Well I must confess, I did not see a torus, I don't know what to call it. Not wanting to be rude, I left it for a later time, after some thought.
Statistics : Posted by JS • on Fri Oct 09, 2009 11:59 am • Replies 15 • Views 791
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Well I must confess, I did not see a torus, I don't know what to call it. Not wanting to be rude, I left it for a later time, after some thought.]]>
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Also, comments will now post--BUT WILL NOT PREVIEW. Seems to be a problem in the wysiwyg editor interface. They are working on it, but do not yet have a solution. In the past, you always had to Preview first... I have disabled this feature so you can SAVE without previewing. This SHOULD post the comment properly.
Statistics : Posted by bperet • on Tue Sep 29, 2009 9:26 am • Replies 2 • Views 209
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Also, comments will now post--BUT WILL NOT PREVIEW. Seems to be a problem in the wysiwyg editor interface. They are working on it, but do not yet have a solution. In the past, you always had to Preview first... I have disabled this feature so you can SAVE without previewing. This SHOULD post the comment properly.]]>
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| Quote: Larson is talking about "fictitious results of relating motions to an arbitrary reference system". He was referring to the currently accepted theories. | |
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wsitze - Thanks for pointing this out. That does make sense. Going back and re-reading the text from Larson that I quoted at the beginning of this thread, the last sentence may answer my original question.
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| Quote: These independent units of motion, as we will call them, are superimposed on the moving reference background in much the same manner as | |
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Statistics : Posted by RAB • on Sun Aug 02, 2009 9:20 am • Replies 10 • Views 434
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| Quote: Larson is talking about "fictitious results of relating motions to an arbitrary reference system". He was referring to the currently accepted theories. | |
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wsitze - Thanks for pointing this out. That does make sense. Going back and re-reading the text from Larson that I quoted at the beginning of this thread, the last sentence may answer my original question.
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| Quote: These independent units of motion, as we will call them, are superimposed on the moving reference background in much the same manner as | |
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Also:
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| Jan Sammer wrote: Since such photons would be travelling along a simple one-dimensional path, capable of being fully represented in a conventional reference system, they would not display any of the interference phenomena and wave-like features of regular photons. They would be polarized, since the one dimension of translational motion open to such photons would always be congruent | |
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Statistics : Posted by Horace • on Thu Mar 12, 2009 7:48 pm • Replies 22 • Views 862
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Also:
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| Jan Sammer wrote: Since such photons would be travelling along a simple one-dimensional path, capable of being fully represented in a conventional reference system, they would not display any of the interference phenomena and wave-like features of regular photons. They would be polarized, since the one dimension of translational motion open to such photons would always be congruent | |
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It appears to be related to pasting MS Word or OpenOffice documents
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Statistics : Posted by bperet • on Sat Jun 20, 2009 11:40 am • Replies 0 • Views 211
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It appears to be related to pasting MS Word or OpenOffice documents
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| davelook wrote: "Why would similarly-charged particles (protons) adhere in the nucleus instead of flying apart? A new fundamental force was invented to keep them together." It's a shame that quarks were unknown, since the ability of protons (plus neutrons) to form stable nuclei is now fully explained by the sub-charges of the quarks. seehttp://www.unclear2nuclear.com/np.php Of course, the strong force is still required to keep the 2 +2/3 up quarks inside the proton. | |
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Statistics : Posted by StevenO • on Sun Aug 05, 2007 8:42 am • Replies 3 • Views 875
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| davelook wrote: "Why would similarly-charged particles (protons) adhere in the nucleus instead of flying apart? A new fundamental force was invented to keep them together." It's a shame that quarks were unknown, since the ability of protons (plus neutrons) to form stable nuclei is now fully explained by the sub-charges of the quarks. seehttp://www.unclear2nuclear.com/np.php Of course, the strong force is still required to keep the 2 +2/3 up quarks inside the proton. | |
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LineLand (current project): A Universe of Motion in ONE dimension. Logic being that it is much easier to understand how ONE scalar dimension creates and manifests in the material and cosmic sectors, than three. It has proved to be a most valuable learning tool.
FlatLand The 2-dimensional Universe of Motion, to be developed after the details
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Statistics : Posted by bperet • on Sat May 09, 2009 2:06 pm • Replies 0 • Views 290
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LineLand (current project): A Universe of Motion in ONE dimension. Logic being that it is much easier to understand how ONE scalar dimension creates and manifests in the material and cosmic sectors, than three. It has proved to be a most valuable learning tool.
FlatLand The 2-dimensional Universe of Motion, to be developed after the details
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The only problem I currently know of is the upload progress bar is broken, for file uploads. Says it is missing a library, but the library is there and readable, so I have no idea what is going on.
A number of
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Statistics : Posted by bperet • on Sun Mar 29, 2009 6:01 pm • Replies 5 • Views 243
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The only problem I currently know of is the upload progress bar is broken, for file uploads. Says it is missing a library, but the library is there and readable, so I have no idea what is going on.
A number of
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If you take the fundamental weak force particles, the W boson (80.398 GeV) and Z boson (91.1876 GeV), you get the cosine of the "weak mixing angle", or, arccos(W/Z)= 28.154687 deg. Let's assume that the weak mixing angle is really just (arctan(3/2))/2=28.154966 (see the arxiv ref)
tan(45-28.154966)=.30277563773
(What's really cool is this is one of the metallic means, because
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Statistics : Posted by davelook • on Mon Dec 29, 2008 9:47 am • Replies 16 • Views 1224
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If you take the fundamental weak force particles, the W boson (80.398 GeV) and Z boson (91.1876 GeV), you get the cosine of the "weak mixing angle", or, arccos(W/Z)= 28.154687 deg. Let's assume that the weak mixing angle is really just (arctan(3/2))/2=28.154966 (see the arxiv ref)
tan(45-28.154966)=.30277563773
(What's really cool is this is one of the metallic means, because
...]]>
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| Quote: One of the long-standing quandary of nuclear physics is why does p-n(proton-neutron or deuterium) form a stable nucleus but the p-p (proton-proton) does not form a stable nucleus. Even more puzzling is why the n-n (neutron-neutron) pair will not form a stable particle. Analyzed from the perspective of the current nuclear models and | |
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Statistics : Posted by davelook • on Sun Mar 22, 2009 12:27 pm • Replies 1 • Views 168
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| Quote: One of the long-standing quandary of nuclear physics is why does p-n(proton-neutron or deuterium) form a stable nucleus but the p-p (proton-proton) does not form a stable nucleus. Even more puzzling is why the n-n (neutron-neutron) pair will not form a stable particle. Analyzed from the perspective of the current nuclear models and | |
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With this release of phpBB, newly registered users will have their first post put on the moderation queue, which will have to be approved by a moderator, before it will be visible to other forum
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Statistics : Posted by bperet • on Sun Mar 22, 2009 12:05 pm • Replies 0 • Views 806
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With this release of phpBB, newly registered users will have their first post put on the moderation queue, which will have to be approved by a moderator, before it will be visible to other forum
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If for some reason you cannot access the site for more than 2 days, please contact me via theRS2 theory forum as a backup.
Things have been updating pretty fast for
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Statistics : Posted by bperet • on Sat Mar 21, 2009 6:34 pm • Replies 0 • Views 195
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If for some reason you cannot access the site for more than 2 days, please contact me via theRS2 theory forum as a backup.
Things have been updating pretty fast for
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For angular velocity, I would suggest that a line segment, as defined by two points, rotating about itself is all that's needed. The point of rotation does not need to be defined, but
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Statistics : Posted by wsitze • on Wed Feb 25, 2009 1:32 pm • Replies 27 • Views 627
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For angular velocity, I would suggest that a line segment, as defined by two points, rotating about itself is all that's needed. The point of rotation does not need to be defined, but
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| ad wrote: Steven, I noted your debating attempt on the BAUT ATM forum. Personally I think such challenges are pointless. Never the less, good luck... PS> I made the comment about locating a reference for Nehru's formula from GR for the precession. PPS> Bruce, I don't seem to be able to find the equivalent of this: http://www.reciprocalsystem.com/rs/cwkv ... etperi.htm on the library pages? | |
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Statistics : Posted by StevenO • on Tue Feb 24, 2009 1:13 am • Replies 3 • Views 182
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| ad wrote: Steven, I noted your debating attempt on the BAUT ATM forum. Personally I think such challenges are pointless. Never the less, good luck... PS> I made the comment about locating a reference for Nehru's formula from GR for the precession. PPS> Bruce, I don't seem to be able to find the equivalent of this: http://www.reciprocalsystem.com/rs/cwkv ... etperi.htm on the library pages? | |
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| Attachment: Fundamentals-Flowchart.gifFundamentals-Flowchart.gifhttp://space.newscientist.com/article/d ... Statistics : Posted by frankm • on Fri Aug 31, 2007 3:44 pm • Replies 0 • Views 851 ]]> The article is available at arXiv. http://www.arxiv.org/PS_cache/arxiv/pdf ... 0908v2.pdf The newscientist has a review of the article. http://space.newscientist.com/article/d ...]]> This letter is a result of an invitation from ISUS via Ron Satz to attend the conference and give an after dinner talk on my memories of Dewey. I regret that I have had to decline both invitations. However, with the possibility there might be some interest in a letter, I will share with you a few memories of my 58 years with my husband and his interest in ISUS. This will not be a speech, it will be an informal chat with you about ... Statistics : Posted by bperet • on Fri Jul 20, 2007 12:17 pm • Replies 0 • Views 763 ]]> This letter is a result of an invitation from ISUS via Ron Satz to attend the conference and give an after dinner talk on my memories of Dewey. I regret that I have had to decline both invitations. However, with the possibility there might be some interest in a letter, I will share with you a few memories of my 58 years with my husband and his interest in ISUS. This will not be a speech, it will be an informal chat with you about ...]]>
This is a good question, Phil. I'll give it a shot. Youwrote:
In the universe of motion, a "Euclidean point" doesn't exist except as motion; that is, the point itself that is "backing ... Statistics : Posted by Douglas Bundy • on Sun Sep 03, 2006 5:08 am • Replies 11 • Views 2196 ]]>
This is a good question, Phil. I'll give it a shot. Youwrote:
In the universe of motion, a "Euclidean point" doesn't exist except as motion; that is, the point itself that is "backing ...]]>
They're theoretical postulates, not laws. Replace "must be Euclidean" with "can be considered in a Euclidean context" and you're got a more valid statement. Statistics : Posted by mwells • on Tue Nov 15, 2005 6:57 pm • Replies 168 • Views 17679 ]]>
They're theoretical postulates, not laws. Replace "must be Euclidean" with "can be considered in a Euclidean context" and you're got a more valid statement.]]>
I would suggest that we designate the RST as the system used to construct the theory, and the RSt as the theory constructed using the RST. That way, we can avoid confusion, I think.
... Statistics : Posted by Douglas Bundy • on Fri Nov 11, 2005 2:24 am • Replies 12 • Views 1262 ]]>
I would suggest that we designate the RST as the system used to construct the theory, and the RSt as the theory constructed using the RST. That way, we can avoid confusion, I think.
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The interesting thing about this whole issue of non-commutative products of vectors, is that it's a natural outcome of vectorial motion in a 3D spatial coordinate system. What this reflects, then, is that the RST is a system OF physical theory, which encompasses ... Statistics : Posted by Douglas Bundy • on Mon Nov 07, 2005 6:14 am • Replies 20 • Views 1810 ]]>
The interesting thing about this whole issue of non-commutative products of vectors, is that it's a natural outcome of vectorial motion in a 3D spatial coordinate system. What this reflects, then, is that the RST is a system OF physical theory, which encompasses ...]]>
It seems that the "shape of now" is not a sphere after all !!! The drawing below has been taken from an exellent paper by D.G.Pavlov, available at: http://hypercomplex.xpsweb.com/articles/151/en/pdf/01-03-e.pdf ... Statistics : Posted by Horace • on Wed Nov 03, 2004 5:50 pm • Replies 9 • Views 1221 ]]>
It seems that the "shape of now" is not a sphere after all !!! The drawing below has been taken from an exellent paper by D.G.Pavlov, available at: http://hypercomplex.xpsweb.com/articles/151/en/pdf/01-03-e.pdf ...]]> Regards David ------------------------------------------------------------------------------------------------
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