Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 14 Nov 2005 08:19 Henri Wilson wrote: > On Sun, 13 Nov 2005 11:52:57 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <HW@..> wrote in message > >news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... > >> > >> George, George, George. > >> > >> I have finally woken up to your (and MY) complete misinterpretation of the > >> problem. > >> > >> We have both been arguing about whether or not the fringes will move > >> during > >> constant angular rotation...and of course they don't. > > > >Actually they do, that's exactly how fibre gyros work. > >The output is proportional to the angular velocity. > >That is why these devices are such important evidence > >and precisely why I have taken the trouble of bringing > >them to your attention. > > George, a few months ago, you (or maybe it was Paul) went to great trouble > to explain to me that fringes shift only during angular acceleration. Gyros > indicate total rotation by continuously integrating the rate of fringe shift > with time. > > Are you now saying this is wrong? I am sticking to what we both agreed. There had been some talk earlier but we revised this around the 11th October. See your message, ID: a4cok15ekbocc95d6ahg68foum9hm082hn(a)4ax.com [Henri wrote:] >[George wrote:] >>[Henri wrote:] >>> George, if a sagnac is rotating at constant angular speed, Do the >>> fringes move continuously or remain steady but offset? > >>The latter, steady but with an offset proportional >>to the speed of rotation. > > (Yes I thought that's what came up before. Just checking). The formula for the Sagnac effect is: delta_t = 4Aw/(c^2-(wR)^2) where where w is the angular speed. For wR << c that simplifies to: delta_t = 4Aw / c^2 so the output is proportional to the angular speed, not the acceleration. > >Acceleration is a more complex subject, involving Doppler > >at the source, Doppler at the receiver and the temporary > >lack of cancellation due to the flight time delay between. > > When acceleration occurs, path lengths vary. > The number of wavelengths in each path changes. > Fringes MOVE. > > It matters not whether the beam moves at c or c+v. > > During constant rotation, including zero, there is no fringe movement. >From page 6 of the comprehensive reference Jerry posted: "Sagnac's polygonal interferometer was mounted on a turntable. It had an area of 0.0860 m^2, a rotation rate of order 2 Hz, and the resulting fractional fringe shift 0.07 +/- 0.01." > >> Actually, you have shown that the BaTh does what it should do. > >> It expects NO fringe shifts under constant rotation. Exactly. > >> BUT!!!! > >> The standard SR explanation says that there WILL BE a continuous fringe > >> shift during steady rotation. > >> > >> Sagnac proves SR to be wrong!!! > > > >If the output was proportional to acceleration then you > >would be right. In fact the output is exactly what SR > >predicts, it is proportional to the angular velocity in > >actual devices and experiments. > > The output is given in degrees rotation from zero angle. That is calculated via > a time integral during acceleration. Angle turned is the integral of angular speed. There is a single integration to get the heading from the raw signal which is proportional to speed, not acceleration. > I think you and your colleagues should learn a few facts. Facts: 1) Ritz predicts no fringe shift for constant angular speed. 2) Sagnac measured a fringe shift of 0.07 at constant rate of 2Hz (120rpm) compared to non-rotating. 3) The experiment measures the speed of light from the moving source and it turns out to be unchanged from that in the non-rotating situation. George
From: George Dishman on 14 Nov 2005 08:37 Black Knight wrote: > > Grandpa puts granddaughter and grandson on a carousel and watches > them walk around it in opposite directions. The pass each other on the > opposite side and meet back at grandpa if the carousel isn't turning, ... Good analogy but the source and detector are on the turntable so Grandpa is on the carousel. You have shown why Ritz says there will be no output, the kids return to Grandpa at the same time regardless of the rate the carousel is turning. George
From: Paul B. Andersen on 14 Nov 2005 09:45 Eric Gisse wrote: > Paul B. Andersen wrote: > > [snip] > > >>I think E = mc^2 follows from Maxwell's equations, >>EM-radiation has momentum. > > > That seemed highly reasonable to me. > > Though I was hoping for Henri to do it instead of you. No point in hoping for the impossible. It is thoroughly documented that Henri couldn't do it. Paul
From: Henri Wilson on 14 Nov 2005 16:18 On 14 Nov 2005 05:19:05 -0800, "George Dishman" <george(a)briar.demon.co.uk> wrote: >Henri Wilson wrote: >> On Sun, 13 Nov 2005 11:52:57 -0000, "George Dishman" <george(a)briar.demon.co.uk> >> wrote: >> >> > >> >"Henri Wilson" <HW@..> wrote in message >> >news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... >> >> >> >> George, George, George. >> >> >> >> I have finally woken up to your (and MY) complete misinterpretation of the >> >> problem. >> >> >> >> We have both been arguing about whether or not the fringes will move >> >> during >> >> constant angular rotation...and of course they don't. >> > >> >Actually they do, that's exactly how fibre gyros work. >> >The output is proportional to the angular velocity. >> >That is why these devices are such important evidence >> >and precisely why I have taken the trouble of bringing >> >them to your attention. >> >> George, a few months ago, you (or maybe it was Paul) went to great trouble >> to explain to me that fringes shift only during angular acceleration. Gyros >> indicate total rotation by continuously integrating the rate of fringe shift >> with time. >> >> Are you now saying this is wrong? > >I am sticking to what we both agreed. There had >been some talk earlier but we revised this around >the 11th October. See your message, ID: > > a4cok15ekbocc95d6ahg68foum9hm082hn(a)4ax.com > >[Henri wrote:] >>[George wrote:] >>>[Henri wrote:] >>>> George, if a sagnac is rotating at constant angular speed, Do the >>>> fringes move continuously or remain steady but offset? >> >>>The latter, steady but with an offset proportional >>>to the speed of rotation. >> >> (Yes I thought that's what came up before. Just checking). > >The formula for the Sagnac effect is: > > delta_t = 4Aw/(c^2-(wR)^2) > >where where w is the angular speed. For wR << c >that simplifies to: > > delta_t = 4Aw / c^2 > >so the output is proportional to the angular speed, >not the acceleration. That's correct. But the fringes only move during acceleration. > >> >Acceleration is a more complex subject, involving Doppler >> >at the source, Doppler at the receiver and the temporary >> >lack of cancellation due to the flight time delay between. >> >> When acceleration occurs, path lengths vary. >> The number of wavelengths in each path changes. >> Fringes MOVE. >> >> It matters not whether the beam moves at c or c+v. >> >> During constant rotation, including zero, there is no fringe movement. > >>From page 6 of the comprehensive reference Jerry >posted: > > "Sagnac's polygonal interferometer was mounted > on a turntable. It had an area of 0.0860 m^2, a > rotation rate of order 2 Hz, and the resulting > fractional fringe shift 0.07 +/- 0.01." > >> >> Actually, you have shown that the BaTh does what it should do. >> >> It expects NO fringe shifts under constant rotation. > >Exactly. > >> >> BUT!!!! >> >> The standard SR explanation says that there WILL BE a continuous fringe >> >> shift during steady rotation. >> >> >> >> Sagnac proves SR to be wrong!!! >> > >> >If the output was proportional to acceleration then you >> >would be right. In fact the output is exactly what SR >> >predicts, it is proportional to the angular velocity in >> >actual devices and experiments. >> >> The output is given in degrees rotation from zero angle. That is calculated via >> a time integral during acceleration. > >Angle turned is the integral of angular speed. There is >a single integration to get the heading from the raw >signal which is proportional to speed, not acceleration. Yes George. The fringe shift is proportional to speed BUT IT OCCURS ONLY DURING ACCELERATION. > >> I think you and your colleagues should learn a few facts. > >Facts: > >1) Ritz predicts no fringe shift for constant angular > speed. > >2) Sagnac measured a fringe shift of 0.07 at constant > rate of 2Hz (120rpm) compared to non-rotating. > >3) The experiment measures the speed of light from the > moving source and it turns out to be unchanged from > that in the non-rotating situation. > >George Sorry George, you have it all wrong. The fringes DO NOT move during constant rotation. They are displaced by a constant amount. You said this yourself on for instance,12th October. """"" >> George, if a sagnac is rotating at constant angular speed, Do the fringes >> move >> continuously or remain steady but offset? > >The latter, steady but with an offset proportional >to the speed of rotation." """"" The displacement arises from the path length change that occurs DURING ACCELERATION. The number of 'wavelengths' in each path changes ONLY during acceleration. The amount of shift signifies a rotation rate. Integrate that over very short time intervals and you have a fairly accurate measurement of the total angle of rotation from zero. You should be able to see now, (from the duck shoot experiment) that virtually the same path length difference occurs whether you use c or c + v.....it differs by only c/(c+v) With a FoG, the path length is greatly increased for higher accuracy. So Sagnac DOES NOT refute the BaTh. HW. www.users.bigpond.com/hewn/index.htm see: www.users.bigpond.com/hewn/variablestars.exe "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong".
From: Henri Wilson on 14 Nov 2005 16:23
On 14 Nov 2005 05:37:52 -0800, "George Dishman" <george(a)briar.demon.co.uk> wrote: >Black Knight wrote: >> >> Grandpa puts granddaughter and grandson on a carousel and watches >> them walk around it in opposite directions. The pass each other on the >> opposite side and meet back at grandpa if the carousel isn't turning, ... > >Good analogy but the source and detector are on >the turntable so Grandpa is on the carousel. You >have shown why Ritz says there will be no output, >the kids return to Grandpa at the same time >regardless of the rate the carousel is turning. Forget it George. I have now completely explained why sagnac does not refute the BaTh. .. >George HW. www.users.bigpond.com/hewn/index.htm see: www.users.bigpond.com/hewn/variablestars.exe "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong". |