Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 3 Dec 2005 06:14 "Henri Wilson" <HW@..> wrote in message news:i087o1h7iv95uoahv80vdv3m9v99l2a2iu(a)4ax.com... > >>> Like I said, I can't find a decent description of FoGs and their design >>> features. >>> ....Nothing much on google. >> >>Well the page I gave you has al the main features. There >>is another but I think the only thing it added was that >>telecomms laser diodes have built-in photodiodes for >>power management which are used instead of a discrete >>device but the principles and equations are as shown. >>What more are you looking for? > > Number of fibre turns, path length and typical fringe displacements. 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. 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. It's not perfect by any means but it should give you a reasonable idea of the magnitude of the numbers. George
From: Henri Wilson on 3 Dec 2005 15:51 On Sat, 3 Dec 2005 10:08:59 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:jklqo1d0l52ei3jv06scq9p7mu9309r8df(a)4ax.com... >> On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: > >> Frankly, I'm not quite sure how circular fringes are formed in either >> interferometer. If the beams were both perfectly parallel and coherent, >> there >> should be no phase difference at any point across the whole viewing area. >> I suppose if one considers that the two uniting wavefronts are spherical >> rather >> than flat, one gets a type 'Newton's rings' pattern...but why aren't the >> wavefronts flat? > >Henri, I just realised you may be forgetting the effect >of the eyepiece. Planes waves passing along the paths >are focussed to a point by that. Otherwise there wouldn't >be anything in the system to determine the centre of the >circular pattern where you have an extended source and >flat mirrors. No I hadn't forgotten the eyepiece, I know it plays a role. In effect, it focusses on the source via both paths. ....but I still cannot see why fringes are formed if both beams are perfectly parallel and coherent. > >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".
From: Henri Wilson on 3 Dec 2005 16:20 On Sat, 3 Dec 2005 11:14:22 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:i087o1h7iv95uoahv80vdv3m9v99l2a2iu(a)4ax.com... >> >>>> Like I said, I can't find a decent description of FoGs and their design >>>> features. >>>> ....Nothing much on google. >>> >>>Well the page I gave you has al the main features. There >>>is another but I think the only thing it added was that >>>telecomms laser diodes have built-in photodiodes for >>>power management which are used instead of a discrete >>>device but the principles and equations are as shown. >>>What more are you looking for? >> >> Number of fibre turns, path length and typical fringe displacements. > >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. 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. >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) that the light speed is c and not c/n or (c+v/root2)/n....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. >It's not perfect by any means but it should give you >a reasonable idea of the magnitude of the numbers. > >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".
From: George Dishman on 4 Dec 2005 12:14 "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 > 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. >>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 Aether theories struggle with Sagnac, SR has no problem at all. > 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). George
From: Jerry on 4 Dec 2005 15:02
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. Jerry |