tag:blogger.com,1999:blog-8282587.post7044613292609690776..comments2024-03-19T01:48:39.709+00:00Comments on Power and Control: Polywell - Adding DetailsM. Simonhttp://www.blogger.com/profile/09508934110558197375noreply@blogger.comBlogger18125tag:blogger.com,1999:blog-8282587.post-1524030634980489102008-02-24T20:34:00.000+00:002008-02-24T20:34:00.000+00:00some misconceptions here: the "quasi-spherical" p...some misconceptions here: the "quasi-spherical" phrase Bussard used refers to electron cloud. Magnetic mirroring (tendency of charged particles with spiraling along magnetic lines to "bounce back" from regions where field strength higher) is used to confine electrons. Particles with angle of velocity to magnetic field greater than a critical value will be reflected.rubycodezhttps://www.blogger.com/profile/08184263858078881047noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-56351550388826751522007-06-02T22:42:00.000+00:002007-06-02T22:42:00.000+00:00The animation I find clearer than the static illus...The animation I find clearer than the static illustration, but it shows up another curious point: To form the well, the electrons have to stick around - at least mostly.<BR/><BR/>In the animation, they come in from all sides, but they don't get out very far. If an electron heads straight out one face, it shouldn't get any fight from the magnetic fields; and since a static magnetic field does not affect kinetic energy, that electron should be heading out with all the energy it started with, which was quite a lot. So why do the electrons in the animation stop and turn around? Just as the well is attracting the positive nuclei, it will be actively pushing away the electrons.<BR/><BR/>Now, maybe there is an ongoing balance between the incoming electrons from the beam and the escaping electrons; but this is not reflected in the animation.<BR/><BR/>Is there any explanation for this?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-17396268296439968192007-06-01T09:49:00.000+00:002007-06-01T09:49:00.000+00:00Neal,The effect is a diamagnetic one. The field li...Neal,<BR/><BR/>The effect is a diamagnetic one. The field lines get compressed.<BR/><BR/>True it is not spherical. It is described as quasi-spherical. <BR/><BR/>Look at Indrek's latest simulation at:<BR/><BR/><A HREF="http://tech.groups.yahoo.com/group/IEC_Fusion/" REL="nofollow">IEC Fusion Newsgroup</A> <BR/><BR/>You can see electron circulation and also what looks like well formation.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-90196038266862880312007-06-01T07:17:00.000+00:002007-06-01T07:17:00.000+00:00M. Simon,That always has the effect of countering ...M. Simon,<BR/><BR/>That always has the effect of countering the local magnetic in which the electron is orbiting: The current loop formed by the electron's orbit produces a field (within the circular region bounded by the orbit) which is opposite to the field which is forcing the electron to move that way. Makes sense: If it worked the other way, the magnetic field would be getting stronger and stronger, which has a perpetual-motion-machine feel to it. <BR/><BR/>As a local effect, it weakens the field wherever it finds it. Therefore: Since the basic symmetry is cubical and not spherical, that will not be changed by this effect. <BR/><BR/>Further, both the fields produced by the device and by the individual electrons satisfy<BR/>div B = 0 , <BR/>so the sum of the fields do as well, because this equation is linear. <BR/><BR/>If you imagine the magnetic field as a flow of liquid, it would mean that overall the flow keeps continuing: If it's coming out of a region, it has to go back into that region. A <I>spherically symmetric</I> flow that satisfies that does not exist; the best you can get is an axially symmetric flow, with the B-field confined to the plane transverse to the axis; and along the axial direction, you could confine it to a region. But the field will still be confined to the transverse plane in that configuration.<BR/><BR/>Yes, I took a class in plasma physics decades ago. An old buddy of mine from grad school is a professor of plasma physics in East Lansing; although his interest is in atmospheric physics rather than fusion.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-28338482139319600032007-06-01T05:21:00.000+00:002007-06-01T05:21:00.000+00:00Neal,You are correct except that the rotation of e...Neal,<BR/><BR/>You are correct except that the rotation of electrons in the magnetic field produces a magnetic field.<BR/><BR/>This magneticfield "circularizes" the magnetic field from the coils.<BR/><BR/>Plasma physics is not easy because everything affects everything.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-72433696522487952352007-06-01T02:53:00.000+00:002007-06-01T02:53:00.000+00:00M. Simon,Even if you use 6 magnets oriented to "po...M. Simon,<BR/><BR/>Even if you use 6 magnets oriented to "point" outward, the field will not be anything like spherical. However much magnetic flux comes out of the faces will have to snake back in at the corners: This is the implication of:<BR/>div B = 0<BR/>in Maxwell's equations.<BR/><BR/>For that reason, using the term "quasi-spherical", as I think even Bussard does in his talk to Google, is profoundly wrong. There will be a symmetry, but that corresponding to a cube, not at all to a sphere. A sphere can be turned in any direction, and to any degree, without a change. Whereas everytime you rotate this cube by 90 degrees, the field at the upper face will convert from outward to inward for as the corner passes, and then back again. Not even roughly spherical.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-37657552664294106222007-05-31T03:16:00.000+00:002007-05-31T03:16:00.000+00:00Neal,I understand English is not your first langua...Neal,<BR/><BR/>I understand English is not your first language. However, if you see something that seems incredible re-read it to make sure you are not jumping to conclusions. <BR/><BR/><I>The magnets can be all North poles facing in or all South poles facing in.</I><BR/><BR/>This is true without recourse to magnetic monopoles.<BR/><BR/>As to your second point. There is no computer today fast enough to model all this in a reasonable amount of time. So build one and take measurements.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-36237046628565269552007-05-30T23:09:00.000+00:002007-05-30T23:09:00.000+00:00Related to the comment above:You cannot have a mag...Related to the comment above:<BR/><BR/>You cannot have a magnetic field that is "all North pole" in all directions: that would violate Maxwell's equations. If the magnetic field lines are envisioned as "coming out" of the faces, then they are "going back" into the corners. So I find the term "quasi-spherical magnetic field" to be rather confusing.<BR/><BR/>With respect to collecting the power: During steady-state operation, nuclei should be dropping into the potential well, and the fusion products should come boiling out. The outward pressure from these exiting alpha particles will combat the in-falling motion of the p and B-11, as well as the electrons. (It's not mechanical pressure: the electromagnetic effect of fast-passing nuclei should be like being hit by photons.) This needs to be taken into account: without the outward flow of the alpha particles, there's no way to extract the energy of the process.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-57960986983864753002007-04-10T15:24:00.000+00:002007-04-10T15:24:00.000+00:00Anon. 30 March,The field lines are going to close ...Anon. 30 March,<BR/><BR/>The field lines are going to close no matter what.<BR/><BR/>The question is what will the field gradient be. Which will determine what the density of ion current will be. Which will in turn determine how the ion current generated magnetic field will interact with the coil generated magnetic fields.<BR/><BR/>It may be that the truncated polyhedron is optimum. <BR/><BR/>I might add that there is also the question of field lines intersecting the adjacent magnet structure. Which is why the spacing Bussard talks about is important.<BR/><BR/>All the interactions get complicated which is why what is going on is so hard to visualise.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-22768330920311780872007-03-30T17:02:00.000+00:002007-03-30T17:02:00.000+00:00I don't think your description of the magnets is r...I don't think your description of the magnets is right. The polyhedral geometry of the WB6 is a <I>truncated</I> cube, not a regular cube. It's true that the coils don't quite fit the shape of the polyhedron, but if they did, they would be diamonds (i.e. squares with corners at top, bottom, left, and right), not squares like the face of a cube. You need someplace for the field lines that go in through the centers of the coils to come back out; that's the corners. The circular coils are a fair approximation of the correct truncated-cube design.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-46915970204351275412007-03-28T07:18:00.000+00:002007-03-28T07:18:00.000+00:00Neutron flux causes loss of superconductivity beca...Neutron flux causes loss of superconductivity because they transmutate the superconductor's atoms, mostly by smashing them apart. Neutron radiation is <I>tremendously</I> destructive, which is why nice, neutron free reactions are so helpful. In the case of p-B11, you also get some high-energy alpha particles of a precise energy, which can be decelerated gracefully and used to produce power. This effect has been used in some radioisotope batteries, and is a fairly mature technology.bbothttps://www.blogger.com/profile/16749016348338381812noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-54267182155042066282007-03-26T21:29:00.000+00:002007-03-26T21:29:00.000+00:00Ah, I understand, thanks. I misunderstood the dir...Ah, I understand, thanks. I misunderstood the direct conversion mechanism. I wish you the best in this work, it's very exciting.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-229542023859756142007-03-26T20:18:00.000+00:002007-03-26T20:18:00.000+00:00The holy grail is to collect the energetic alphas ...The holy grail is to collect the energetic alphas in a 2 million volt (or so) grid.<BR/><BR/>Direct conversion so to speak.<BR/><BR/>The alternative is to figure some way to get the alphas to boil water.<BR/><BR/>Neutrons are easier. However, neutron fluxes are not a good idea for a number of reasons. One is that if you are using superconducting magnet coils they cause loss of superconductivity.<BR/><BR/>This idea is a long way from practical application. It needs lots of work.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-52646765689494663492007-03-26T19:55:00.000+00:002007-03-26T19:55:00.000+00:00So how does one get power out of a Bussard reactor...So how does one get power out of a Bussard reactor? I'm still confused.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-64238870346804760962007-03-26T16:17:00.000+00:002007-03-26T16:17:00.000+00:00neil,The spare electrons are injected by an electr...neil,<BR/><BR/>The spare electrons are injected by an electron gun.<BR/><BR/>The Bussard reactor = Polywell.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-22898678479213582472007-03-26T16:11:00.000+00:002007-03-26T16:11:00.000+00:00As I understand it, the Bussard reactor produces a...As I understand it, the Bussard reactor produces a potential (and current) as a direct product of fusion. I realize the models being described are just test versions, but once break-even fusion is achieved, where do the spare electrons come from?<BR/><BR/>Or maybe that's the obvious point I'm missing? The Bussard reactor is different from the Polywell models?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8282587.post-47402464932792066432007-03-25T22:47:00.000+00:002007-03-25T22:47:00.000+00:00I don't understand your question.The object of the...I don't understand your question.<BR/><BR/>The object of the device is not current production. It is fusion.M. Simonhttps://www.blogger.com/profile/09508934110558197375noreply@blogger.comtag:blogger.com,1999:blog-8282587.post-30932974453108604402007-03-25T21:33:00.000+00:002007-03-25T21:33:00.000+00:00It's possible I'm missing something obvious, but w...It's possible I'm missing something obvious, but which parts form the usable, current-producing potential?Anonymousnoreply@blogger.com