Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 4 Dec 2005 16:33 On Sun, 4 Dec 2005 17:14:26 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:3t14p19pfdumiaijpi8rq2e4c25vbe5qqk(a)4ax.com... >> On Sat, 3 Dec 2005 11:14:22 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >... >>>I can't give you a definitive answer but we can get >>>an order-of-magnitude estimate by inference. >>> >>>This explains that technique and also gives a frequency >>>which is probably typical. For stability they commercial >>>product probably uses an AT cut crystal and they typically >>>run in the low megahertz range: >>> >>>http://www.physik.fu-berlin.de/~bauer/habil_online/node11.html#phaseshift >>> >>> "The modulation is most effective if tau is half of >>> the period time of the modulation ..." >>> >>>Tau is the light time round the fibre. >>> >>> "in our experiments this setting is 5.05 MHz." >>> >>>A frequency of 5.05MHz is a wavelength of 59.4m so >>>their loop had a total length of about 29.7m. >>> >>>http://www.kvh.com/pdf/DSP3000_5.04.pdf >>> >>>The DSP-3000 has the dimensions 3.5" x 2.3" x 1.3" >>>so the fibre coil must have a diameter of less than >>>3cm allowing 1.5mm for the box thickness, that's a >>>circumference of about 9.4cm. >>> >>>The two equipments are different but if the device >>>used 5.05MHz modulation, that would imply 315 turns. >>> >>>Each turn has area of 7.1 cm^2 so 315 turns gives a >>>total area of 2227 cm^2. It is equivalent to a single >>>loop of diameter 53.3 cm or a square set of mirrors >>>at the periphery of a turntable of radius 66.7cm. >> >> I cannot see why you consider loop area to be the important factor. > >Because the output is proportional to the >enclosed area: > >http://www.mathpages.com/rr/s2-07/2-07.htm according to your theory, OK. > >> It is fibre >> length that you need to know. That is about 10 metres....equivalent to a >> four >> mirror sagnac with 2.0 metre diagonals. > >Imagine a fibre folded in half and laid in a straight >line. Obviously there would be no difference. It's not >just length that matters but the integral of a factor >involving the radius along the length. It works out to >be the area. according to your theory, OK. > >>>The KVH spec says the maximum rotation rate is >>>100 degrees/second for the analog output. I'll let >>>you work out whether that is multiple fringes or a >>>fraction of a fringe. >> >> That will depend on which theory I use. >> The peripheral speed is around 3 m/sec. >> >> Using your aether theory (alias SR) > >Very funny Henri. It always amuses me that you >can't resist showing that you haven't a clue >about SR. I suggest you read the last three >paragraphs of this page: > >http://www.mathpages.com/home/kmath169/kmath169.htm Obviously written by a relativist. > >Aether theories struggle with Sagnac, SR >has no problem at all. Except that it would result in NO fringes at all. The whole image would be uniform in colour. Incidentally, your theory say 'light moves at c in a vacuum'. In the SR analysis of sagnac, what is the reference for that speed? Is it the "point in absolute space where the source was when the light was emitted?" ....it's quite amusing really. You people have no idea that you are merely preaching LET. >> that the light speed is c and not c/n or >> (c+v/root2)/n.... > >According to SR, the speed is c/n in the >rotating frame or c/n + v(1-n^2)/sqrt(2) in >the lab frame. > >On the above page you'll find the maths that >shows the effect is proportional to area for any >arbitrary polygon of mirrors and a less detailed >explanation of why the time difference is >independent of the refractive index. > >> the travel time around the path is 3.3E-8 secs, during which >> the periphery moves 10^-7 m.....now double that for the two paths. >> = 2E-7 m or 0.2 microns. >> ~1/3 of a wavelength. > >Unless I have an error in my arithmetic, the time >difference between the two paths is: > > dt = 4Aw/c^2 ~ 1.73*10^-17s > > c dt = 5.18 nm > >For 600nm light which you assumed, the fraction >is 0.0086 of a wavelength. For such low values, >sin(x) ~ x so the output of the detector would be >proportional to the angular speed (according to >SR and in reality, proportional to the angular >acceleration according to Ritz). But of course we know differently now, don't we George. We know that the rays which take the same travel time according to Ritz, do NOT end up on the same point on the image. Therefore they are not the ones which determine the state of the interference pattern at all. > >George > HW. www.users.bigpond.com/hewn/index.htm
From: Henri Wilson on 4 Dec 2005 21:35 On 4 Dec 2005 12:02:40 -0800, "Jerry" <Cephalobus_alienus(a)comcast.net> wrote: >Henri Wilson wrote: > >> I cannot see why you consider loop area to be the important factor. >> It is fibre length that you need to know. That is about 10 metres.... >> equivalent to a four mirror sagnac with 2.0 metre diagonals. > >Er, no. The shape of the ring is extremely important. Stretch out the >ring so that the fibers are parallel, and you will have destroyed the >sensitivity of the device. Mathematically, the sensitivity the device >turns out to be directly proportional to the area. Only because v=w.r. You still multiply by the number of turns. > >Jerry HW. www.users.bigpond.com/hewn/index.htm
From: Jerry on 5 Dec 2005 08:26 Henri Wilson wrote: > On Sun, 4 Dec 2005 17:14:26 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <HW@..> wrote in message > >news:3t14p19pfdumiaijpi8rq2e4c25vbe5qqk(a)4ax.com... > >> I cannot see why you consider loop area to be the important factor. > > > >Because the output is proportional to the > >enclosed area: > > > >http://www.mathpages.com/rr/s2-07/2-07.htm > > according to your theory, OK. > > > > >> It is fibre > >> length that you need to know. That is about 10 metres....equivalent to a > >> four > >> mirror sagnac with 2.0 metre diagonals. > > > >Imagine a fibre folded in half and laid in a straight > >line. Obviously there would be no difference. It's not > >just length that matters but the integral of a factor > >involving the radius along the length. It works out to > >be the area. > > according to your theory, OK. Henri, re-work your Sagnac simulation so that the light path is an elongated rectangle of zero width, zero area, i.e. |=====================================| You have converted the Sagnac interferometer into an MMX interferometer, which is not sensitive to rotation by ANY theory, Lorentz, SR, aetheric, BaT... Jerry
From: George Dishman on 5 Dec 2005 09:19 Henri Wilson wrote: > On Sun, 4 Dec 2005 17:14:26 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <HW@..> wrote in message > >news:3t14p19pfdumiaijpi8rq2e4c25vbe5qqk(a)4ax.com... > >> On Sat, 3 Dec 2005 11:14:22 -0000, "George Dishman" > >> <george(a)briar.demon.co.uk> > >> wrote: > >... > >>>I can't give you a definitive answer but we can get > >>>an order-of-magnitude estimate by inference. > >>> > >>>This explains that technique and also gives a frequency > >>>which is probably typical. For stability they commercial > >>>product probably uses an AT cut crystal and they typically > >>>run in the low megahertz range: > >>> > >>>http://www.physik.fu-berlin.de/~bauer/habil_online/node11.html#phaseshift > >>> > >>> "The modulation is most effective if tau is half of > >>> the period time of the modulation ..." > >>> > >>>Tau is the light time round the fibre. > >>> > >>> "in our experiments this setting is 5.05 MHz." > >>> > >>>A frequency of 5.05MHz is a wavelength of 59.4m so > >>>their loop had a total length of about 29.7m. > >>> > >>>http://www.kvh.com/pdf/DSP3000_5.04.pdf > >>> > >>>The DSP-3000 has the dimensions 3.5" x 2.3" x 1.3" > >>>so the fibre coil must have a diameter of less than > >>>3cm allowing 1.5mm for the box thickness, that's a > >>>circumference of about 9.4cm. > >>> > >>>The two equipments are different but if the device > >>>used 5.05MHz modulation, that would imply 315 turns. > >>> > >>>Each turn has area of 7.1 cm^2 so 315 turns gives a > >>>total area of 2227 cm^2. It is equivalent to a single > >>>loop of diameter 53.3 cm or a square set of mirrors > >>>at the periphery of a turntable of radius 66.7cm. > >> > >> I cannot see why you consider loop area to be the important factor. > > > >Because the output is proportional to the > >enclosed area: > > > >http://www.mathpages.com/rr/s2-07/2-07.htm > > according to your theory, OK. True, but according to your theory dt=0 for constant speed. The formula on that page is empirically proven too and I suspect if you work through Ritz you will find any effect is proportional to the area too. In fact you explain why in another post: "Henri Wilson" <HW@..> wrote in message news:rn97p1lh5hom3b79j4jh64oqec9d1ct7jh(a)4ax.com... > On 4 Dec 2005 12:02:40 -0800, "Jerry" <Cephalobus_alienus(a)comcast.net> wrote: > >... Mathematically, the sensitivity the device > >turns out to be directly proportional to the area. > > Only because v=w.r. Exactly. The speed depends on r and so does the length of the path (circumference of a circular loop or perimeter of an inscribed square) so the overall effect is expected to go as the square of the radius. > You still multiply by the number of turns. That's the direct way to do it of course, work out dt for a single turn of radius 15mm and then multiple by 315 turns. I expect you to get the same as for a single loop of radius 15*sqrt(315) mm but you can check that for yourself. > >> It is fibre > >> length that you need to know. That is about 10 metres....equivalent to a > >> four > >> mirror sagnac with 2.0 metre diagonals. > > > >Imagine a fibre folded in half and laid in a straight > >line. Obviously there would be no difference. It's not > >just length that matters but the integral of a factor > >involving the radius along the length. It works out to > >be the area. > > according to your theory, OK. Indeed, but as we have seen, Ritz says dt depends on the acceleration. Other than that I expect it still to be proportional to the area for the reasons given above. > >>>The KVH spec says the maximum rotation rate is > >>>100 degrees/second for the analog output. I'll let > >>>you work out whether that is multiple fringes or a > >>>fraction of a fringe. > >> > >> That will depend on which theory I use. > >> The peripheral speed is around 3 m/sec. For a single loop of a 30mm diameter coil, v=26mm/s. > >> Using your aether theory (alias SR) > > > >Very funny Henri. It always amuses me that you > >can't resist showing that you haven't a clue > >about SR. I suggest you read the last three > >paragraphs of this page: > > > >http://www.mathpages.com/home/kmath169/kmath169.htm > > Obviously written by a relativist. > > > > >Aether theories struggle with Sagnac, SR > >has no problem at all. > > Except that it would result in NO fringes at all. The whole image would be > uniform in colour. Well done, I kept telling you there were no fringes in an iFOG. A photodiode measures the intensity which varies because dt produces a phase difference. When you add two sine waves, you get an amplitude that depends on the phase difference. I'll explain how the fringes are formed in the lab version separately, I'm out of time now. > Incidentally, your theory say 'light moves at c in a vacuum'. In the SR > analysis of sagnac, what is the reference for that speed? Is it the "point in > absolute space where the source was when the light was emitted?" There's no point wasting time on asides Henri, if you want to understand SR, you'll need to make the effort. > ...it's quite amusing really. > > You people have no idea that you are merely preaching LET. It is amusing, you have no concept of anything other than LET and Ritz, just like Sagnac. > >> that the light speed is c and not c/n or > >> (c+v/root2)/n.... > > > >According to SR, the speed is c/n in the > >rotating frame or c/n + v(1-n^2)/sqrt(2) in > >the lab frame. > > > >On the above page you'll find the maths that > >shows the effect is proportional to area for any > >arbitrary polygon of mirrors and a less detailed > >explanation of why the time difference is > >independent of the refractive index. > > > >> the travel time around the path is 3.3E-8 secs, during which > >> the periphery moves 10^-7 m.....now double that for the two paths. > >> = 2E-7 m or 0.2 microns. > >> ~1/3 of a wavelength. > > > >Unless I have an error in my arithmetic, the time > >difference between the two paths is: > > > > dt = 4Aw/c^2 ~ 1.73*10^-17s > > > > c dt = 5.18 nm > > > >For 600nm light which you assumed, the fraction > >is 0.0086 of a wavelength. For such low values, > >sin(x) ~ x so the output of the detector would be > >proportional to the angular speed (according to > >SR and in reality, proportional to the angular > >acceleration according to Ritz). > > But of course we know differently now, don't we George. The difference beteen your figure and mine is roughly sqrt(315) which is explained by the bit about area above so I think in general we will agree once you work out a single loop and multiply by 315. Bottom line is that we both think dt is only a fraction of a cycle ("fringe") so counting dark/light transitions is not the way the devices work. In fact the analogue output is just an amplified version of the demodulator signal. > We know that the rays which take the same travel time according to Ritz, do NOT > end up on the same point on the image. > Therefore they are not the ones which determine the state of the interference > pattern at all. I think you missed some of my replies at the end of last week, I already pointed out that your diagram that we used to derive t'=t includes that effect: "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:dmidot$u6p$1(a)news.freedom2surf.net... > > "Henri Wilson" <HW@..> wrote in message > news:aa4no19bd927eqmskbfgrscmro3vculefs(a)4ax.com... > > > Yes OK you were right. > > > > Trouble is, the rays that have the same travel times aren't > > the ones that meet at the same points. > > We must only consider those that DO reunite at the same points. > > Your diagram already copes with that. I have added > two wavefronts in colour to illustrate the point: > > http://www.briar.demon.co.uk/Henri/sagnac.gif > > Think of them as arcs centred on A or farther away > for a laser source. > > The red line C-C' is the wavefront when it hits the > detector at point C when the table is not turning. > The blue line D-D' is the wavefront when it hits the > detector at point D when the table is turning at > constant speed. Note that your existing lines show > the two rays, A-C and A-D. > > Consider the constant speed case: the light that > would have hit the detector at C is at C' when > the wavefront reaches D and misses the detector. > In the non-rotating case photons moving along the > A-D ray miss the detector. George
From: Jerry on 5 Dec 2005 09:50
Jerry wrote: > Henri Wilson wrote: > > On Sun, 4 Dec 2005 17:14:26 -0000, "George Dishman" <george(a)briar.demon.co.uk> > > wrote: > > > > > > > >"Henri Wilson" <HW@..> wrote in message > > >news:3t14p19pfdumiaijpi8rq2e4c25vbe5qqk(a)4ax.com... > > > >> I cannot see why you consider loop area to be the important factor. > > > > > >Because the output is proportional to the > > >enclosed area: > > > > > >http://www.mathpages.com/rr/s2-07/2-07.htm > > > > according to your theory, OK. > > > > > > > >> It is fibre > > >> length that you need to know. That is about 10 metres....equivalent to a > > >> four > > >> mirror sagnac with 2.0 metre diagonals. > > > > > >Imagine a fibre folded in half and laid in a straight > > >line. Obviously there would be no difference. It's not > > >just length that matters but the integral of a factor > > >involving the radius along the length. It works out to > > >be the area. > > > > according to your theory, OK. > > Henri, re-work your Sagnac simulation so that the light path > is an elongated rectangle of zero width, zero area, i.e. > > |=====================================| > > You have converted the Sagnac interferometer into HALF of an MMX interferometer > an MMX interferometer, which is not sensitive to rotation > by ANY theory, Lorentz, SR, aetheric, BaT... Sorry, typo. Jerry |