[Vo]:steorn 2009 patent: System and method for measuring energy in magnetic interactions
(gleaned from overunity.com) US 2009/0009157 A1 : SYSTEM AND METHOD FOR MEASURING ENERGY IN MAGNETIC INTERACTIONS : An apparatus and method is provided for measuring magnetic force response time due to the magnetic viscosity of materials and for measuring total energy exchanged due to relative motion of magnetic materials. Voltage and current versus time through an electromagnet is measured and recorded Inventors: Sean David McCarthy, Alan Simpson, Martin Flood, Maxime Sorin and http://sites.google.com/site/steornlab/home
Re: [Vo]:steorn 2009 patent: System and method for measuring energy in magnetic interactions
http://www.google.com/patents?id=1dOyEBAJzoom=4pg=PA1#v=onepageq=f=true On Tue, Dec 29, 2009 at 12:11 PM, Esa Ruoho esaru...@gmail.com wrote: (gleaned from overunity.com) US 2009/0009157 A1 : SYSTEM AND METHOD FOR MEASURING ENERGY IN MAGNETIC INTERACTIONS : An apparatus and method is provided for measuring magnetic force response time due to the magnetic viscosity of materials and for measuring total energy exchanged due to relative motion of magnetic materials. Voltage and current versus time through an electromagnet is measured and recorded Inventors: Sean David McCarthy, Alan Simpson, Martin Flood, Maxime Sorin and http://sites.google.com/site/steornlab/home
Re: [Vo]:steorn 2009 patent: System and method for measuring energy in magnetic interactions
You might note that these are patent applications. The patents are pending. Terry On Tue, Dec 29, 2009 at 5:27 AM, Esa Ruoho esaru...@gmail.com wrote: http://www.google.com/patents?id=1dOyEBAJzoom=4pg=PA1#v=onepageq=f=true
RE: [Vo]:OT: Change Metal Surface to Reflct any Color (Not plating or Paint)
This is great stuff! Amazing. Summary: they are changing the color of metals by changing the surface structure. - Jed
RE: [Vo]:Significant Implications - Kitamura
On Mon, 28 Dec 2009 10:22:14 Jones Beene said [snip] Others here on Vo - have mentioned or debated the fact that the gravity force must grow exponentially at close dimensions - IF - grand unification is accurate. I think it is accurate. Dufour puts some numbers to that hypothesis. He may be onto something.[End Snip] [snip] But here is an irony. We have often asked the rhetorical question: if the Casimir 'force' is essentially negative, then how can it produce a net energy gain? And now, with pico-gravity in the picture, we seems to have a tantalizing clue, in a reversed solution, so to speak. [End Snip] [reply] How about if the gravity is also a relativistic effect? If Naudts is correct then the hydrino can be explained relativistically. The Casimir plates are a negative energy sink reshaping longer vacuum fluctuations to fit between the plates meaning the Casimir cavity represents a different inertial frame. Any matter diffused inside the cavity is redrawn on these reshaped vacuum fluctuations which also modify gravity from our perspective outside the cavity because gravity is defined as distance/time^2. Gamma is changing inside the cavity in the same way as the Twin approaching C see his twin back on earth except the boundary is abrupt and the accumulating dv results from a difference in equivalent accelerations between the ambient gravitational field outside the cavity and reduced field inside. The cavity maintains the zones spatially stationary to each other but the reshaped / restricted vacuum flux open a novel relativistic solution for the dv. We pull away in our ambient inertial frame while the cavity contents fall behind in time proportional to the Casimir force/ plate spacing at their locality, different inertial frames forming a gradient for different spacing until finally reaching the plate boundaries and restoring the normal ambient energy levels for vacuum fluctuations. The gradient represents different levels of deceleration (or negative energy/sink), the energy would be conservative upon exiting the cavity unless we somehow pin the reshaped atoms into their new shape making the vacuum flux do work to restore the atoms on the way out such as forming a compound or molecule. I don't think the world will ever see a steep fractional hydrogen outside of a Casimir cavity / skeletal catalyst and I think this property can be exploited in conjunction with natures preference for diatomic states. The black light plasma could well be decelerated hydrogen oscillating between H1/H2 as it is drug back up to speed exiting the cavity.[end reply] Fran animation - long vacuum fluctuations reshaped instead of displaced http://www.byzipp.com/finished2.swf
Re: [Vo]:Significant Implications - Kitamura
2009/12/28 Jones Beene jone...@pacbell.net: - but the 2 eV available from loading alone without deuterium (contrast that to about .5 eV if the hydrogen were burned in air) is a huge surprise - Jones, where did you get that .5 eV figure? I did the maths and found about 1.5 eV instead, here is the Google calculator result; ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts 294.6 kJ/mol is the energy released per mole of D2O formed (=minus the enthalpy of formation of D2O), which I divided by 2 (2 D per D2O) and by Avogadro's number and then converted to eV to find the burning energy in eV per D atom. Did I get it wrong? Michel
RE: [Vo]:Significant Implications - Kitamura
-Original Message- From: Michel Jullian - but the 2 eV available from loading alone without deuterium (contrast that to about .5 eV if the hydrogen were burned in air) is a huge surprise - MJ: Jones, where did you get that .5 eV figure? I did the maths and found about 1.5 eV instead, here is the Google calculator result; ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts Michel, the half-eV figure is the common 'real world' estimate based on the maximum average temperature of the resultant steam - but even so, it appears you did not first deduct the dissociation energy of O2 and H2 and then later deduct the parasitic losses of NOx, peroxides etc. and the other losses that are expected in actual practice, for combustion in air? IOW there are lies, damn lies, and theoretical calculations ;) when trying to go from 'paper numbers' to actual practice. Kitamura's numbers were indicated to be actual practice (if they can be trusted) so it is fair to contrast those numbers with that which would happen if one were to actually burn H2 in air - and .5 eV is a fair estimate even if you discount the 80% of air which is nearly inert. Since water can be split into H2 and O2 with 1.23 volts - does it stand to reason that one could get 1.5 eV in return ? That was rhetorical; and of course this one of nature's built-in cases of systemic overunity - ... except for the damn lie that it simply does not work out that way in practice - but it does serve to contrast the large disparity of the actual with the calculated. Did I get it wrong? Well, let's say that you got it partly right and mostly wrong - if your intent was to suggest that hydrogen can be burned in air with resultant steam being formed at about 17,000 degrees K. Jones
Re: [Vo]:Significant Implications - Kitamura
On 12/29/2009 11:19 AM, Jones Beene wrote: -Original Message- From: Michel Jullian - but the 2 eV available from loading alone without deuterium (contrast that to about .5 eV if the hydrogen were burned in air) is a huge surprise - MJ: Jones, where did you get that .5 eV figure? I did the maths and found about 1.5 eV instead, here is the Google calculator result; ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts Michel, the half-eV figure is the common 'real world' estimate based on the maximum average temperature of the resultant steam Isn't combustion of hydrogen in air rather different from the situation we've got here? - but even so, it appears you did not first deduct the dissociation energy of O2 and H2 and then later deduct the parasitic losses of NOx, peroxides etc. and the other losses that are expected in actual practice, for combustion in air? Parasitic losses, in particular, would not seem to apply in the present case. IOW there are lies, damn lies, and theoretical calculations ;) when trying to go from 'paper numbers' to actual practice. Kitamura's numbers were indicated to be actual practice (if they can be trusted) so it is fair to contrast those numbers with that which would happen if one were to actually burn H2 in air - and .5 eV is a fair estimate even if you discount the 80% of air which is nearly inert. Since water can be split into H2 and O2 with 1.23 volts - does it stand to reason that one could get 1.5 eV in return ? That was rhetorical; and of course this one of nature's built-in cases of systemic overunity - Now you're neglecting the splitting cost of H2-2H and O2-2H. ... except for the damn lie that it simply does not work out that way in practice - but it does serve to contrast the large disparity of the actual with the calculated. Did I get it wrong? Well, let's say that you got it partly right and mostly wrong - if your intent was to suggest that hydrogen can be burned in air with resultant steam being formed at about 17,000 degrees K. Jones
Re: [Vo]:Significant Implications - Kitamura
2009/12/29 Jones Beene jone...@pacbell.net: -Original Message- From: Michel Jullian - but the 2 eV available from loading alone without deuterium (contrast that to about .5 eV if the hydrogen were burned in air) is a huge surprise - MJ: Jones, where did you get that .5 eV figure? I did the maths and found about 1.5 eV instead, here is the Google calculator result; ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts Michel, the half-eV figure is the common 'real world' estimate based on the maximum average temperature of the resultant steam - but even so, it appears you did not first deduct the dissociation energy of O2 and H2 Their formation enthalpy is zero, by convention and then later deduct the parasitic losses of NOx, peroxides etc. and the other losses that are expected in actual practice, for combustion in air? Negligible IOW there are lies, damn lies, and theoretical calculations ;) when trying to go from 'paper numbers' to actual practice. Kitamura's numbers were indicated to be actual practice (if they can be trusted) so it is fair to contrast those numbers with that which would happen if one were to actually burn H2 in air - and .5 eV is a fair estimate No (see below) even if you discount the 80% of air which is nearly inert. why would you not discount them??? Since water can be split into H2 and O2 with 1.23 volts - does it stand to reason that one could get 1.5 eV in return ? That was rhetorical; and of course this one of nature's built-in cases of systemic overunity - This was not rhetorical at all actually, I hadn't made the connexion but yes, the combustion energy per D atom in eV should be, of course, exactly equal to the thermoneutral electrolysis voltage... and it is, as a matter of fact: the thermoneutral voltage for electrolysis of D2O is 1.54V, which confirms my 1.53V calculation. And BTW, it's 1.48V for H2O, not 1.23V. ... except for the damn lie that it simply does not work out that way in practice - but it does serve to contrast the large disparity of the actual with the calculated. Did I get it wrong? Well, let's say that you got it partly right and mostly wrong Or rather, as it turns out, exactly right. Physics works, contrary to your suggestions :) Besides, you don't have to take my word, see http://en.wikipedia.org/wiki/Heat_of_combustion Hydrogen: 140 kJ/g, which is about 1.5eV per atom. The important result here is that the 2 eV you get by letting an hydrogen atom bond to the _surface_ of a Pd nanoparticle are comparable with the chemical energy you get by letting it bond to an oxygen atom (starting from molecular gas phase in both cases) Michel
[Vo]:Personal:Little help with UFO sighting?
On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote: Saw an orange fire colored UFO last night just after nightfall. The path was that of an object flying in a curved path at high altitude (a u-turn, basically), definitely not a satellite, and a bit brighter than a good space station sighting. Even through 8x binoculars it appeared as a point source. The time of day, the sighting angle, and the variable characteristics of the light over the two minutes or so the object was visible suggests it was reflected light from the setting sun on the lower surface of a solid object, but given that the sky was almost completely dark at my location at that point, my guess is that its altitude could have been above the atmosphere. I’ve seen conventional aircraft reflecting sunset’s light after local sunset, and though the appearance was similar, in those cases it was much closer to sunset and sky was still quite light. Once it gets dark, I think such reflections tend to be in satellite territory. Anyway, since I can find the sight angle because of its passage near identifiable stars, the time, and my location on that date, it should be possible to calculate the earth’s shadow line from the setting sun and see where my sight line crosses it. That would tell me if it was just an airplane at very high altitude, or something maneuvering up a bit higher than conventional aircraft can reach. Anybody have an idea how I would go about that (umbra?) line? Thanks, - Rick Hi Rick, Coincidentally, I saw something similar yesterday (Dec 28, 2009) around noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW. It was one small finger width at arms length above the horizon. It had a periodic (about 10 second) flash to it, so I assumed it might be a booster, but strange it was heading SW, not SE or NE, or just S. Of course a U-turn is not a typical satellite maneuver, nor did I see that! The altitude h to the directly overhead sun midline is given by: h = r_earth * ( SQRT(1 + sin^2 theta) -1) Given time after sunset t we have: theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers: t (min) theta (radians) h (miles) 1 0.00436331944 0.03760073165 5 0.02181659722 0.93976780755 10 0.04363319444 3.75594358 20 0.08726638889 14.973936498 30 0.13089958333 33.506081478 60 0.26179916667 130.1553394 90 0.39269875 279.3533269 The above only provides mid-line information. The angular radius of the sun is 31.65 minutes, so the total sunset line follows the above by about 31 minutes and 39 seconds. Even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time passed from sunset to rule out an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Personal:Little help with UFO sighting?
On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote: Saw an orange fire colored UFO last night just after nightfall. The path was that of an object flying in a curved path at high altitude (a u-turn, basically), definitely not a satellite, and a bit brighter than a good space station sighting. Even through 8x binoculars it appeared as a point source. The time of day, the sighting angle, and the variable characteristics of the light over the two minutes or so the object was visible suggests it was reflected light from the setting sun on the lower surface of a solid object, but given that the sky was almost completely dark at my location at that point, my guess is that its altitude could have been above the atmosphere. I’ve seen conventional aircraft reflecting sunset’s light after local sunset, and though the appearance was similar, in those cases it was much closer to sunset and sky was still quite light. Once it gets dark, I think such reflections tend to be in satellite territory. Anyway, since I can find the sight angle because of its passage near identifiable stars, the time, and my location on that date, it should be possible to calculate the earth’s shadow line from the setting sun and see where my sight line crosses it. That would tell me if it was just an airplane at very high altitude, or something maneuvering up a bit higher than conventional aircraft can reach. Anybody have an idea how I would go about that (umbra?) line? Thanks, - Rick Hi Rick, Coincidentally, I saw something similar yesterday (Dec 28, 2009) around noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW. It was one small finger width at arms length above the horizon. It had a periodic (about 10 second) flash to it, so I assumed it might be a booster, but strange it was heading SW, not SE or NE, or just S. Of course a U-turn is not a typical satellite maneuver, nor did I see that! The altitude h to the directly overhead sun midline is given by: h = r_earth * ( SQRT(1 + sin^2 theta) -1) Given time after sunset t we have: theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers: t (min) theta (radians) h (miles) 1 0.00436331944 0.03760073165 5 0.02181659722 0.93976780755 10 0.04363319444 3.75594358 20 0.08726638889 14.973936498 30 0.13089958333 33.506081478 60 0.26179916667 130.1553394 90 0.39269875 279.3533269 The above only provides mid-line information. The angular radius of the sun is 31.65 minutes, so the total sunset line follows the above by about 15 minutes and 50 seconds. Even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time passed from sunset to rule out an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Significant Implications - Kitamura
2009/12/29 Stephen A. Lawrence sa...@pobox.com: On 12/29/2009 11:19 AM, Jones Beene wrote: Since water can be split into H2 and O2 with 1.23 volts - does it stand to reason that one could get 1.5 eV in return ? That was rhetorical; and of course this one of nature's built-in cases of systemic overunity - Now you're neglecting the splitting cost of H2-2H and O2-2H. No he isn't, that's comprised in the price (if you use the correct value of 1.48V that is). What's the energy needed to go from water to the gases? 1.48V, times the charge of the transferred electrons (1 electron per hydrogen atom). Of course, you get the same energy when going the other way. Michel
[Vo]:Personal:Little help with UFO sighting?
On Dec 28, 2009, at 2:09 PM, Rick Monteverde wrote: Saw an orange fire colored UFO last night just after nightfall. The path was that of an object flying in a curved path at high altitude (a u-turn, basically), definitely not a satellite, and a bit brighter than a good space station sighting. Even through 8x binoculars it appeared as a point source. The time of day, the sighting angle, and the variable characteristics of the light over the two minutes or so the object was visible suggests it was reflected light from the setting sun on the lower surface of a solid object, but given that the sky was almost completely dark at my location at that point, my guess is that its altitude could have been above the atmosphere. I’ve seen conventional aircraft reflecting sunset’s light after local sunset, and though the appearance was similar, in those cases it was much closer to sunset and sky was still quite light. Once it gets dark, I think such reflections tend to be in satellite territory. Anyway, since I can find the sight angle because of its passage near identifiable stars, the time, and my location on that date, it should be possible to calculate the earth’s shadow line from the setting sun and see where my sight line crosses it. That would tell me if it was just an airplane at very high altitude, or something maneuvering up a bit higher than conventional aircraft can reach. Anybody have an idea how I would go about that (umbra?) line? Thanks, - Rick Hi Rick, Coincidentally, I saw something similar yesterday (Dec 28, 2009) around noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW. It was one small finger width at arms length above the horizon. It had a periodic (about 10 second) flash to it, so I assumed it might be a booster, but strange it was heading SW, not SE or NE, or just S. Of course a U-turn is not a typical satellite maneuver, nor did I see that! The altitude h to the directly overhead sun midline is given by: h = r_earth * ( SQRT(1 + sin^2 theta) -1) Given time after sunset t we have: theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers: t (min) theta (radians) h (miles) 1 0.00436331944 0.03760073165 5 0.02181659722 0.93976780755 10 0.04363319444 3.75594358 20 0.08726638889 14.973936498 30 0.13089958333 33.506081478 60 0.26179916667 130.1553394 90 0.39269875 279.3533269 Since the above is time after total sunset, you don't have to correct for the angular width of the sun. However, even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time, i.e. shortly after sunset, passed to rule out an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Personal:Little help with UFO sighting?
In case there is any doubt, the following is my final answer - unless of course I find other mistakes! 8^) On Dec 29, 2009, at 8:44 AM, Horace Heffner wrote: Hi Rick, Coincidentally, I saw something similar yesterday (Dec 28, 2009) around noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW. It was one small finger width at arms length above the horizon. It had a periodic (about 10 second) flash to it, so I assumed it might be a booster, but strange it was heading SW, not SE or NE, or just S. Of course a U-turn is not a typical satellite maneuver, nor did I see that! The altitude h to the directly overhead sun midline is given by: h = r_earth * ( SQRT(1 + sin^2 theta) -1) Given time after sunset t we have: theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers: t (min) theta (radians) h (miles) 1 0.00436331944 0.03760073165 5 0.02181659722 0.93976780755 10 0.04363319444 3.75594358 20 0.08726638889 14.973936498 30 0.13089958333 33.506081478 60 0.26179916667 130.1553394 90 0.39269875 279.3533269 Since the above is time after total sunset, you don't have to correct for the angular width of the sun. However, even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time, i.e. shortly after sunset, passed to rule out an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:Significant Implications - Kitamura
-Original Message- From: Michel Jullian [ Or rather, as it turns out, exactly right. Physics works, contrary to your suggestions :) It works of course, but not as perfectly as you suggest, in real world applications. Besides, you don't have to take my word, see http://en.wikipedia.org/wiki/Heat_of_combustion Hydrogen: 140 kJ/g, which is about 1.5eV per atom. Yes, but once again your reference is NOT to burning hydrogen in air. At the very top of the page you site, it clearly says The heat of combustion is the energy released as heat when one mole of a compound undergoes complete combustion with oxygen Burning H2 in air is not complete combustion with oxygen and in fact H2 can be leaned-out sufficiently in air so that it will not burn at all. Contrary to what you state, parasitic loses cannot be ignored - unless you are merely trying to prove a pedantic point, which seems to be the case. The important result here is that the 2 eV you get by letting an hydrogen atom bond to the _surface_ of a Pd nanoparticle are comparable with the chemical energy you get by letting it bond to an oxygen atom (starting from molecular gas phase in both cases) NO! Absolutely not a relevant comparison, nor an accurate value. Otherwise metal hydrides could not be used for hydrogen storage, and palladium could not be used as a filter to separate H2 from other gases, both of which applications are common. Imagine having to apply 2 eV of thermal energy to a metal hydride in order to release the stored hydrogen gas for use in an engine. That is absurd. Jones
[Vo]:Personal:Little help with UFO sighting?
For background information, the attached is a graphic showing my derivation of the formula used below. The umbra line, shown as horizontal, and the line segment x directly above the viewer is roughly the same as the altitude to an object on the umbra line, depending on the latitudinal distance from the viewer. The altitude h to the directly overhead sun midline, with r factored out, is given by: h = r_earth * ( SQRT(1 + sin^2 theta) -1) Given time after sunset t we have: theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers: t (min) theta (radians) h (miles) 1 0.00436331944 0.03760073165 5 0.02181659722 0.93976780755 10 0.04363319444 3.75594358 20 0.08726638889 14.973936498 30 0.13089958333 33.506081478 60 0.26179916667 130.1553394 90 0.39269875 279.3533269 Since the above is time after total sunset, you don't have to correct for the angular width of the sun. However, even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time, i.e. shortly after sunset, passed to rule out an airplane. inline: Umbra.jpg Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:Personal:Little help with UFO sighting?
Horace - My sighting wasn't just after sunset, it was just after nightfall - total darkness. There was just a vague hint of fading light on the horizon, but the sky surrounding the object, which was relatively low in the southwest, was already black. I did find something on the after-sunset atmospheric distortion - they say add 6 arc minutes to the apparent semidiameter of the sun. I'll try to muddle through your figures in a little while. I sure appreciate the help, thanks. - R. From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Tuesday, December 29, 2009 8:08 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Personal:Little help with UFO sighting? In case there is any doubt, the following is my final answer - unless of course I find other mistakes! 8^) On Dec 29, 2009, at 8:44 AM, Horace Heffner wrote: Hi Rick, Coincidentally, I saw something similar yesterday (Dec 28, 2009) around noon AKST, (about 11 orbits later) west of Palmer AK, but heading SW. It was one small finger width at arms length above the horizon. It had a periodic (about 10 second) flash to it, so I assumed it might be a booster, but strange it was heading SW, not SE or NE, or just S. Of course a U-turn is not a typical satellite maneuver, nor did I see that! The altitude h to the directly overhead sun midline is given by: h = r_earth * ( SQRT(1 + sin^2 theta) -1) Given time after sunset t we have: theta = (t/(8.64x10^4 s))*(2*Pi) radians = (t/(1440 min))*(2*Pi) radians Earth radius, r_earth, at Hawaii is about 3951 mi. Here are some numbers: t (min) theta (radians) h (miles) 1 0.00436331944 0.03760073165 5 0.02181659722 0.93976780755 100.04363319444 3.75594358 200.08726638889 14.973936498 300.13089958333 33.506081478 600.26179916667 130.1553394 900.39269875 279.3533269 Since the above is time after total sunset, you don't have to correct for the angular width of the sun. However, even total sunset is not good enough to black out an object though, due to light diffraction. Clearly not enough time, i.e. shortly after sunset, passed to rule out an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:Personal:Little help with UFO sighting?
Sunset at my location in Honolulu that day was 5:59 PM, sighting was at 6:58 PM. I'll look for a star chart to get the sighting angle for the object. - R.
Re: [Vo]:OT: Change Metal Surface to Reflct any Color (Not plating or Paint)
Heat from welding can tint steel too, but I guess with lasers the effect can be controlled. This topic reminds of Chromoskedasic printing, where blackwhite photographic paper is chemically treated so that it will scatter light and generate abstract/psychedelic colour effects. http://chemigramist.com/chromoskedasic.html example http://www.loreal.com/_en/_ww/loreal-art-science/bronzefigure-2005.aspx Harry - Original Message From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com Sent: Tue, December 29, 2009 9:24:51 AM Subject: RE: [Vo]:OT: Change Metal Surface to Reflct any Color (Not plating or Paint) This is great stuff! Amazing. Summary: they are changing the color of metals by changing the surface structure. - Jed __ Get a sneak peak at messages with a handy reading pane with All new Yahoo! Mail: http://ca.promos.yahoo.com/newmail/overview2/
Re: [Vo]:Personal:Little help with UFO sighting?
On Dec 29, 2009, at 10:02 AM, Rick Monteverde wrote: Horace – My sighting wasn’t just after sunset, it was just after nightfall - total darkness. There was just a vague hint of fading light on the horizon, but the sky surrounding the object, which was relatively low in the southwest, was already black. I did find something on the after-sunset atmospheric distortion – they say add 6 arc minutes to the apparent semidiameter of the sun. I’ll try to muddle through your figures in a little while. I sure appreciate the help, thanks. Yes, that accounts for distortion of the image. There is also atmospheric scatter involved, and that might make an object look orange or red. It makes snowy mountains here take on hot pastel colors - called Alpenglow. This time of year the sun sets almost due west there. If the object was ever in a direction approximately due north or south of you, i.e. on a line perpendicular to the sunset location, then the altitude h I provided fairly closely applies to the object for that time t in the table. If it was mainly east or west then another calculation is needed. I would say anything above 100,000 feet, or 18.9 miles, was probably not a military jet, and certainly not a passenger jet. That altitude h corresponds to about 22 minutes after surface darkness - to whatever degree such darkness needs to be defined. From experience there, I know it gets dark pretty fast in Hawaii after sunset - especially compared to here - where sunsets can take a very long time. 8^) If you observed the object an hour after sunset then I'd say it was well past the 22 minutes after darkness mark. A general compass direction thus may be sufficient information for a definitive answer. That far after sunset, an hour, taken even alone, is a pretty strong indication it was not an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:Significant Implications - Kitamura
OK, vorticians. This is could be an important paper and topic, so let me add one more point of clarification to Michel Jullian's point about the heat of combustion of hydrogen, compared to the anomalous loading heat of Kitamura's claim. Michel correctly finds that if you only look at one-half of the reaction, and ignore the mass of the end product, then what we have is: (294.6 / 2) / 6.02e23) * kJ = ~1.5 electron volts/amu based on hydrogen This is the energy released relative to initial hydrogen mass, but that might assume that oxygen is unnecessary, if you leave it out. One should take the mass of O2 into consideration for the comparison with reversible hydride loading. ERGO. It would have been clearer back a few posts ago - if I had broken the comparison down this way. The steam from hydrogen combustion will have a molecular wt of 18 amu per hot molecule. The heat of combustion of the two hydrogen atoms is ~3+ eV in total. The resultant energy per amu of the steam, therefore, is 3/18 or .16 eV per amu of combustion end product. When we compare that energy per mass of combustion product - with the Kitamura reaction of hydrogen which has been reversibly loaded into a metal matrix, and then released, then we find that the amu of the end product is still about one since there is/was no permanent bond. The thermal energy released, according to Kitamura is ~2 eV, so the eV per amu is about a *ten to one ratio,* when the energy of the hydride bond is deducted - compared to hydrogen combustion (by mass of all non-renewable reactants). Next big issue. What is the real hydride bond energy for Pd? There is a chart here (Fig 3): http://www.iop.org/EJ/article/1742-6596/79/1/012028/jpconf7_79_012028.pdf?re quest-id=e4195775-a6d5-4d5f-83b9-da98912aa8c1 It appears that the bond energy for Pd varies between .9 eV and a negative value, depending of a number of variables. The bond is field influenced, which could be important. From the chart - an average value appears to be less than .5 eV. However, the indication is that it could be much lower. Therefore, if Kitamura were correct on the heat energy (which I am beginning to doubt), then this kind of iterative recycling of hydrogen would be a window of opportunity for gainfulness, since the spread is very large. This is too simple and robust to be real, no? This looks like a COP of close to three. For an accurate cross-comparison based on all reactants - it is fair to say that we are looking an initial gain of almost ten to one over combustion; moreover it is an infinite gain if based on the renewability of the hydrogen, that is: if the COP~3 allows that to happen, after the conversion losses of heat back into electricity. Before we can arrive at an accurate final appraisal for the usefulness of the process, we must consider the net energy necessary to release the hydrogen from the matrix. If that were to be .5 eV as the IOP paper suggests (or less with an electric field) - then there is a huge potential for net gain from recycling the hydrogen. IF of course, Kitamura got the 2 eV thermal number correct. Doubts remain on that issue. The big if. Jones
[Vo]:Horrace help
How do you calculate the phonon frequency in the solid. I get 10 exp 17 hertz..the number should be about 10 exp 12 hertz. Is there an easy way. Frank Z
Re: [Vo]:JL-naudin replicates current Steorn Orbo (Dec) demo
At 06:00 AM 12/28/2009, William Beaty wrote: Rather than focusing on some perhaps-unexpected measurement, just close the loop. Ditch the battery. Make a perpetual wheel. Close the loop. If it's real, then closing the loop should be easy. If it's an artifact which misleads FE-enthusiasts, then closing the loop will be impossible. But there is the tantalizing middle. They find that they almost close the loop. So they think that they are actually over unity, but with losses that maybe with better engineering they can fix. All it takes is more money. But this is the real and present tipoff: their development of extremely low-friction bearings. That is an abandonment of over-unity and indicates a desire to become ever more and more sensitive, allowing more spectacular demonstrations where a tiny effect is accumulated. But given so much energy being dumped into heat, in the end, it only takes a tiny, tiny fraction of that to be coupled into rotor motion instead, very difficult to detect, if you have a very low-friction rotor which won't lose heat there. So much, though, for actually generating power, which will immediately dump much rotor energy into heat again. Calorimetry would show the overall problem, but, of course, doing really accurate calorimetry is difficult. Much easier to make a roter spin fast and claim that the energy for that is free, that the battery is only generating heat, that none of this cycling of the magnetic field is accelerating the rotor. Though it obviously is. They claim there is no energy going there, but that hasn't actually been shown except by a gross and coarse display that would completely miss the tiny amount of energy expenditure necessary to make that rotor accumulate angular momentum.
[Vo]:Low-Energy Nuclear Reactions Sourcebook Volume 2 abstracts
The abstracts for each chapter are available here: http://pubs.acs.org/isbn/9780841224544 - Jed
Re: [Vo]:Personal:Little help with UFO sighting?
Sounds like a Fastwalker. Terry On Tue, Dec 29, 2009 at 3:57 PM, Horace Heffner hheff...@mtaonline.net wrote: On Dec 29, 2009, at 10:02 AM, Rick Monteverde wrote: Horace – My sighting wasn’t just after sunset, it was just after nightfall - total darkness. There was just a vague hint of fading light on the horizon, but the sky surrounding the object, which was relatively low in the southwest, was already black. I did find something on the after-sunset atmospheric distortion – they say add 6 arc minutes to the apparent semidiameter of the sun. I’ll try to muddle through your figures in a little while. I sure appreciate the help, thanks. Yes, that accounts for distortion of the image. There is also atmospheric scatter involved, and that might make an object look orange or red. It makes snowy mountains here take on hot pastel colors - called Alpenglow. This time of year the sun sets almost due west there. If the object was ever in a direction approximately due north or south of you, i.e. on a line perpendicular to the sunset location, then the altitude h I provided fairly closely applies to the object for that time t in the table. If it was mainly east or west then another calculation is needed. I would say anything above 100,000 feet, or 18.9 miles, was probably not a military jet, and certainly not a passenger jet. That altitude h corresponds to about 22 minutes after surface darkness - to whatever degree such darkness needs to be defined. From experience there, I know it gets dark pretty fast in Hawaii after sunset - especially compared to here - where sunsets can take a very long time. 8^) If you observed the object an hour after sunset then I'd say it was well past the 22 minutes after darkness mark. A general compass direction thus may be sufficient information for a definitive answer. That far after sunset, an hour, taken even alone, is a pretty strong indication it was not an airplane. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Horrace help
On Dec 29, 2009, at 12:08 PM, fznidar...@aol.com wrote: How do you calculate the phonon frequency in the solid. I get 10 exp 17 hertz..the number should be about 10 exp 12 hertz. Is there an easy way. Frank Z Yes, from the very little information provided above I suspect there might be. I suspect you are using c as the speed of light instead of the speed of sound in the medium. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Significant Implications - Kitamura
So, how does this compare to the recombination energy of atomic hydrogen? Here's a reference by a dubious source: http://www.cheniere.org/misc/a_h%20reaction.htm :-) Terry On Tue, Dec 29, 2009 at 4:04 PM, Jones Beene jone...@pacbell.net wrote: OK, vorticians. This is could be an important paper and topic, so let me add one more point of clarification to Michel Jullian's point about the heat of combustion of hydrogen, compared to the anomalous loading heat of Kitamura's claim. Michel correctly finds that if you only look at one-half of the reaction, and ignore the mass of the end product, then what we have is: (294.6 / 2) / 6.02e23) * kJ = ~1.5 electron volts/amu based on hydrogen This is the energy released relative to initial hydrogen mass, but that might assume that oxygen is unnecessary, if you leave it out. One should take the mass of O2 into consideration for the comparison with reversible hydride loading. ERGO. It would have been clearer back a few posts ago - if I had broken the comparison down this way. The steam from hydrogen combustion will have a molecular wt of 18 amu per hot molecule. The heat of combustion of the two hydrogen atoms is ~3+ eV in total. The resultant energy per amu of the steam, therefore, is 3/18 or .16 eV per amu of combustion end product. When we compare that energy per mass of combustion product - with the Kitamura reaction of hydrogen which has been reversibly loaded into a metal matrix, and then released, then we find that the amu of the end product is still about one since there is/was no permanent bond. The thermal energy released, according to Kitamura is ~2 eV, so the eV per amu is about a *ten to one ratio,* when the energy of the hydride bond is deducted - compared to hydrogen combustion (by mass of all non-renewable reactants). Next big issue. What is the real hydride bond energy for Pd? There is a chart here (Fig 3): http://www.iop.org/EJ/article/1742-6596/79/1/012028/jpconf7_79_012028.pdf?re quest-id=e4195775-a6d5-4d5f-83b9-da98912aa8c1 It appears that the bond energy for Pd varies between .9 eV and a negative value, depending of a number of variables. The bond is field influenced, which could be important. From the chart - an average value appears to be less than .5 eV. However, the indication is that it could be much lower. Therefore, if Kitamura were correct on the heat energy (which I am beginning to doubt), then this kind of iterative recycling of hydrogen would be a window of opportunity for gainfulness, since the spread is very large. This is too simple and robust to be real, no? This looks like a COP of close to three. For an accurate cross-comparison based on all reactants - it is fair to say that we are looking an initial gain of almost ten to one over combustion; moreover it is an infinite gain if based on the renewability of the hydrogen, that is: if the COP~3 allows that to happen, after the conversion losses of heat back into electricity. Before we can arrive at an accurate final appraisal for the usefulness of the process, we must consider the net energy necessary to release the hydrogen from the matrix. If that were to be .5 eV as the IOP paper suggests (or less with an electric field) - then there is a huge potential for net gain from recycling the hydrogen. IF of course, Kitamura got the 2 eV thermal number correct. Doubts remain on that issue. The big if. Jones
Re: [Vo]:JL-naudin replicates current Steorn Orbo (Dec) demo
On Tue, 29 Dec 2009, Abd ul-Rahman Lomax wrote: But there is the tantalizing middle. They find that they almost close the loop. You're giving them the benefit of the doubt. Count how many times you have to do that! It's very telling. Their acting very Newman-esque and using a battery? Rather than using a five dollar supercapacitor? They're either insane, or they're scammers. So they think that they are actually over unity, but with losses that maybe with better engineering they can fix. All it takes is more money. If they're insane, then they'll talk themselves into using a battery and never actually try a supercap, even in private. They'll have all sorts of important reasons why they cannot ever try a supercap. Oh, and by insane, I mean the same as fooling themselves. There is a threshold past which the self-fooing becomes a complete break with reality. Were you here when Doctor Stiffler was presenting his LED overunity device? One of his odd behaviors was, rather than just sitting down and honestly demonstrating his claims, and always sticking with straight un-twisted discussions, he claimed to be making youtube postings to mess with the heads of skeptics. In that case, nobody knew which of his videos were hoaxes intended to mislead skeptics, and which were honest experiments. Steorn mentioned doing something similar. But this is the real and present tipoff: their development of extremely low-friction bearings. That is an abandonment of over-unity and indicates a desire to become ever more and more sensitive, allowing more spectacular demonstrations where a tiny effect is accumulated. Definitely! That's the Newman fallacy: pretending that a whirling massive flywheel represents a huge energy output. With low-friction bearings, you can spin a fairly large wheel for months using just a few 10s of cc of battery volume. That's how the fake PM machine sculpture built by David Jones of Nature journal accomplished its feat. (I replaced those hidden batteries myself more than once over the years.) Though it obviously is. They claim there is no energy going there, but that hasn't actually been shown except by a gross and coarse display that would completely miss the tiny amount of energy expenditure necessary to make that rotor accumulate angular momentum. Why not just use a supercap and remove the whole battery problem? Watch the capacitor voltage rise slowly as excess energy comes from nowhere? Stick a Zener across it to keep it from overvoltage. There's really no sensible excuse for their bizarre setup, unless it's obfuscation. Their setup looks sensible unless one realizes what the lack of a supercap implies about their collective mental state. (( ( ( ( ((O)) ) ) ) ))) William J. BeatySCIENCE HOBBYIST website billb at amasci com http://amasci.com EE/programmer/sci-exhibits amateur science, hobby projects, sci fair Seattle, WA 206-762-3818unusual phenomena, tesla coils, weird sci
Re: [Vo]:JL-naudin replicates current Steorn Orbo (Dec) demo
At 08:58 PM 12/29/2009, William Beaty wrote: On Tue, 29 Dec 2009, Abd ul-Rahman Lomax wrote: But there is the tantalizing middle. They find that they almost close the loop. You're giving them the benefit of the doubt. Count how many times you have to do that! It's very telling. Their acting very Newman-esque and using a battery? Rather than using a five dollar supercapacitor? They're either insane, or they're scammers. I've already concluded that they are a variation on the latter. But there is a *possibility*, I'm pointing out, that they are sincere. Still unethical, but that they believe they just need to get some more money to fix this or that, and that this justifies withholding the critical information. The critical information is *why* they believe they have overunity, or, in fact, why they believe that they have any evidence at all of excess energy. What they showed us, quite simply, didn't reveal that. Come on, they looked at some oscilloscope traces and they looked okay? The amount of energy that would need to be dumped to rotation would be quite small compared to the heat, as if the toroids were resistive loads. But, as I recall, I saw some ringing. So they think that they are actually over unity, but with losses that maybe with better engineering they can fix. All it takes is more money. If they're insane, then they'll talk themselves into using a battery and never actually try a supercap, even in private. They'll have all sorts of important reasons why they cannot ever try a supercap. Oh, and by insane, I mean the same as fooling themselves. There is a threshold past which the self-fooing becomes a complete break with reality. I don't think they are simply fooling themselves, I think they got led into a situation where they needed to fool others. Do they know that the whole thing is bogus? How could they *know* that? They'd have to do much more careful work, and they are too busy marketing what they have: a concept, not engineering to *actually work*, just an idea that there is some anomaly here, and they want to see you the anomaly. You can figure out how to use it, not their business, they are in the business of selling you the idea and some of the equipment you'd use to test it. That way, they make money whether there is anything real here or not. Quite a business concept, actually. I'm even doing something a *little* like it, except that I'm fully disclosing everything. I don't have any supersecret idea, I'm trying to sell kits to replicate a SPAWAR experiment. In theory, I could make money even if SPAWAR is bogus, though it would be more difficult. I could sell you the kits to show that it doesn't work. (But the problem is, how would I know that my kit wasn't missing some critical feature, some parameter that I varied, perhaps without realizing it?) I can say this: if I can't get the kits to work, i.e., to show radiation evidence, I might still sell them, but with that disclosure and all the associated caveats. Maybe somebody else could figure out the missing link. Quite simply, I have a few thousand dollars in this, and I could get most of it back by selling my stuff for other applications. I have no intention of putting myself in a position where I'd have to lie or deceive in order to escape with my shirt on. I'd rather eke it out on social security, I'd sleep better.) Were you here when Doctor Stiffler was presenting his LED overunity device? One of his odd behaviors was, rather than just sitting down and honestly demonstrating his claims, and always sticking with straight un-twisted discussions, he claimed to be making youtube postings to mess with the heads of skeptics. Steorn made a claim like that about one of their prior announcements. It was to lead the Men in Black astray. In that case, nobody knew which of his videos were hoaxes intended to mislead skeptics, and which were honest experiments. Steorn mentioned doing something similar. You noticed. But this is the real and present tipoff: their development of extremely low-friction bearings. That is an abandonment of over-unity and indicates a desire to become ever more and more sensitive, allowing more spectacular demonstrations where a tiny effect is accumulated. Definitely! That's the Newman fallacy: pretending that a whirling massive flywheel represents a huge energy output. With low-friction bearings, you can spin a fairly large wheel for months using just a few 10s of cc of battery volume. That's how the fake PM machine sculpture built by David Jones of Nature journal accomplished its feat. (I replaced those hidden batteries myself more than once over the years.) Yeah. Classic. I've been reading Park's Voodoo Science. He makes, of course, some crucial errors, he fails to understand and apply his own advice. But he's also right about some stuff. Some of the scams he reports on were truly cheeky. And he seems to
Re: [Vo]:Horrace help
In a message dated 12/29/2009 7:33:04 PM Eastern Standard Time, hheff...@mtaonline.net writes: Yes, from the very little information provided above I suspect there might be. I suspect you are using c as the speed of light instead of the speed of sound in the medium. Best regards, Horace Heffner _http://www.mtaonline.net/~hheffner/_ (http://www.mtaonline.net/~hheffner/) No. I tried to get an estimate of phonon frequency from the resonant freq formula. It is the sq root of K/M times 1/rwo pie For M I used 2* 1836 *9.1 *10 exp -31 KG For K I used 1.45 exp 11 Newtons / meter What is wrong? Frank Z
Re: [Vo]:Horrace help
On Dec 29, 2009, at 6:42 PM, fznidar...@aol.com wrote: In a message dated 12/29/2009 7:33:04 PM Eastern Standard Time, hheff...@mtaonline.net writes: Yes, from the very little information provided above I suspect there might be. I suspect you are using c as the speed of light instead of the speed of sound in the medium. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ No. I tried to get an estimate of phonon frequency from the resonant freq formula. It is the sq root of K/M times 1/rwo pie By the above I take it you mean frequency f is given by: f = (1/(2*Pi)) (k/M)^0.5 which is the resonant frequency of a mechanical system. For M I used 2* 1836 *9.1 *10 exp -31 KG The mass of the deuteron is 3.3444941 x 10^-27 kg. For K I used 1.45 exp 11 Newtons / meter What is wrong? Frank Z The above k is the spring constant, given by Hooke's law: F = -k/x where F is the displacement force and x is the displacement. It assumes a linear relationship, which does not exist with respect to Coulomb's law. But, let's ignore that for now. Palladium has a room temperature lattice constant of about 0.389 nm, or 3.89x10^-10 m. Let the deuteron move 1/100 of that distance from the zero force location, or let x = 3.89x10^-12 m. Assume the deuteron is vibrating at a displacement +- x from the neutral force point. From Coulomb's law: F1 = -C_k * q^2/(r1)^2 = -C_k * q^2/(3.89x10^-10 m - 3.89x10^-12 m )^2 F2 = +C_k * q^2/(r1)^2 = -C_k * q^2/(3.89x10^-10 m + 3.89x10^-12 m )^2 where C_k * q^2 = 2.30708 J/m, and: F1 = -1.8x10^-9 N F2 = 1.49459x10^-9 N F = F1 + F2 = -6.099724x10^-11 N k = -F/x = (6.099724x10^-11 N)/(3.89x10^-12 m) = 1.568 N/m and we obtain the frequency by: f = (1/(2*Pi)) (k/M)^0.5 = (1/(2*Pi)) ((1.568 N/m)/(3.3444941 x 10^-27 kg))^0.5 f = 3.459 x 10^12 Hz Now that is resolved, there is another problem. The deuterons actually are screened from each other by lattice electrons. They exist in lattice potential wells. They essentially move at lattice speeds due to lattice vibrations, except when tunneling between lattice sites. These lattice vibrations might be considered to achieve resonance when lambda = N * 2 * (lattice constant), which for Pd is 7.78x10^-10 m. The resonant frequency is given by: f = v / lambda where v is the speed of sound in the medium, or 3070 m/s, so: f = ( 3070 m/s) / (7.78x10^-10 m) = 3.946 x 10^12 Hz which is not far off from the other frequency I gave, which was computed from the deuteron 1 dimensional mechanical resonance. In all cases the phonon energy E is quantized to: E = (N + 1/2)*h*f where (1/2)*h*f is the zero point energy, and N is an integer. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
