Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 28 Nov 2005 12:29 "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
From: George Dishman on 28 Nov 2005 15:07 "Henri Wilson" <HW@..> wrote in message news:06c7o1ptp360adjulvbpol9ved8io57jk8(a)4ax.com... > 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. 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. 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 Obviously t" = L"/c" 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. 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. 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
From: George Dishman on 28 Nov 2005 15:16 <jgreen(a)seol.net.au> wrote in message news:1132819312.828960.57210(a)f14g2000cwb.googlegroups.com... > > Henri Wilson wrote: >> On 22 Nov 2005 21:27:55 -0800, jgreen(a)seol.net.au wrote: >> >> > >> >George Dishman wrote: >> >> "Henri Wilson" <HW@..> wrote in message >> >> news:2d67o1pq903b4eikhdbu2qik79ptjggotf(a)4ax.com... >> >> > On Tue, 22 Nov 2005 21:00:01 -0000, "George Dishman" >> >> > <george(a)briar.demon.co.uk> >> >> > wrote: >> >> >> >> >> 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'. >> >> >> >> If Jim used 'fringe displacement' it would clear up what >> >> he meant. >> >> >> >> George >> > >> >Verb or noun?Displacement indicating motion, or position (stationary >> >after motion)? >> >There is "displacement" occurring while the airframe rotates, and the >> >(new position) "displacement" (eg 2 o'clock). >> >It seems that your sagnac has additional electronic gadgetry which >> >defaults the 2 o'clock to 12 o'clock, once the new heading stabilises. >> > >> >Jim G >> >c'=c+v >> >> Yes Jim, we must get our definitions right or we talk at cross purposes. >> >> I suggest we use: >> OUTPUT: The final reading on some kind of instrument of current rotation >> angle >> from the calibrated zero. (static or dynamic) >> DISPLACEMENT or SHIFT: the distance between current fringe position and >> the >> zero position. (static) >> FRINGE MOVEMENT: the actual transient process when fringes are changing >> displacement. (dynamic) >> >> The fact is, fringes remain displaced but static during periods of >> constant >> rotation speeds. The amount of displacement is proportional to rotation >> speed. >> Movement occurs only during periods of acceleration. > > Sounds overly complicated to me. It came from a post where Henri used "fringe shift" to mean the static displacement at the beginning of a sentence and later meant shift as in movement of the fringes. It took a few days to sort out the confusion and using "fringe displacement" avoids the ambiguity. > The addition of the rotational velocity of an airframe to c is VERY > small. The rotation of the sagnac (650rpm whatever), which is enhanced > by the mirroring, magnifies this difference, resulting in a situation > where the position where the beams interdict becomes discernible. > Sagnac WORKS! The output you get from iFOG devices is also well documented, they give a signal that tells you the rate of turn, not the present heading. You need an extra bit if electronics called an integrator to figure out the heading. > It is the interpretation of WHY it works is the issue. Exactly, but you need to understand WHAT it does before you can argue about WHY it does it. That's where you have a little catching up to do. George
From: George Dishman on 28 Nov 2005 15:31 "Henri Wilson" <HW@..> wrote in message news:9c6ko1lra1up33csccovdvcv9blr34kcuk(a)4ax.com... > 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 > > 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. Nah, holiday party season has started :-) This is my first night in since Wednesday. >>What I could never pin down, were the FIXED; airframe, earth, or >>hirdygirdy. Jim, I gave you links to a photo of the board in one of the KVH devices a year or so ago. It's since gone obsolete and I can't see an equivalent now but the construction is very simple. The light source and the detector are a single silicon device and it is connected to the end of a coil of optical fibre. Imagine taking a light bulb (laser diode actually) and gluing a photo sensor on one side and both ends of a loop of fishing line on the other. Half the light goes into each end of the line and comes out the other, travels across the bulb and hits the sensor. Now roll up the loop of line, put the whole lot in a box and it looks like this http://www.kvh.com/pdf/DSP-4000_6.04.pdf The diagram on page two shows the parts in more detail: http://www.kvh.com/pdf/ECoreTech.pdf The "hirdygirdy" or "carousel" is the airframe, the box is bolted to it and the light source, fibre and sensor are all fixed inside the box so it has "no moving parts" in the conventional sense. George
From: George Dishman on 28 Nov 2005 15:34
<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. George |