Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 22 Nov 2005 18:12 "Henri Wilson" <HW@..> wrote in message news:tt67o1psr5pot6bmgq5nrap86l76kmbup6(a)4ax.com... > On Tue, 22 Nov 2005 21:00:21 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:68v6o11dhue4eib2fmbm68qbmcnikv832h(a)4ax.com... >>> On Tue, 22 Nov 2005 09:51:18 -0000, "George Dishman" > >>>> >>>>> Fringes move only during acceleration periods. >>>>> If a +ve acceleration is followed by an identical -ve one, the >>>>> rotation >>>>> speed >>>>> is back to where it was....and so is fringe displacement. >>>>> >>>>>>>>There is no method to provide physical integration >>>>>>>>because the number of wavelengths in the path does >>>>>>>>not affect the time difference between wavefront >>>>>>>>arrivals in the two beams which is what produces >>>>>>>>the output. >>>>>>> >>>>>>> The 'change in fringe displacement' is effectively an integration of >>>>>>> the >>>>>>> path >>>>>>> length increase during the acceleration period. >>>>>> >>>>>>No, the fringe displacement is a t^2 / 2 as you show >>>>>>in your diagram. >>>>> >>>>> Yes, I suppose that its right. It is proportional to the path length >>>>> difference of the two beams.. >>> >>> Well, George, it certainly isn't obvious that the fringe displacement is >>> at^2/2. >> >>Then why say "Yes, I suppose that its right." when >>I said " the fringe displacement is a t^2 / 2 ..." >>and why are you now disputing what your own sketch >>shows? > > I'm wasn't saying it was wrong. I was just pointing out that it was not > OBVIOUS. OK. > Anyway it IS wrong. Look at the diagram again. > The path length change is AE-AD which, for small angles is at^2/(2root2) OK. That is still proportional to the acceleration. dL = k * a where k = t^2 / (2 * root(2)) >>> In fact it probably is NOT. You can forget the '/2' anyway because the >>> effect is doubled. >> >>Sure, a fixed factor of 2 isn't important when we >>are both using the term "proportional to". I was >>just trying to keep faithful to the text on your >>diagram so you could see where I got it from. >> >>In fact it is slightly more complex because the >>speed change that cancels the "vt" term also >>divides the "at^2" part but again it's a small >>factor of the order of 1/(1+L*sqrt(1/2)/vt), but >>this is really unnecessary nitpicking. > > Anyway, you left out the 1/root2. I quoted your diagram, either way it remains proportional which is what matters, when a=0 the term is zero regardless of the sqrt(2). >>>>>>>>Actually I think your diagram is oversimplified but >>>>>>>>we can go with it for the moment, it is close enough. >>>>>>> >>>>>>> It shows what happens during a constant acceleration. In practice, >>>>>>> acceleration >>>>>>> would vary with time. >>>>>> >>>>>>Indeed. I can't show a quadratic start and end to >>>>>>each period of acceleration but if I could the >>>>>>resulting output would look like this: >>>>>> >>>>>> +ve __ >>>>>> / \ >>>>>> / \ >>>>>> / \ >>>>>> 0 ___/ \_______ _____ >>>>>> \ / >>>>>> \ / >>>>>> \ / >>>>>> -ve \__/ >>>>>> >>>>>>Since the acceleration is changing, you now have to >>>>>>understand the 'a' in your diagram to be the mean >>>>>>acceleration during the flight time. >>>>> >>>>> yes, something like that. >>> >>> 'a' IS the acceleration during flight time. Where is the problem? >> >>No problem at all, I agree, that would be the output >>from the device if Ritz were correct, hence Ritz is >>falsified. >> >>Henri, I really think you need to get clear in your >>mind how you think your diagram gives something >>other than acceleration as the cause of fringe >>displacement. > > What are you talking about? Your diagram shows that, if a=0, the path length is only changed by the vt term. That is what I am saying > Read what Jim Greenfielfd said. I read him as saying the fringes stay displaced when moving on a constant heading of north-east which doesn't match either of our views. > 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. > If acceleration > varies with time, the fringe movement automatically integrates that. There is no integrator. > It fringes stay where they are when the acceleration ceases. Not according to your diagram. > 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. George
From: George Dishman on 22 Nov 2005 18:21 "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: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:d107o15053t20qmtjmmrs36po2q1853c5f(a)4ax.com... >>> On Tue, 22 Nov 2005 10:08:49 -0000, "George Dishman" > >>>>This isn't bad. I've seen a more detiled one but didn't >>>>bookmark it. >>>> >>>>http://www.physik.fu-berlin.de/~bauer/habil_online/node11.html#sagnac1 >>> >>> Doesn't tell us anything about path lengths. >> >>No, it exlains the use of the modulator which is what >>we were talking about when you said you couldn't find >>any description of how it worked. > > Yes, that's quite a neat idea. > >>>>I'm pushed for time at the moment so I'll answer more >>>>fully tonight. >>>> >>>>> 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. >>>> >>>>Yes. >>>> >>>>I'll just answer one other point because it's quick and >>>>you might need some time to consider: >>>> >>>>>>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. >>>> >>>>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? George
From: Henri Wilson on 22 Nov 2005 18:29 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... >> On 21 Nov 2005 06:40:55 -0800, "George Dishman" <george(a)briar.demon.co.uk> >> wrote: >>>Henri Wilson wrote: >... >>>> It counts the light and dark pulses. It probably also senses 'fractions >>>> of >>>> fringe.' >>> >>>My understanding is that they always work in the >>>fraction of a fringe regime. The number of turns >>>needed to work the way you suggest at low rates >>>would be in the billions. >> >> I cannot see how any kind of acceptable accuracy would be achieved if only >> fractions of fringes were measured. I don't know how the source would be >> kept >> sufficiently stable for one thing. > >The source is a semiconductor laser so highly >stable in the short term and slow variation >has negligible impact since the one laser is >split to create the two beams. > >> Also the detector would be quite temperature >> sensitive. > >Temperature drift affects leakage current so would be >a slow DC shift in the measured signal. If you look at >the page on the modulator, it uses a "phase-sensitive >lock-in amplifier set to the first harmonic of the >modulation frequency" so is unaffected by DC, they >are measuring the amplitude of an AC component. > >>>> Wait I think we are talking about different things here George. >>> >>>Sort of. >>> >>>> The integration of fringe displacement with time gives total angle moved >>>> during >>>> that time. >>> >>>That's right, but also the integration of acceleration would >>>give the speed. Where we disagree, perhaps, is whether >>>the fringe displacement is a function of the speed, the >>>acceleration or a bit of both. Again, this will become >>>clearer later. >> >> I say the fringe displacement (noun) is a function of speed. >> Changes in displacement occur DURING acceleration. >> The rate of change is a function of acceleration. > >That conflicts with your diagram as I read it but >as I pointed out when you first posted it, it only >gives the path lengths and doesn't take account of >the c+kv speed part which also influences path times: > >http://www.users.bigpond.com/hewn/sagnac.jpg > >>>> Each rotation speed has a different path length. You are ignotring the >>>> 'vt' >>>> term. >>> >>>Not ignoring it, the path length changes as you say but the >>>your diagram is drawn in the lab frame so the speed is not >>>c, it is also modified. That was the point I have been making >>>for months, the speed and path length changes to the two >>>paths are equal but opposite (+vt versus -vt, c+kv versus c-kv) >>>so that factor cancels. >> >> 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... one path increases to 100000000.1 wavelengths...the other decreases to 99999999.9....fringes have moved 0.2 fringe widths. They don't go back to zero just because the acceleration ceases. You see to think they do. This also proves the travel times around each path cannot be equal under constant c+v, as you claim. >>>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. > >>>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. During that time, the path lengths diverge by at^2/root2 (for constant a otherwise (integral a(t)dt)^2/root2). > >> 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. The fringe displacement is proportional to v, not dv/dt. 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. 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. > >>>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. >>> >>>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. > >>>> 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. >>> >>>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. >> >> 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. 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?. > >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 22 Nov 2005 18:57 On Tue, 22 Nov 2005 23:00:08 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:2d67o1pq903b4eikhdbu2qik79ptjggotf(a)4ax.com... >>>> That's exactly what I am saying, too. >>> >>>What???? Are you now saying the output is proportional >>>to the change in heading? >> >> George, you are becoming progressiveley more confused. >> The output is proportional to rotation speed. >> >>>That means no integrators are >>>needed where yesterday we were arguing whether it was >>>one or two! Are you sure you haven't misread what Jim >>>said? >> >> You have misread both Jim and me. >> I agree with what Jim said. >> The fringe displacement changes only during an acceleration period. > >Jim said "Now the observed fringe shift will REMAIN >at 2 o'clock, UNLESS a reverse rotation takes place." > >If the output is proportional to rotation speed, the >fringe displacement is zero for any constant heading. Oh! OK. I don't agree with Jim on that one. I didn't realize that Jim's plane had experienced both a positive and a negative acceleration. The fringe shift will indeed go back to zero during no rotation periods. >>>> George, out of pure desperation, tried to make he fringe movement >>>> proportional to da/dt. >>> >>>"fringe displacement proportional to dv/dt (=a)" please >>>Henri. Your phrasing isn't wrong but let's not get >>>confused over verbs and nouns again. >> >> George, according to your diagrams, you believe fringe movement is >> proportional >> to da/dt and not a.. > >Which is the same as we both said above, I only changed >the wording to use the term "displacement" as you >suggested. > >> In other words, you believe the fringe moves back to zero every time the >> RATE >> OF acceleration decreases. > >No, I am saying Ritz predicts displacement is proportional >to angular acceleration which is what you said above, >"fringe movement proportional to da/dt" assuming you mean >'change of displacement' when you say 'movement'. No you are confusing da/dt with dv/dt >If Jim used 'fringe displacement' it would clear up what >he meant. No I think Jim's terminology was wrong there.. When the plane stays at 2 o'clock, the fringe displacement is zero. However the FINAL output will read a constant rotation angle, determined by electronic integration (with time) of the previous rotation speed history. That's probably what Jim meant. 'Output' should refer to the actual angular displacement from zero, as finally indicated on some kind of dial. 'Displacement' should always be use as a noun, the passive situation. Thus 'fringe displacement' is the position of the fringes in relation to the calibrated zero. 'Fringe movement' describes how fringe displacement is in the process of changing value. > >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 22 Nov 2005 19:07
On Tue, 22 Nov 2005 23:12:16 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:tt67o1psr5pot6bmgq5nrap86l76kmbup6(a)4ax.com... >> On Tue, 22 Nov 2005 21:00:21 -0000, "George Dishman" >>>> >>>> Well, George, it certainly isn't obvious that the fringe displacement is >>>> at^2/2. >>> >>>Then why say "Yes, I suppose that its right." when >>>I said " the fringe displacement is a t^2 / 2 ..." >>>and why are you now disputing what your own sketch >>>shows? >> >> I'm wasn't saying it was wrong. I was just pointing out that it was not >> OBVIOUS. > >OK. > >> Anyway it IS wrong. Look at the diagram again. >> The path length change is AE-AD which, for small angles is at^2/(2root2) > >OK. That is still proportional to the acceleration. > > dL = k * a where k = t^2 / (2 * root(2)) > >>>> In fact it probably is NOT. You can forget the '/2' anyway because the >>>> effect is doubled. >>> >>>Sure, a fixed factor of 2 isn't important when we >>>are both using the term "proportional to". I was >>>just trying to keep faithful to the text on your >>>diagram so you could see where I got it from. >>> >>>In fact it is slightly more complex because the >>>speed change that cancels the "vt" term also >>>divides the "at^2" part but again it's a small >>>factor of the order of 1/(1+L*sqrt(1/2)/vt), but >>>this is really unnecessary nitpicking. >> >> Anyway, you left out the 1/root2. > >I quoted your diagram, either way it remains >proportional which is what matters, when a=0 the >term is zero regardless of the sqrt(2). OK > >>>>>>>>>Actually I think your diagram is oversimplified but >>>>>>>>>we can go with it for the moment, it is close enough. >>>>>>>> >>>>>>>> It shows what happens during a constant acceleration. In practice, >>>>>>>> acceleration >>>>>>>> would vary with time. >>>>>>> >>>>>>>Indeed. I can't show a quadratic start and end to >>>>>>>each period of acceleration but if I could the >>>>>>>resulting output would look like this: >>>>>>> >>>>>>> +ve __ >>>>>>> / \ >>>>>>> / \ >>>>>>> / \ >>>>>>> 0 ___/ \_______ _____ >>>>>>> \ / >>>>>>> \ / >>>>>>> \ / >>>>>>> -ve \__/ >>>>>>> >>>>>>>Since the acceleration is changing, you now have to >>>>>>>understand the 'a' in your diagram to be the mean >>>>>>>acceleration during the flight time. >>>>>> >>>>>> yes, something like that. >>>> >>>> 'a' IS the acceleration during flight time. Where is the problem? >>> >>>No problem at all, I agree, that would be the output >>>from the device if Ritz were correct, hence Ritz is >>>falsified. >>> >>>Henri, I really think you need to get clear in your >>>mind how you think your diagram gives something >>>other than acceleration as the cause of fringe >>>displacement. >> >> What are you talking about? > >Your diagram shows that, if a=0, the path length is >only changed by the vt term. That is what I am saying OK > >> Read what Jim Greenfielfd said. > >I read him as saying the fringes stay displaced when >moving on a constant heading of north-east which doesn't >match either of our views. 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. (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'. When acceleration ceases, v is larger than it was and vt is therefore longer than it was. > >> 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. > >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". |