Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 28 Nov 2005 16:08 "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 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. >> >>> The number of wavelengths in each path is different (you claim only >>> fractionally, no matter) >> >>Erm, no I have never mentioned the number of waves >>in the path, it has almost no influence. The output >>depends on the time difference between the two halves >>of the signal getting to the detector and not the >>actual time taken from their emission. > > Consider this. > No rotation. ..path lengths equal...100000000 wavlengths in each path > (let's > say). > Short period of acceleration... I'm assuming you next talk about the time after the speed has stabilised at the new value: > one path increases to 100000000.1 > wavelengths...the other decreases to 99999999.9.... OK. > fringes have moved 0.2 fringe widths. No. One path has increased in length but the light is moving faster due to Ritz so the time taken is as before. The other path has shortened but the light is moving slower and again the time taken is as before. The two beams arrive in sync so there is no fringe displacement. > They don't go back to zero just because the acceleration ceases. You see > to > think they do. See my other post where I show that your diagram requires that they do. > This also proves the travel times around each path cannot be equal under > constant c+v, as you claim. Again see the more detailed post where I show that t' = t. >>>>As the acceleration build up to the constant value, there >>>>is a beat which is counted, the fractional part being added. >>>>(The counter would increment each time the fractional part >>>>rolled over.) While the acceleration is constant, the >>>>displacement is constant so the counter output is constant. >>> >>> No George. >>> You are confusing constant acceleration with constant speed. >> >>No, I'm not confusing them, I am saying your words are >>wrong but your diagram is right. The diagram shows what >>I say above. > > It matters not whether the acceleration is constant or not. > If it is +ve, the path lengths will diverge, if -ve, they come together. According to Ritz, the displacement is proportional to the acceleration at all times whether it is constant or not (for reasonable values, i.e. where it isn't fast enough to create an event horizon). >>>>While the acceleration decreases to zero, the fringes would >>>>pass in the other direction so the couter would need to >>>>decrease until it should be back at zero when the rotational >>>>reaches zero. >>> >>> No that is not right. If the acceleration is +ve, the fringes move one >>> way. >> >>Not according to your diagram. It says the path length is >>increased by vt+1/2at^2 and we know the vt part is cancelled >>by the Ritzian speed change (c+kv) so the path length is >>altered by 1/2at^2 which is constant, and the path length >>determines the fringe displacement. > > Why do you say "1/2at^2 is constant"? > The value of t signifies the end of the acceleration period. No, that value is the time for wich the light is travelling. The speed of the light is set according to Ritz at the moment it is emitted. The extra path length (compared to the situation of constant speed at the value the table had when the light was emitted) is dependent on how much the table accelerates while the light is "in the air". It increases by dv = at. Later emissions start at progressively faster speeds as the table accelerates but that speeds up the light too. Only the Vt term in the path length depends on the actual speed, and that is cancelled out by the Ritzian boost. > During that time, the path lengths diverge by at^2/root2 (for constant a > otherwise (integral a(t)dt)^2/root2). That's about right though see my other post which shows why the root(2) also gets cancelled. >>> It >>> matters not whether the magnitude of that acceleration is increasing or >>> decreasing. >>> 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. > > George please try to understand the difference between fringe > 'displacement' > and fringe 'movement'. The former is the static situation, measured from > zero, > the latter the process of actually moving. That is the sense in which I am using them. > The fringe displacement is proportional to v, not dv/dt. In reality, yes. According to SR, yes. According to Ritz, no. Ritz says the fringe displacement should be proportional to the acceleration. Ritz is wrong. > That means they DO NOT move during constant v. > > The rate of fringe movement is proportional to a (and maybe a secondary > effect) > The change in fringe displacement after an acceleration period is > proprtional > to the change in v. > >>I have already spent months showing you the "vt" >>part is cancelled by the speed change, or think of the >>carousel analogy which shows it nicely, or do "the duck, >>the car and the goose" diagram. All of them prove it. > > You haven't proved that at all. You forgot the root2 again. See my other post where I do all the math. The sqrt(2) cancels out and the speed boost exactly cancels the path length increase for the constant speed situation. > I say the travel times around the two paths are 'rotation speed' > dependent. > they have to be.... because the number of wavelengths in each path is > speed > dependent and wavelength doesn't change under BaTh. Don't just "say" Henri, opinion carries no weight. Show your math as I have done. >>>>The output of that counter would therefore be a digital >>>>indication of the acceleration, not the speed. >>>> >>>>One point Henri, in this part you seem to have lost the "vt" >>>>term which you mention above. I think you need to consider >>>>just which of the terms is responsible for the output here, >>>>it's quite fundamental. >>> >>> The at^2/2 is responsible for fringe movement. >>> The vt is responsible for fringe displacement during constant rotation. >> >>See my earlier post on that. > > It was very confused. Too much history and back quoting perhaps. I have redone it froma scratch. >>>>Both integrators need to have an initial value placed in them. >>> >>> yes >> >>Bu&*&&*r. Now you are back to agreeing with me. > > All integrators need an initial reference value. I know, but you are trying to convince me that Ritz only calls for ONE intgrator because it gives an output proportional to speed while I an saying it needs TWO to determine heading because the output is proportional to acceleration, and you just said "both" >>>>> A second continuous integration of instantaneous displacement with >>>>> time >>>>> gives >>>>> the rotation angle from zero. >>>> >>>>Exactly. >>>> >>>>> >You could imagine that this is how commercial units >>>>> >work (it isn't) but that wouldn't apply to the lab >>>>> >experiment. Remember in the original experiment of >>>>> >Sagnac, he saw a shift of 7% of a fringe while the >>>>> >table was turning at 120rpm. What you are describing >>>>> >would be that he counted 0.07 fringes during the >>>>> >acceleration phase and the displacement returned to >>>>> >zero once constant speed was achieved. >>>>> >>>>> I dont think that would be accurate enough for any practical purpose. >>>>> I'm sure fringes move a lot more in multi turn FoGs. >>>> >>>>Well see the numbers above. >>> >>> 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. Number of turns and path length is too detailed for ordinary documents (I already wrote this in another reply). "Fringe displacements" won't be mentioned since there are no fringes, there is a synchronous demodulator measuring the amplitude of the signal from the optical modulator. >>>>That's not what you described above where the fringe >>>>counter is supposedly doing the integration. >>> >>> It registers the answer. >> >>OK, that would be correct but now look at the description >>on the page I cited. There is no counter and no fringes, >>the signal is being measured as the amplitude of a 5.05MHz >>signal used in the modulator. > > Yes, I'm not sure how that works. Which part, the synchronous demodulator? >>> 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. George
From: Henri Wilson on 28 Nov 2005 18:15 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... >> >> Yes I accept he was wrong for the reasons stated in the other message. >> >>>> The fringe displacement only changes during an acceleration. >>> >>>The fringe displacement is determined by the path length >>>difference. Your diagram shows that termm as 1/2at^2 so >>>if a=0 then path length difference=0 therefore fringe >>>displacement is zero. >> >> yes. If v =0 then displacement is zero. > >I said no displacement when a=0, not when v=0. >You said "OK" to the reason behind this above >but this is just for the record, I'll come back >to this later as it needs a clean start as you >say at the bottom. > >> (But the electronically integrated output is not. It shows angle turned).. >> >>>> If acceleration >>>> varies with time, the fringe movement automatically integrates that. >>> >>>There is no integrator. >> >> Path length is the integrated result. >> >>>> The fringes stay where they are when the acceleration ceases. >>> >>>Not according to your diagram. >> >> You are confusing 'acceleration ceases' with 'rotation ceases'. > >No confusion, I have just reached a different >conclusion from you. All the details are below. > >> When acceleration ceases, v is larger than it was and vt is therefore >> longer >> than it was. > >I know but since v is larger, the fact the path >length is longer need not mean that the time of >flight is greater. In fact it remains the same. > >>>> That means the output is constant during constant rotation and is >>>> determined by >>>> that rotation speed. >>>> >>>> Your logic has gone astray George.... maybe you are working too hard. >>> >>>There is no logic involved at all Henri, I am just >>>quoting your own page back at you. If you don't like >>>it, redraw it. >> >> 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) > >Obviously > > t" = L"/c" (To first order only. L is not constant... but the 2nd order effect is very small.) > 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) > >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. >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 |__| > > >George > That's all very well George. ..but quite irrelevant in light of my latest findings. 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. 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. 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. 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 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 28 Nov 2005 18:21 On Mon, 28 Nov 2005 17:29:37 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:nnc7o154bd4j6v0som2prparadf6meijsp(a)4ax.com... >> On Tue, 22 Nov 2005 23:21:43 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >> >>> >>>"Henri Wilson" <HW@..> wrote in message >>>news:un77o1d04j3mucd353ud969ohinls6gc5c(a)4ax.com... >>>> On Tue, 22 Nov 2005 21:04:50 -0000, "George Dishman" >>>> <george(a)briar.demon.co.uk> >>>> wrote: >> >>>>>>> >>>>>>>Consider: if "vt is responsible for fringe displacement >>>>>>>during constant rotation" then the change of v with >>>>>>>time already provides the movement, the a(t^2)/2 factor >>>>>>>is an _additional_ offset on top of that which would be >>>>>>>_constant_ during constant acceleration. >>>>>>> >>>>>>>I won't be able to do that in ASCII since it involves >>>>>>>two different slopes so try it for yourself for this >>>>>>>speed profile: >>>>>>> >>>>>>> >>>>>>> _________ >>>>>>> / \ >>>>>>> / \ ^ >>>>>>> __________/ \_________ | speed >>>>>>> | >>>>>>> __________________________________ >>>>>>> >>>>>>> ------> >>>>>>> time >>>>>>> >>>>>>> >>>>>>> >>>>>>>George >>>>>> >>>>>> I think you need more time. >>>>>> You are becoming MORE confused than ever. >>>>> >>>>>I only quoted your own words Henri. Plot them and >>>>>see what they produce because you clearly haven't >>>>>fully realised the consequences yet. >>>> >>>> You are totally confused about what I said. >>> >>> >>>http://www.users.bigpond.com/hewn/sagnac.jpg >>> >>>If you think I am confused by your drawing, answer this >>>question and it will clear it up. >>> >>>Your diagram shows the length from C to E for constant >>>acceleration as "vt + 1/2 a t^2", right? Apply that to >>>the example speed profile and what do you get? >> >> Your speed profile has identical + and - accelerations. > >Yes. > >> You end up at the starting speed. > >Yes. > >> The fringe displacement returns to the original, determined by v. > >Not quite, the graph is taken from your diagram which >shows path length, not fringe displacement, so you >would be right if you said "The path length returns >to the original, determined by v." > >However, the interesting part is what happens _during_ >the acceleration. Here is the answer assuming infinite >jerk (instantaneous change of acceleration, I can't do >a realistic one in ASCII): > > . > /| > / | > / |_______ > | | > | | ^ > __________| | _________ | path length > \ | | > ______________________\ |_________ > \| > ------> ' > time > >Check for yourself, it is just a plot of what is on >your diagram. Remember "t" is the time it takes a >wavefront to travel from source to detector so is >essentially a constant, not the usual time coordinate. >I tend to use capitals for specific values but I've >stayed with the notation from your drawing. > >George > All this is now insignificant in light of my latest findings. 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 28 Nov 2005 18:24 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 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. >>> >>>> The number of wavelengths in each path is different (you claim only >>>> fractionally, no matter) >>> >>>Erm, no I have never mentioned the number of waves >>>in the path, it has almost no influence. The output >>>depends on the time difference between the two halves >>>of the signal getting to the detector and not the >>>actual time taken from their emission. >> >> Consider this. >> No rotation. ..path lengths equal...100000000 wavlengths in each path >> (let's >> say). >> Short period of acceleration... > >I'm assuming you next talk about the time after the >speed has stabilised at the new value: > >> one path increases to 100000000.1 >> wavelengths...the other decreases to 99999999.9.... > >OK. > >> fringes have moved 0.2 fringe widths. > >No. One path has increased in length but the light >is moving faster due to Ritz so the time taken is >as before. The other path has shortened but the >light is moving slower and again the time taken >is as before. The two beams arrive in sync so >there is no fringe displacement. > >> They don't go back to zero just because the acceleration ceases. You see >> to >> think they do. > >See my other post where I show that your diagram >requires that they do. > >> This also proves the travel times around each path cannot be equal under >> constant c+v, as you claim. > >Again see the more detailed post where I show >that t' = t. > >>>>>As the acceleration build up to the constant value, there >>>>>is a beat which is counted, the fractional part being added. >>>>>(The counter would increment each time the fractional part >>>>>rolled over.) While the acceleration is constant, the >>>>>displacement is constant so the counter output is constant. >>>> >>>> No George. >>>> You are confusing constant acceleration with constant speed. >>> >>>No, I'm not confusing them, I am saying your words are >>>wrong but your diagram is right. The diagram shows what >>>I say above. >> >> It matters not whether the acceleration is constant or not. >> If it is +ve, the path lengths will diverge, if -ve, they come together. > >According to Ritz, the displacement is proportional to >the acceleration at all times whether it is constant >or not (for reasonable values, i.e. where it isn't >fast enough to create an event horizon). > >>>>>While the acceleration decreases to zero, the fringes would >>>>>pass in the other direction so the couter would need to >>>>>decrease until it should be back at zero when the rotational >>>>>reaches zero. >>>> >>>> No that is not right. If the acceleration is +ve, the fringes move one >>>> way. >>> >>>Not according to your diagram. It says the path length is >>>increased by vt+1/2at^2 and we know the vt part is cancelled >>>by the Ritzian speed change (c+kv) so the path length is >>>altered by 1/2at^2 which is constant, and the path length >>>determines the fringe displacement. >> >> Why do you say "1/2at^2 is constant"? >> The value of t signifies the end of the acceleration period. > >No, that value is the time for wich the light is >travelling. The speed of the light is set according >to Ritz at the moment it is emitted. The extra >path length (compared to the situation of constant >speed at the value the table had when the light was >emitted) is dependent on how much the table accelerates >while the light is "in the air". It increases by dv = at. > >Later emissions start at progressively faster speeds as >the table accelerates but that speeds up the light too. >Only the Vt term in the path length depends on the actual >speed, and that is cancelled out by the Ritzian boost. > >> During that time, the path lengths diverge by at^2/root2 (for constant a >> otherwise (integral a(t)dt)^2/root2). > >That's about right though see my other post which >shows why the root(2) also gets cancelled. > >>>> It >>>> matters not whether the magnitude of that acceleration is increasing or >>>> decreasing. >>>> 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. >> >> George please try to understand the difference between fringe >> 'displacement' >> and fringe 'movement'. The former is the static situation, measured from >> zero, >> the latter the process of actually moving. > >That is the sense in which I am using them. > >> The fringe displacement is proportional to v, not dv/dt. > >In reality, yes. According to SR, yes. According >to Ritz, no. Ritz says the fringe displacement >should be proportional to the acceleration. Ritz >is wrong. > >> That means they DO NOT move during constant v. >> >> The rate of fringe movement is proportional to a (and maybe a secondary >> effect) >> The change in fringe displacement after an acceleration period is >> proprtional >> to the change in v. >> >>>I have already spent months showing you the "vt" >>>part is cancelled by the speed change, or think of the >>>carousel analogy which shows it nicely, or do "the duck, >>>the car and the goose" diagram. All of them prove it. >> >> You haven't proved that at all. You forgot the root2 again. > >See my other post where I do all the math. The >sqrt(2) cancels out and the speed boost exactly >cancels the path length increase for the constant >speed situation. > >> I say the travel times around the two paths are 'rotation speed' >> dependent. >> they have to be.... because the number of wavelengths in each path is >> speed >> dependent and wavelength doesn't change under BaTh. > >Don't just "say" Henri, opinion carries no weight. >Show your math as I have done. > >>>>>The output of that counter would therefore be a digital >>>>>indication of the acceleration, not the speed. >>>>> >>>>>One point Henri, in this part you seem to have lost the "vt" >>>>>term which you mention above. I think you need to consider >>>>>just which of the terms is responsible for the output here, >>>>>it's quite fundamental. >>>> >>>> The at^2/2 is responsible for fringe movement. >>>> The vt is responsible for fringe displacement during constant rotation. >>> >>>See my earlier post on that. >> >> It was very confused. > >Too much history and back quoting perhaps. I have >redone it froma scratch. > >>>>>Both integrators need to have an initial value placed in them. >>>> >>>> yes >>> >>>Bu&*&&*r. Now you are back to agreeing with me. >> >> All integrators need an initial reference value. > >I know, but you are trying to convince me that Ritz >only calls for ONE intgrator because it gives an >output proportional to speed while I an saying it >needs TWO to determine heading because the output is >proportional to acceleration, and you just said "both" > >>>>>> A second continuous integration of instantaneous displacement with >>>>>> time >>>>>> gives >>>>>> the rotation angle from zero. >>>>> >>>>>Exactly. >>>>> >>>>>> >You could imagine that this is how commercial units >>>>>> >work (it isn't) but that wouldn't apply to the lab >>>>>> >experiment. Remember in the original experiment of >>>>>> >Sagnac, he saw a shift of 7% of a fringe while the >>>>>> >table was turning at 120rpm. What you are describing >>>>>> >would be that he counted 0.07 fringes during the >>>>>> >acceleration phase and the displacement returned to >>>>>> >zero once constant speed was achieved. >>>>>> >>>>>> I dont think that would be accurate enough for any practical purpose. >>>>>> I'm sure fringes move a lot more in multi turn FoGs. >>>>> >>>>>Well see the numbers above. >>>> >>>> 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. > >Number of turns and path length is too detailed >for ordinary documents (I already wrote this in >another reply). "Fringe displacements" won't be >mentioned since there are no fringes, there is >a synchronous demodulator measuring the amplitude >of the signal from the optical modulator. > >>>>>That's not what you described above where the fringe >>>>>counter is supposedly doing the integration. >>>> >>>> It registers the answer. >>> >>>OK, that would be correct but now look at the description >>>on the page I cited. There is no counter and no fringes, >>>the signal is being measured as the amplitude of a 5.05MHz >>>signal used in the modulator. >> >> Yes, I'm not sure how that works. > >Which part, the synchronous demodulator? > >>>> 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. > >George > 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. 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: jgreen on 28 Nov 2005 21:09
George Dishman wrote: > <jgreen(a)seol.net.au> wrote in message > news:1133131335.712029.264010(a)g43g2000cwa.googlegroups.com... > > > > Henri Wilson wrote: > >> On 27 Nov 2005 00:13:23 -0800, jgreen(a)seol.net.au wrote: > >> > >> > > >> >Henri Wilson wrote: > >> >> On 24 Nov 2005 00:01:52 -0800, jgreen(a)seol.net.au wrote: > >> >> > >> > >> >> > >> >> You are a bit behond us Jim. > >> >> It is turning out to be a very complicated problem. > >> > > >> >Is "behond" a Wilsonism, or a typo? > >> >Is it "behind" or "beyond" > >> > >> Maybe you are 'behind' in some areas and 'beyond' in others. > >> > >> >Get back to basics. I presume, after all the argi between you and > >> >George, that he would have sent you his sagnac animation (demo of why > >> >it works) > >> > >> Actually I sent it to him. www.users.bigpond.com/hewn/sagnac1.exe > > > > Explorer blocks it from me (citing secrity) > >> > >> George seems to have disappeared since I sent him my latest sagnac > >> animation. > >> That's not surprising. It shows clearly why the sagnac DOES NOT refute > >> the > >> BaTh. > >> > >> >What I could never pin down, were the FIXED; airframe, earth, or > >> >hirdygirdy. > >> >The reference point in time AND position, could not be determined, from > >> >the information > >> >(as regards the emission and receival of the light) > >> > >> It is a very complicated problem The SR (actually LET) analysis is vastly > >> oversimplified and wrong. > >> > >> What my demo shows is that two opposite beams which leave the first 45 > >> mirror > >> at right angles DO NOT meet up at the same point even though the BaTh > >> says > >> their travel times are equal (in the opoosite directions). > > > > Neither would bullets fired from identical guns. > > See Henri's analogy about the car and the duck. > If you want to hit the duck, you aim ahead. The > duck doesn't care about near misses and the > detector isn't affected by light that doesn't > hit it. > > > 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 ;-( 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................ ref sagnac: the whole shebang constitutes the equipment; airframe, opticals, and spinning turntable. It is the relationship between these and the STARTING rotation of the lot, and the observed fringe displacements, which are at issue. 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. Jim G c'=c+v |