A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: Inertial on 15 Feb 2010 17:31 "Henry Wilson DSc" <..@..> wrote in message news:qihjn5d4hapu9j5qu1kih5gi09m6bengf1(a)4ax.com... > The numbers arriving from each direction between the emission of pulse X > and > its arrival at the detector differ by 124 in my example. So, according to you in some fixed period of time, there are 62 more pulses arriving in one direction than were emitted in that same period of time. So if you leave it running for a while, you end up with more pulses coming out of the device than have gone into it. You've created free energy. Congratulations.
From: Henry Wilson DSc on 16 Feb 2010 01:00 On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Feb 15, 4:20�pm, ..@..(Henry Wilson DSc) wrote: >> On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry >> > >Your "rayphases" model has been refuted many times over, by me >and by many others employing different arguments and methods of >analysis. It is self-contradictory and predicts effects that have >never been seen. In this thread alone, I've seen you squirm and >deny assertions that you made mere hours previously. Your model >is HOPELESS. >http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm Well, this experiment requires no light rays or interferometers. It is simple. Just wrap two optical fibres around the equator and send time labelled light pulses every 3.33 nanoseconds down each one in opposite directions. The pulses will be 1 metre apart in both fibres. According to BaTh, the Earth rotates by 62 metres during the time it takes for a pulse to travel around the ring. So one light path will be 4000062 metres long and the other 3999938. Therefore, between the time pulse N is emitted and received, 4000062 pulses arrive from one fibre and only 3999938 from the other. All you need is a counter. >Jerry Henry Wilson... ........provider of free physics lessons
From: Inertial on 16 Feb 2010 01:31 "Henry Wilson DSc" <..@..> wrote in message news:ne2kn59rbmokiv1revbefkn1m3npnbq3k7(a)4ax.com... > On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry > <Cephalobus_alienus(a)comcast.net> wrote: > >>On Feb 15, 4:20 pm, ..@..(Henry Wilson DSc) wrote: >>> On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry >>> > >> >>Your "rayphases" model has been refuted many times over, by me >>and by many others employing different arguments and methods of >>analysis. It is self-contradictory and predicts effects that have >>never been seen. In this thread alone, I've seen you squirm and >>deny assertions that you made mere hours previously. Your model >>is HOPELESS. >>http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm > > Well, this experiment requires no light rays or interferometers. So a light pulse isn't just a short light ray? > It is simple. Just wrap two optical fibres around the equator and send > time > labelled light pulses every 3.33 nanoseconds down each one in opposite > directions. The pulses will be 1 metre apart in both fibres. OK > According to BaTh, the Earth rotates by 62 metres during the time it takes > for > a pulse to travel around the ring. So one light path will be 4000062 > metres > long and the other 3999938. But it will travel the longer path quicker, so they arrive at the same time with the same arrival rate > Therefore, between the time pulse N is emitted and received, 4000062 > pulses > arrive from one fibre and only 3999938 from the other. WRONG. If the pulses leave in pairs (as you imply), and take the same time to travel all the way around their path and arrive back again (as emission theory says), then they MUST arrive back in the same pairs they were emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path in any given time interval. You cannot determine the rotation of the device (according to emission theory) by looking at the arrival times and rates of the pulses. And the fact that your CAN in reality (with a real sagnac device) is resounding and refutation of emission theory. Why do you keep lying, Henry? > All you need is a > counter. All you need is SR .. which DOES let you determine the absolute rotation.
From: Henry Wilson DSc on 16 Feb 2010 05:02 On Tue, 16 Feb 2010 17:31:51 +1100, "Inertial" <relatively(a)rest.com> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >news:ne2kn59rbmokiv1revbefkn1m3npnbq3k7(a)4ax.com... >> On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry >> <Cephalobus_alienus(a)comcast.net> wrote: >> >>>On Feb 15, 4:20 pm, ..@..(Henry Wilson DSc) wrote: >>>> On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry >>>> >> >>> >>>Your "rayphases" model has been refuted many times over, by me >>>and by many others employing different arguments and methods of >>>analysis. It is self-contradictory and predicts effects that have >>>never been seen. In this thread alone, I've seen you squirm and >>>deny assertions that you made mere hours previously. Your model >>>is HOPELESS. >>>http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm >> >> Well, this experiment requires no light rays or interferometers. > >So a light pulse isn't just a short light ray? > >> It is simple. Just wrap two optical fibres around the equator and send >> time >> labelled light pulses every 3.33 nanoseconds down each one in opposite >> directions. The pulses will be 1 metre apart in both fibres. > >OK > >> According to BaTh, the Earth rotates by 62 metres during the time it takes >> for >> a pulse to travel around the ring. So one light path will be 4000062 >> metres >> long and the other 3999938. > >But it will travel the longer path quicker, so they arrive at the same time >with the same arrival rate > >> Therefore, between the time pulse N is emitted and received, 4000062 >> pulses >> arrive from one fibre and only 3999938 from the other. > >WRONG. > >If the pulses leave in pairs (as you imply), and take the same time to >travel all the way around their path and arrive back again (as emission >theory says), then they MUST arrive back in the same pairs they were >emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path >in any given time interval. > >You cannot determine the rotation of the device (according to emission >theory) by looking at the arrival times and rates of the pulses. And the >fact that your CAN in reality (with a real sagnac device) is resounding and >refutation of emission theory. > >Why do you keep lying, Henry? > >> All you need is a >> counter. > >All you need is SR .. which DOES let you determine the absolute rotation. You will remain plonked until your brain decides to function Henry Wilson... ........provider of free physics lessons
From: Henry Wilson DSc on 16 Feb 2010 05:26
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Feb 15, 4:20�pm, ..@..(Henry Wilson DSc) wrote: >Your "rayphases" model has been refuted many times over, by me >and by many others employing different arguments and methods of >analysis. It is self-contradictory and predicts effects that have >never been seen. In this thread alone, I've seen you squirm and >deny assertions that you made mere hours previously. Your model >is HOPELESS. >http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm It is true that I made a late addition to that program which is rather misleading...but that matters not. The reason you are having trouble understanding the Bath explanation is that three separate oscillation periods are involved. There is the the rotation of the Earth, the rotation of the pulses around the ring and the oscillator that causes pulses to be emitted at regular intervals. The fact that movement is circular rather than linear further complicates the issue. Only those with superior intelligence should even try to fathom what is actually happening here. Certainly, neither a raw graduate nor anyone with early dementia would, have a hope. >Jerry Henry Wilson... ........provider of free physics lessons |