its self sustained reaction falls down.(on 65% time,off 35%)
The cop of the E-cat is calculated with the ratio between its energy output and the input of the activator more or less because the cop of activator is near 1(in 100% of the time).
To have a steady output of the machine many of these rat-tiger works together .
Is correct?
Regards G G
Estimating Average COP for the Activator + Tiger
Assumptions:
Tiger output: 100 kW
Activator output: 25% of Tiger = 25kW
Tiger COP: 150
Activator COP: 1.1
Activator Operation: 35%
Tiger Alone Operation: 65%
Analysis:
After start-up, when Activator is operating, input to Tiger is zero. Activator input power = 25kW / 1.1 = 22.7 kW
Tiger input power (Activator is off) = 100kW / 150 = 0.67 kW
Average input power = 0.35 * 22.7 kW + 0.65 * 0.67 kW
= 8 kW + 0.44 kW = 8.44 kW
Average COP = 100 kW / 8.44 = 12
Joe wrote in May 10th, 2013 at 5:03 AM
Wladimir,
I do not know all the nuclear models that are in existence. Maybe there are some that could perhaps explain the excited 6C12 but have yet to be acknowledged or recognized by the general establishment.
COMMENT:
Dear Joe,
from the current principles of the Standard Nuclear Theory it’s IMPOSSIBLE to explain the behavior of the excited 6C12
If there is a new nuclear model some that could perhaps explain the excited 6C12 (to be acknowledged or recognized by the general establishment), it CANNOT WORK BY KEEPING the foundations of the Standard Nuclear Theory.
Such a new nuclear model needs to work with some principles no considered in the current nuclear theory.
.
This would include QRT. Even if the most accepted current model is deficient in this respect, that does not mean that it needs to be COMPLETELY rethought, since it has already demonstrated its predictive potential.
COMMENT:
You are wrong.
It seems you did not understand the words of Dr. Wilfried Nörtershäuse in the abstract of my paper.
The predictive potential of the current nuclear models (when they get good predictions) is concerning the heavy nuclei.
Because the heavy nuclei behave by following statistical tendency (because there is a big quantity of protons and neutrons).
Therefore, for the heavy nuclei, there is no need to apply the true nuclear laws existing in the nuclear structure existing in the nature.
For the light nuclei (as emphasized by Dr. Wilfried Nörtershäuse), the Standard Nuclear Physics fails completely, because (as the quantity of nucleons is small), statistical tendencies cannot be applied.
For light nuclei there is need to apply the true nuclear laws existing in nature. That’s why the Standard Nuclear Physics fails completely for light nuclei.
.
It would only need small, well-thought alterations in order to conform to new experimental realities.
COMMENT:
Again, you did not understand the point. The statistical considerations applied for heavy nuclei cannot be applied for light nuclei.
In heavy nuclei there is preponderance of statistics tendencies (which mask the true nuclear laws).
In light nuclei the true laws cannot be masked by statistical improvements.
.
As far as Nature is concerned, it does not follow a model. Models are abstractions of Nature and therefore LESS than Nature.
COMMENT:
You are wrong.
Such a belief was created by the scientists, because they had failed in their attempt of finding the true PHYSICAL models existing in nature.
Nature works by PHYSICAL models.
Because the nuclear physicists did not succeed in their enterprise, it does not mean that nature does not work with a model.
Such sort of argument is used in general by the scientific community in order to protect their theories.
As they did not succeed to discover the physical models used by nature, the theorists had developed mathematical models, where some physical principles [ as for instance the flux n(o) ] are missing.
In such an attempt, they were obliged to consider incompatible models, in order to explain the phenomena.
That’s why they claim that “Models are abstractions of Nature and therefore LESS than Nature.
Of course mathematical models (where physical are missing) must be abstractions
.
A multitude of models will always exist for this very reason.
COMMENT:
Yes, because in the mathematical abstractions of the nuclear theorists are missing the physical laws existing in the nuclear model of the nature.
Therefore a theoretical model A explains a nuclear phenomenon X, but it is incompatible with another phenomenon Y. Then there is need to conceive other theoretical model B, incompatible with A, so that to explain the phenomenon Y.
In this way, of course the theorists need to claim that “Models are abstractions of Nature and therefore LESS than Nature.
.
There will always be new questions and therefore new explanations needed. And these new explanations might not all fit into one model.
COMMENT:
If the model is wrong (because it is missing some physical laws of nature in the model), of course it cannot explain all the nuclear phenomena.
Then there is need to propose a second wrong model, so that to explain some phenomena not explained by the first model.
But as the second model is wrong, it cannot explain other phenomena.
Then there is need to proposed a third wrong model.
And so on…
By this, by proposing wrong models (which do not work with the true physical laws of the nature), of course There will always be new questions and therefore new explanations needed.
.
While it is true that QRT predicted a non-spherical shape for light nuclei with an even number of protons and neutrons, the total number of options for shape is really just two: spherical and non-spherical.
Mainstream scientists expect more than just a game of chance.
COMMENT:
You are wrong.
The mainstream scientists had believed (along 80 years) that light nuclei with Z=N=pair have spherical shape.
This is not a question of “a game of chance”.
They had wrongly predicted that light nuclei with Z=N=pair have spherical shape BECAUSE IT IS IMPOSSIBLE, (from the current principle of the Standard Nuclear Physics) to explain why they have non-spherical shape.
From the principles of the Standard Nucler Theory the light nuclei with Z=N=pair have to have SPHERICAL SHAPE.
I had predicted correctly that light nuclei with Z=N=pair have non-spherical shape because from the principles of Quantum Ring Theory they CANNOT have spherical shape.
Joe,
think about the following:
1- My nuclear model works by laws missing in the the current nuclear theory
2- Therefore it is very hard for a mainstream scientist to accept my new nuclear model.
3- Then imagine my problem when I had proposed that light nuclei with Z=N=pair have non-spherical shape.
That prediction of mine had been a big argument against my nuclear model. Indeed, any nuclear theorist, looking to my model, could claim:
“What a stupid nuclear model. Any nuclear theorists knows that light nuclei with Z=N=pair have spherical shape, because from the principles of current nuclear physics it’s IMPOSSIBLE that those light nuclei may have non-spherical shape.
When my book had been published in 2006, I had never imagined that new experiments could detect and confirm my prediction.
Therefore, so that to increase the chance of getting my theory to be accepted by the nuclear theorists, it should be better if I had NEVER mentioned that those nuclei have non-spherical shape.
As you can see, it is not a game of chance, as you claim.
The laws of my new nuclear model do not allow to the light nuclei with Z=N=pair to have spherical shape.
.
And the same logic applies to the agglutination of nucleons.
They are either tightly set, or they are not. And it is not a stretch of the imagination to envisage nucleons at slightly greater distances from each other, since the concepts of clustering, shells, and mean free paths were already part of the vocabulary of nuclear scientists for quite some time. And again, this does not mean that it is just a mere coincidence that QRT called it right. But, scientists expect much more, especially in terms of quantitative predictions with great accuracy – and many of them.
.
COMMENT:
Joe,
the facts are the following:
1- Strong nuclear force acts in a maximum distance of 3fm
2- The halo neutron in 4Be11 is a distance of 7fm
So, the strong force cannot keep that neutron.
Of course it more confortable to try to save the current nuclear models, by thinking about desperate solutions, as Wilfried Nörtershäuse and others theorists are trying to do.
rssss
.
Scientists will always be wrong because a model is never Nature itself.
COMMENT:
This is the philosophy proposed by Heisenberg. The mainstream scientists followed his philosophy. And now the Theoretical Physics is living its worst crisis of the whole time.
Heisenberg had believed that some physical models cannot be considered seriously (because they are speculations). For instance, he rejected the helical trajectory.
But suppose that nature uses the helical trajectory.
Do you think that such philosophy proposed by Heisenberg can lead the Theoretical Physics to a good way ? (in the case they continue to keep Heisenberg philosophy and continue rejecting the helical trajectory)
According to Heisenberg philosophy, the flux n(o) is a speculation, not unacceptable from the scientific criterium.
But suppose the flux n(o) exists in the nature, within the nuclei.
Do you think that from Heisenberg philosophy the nuclear theorist can get a satisfactory model, by neglecting the flux n(o)?
.
That is why science is progress. Science only ends when we stop asking ‘why?’ – and not because we have reached the ultimate model.
COMMENT:
Of course such philosophy is valid while the ultimate model is not discovered
rssss
regards
wlad
WARNING:
Today for a wrong move many comments have been erroneously spammed: all our Readers who did not see their comment published are invited to send it again.
JoNP
Dear Barty:
This does not depend on me, but mass production is for sure in preparation.
Warm Regards,
A.R.
Dear Hank Mills:
1- the only source of external energy is the resistance of the Activator
2- confidential
3- no
4- confidential
Warm Regards,
A.R.
Dear John:
No, there are no other energy sources.
Warm Regards,
A.R.
Dear Giovanni Guerrini:
Activator and E-Cat never go at the same time.
Warm Regards,
A.R.
…but,I suppose that when is running the activator (35% of time) the e-cat runs in ssm and the calculation is less simple .
Regards G G