Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 29 Nov 2005 08:43 jgreen(a)seol.net.au wrote: > George Dishman wrote: > > <jgreen(a)seol.net.au> wrote in message > > news:1133131335.712029.264010(a)g43g2000cwa.googlegroups.com... > > > > > > DHR's get so involed > > > with the magic of "c", that the don't realise that they USE magic, in > > > order to SHOW it! > > > > Keep wishing Jim, Santa's coming. > > (sigh) He doesn't love me ;-( I know. In this Sagnac discussion, you can see the speed is determined without first assuming it. > For years I've been asking for an explanation on HOW light from the > Xmas candles strike me at the SAME speed, when I walk from one toward > another. Disappointment year after year................... > .............and so little to ask for................ The trouble is that you look at the wrapping paper and don't like the pattern so you never open the box. I've tried to explain it to you a few times but you never try to understand, you just nitpick. Nobody can force understanding onto you Jim, you have to make the effort - open the present and look inside. If you want to discuss it again I'll be happy to do that but if you aren't prepared to try, there's no point. > ref sagnac: the whole shebang constitutes the equipment; airframe, > opticals, and spinning turntable. I think perhaps you are confusing two things. Sometimes Henri and I talk about "the Sagnac experiment". In that, there is a turntable in a lab and the light source and detector are mounted on the table. Some optics allows the fringes to be seen in the lab but after they have been formed. In the other case, we are discussing the iFOG devices you and I looked at last year. These can be mounted in vehicles or planes but there is no turntable as such. The vehicle or airframe itself plays the part of the turntable. The components are rigidly fixed to the chassis. > It is the relationship between these > and the STARTING rotation of the lot, and the observed fringe > displacements, which are at issue. Not really, we know that the relationship is that the fringe displacement is a measure of the rotation rate, not the heading. There is another type of device called a ring laser gyro where the fringes act more like the needle on a compass and indicate heading but we aren't discussing that type of device at all. The question is the mechanism that produces the shift in your Ritzian model when applied to iFOGs because a straightforward analysis says the fringes should NOT be displaced when turning at constant rate and we know they are. > nb: Starting rotation being the airframe on the tarmac, sagnac spinning > at an agreed rate/sec which will not alter according to the onboard > clock checking rotations/sec, and the position of the fringe marked > zero. No, when the aircraft is sitting on the tarmac, it is rotating once per day along with the Earth. There is no turntable in the aircraft, the device is bolted straight onto the airframe. George
From: George Dishman on 29 Nov 2005 15:38 "Henri Wilson" <HW@..> wrote in message news:aa4no19bd927eqmskbfgrscmro3vculefs(a)4ax.com... > On Mon, 28 Nov 2005 21:08:30 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: >>"Henri Wilson" <HW@..> wrote in message >>news:i087o1h7iv95uoahv80vdv3m9v99l2a2iu(a)4ax.com... >>> On Tue, 22 Nov 2005 21:44:45 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>>>"Henri Wilson" <HW@..> wrote in message >>>>news:sli4o19bq13po9n86sold869obkic3gka0(a)4ax.com... I think I've snipped to get the essential points, my apologies if you think it is inaccurate, there was far too much history. >>>>> I think you are trying to say that even though the path lengths of the >>>>> two beam change during acceleration and remain changed by a constant >>>>> amount >>>>> during constant rotation, the travel time of light in each beam is >>>>> always the >>>>> same. >>>>> >>>>> I say it is slightly different but constant.... and so there is a no >>>>> fringe movement during constant rotation. >>>> >>>>OK but that's what we have been arguing over for >>>>months, you are now going back to what we were >>>>saying before you brought the question of >>>>acceleration into it. .... >>>>> You are claiming the fringe ''''movement'''' is a function of da/dt. >>>>> I would like to see your proof. >>>> >>>>Let's stick with the term we agreed, I am saying fringe >>>>displacement is proportional to dv/dt. My proof is your >>>>diagram where dv/dt is called "a" and is in the "1/2at^2" >>>>term. .... >>>>> See my comment above about da/dt >>>>> Let's clear that up. >>>> >>>>Yes please, that is fundamental. Please consider >>>>what I said in the earlier post, try plotting your >>>>"vt+1/2at^2" from the diagram using this profile. >>>> >>>>> _________ >>>>> / \ >>>>> / \ ^ >>>>> __________/ \_________ | speed >>>>> | >>>>> __________________________________ >>>>> >>>>> ------> >>>>> time >>>> >>>>Thirty seconds with the back of an envelope should >>>>show you what I am saying. I can't draw it easily so >>>>help me out here Henri. >>> >>> All you have drawn is an increase in speed to a constant followed by an >>> identical decrease which bring it back to the starting point. >>> the output is zero becasue there has been no net rotation. >>> >>> During the diagonal sections, the fringes will move the same amount in >>> opposite directions. >> >>The "vt" part does that but the "at^2/2" part >>adds a further fixed offset. >> >>> During the constant v section in the middle, the fringes will be >>> displaced >>> and the output will indicate the rate of rotation. >>> >>> Where is your problem?. >> >>I wanted you to consider how the value compares >>during the periods of acceleration. Too late, >>I gave you the answer in an earlier reply. > > 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: donstockbauer on 29 Nov 2005 15:49 Going for the gold: this post ups the count to 1162. Have a nice day. What's a troll?????????
From: George Dishman on 29 Nov 2005 16:46 "Henri Wilson" <HW@..> wrote in message news:pjumo1p9d978pli6l3llvkvaq5edjaprrr(a)4ax.com... > On Mon, 28 Nov 2005 20:07:21 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: >>"Henri Wilson" <HW@..> wrote in message >>news:06c7o1ptp360adjulvbpol9ved8io57jk8(a)4ax.com... .... >>> Maybe we should start all over again. >> >>I'll see if I can start from the beginning then, >>maybe some aspect is slipping past without you >>noticing. I am still just trying to get you to >>see the consequences of your own drawing though, >>I'm not introducing anything new. >> >> http://www.users.bigpond.com/hewn/sagnac.jpg >> >>Path length at rest is distance L from point A to >>point C in time t. The light moves at speed c so >> >> t = L/c. >> >>When the table is rotating at constant speed, light >>is emitted at point A but the table moves a distance >>vt from point C to point D while the light is in >>transit. I'll add a single quote to indicate the >>values for this case where they differ from the >>static situation. >> >>The speed of the light in the direction you have >>shown has been increased by Ritz to >> >> c' = c + v/sqrt(2) >> >>where v is the tangential speed of the source. The >>path length increases to >> >> L' = L + vt'/sqrt(2) >> >>and of course >> >> t' = L'/c' >> >>Remember we are interested in the difference between >>the two paths and the other has been shortened but >>the light slowed. The results are symmetrical so I'll >>only do the maths for one. (Just use -v instead of v >>for the other.) >> >>To save typing, note that v appears with sqrt(2) in >>both so let >> >> V = v/sqrt(2) >> >>so: >> >> c' = c + V >> >> L' = L + Vt' >> >>Solve for t': >> >> t' = L'/c' >> >> c't' = L' >> >>Substituting c' and L' we get: >> >> (c + V)t' = (L + Vt') >> >> ct' + Vt' = L + Vt' >> >> ct' = L >> >> t' = L/c >> >> t' = t >> >>So constant speed rotation produces no change to the >>time taken for the light to travel the increased path >>length, the increased speed compensates. That means >>no fringe displacement because the displacement is >>determined by the time difference between the forward >>and backward paths. They are both equal (at t) when >>the table is stationary so the are still equal when >>it is rotating at constant speed. >> >>Notice also that the sqrt(2) factor affects both the >>path length and the boost in speed through the geometry >>so a different number of mirrors would give a different >>factor but it would still cancel out. > > Yes I have now done these calculations myself and I think you are right, > if we > ignore second order effects and make certain assumptions about the way > light > reflects from each mirror. > >>Now consider the case with acceleration. Definitions >>are critical here because v is now a function of time, >>it is increasing linearly. I'll add double quotes for >>this case. >> >>At some instant, a wavefront is emitted from the >>source. At that time, the tangential speed of the >>source is v. The trick here is that the light keeps >>the same speed once it is emitted while the table >>is accelerating hence as before >> >> c" = c + V >> >>so >> >> c" = c' >> >>When the light is in transit for time t", the table >>speeds up by a factor of >> >> dv = a t" >> >>The average increase during t" is half that so the >>distance by which the table moves more than the >>constant speed case is >> >> dL = (a t" / 2) t" >> >>The path length shown on the diagram is given by >> >> L" = L + Vt" + a t"^2 / 2 > > (How about L" = L + Vt" + A t"^2 / 2, where A=a/sqrt2) Yes, you are correct. It reduces the amplitude of the signal you would get according to Ritz but doesn't change the logic, the predicted result is still proportional to acceleration, not speed. >>Obviously >> >> t" = L"/c" > > (To first order only. L is not constant... but the 2nd order effect is > very > small.) No, it is exact. L" and c" are specific values for any particular wavefront so the time taken is the distance divided by the speed. The values of L" and c" for the the next wavefront will both be greater but these equations hold for each individually. It is not until later that it is shown that t" = t and then we can conclude that t" is the same for all wavefronts. At this stage it might have a different value in each case. >> t" = (L' + dL) / c' >> >> t" = L'/c' + dL/c' >> >> t" = t + dt >> >>where >> >> dt = (a t"^2 / 2) / c' >> >>That is what you should expect when you realise that, >>if a=0, we are back to the previous situation and the >>Ritzian change to the speed of the light cancels the >>effect of the Vt" term as it did before. > > OK so far. (apart from the a/sqrt2) Excellent, we agree entirely. >>For the other path however, the acceleration _decreases_ >>the path length so this time we do not get cancellation >>and there is going to be a fringe displacement that >>depends on the acceleration. > > I don't think that is right George. The V terms still cancel but the A > terms > for both directions still add...so we get double the effect. > > Maybe I'm missing something here. You misread me somehow, what I said was the same, there IS going to be a fringe displacement. Maybe I wasn't clear enough but we are in complete agreement. >>The displacement as a >>fraction of a fringe is the ratio of the time difference >>to the period. If the frequency of the source light is F >>then the period is P = 1/F and the displacement is given >>by >> >> D = 2 dt / P >> >> D = 2 F dt >> >> D = [2 F t"^2 / c'] a >> >> >>For v << c, dt << t so the result is proportional to the >>acceleration. >> >>For this speed profile: >> >> _________ >> / \ >> / \ ^ >> __________/ \_________ | speed >> | >> __________________________________ >> >> ------> >> time >> >> >>Ritz predicts this fringe displacement: >> >> __ >> +ve | | ^ >> | | | displacement >> | | | >> 0 _______| |_______ ________ >> | | >> ------> | | >> time | | >> -ve |__| > > That's all very well George. ..but quite irrelevant in light of my latest > findings. Perhaps, but it's taken months to get to the point where we agree it so I want to make sure it is noted. I don't want to have to come back and do this all again next year. Having arrived at the point where we share a view on the path length analysis, I'm happy to move on to look at the next point you raise. > We have to go right back to basics about why fringes are formed at all. > > You know how to make 'line fringes' rather than circular ones using an > 'optical > wedge'.. ..so let's use line fringes. They aren't really the same Henri, lines are produced by a lateral displacement for sources at equal path length whereas in the Sagnac the circular fringes are produced by sources in line at different distances from the screen. > If the beams are perfectly parallel and coherent, they will alternatively > reinforce or destructively interfere when they unite at the final surface > and > lines will appear. For line fringes, if they are perfectly parallel, you get no fringes, the signals maintain constant phase across the screen. I think it would be better to stick with the actual Sagnac configuration where you do get fringes if the beams are parallel but from different distances. I'll maybe sketch that if I get some time later in the week. > Now, what I have shown with my demo: > www.users.bigpond.com/hewn/sagnac1.exe > is that rays that start out at exactly 90 apart, do not reunite at the > same > point on the final mirror surface. Yet the fringe pattern is caused by > rays > that DO unite at the same points. > > So what is actually happening is that rays that DO reunite at the same > point > are NOT those which started out exactly 90degrees apart. The source beam > is of > course not perfectly parallel and has rays going in all possible diagonal > directions within its confines. > > So we have to calculate the difference in travel times of these very > slightly > diagonal rays that DO meet. Yes, again I agree that. In fact that was the important point I was making at the beginning of the month: "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:dk8epa$1kn$1(a)news.freedom2surf.net... >>In Ritzian theory, the light is emitted at some speed greater >>than c from the source. The speed can be found by taking the >>magnitude of the vector sum of the mirror velocity and a >>vector of magnitude c whose direction is such that the light >>eventually reaches the detector. ^^^^^^^^^^^^^^^^^^^^ > This is NOT the same as simply moving one beam > sideways at exactly the same angle. > We can then see how a relative sideways displacement will in fact result > in a > velocity dependent fringe displacement. > > What I am saying is pretty complicated and difficult to analyse. > > I hope I have described it adequately You have described it very well indeed. There are two aspects, the first, the effect of the slight offset itself, is already handled by your diagram but I've marked up a copy to make it more obvious: http://www.briar.demon.co.uk/Henri/sagnac.gif When the table is turning at constant speed, it is light emitted in the A-D direction that hits the detector at D instead of light emitted in the A-C direction when the table is stationary. The consequence is the vt term which we have agreed above is cancelled by the effect of the boost in speed from c to c'. The other aspect is a possible change in the angle between the beams. To sort that out, start by thinking about the car, the duck and the goose: > .. I like your > analogy for the four mirror case so let's look in a > bit more detail. First add a goose following the car > and another hunter shooting back at it: > > > <- Car > * > / \guns > / | \ > / | \ > / | \ > / 44 | 46 \ > / | \ > | / | \ ^ > v * + * | > Duck Goose > > > The "+" in the middle indicates the centre of the > roundabout and all are at equal distance R form > that and moving at the same speed, V, in the > directions indicated by the arrows. > > Now add a large sheet of paper stretched between > the duck, the car and the goose. The guns are fired > simultaneously and the lead bullets just skim the > surface of the paper leaving a drawn line. When nothing is moving, the guns shoot at 45 degrees to a radial line from the centre. When all are moving anti-clockwise as shown, the leading gun must aim ahead of the duck so shoots at less than 45 degrees (shown as "44") while the other gun must shoot at more than 45 degrees, shown as 46. The key here is that the angle between the guns remains 90 degrees because one increases while the other decreases. Now look at your animation www.users.bigpond.com/hewn/sagnac1.exe When it finishes, the enlargement shows that both beams must move slightly anti-clockwise, and by the same amount, to hit the detector. That means that angle between them will stay the same. Once re-combined by the beam splitter, if they started parallel when the table isn't turning then they remain parallel when it is. The fringes must be purely a result of the changed path lengths and Ritz cannot explain that. Remember also that in the iFOG case, we are measuring the actual time delay of the modulating waveform on the signal. There is no fringe pattern produced, the sensor only ever measures the intensity at the centre. George
From: Henri Wilson on 30 Nov 2005 04:09
On Tue, 29 Nov 2005 21:46:57 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:pjumo1p9d978pli6l3llvkvaq5edjaprrr(a)4ax.com... >> On Mon, 28 Nov 2005 20:07:21 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >>>"Henri Wilson" <HW@..> wrote in message >>>news:06c7o1ptp360adjulvbpol9ved8io57jk8(a)4ax.com... >... >>>> Maybe we should start all over again. >>> >>>I'll see if I can start from the beginning then, >>>maybe some aspect is slipping past without you >>>noticing. I am still just trying to get you to >>>see the consequences of your own drawing though, >>>I'm not introducing anything new. >>> >>> http://www.users.bigpond.com/hewn/sagnac.jpg >>> >>>Path length at rest is distance L from point A to >>>point C in time t. The light moves at speed c so >>> >>> t = L/c. >>> >>>When the table is rotating at constant speed, light >>>is emitted at point A but the table moves a distance >>>vt from point C to point D while the light is in >>>transit. I'll add a single quote to indicate the >>>values for this case where they differ from the >>>static situation. >>> >>>The speed of the light in the direction you have >>>shown has been increased by Ritz to >>> >>> c' = c + v/sqrt(2) >>> >>>where v is the tangential speed of the source. The >>>path length increases to >>> >>> L' = L + vt'/sqrt(2) >>> >>>and of course >>> >>> t' = L'/c' >>> >>>Remember we are interested in the difference between >>>the two paths and the other has been shortened but >>>the light slowed. The results are symmetrical so I'll >>>only do the maths for one. (Just use -v instead of v >>>for the other.) >>> >>>To save typing, note that v appears with sqrt(2) in >>>both so let >>> >>> V = v/sqrt(2) >>> >>>so: >>> >>> c' = c + V >>> >>> L' = L + Vt' >>> >>>Solve for t': >>> >>> t' = L'/c' >>> >>> c't' = L' >>> >>>Substituting c' and L' we get: >>> >>> (c + V)t' = (L + Vt') >>> >>> ct' + Vt' = L + Vt' >>> >>> ct' = L >>> >>> t' = L/c >>> >>> t' = t >>> >>>So constant speed rotation produces no change to the >>>time taken for the light to travel the increased path >>>length, the increased speed compensates. That means >>>no fringe displacement because the displacement is >>>determined by the time difference between the forward >>>and backward paths. They are both equal (at t) when >>>the table is stationary so the are still equal when >>>it is rotating at constant speed. >>> >>>Notice also that the sqrt(2) factor affects both the >>>path length and the boost in speed through the geometry >>>so a different number of mirrors would give a different >>>factor but it would still cancel out. >> >> Yes I have now done these calculations myself and I think you are right, >> if we >> ignore second order effects and make certain assumptions about the way >> light >> reflects from each mirror. >> >>>Now consider the case with acceleration. Definitions >>>are critical here because v is now a function of time, >>>it is increasing linearly. I'll add double quotes for >>>this case. >>> >>>At some instant, a wavefront is emitted from the >>>source. At that time, the tangential speed of the >>>source is v. The trick here is that the light keeps >>>the same speed once it is emitted while the table >>>is accelerating hence as before >>> >>> c" = c + V >>> >>>so >>> >>> c" = c' >>> >>>When the light is in transit for time t", the table >>>speeds up by a factor of >>> >>> dv = a t" >>> >>>The average increase during t" is half that so the >>>distance by which the table moves more than the >>>constant speed case is >>> >>> dL = (a t" / 2) t" >>> >>>The path length shown on the diagram is given by >>> >>> L" = L + Vt" + a t"^2 / 2 >> >> (How about L" = L + Vt" + A t"^2 / 2, where A=a/sqrt2) > >Yes, you are correct. It reduces the amplitude >of the signal you would get according to Ritz >but doesn't change the logic, the predicted >result is still proportional to acceleration, >not speed. > >>>Obviously >>> >>> t" = L"/c" >> >> (To first order only. L is not constant... but the 2nd order effect is >> very >> small.) > >No, it is exact. L" and c" are specific values for >any particular wavefront so the time taken is the >distance divided by the speed. > >The values of L" and c" for the the next wavefront >will both be greater but these equations hold for >each individually. It is not until later that it >is shown that t" = t and then we can conclude that >t" is the same for all wavefronts. At this stage >it might have a different value in each case. > >>> t" = (L' + dL) / c' >>> >>> t" = L'/c' + dL/c' >>> >>> t" = t + dt >>> >>>where >>> >>> dt = (a t"^2 / 2) / c' >>> >>>That is what you should expect when you realise that, >>>if a=0, we are back to the previous situation and the >>>Ritzian change to the speed of the light cancels the >>>effect of the Vt" term as it did before. >> >> OK so far. (apart from the a/sqrt2) > >Excellent, we agree entirely. > >>>For the other path however, the acceleration _decreases_ >>>the path length so this time we do not get cancellation >>>and there is going to be a fringe displacement that >>>depends on the acceleration. >> >> I don't think that is right George. The V terms still cancel but the A >> terms >> for both directions still add...so we get double the effect. >> >> Maybe I'm missing something here. > >You misread me somehow, what I said was the same, >there IS going to be a fringe displacement. Maybe >I wasn't clear enough but we are in complete >agreement. > >>>The displacement as a >>>fraction of a fringe is the ratio of the time difference >>>to the period. If the frequency of the source light is F >>>then the period is P = 1/F and the displacement is given >>>by >>> >>> D = 2 dt / P >>> >>> D = 2 F dt >>> >>> D = [2 F t"^2 / c'] a >>> >>> >>>For v << c, dt << t so the result is proportional to the >>>acceleration. >>> >>>For this speed profile: >>> >>> _________ >>> / \ >>> / \ ^ >>> __________/ \_________ | speed >>> | >>> __________________________________ >>> >>> ------> >>> time >>> >>> >>>Ritz predicts this fringe displacement: >>> >>> __ >>> +ve | | ^ >>> | | | displacement >>> | | | >>> 0 _______| |_______ ________ >>> | | >>> ------> | | >>> time | | >>> -ve |__| >> >> That's all very well George. ..but quite irrelevant in light of my latest >> findings. > >Perhaps, but it's taken months to get to the point >where we agree it so I want to make sure it is noted. >I don't want to have to come back and do this all >again next year. I will have a good look at this before I agree 100%. There seems to be something logically wrong here. > >Having arrived at the point where we share a view on >the path length analysis, I'm happy to move on to look >at the next point you raise. > >> We have to go right back to basics about why fringes are formed at all. >> >> You know how to make 'line fringes' rather than circular ones using an >> 'optical >> wedge'.. ..so let's use line fringes. > >They aren't really the same Henri, lines are produced >by a lateral displacement for sources at equal path >length whereas in the Sagnac the circular fringes are >produced by sources in line at different distances >from the screen. I have never seen the fringes in a sagnac interferometer. are you certain that circular fringes are used. It doesn't matter all that much because whatever method is used to measure fringe displacement will focus on a fairly straight section anyway. >> If the beams are perfectly parallel and coherent, they will alternatively >> reinforce or destructively interfere when they unite at the final surface >> and >> lines will appear. > >For line fringes, if they are perfectly parallel, you >get no fringes, the signals maintain constant phase >across the screen. That's right. That's why in the MMX, one mirror can be slightly tilted to make an optical wedge. In that case, the beams meet with sinusoidally varying phase differences across the final screen. 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? >I think it would be better to stick >with the actual Sagnac configuration where you do get >fringes if the beams are parallel but from different >distances. I'll maybe sketch that if I get some time >later in the week. If you are going to use spherical wavefronts with different diameters, I already don't like it. I was under the impression that the two sagnac paths were equal. >> Now, what I have shown with my demo: >> www.users.bigpond.com/hewn/sagnac1.exe >> is that rays that start out at exactly 90 apart, do not reunite at the >> same >> point on the final mirror surface. Yet the fringe pattern is caused by >> rays >> that DO unite at the same points. >> >> So what is actually happening is that rays that DO reunite at the same >> point >> are NOT those which started out exactly 90degrees apart. The source beam >> is of >> course not perfectly parallel and has rays going in all possible diagonal >> directions within its confines. >> >> So we have to calculate the difference in travel times of these very >> slightly >> diagonal rays that DO meet. > >Yes, again I agree that. In fact that was the important >point I was making at the beginning of the month: > >"George Dishman" <george(a)briar.demon.co.uk> wrote in message >news:dk8epa$1kn$1(a)news.freedom2surf.net... >>>In Ritzian theory, the light is emitted at some speed greater >>>than c from the source. The speed can be found by taking the >>>magnitude of the vector sum of the mirror velocity and a >>>vector of magnitude c whose direction is such that the light >>>eventually reaches the detector. > ^^^^^^^^^^^^^^^^^^^^ > >> This is NOT the same as simply moving one beam >> sideways at exactly the same angle. >> We can then see how a relative sideways displacement will in fact result >> in a >> velocity dependent fringe displacement. >> >> What I am saying is pretty complicated and difficult to analyse. >> >> I hope I have described it adequately > >You have described it very well indeed. > >There are two aspects, the first, the effect of the >slight offset itself, is already handled by your >diagram but I've marked up a copy to make it more >obvious: > > http://www.briar.demon.co.uk/Henri/sagnac.gif > >When the table is turning at constant speed, it is >light emitted in the A-D direction that hits the >detector at D instead of light emitted in the A-C >direction when the table is stationary. The >consequence is the vt term which we have agreed >above is cancelled by the effect of the boost in >speed from c to c'. It is for path from the source to the first mirror anyway. I'm not completely convinced about what happens to the subsequent sections. > >The other aspect is a possible change in the angle >between the beams. To sort that out, start by >thinking about the car, the duck and the goose: > >> .. I like your >> analogy for the four mirror case so let's look in a >> bit more detail. First add a goose following the car >> and another hunter shooting back at it: >> >> >> <- Car >> * >> / \guns >> / | \ >> / | \ >> / | \ >> / 44 | 46 \ >> / | \ >> | / | \ ^ >> v * + * | >> Duck Goose >> >> >> The "+" in the middle indicates the centre of the >> roundabout and all are at equal distance R form >> that and moving at the same speed, V, in the >> directions indicated by the arrows. >> >> Now add a large sheet of paper stretched between >> the duck, the car and the goose. The guns are fired >> simultaneously and the lead bullets just skim the >> surface of the paper leaving a drawn line. > >When nothing is moving, the guns shoot at 45 degrees >to a radial line from the centre. > >When all are moving anti-clockwise as shown, the >leading gun must aim ahead of the duck so shoots >at less than 45 degrees (shown as "44") while the >other gun must shoot at more than 45 degrees, shown >as 46. > >The key here is that the angle between the guns >remains 90 degrees because one increases while the >other decreases. There is a bit of a trap here. You have to consider the velocity of the gun as the bullet is fired. The two angles are not quite symetrical about 45 deg. In my diagram, I have represented those components by F-D and G-E. If you draw them on your diagram in the two directions, you will see what I mean. So the angle does not remain at exactly 90. > >Now look at your animation > > www.users.bigpond.com/hewn/sagnac1.exe > >When it finishes, the enlargement shows that both >beams must move slightly anti-clockwise, and by the >same amount, to hit the detector. That means that >angle between them will stay the same. > >Once re-combined by the beam splitter, if they >started parallel when the table isn't turning then >they remain parallel when it is. The fringes must >be purely a result of the changed path lengths and >Ritz cannot explain that. > >Remember also that in the iFOG case, we are measuring >the actual time delay of the modulating waveform on >the signal. There is no fringe pattern produced, the >sensor only ever measures the intensity at the centre. > >George George, I'll try to draw what I'm trying to say. Use fixed pitch fonts. Consider two beams meeting each other as in a sagnac. They alternatively reinforce and destruct across the screen where they meet. The +'s mark the points where the beams reinforce and bright fringes appear. (It could be a section of a circular fringe pattern) + + + + + | | | | | -> | | | | | _____|_|_|_|_| _____|_|_|_|/| _____|_|_|_/|| _____|_|_|/||| _____|_|_/||||^ _____|_|/|||||| _____|_/|||||| _____|/||||||| ||||||||| ||||||||| Q1) Why do fringes form at all? Q2) If one beam is moved sideways by half a fringe, what happens to the fringe pattern? Q3) If the two beams reunite at not quite 90 degrees and their path lengths are slightly different, what happens to the fringe pattern. Q4) What other alternatives are there? 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". |