Ballistic Theory, Progress report...Suitable for 5yo Kids
in [Relativity]
|
Prev: OWLS is not equal to c
Next: Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
From: Jerry on 5 Dec 2005 09:51 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
From: Henri Wilson on 5 Dec 2005 15:16 On 5 Dec 2005 05:26:22 -0800, "Jerry" <Cephalobus_alienus(a)comcast.net> 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 an >MMX interferometer, which is not sensitive to rotation >by ANY theory, Lorentz, SR, aetheric, BaT... Are you trying to tell me that the number of turns of the optic fibre is not important in the result? I don't think you know anything about the sagnac effect. > >Jerry HW. www.users.bigpond.com/hewn/index.htm
From: Jerry on 5 Dec 2005 15:35 Henri Wilson wrote: > Are you trying to tell me that the number of turns of the optic fibre is not > important in the result? > > I don't think you know anything about the sagnac effect. N turns allows you to enclose N times the effective area in the same limited enclosure. But the math is very clear about the proportionality between area and magnitude of observed effect. Why did Michelson enclose a 1100 by 2100 foot field for his demo in the Michelson-Gale experiment, rather than simply bounce the light in a narrow closed loop of, say, 1 x 3200 feet? Jerry
From: Henri Wilson on 5 Dec 2005 16:01 On 5 Dec 2005 06:19:18 -0800, "George Dishman" <george(a)briar.demon.co.uk> wrote: >Henri Wilson wrote: >> On Sun, 4 Dec 2005 17:14:26 -0000, "George Dishman" <george(a)briar.demon.co.uk> >> wrote: >> >> >>>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. I think you and the standard 'SR' explanation are completely wrong. I go back to my earlier claim that light has a 'built-in gyro' in the form of an axis, which doesn't particulalrly want to change angle. > >> >>>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. At least we agree on something.... ....but what about a four mirror system? > >I'll explain how the fringes are formed in the lab version >separately, I'm out of time now. I'll bet you can't even explain why fringes are formed even in the MMX. >> 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. Don't run for cover with your tail between your legs George. According to aether theories all observers will measure OWLS to be c in their respective frames. It will actually move at c wrt the point in the absolute aether where he happens to be at the instant of emission...but to him it moves at c wrt himself. You use that exact concept in your analysis...that there exist 'absolute points in space'. >> ...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. Yoiu don't like it when the truth about SR is revealed do you George. It is just rehashed aether theory. >> >> 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. I still can't see how the use of a modulated beam provides a more sensitive system...but no matter. >> 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. If the beam is exactly parallel, why are there beams traveling in different directions anyway? That is the important question...along with the relative travel times of those which DO meet at the same points. This is a rather difficult problem. My animation 'sagnac.exe' is wrong in that it shows the rays heading for the centres of the next mirrors. They should be aimed at the points where the centres will be when the rays arrive...not easy to animate...but I will try. > >George HW. www.users.bigpond.com/hewn/index.htm
From: Henri Wilson on 5 Dec 2005 18:11
On 5 Dec 2005 12:35:06 -0800, "Jerry" <Cephalobus_alienus(a)comcast.net> wrote: >Henri Wilson wrote: > >> Are you trying to tell me that the number of turns of the optic fibre is not >> important in the result? >> >> I don't think you know anything about the sagnac effect. > >N turns allows you to enclose N times the effective area in the same >limited enclosure. But the math is very clear about the proportionality >between area and magnitude of observed effect. Like I said, only because v=w.r > >Why did Michelson enclose a 1100 by 2100 foot field for his demo >in the Michelson-Gale experiment, rather than simply bounce the >light in a narrow closed loop of, say, 1 x 3200 feet? see http://renshaw.teleinc.com/papers/fizeau/fizeau.stm > >Jerry HW. www.users.bigpond.com/hewn/index.htm |