Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 16 Nov 2005 16:45 On Wed, 16 Nov 2005 19:41:40 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:1g0ln19apsuph4suno5q8b1dodhcc4knut(a)4ax.com... >> >> Maybe we settle on using 'displacement' instead of 'shift' the noun... and >> 'fringe movement' instead of 'shift' the verb. > >Better but I think something like "change of displacement" >than 'fringe movement' would keep it consistent. > >Henri, I think much of the last few days discussion has been >a result of the confusion over these terms. I'll repost your >message of the 12th and then retype it as I understood your >meaning using "displacement". See what you think. I'll >address your specific points separately. > > >-- you typed ... -- > >"Henri Wilson" <HW@..> wrote in message >news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... >> On Sat, 12 Nov 2005 10:35:04 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >> >>>"Henri Wilson" <HW@..> wrote in message >>>news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... >> >> George, George, George. >> >> I have finally woken up to your (and MY) complete misinterpretation of the >> problem. >> >> We have both been arguing about whether or not the fringes will move >> during >> constant angular rotation...and of course they don't. >> >> We should only be considering what happens during angular ACCELERATION >> !!!!!! >> >> That is when the two path lengths change. >> That is when more 'wavelengths' fit into one path than the other. >> That is when fringes move. >> >> Path lengths chaneg because each mirror accelerates slightly as light from >> the >> previous one is in flight. Small second order effect, you say. No way! It >> is >> the whole basis of operation. >> >> Actually, you have shown that the BaTh does what it should do. It expects >> NO >> fringe shifts under constant rotation. >> >> BUT!!!! >> The standard SR explanation says that there WILL BE a continuous fringe >> shift >> during steady rotation. >> >> Sagnac proves SR to be wrong!!! > > >-- I took that to mean (edited lines marked with asterisks) ... -- > > >"Henri Wilson" <HW@..> wrote in message >news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... >> On Sat, 12 Nov 2005 10:35:04 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >> >>>"Henri Wilson" <HW@..> wrote in message >>>news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... >> >> George, George, George. >> >> I have finally woken up to your (and MY) complete misinterpretation of the >> problem. >> >* We have both been arguing about whether or not the fringes will BE >DISPLACED >> during constant angular rotation...and of course they don't. >> >> We should only be considering what happens during angular ACCELERATION >> !!!!!! >> >> That is when the two path lengths change. >> That is when more 'wavelengths' fit into one path than the other. >* That is when fringes ARE DISPLACED. >> >> Path lengths chaneg because each mirror accelerates slightly as light from >> the >> previous one is in flight. Small second order effect, you say. No way! It >> is >> the whole basis of operation. >> >> Actually, you have shown that the BaTh does what it should do. It expects >> NO >* fringe DISPLACEMENT under constant rotation. >> >> BUT!!!! >* The standard SR explanation says that there WILL BE a CONSTANT fringe >* DISPLACEMENT during steady rotation. >> >> Sagnac proves SR to be wrong!!! > > >George > Sorry George, you cannot win an argument by confusion. What I said originally was what I meant. Your 'translation' is completely wrong. Thsi must be very hard for you to take, I know. 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 16 Nov 2005 16:59 On Wed, 16 Nov 2005 20:10:04 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:1g0ln19apsuph4suno5q8b1dodhcc4knut(a)4ax.com... >>>>>> George, if a sagnac is rotating at constant angular speed, Do the >>>>>> fringes >>>>>> move continuously or remain steady but offset? >>>>> >>>>>The latter, steady but with an offset proportional >>>>>to the speed of rotation." >>>> """"" >>> >>>Yep, and you seem to agree with that above. As I say, >>>no difference in views so far. >> >> except what you said in 1) above. >> >> Ritz predicts a constant fringe displacement but no fringe movement when >> there >> is constant rotational speed. > >Well we seem to be back to where we were last week before >you said "I have finally woken up to your (and MY) complete >misinterpretation of the problem." > >I have been showing you for months that Ritz predicts >no fringe displacement at constant angular speed while >you have been trying to explain how it did. I fail to >see what you think your sudden insight was. These bright ideas come to me in my sleep George. >>>Yep, again that is entirely correct and exactly what I >>>said earlier, still no difference in views. However ... >>> >>>> The displacement arises from the path length change that occurs DURING >>>> ACCELERATION. >>> >>>You need to explain that Henri. If the displacement occurs >>>only during acceleration then there should be no displacement >>>whenrunning at constant speed, but you just agreed there was >>>a constant (non-zero) displacement at constant speed. >> >> The 'current displacement' is a reflection of the integrated instantaneous >> path >> length changes during period of acceleration, no matter how small. > >There is no physical mechanism to perform such an >integration. There is an integrator in devices to >change the angular speed returned as the output >into a change of heading but that requires a speed >related output to start with. The former is an automatic phyical integration. (number of wavelengths in the path) The latter is electronic. > >To avoid further confusion, suppose the table is >turning at one speed, then for a while it accelerates >and finally it runs at a constant but higher speed. >Draw a graph of the angular speed and we get this: > > _________ > / > / ^ > __________/ | speed > | > ______________________ > > ------> > time > > >> Path length vary ONLY during acceleration. So does the number of >> wavelengths in >> each beam. > >Careful with the wording. Do you mean the path lengths >differ from each other or from the non-rotating value? >Do you mean path lengths only CHANGE during acceleration? Yes, the latter... each path length changes only during acceleration. You should be able to work that out yourself. > >>>> The number of 'wavelengths' in each path changes ONLY during >>>> acceleration. >>> >>>I'm less concerned about your ideas on the mechanism at >>>this stage, we might get to that later. >> >> Read what I said to Paul A and study my duck shooting diagram: >> >> http://www.users.bigpond.com/hewn/sagnac.jpg > >It is a bit too simplistic because you have drawn >straight lines so it must be in the inertial frame. >In that frame you have c'=c+kv so the time is a >combination of factors that you cannot see from >that diagram alone. It is the non-rotating frame. Naturally I have shown 'vt' to be a lot longer than it would be in practice. For small time intervals and minute velocities, you can ignore the curvature. > >> PS: I don't advocate the shooting of ducks. > >Christmas is coming, how about geese? 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 > > > > > | / \ ^ > 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. > >Obviously if V = 0, both lines are straight and of >length R * sqrt(2). The bullets leave the guns at >the same speed relative to the guns so they take >equal times to reach their targets, hitting >simultaneously. > >When V > 0 and constant, tell me what the lines on >the paper will look like, are they straight or >curved? If curved are they farther from the centre >or closer to it at the midway point (I think one of >each)? The bullets still leave the guns at the same >speed they did before but do they still hit the >birds simultaneously? > >I told you my answers to this some months ago but >I'm interested to see how you now relate this to >your comments on acceleration. > >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 16 Nov 2005 17:01 On Wed, 16 Nov 2005 20:10:04 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:1g0ln19apsuph4suno5q8b1dodhcc4knut(a)4ax.com... > >Christmas is coming, how about geese? 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 > > > > > | / \ ^ > 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. > >Obviously if V = 0, both lines are straight and of >length R * sqrt(2). The bullets leave the guns at >the same speed relative to the guns so they take >equal times to reach their targets, hitting >simultaneously. > >When V > 0 and constant, tell me what the lines on >the paper will look like, are they straight or >curved? If curved are they farther from the centre >or closer to it at the midway point (I think one of >each)? The bullets still leave the guns at the same >speed they did before but do they still hit the >birds simultaneously? > >I told you my answers to this some months ago but >I'm interested to see how you now relate this to >your comments on acceleration. > >George > George, my diagram shows all. the BaTh is fully supported by sagnac. 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 16 Nov 2005 17:05 On Wed, 16 Nov 2005 20:14:17 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:dufln1ptkv48kelkvv45pukcaeo868gkjc(a)4ax.com... >> >> Poor old George has spent years proving that according to the BaTh, >> fringes >> will not move during constant rotation. That is of course what happens. > >No, George has been patiently explaining to you why >Ritz predicts that the fringes will not be displaced >from the non-rotating pattern by rotation at constant >speed. No you haven't. You have shown that the fringes shold not MOVE during constant rotation, according to the BaTh. That is what happens. You have proved the BaTh to be consistent with sagnac. > >You knew that a week ago but this whole acceleration >sidetrack seems to have confused you. Anyway, have a >look at the car, duck and goose and see what you make >of it. It's a diversion. You know that the fringes only change their displacement during acceleration. My diagram shows why. The path length change only during acceleration, not during constant rotation. The fringe 'displacement' at any instant is the integrated effect of all previous ACCELERATIONS. > >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: Paul B. Andersen on 16 Nov 2005 17:34
Henri Wilson wrote: > On Tue, 15 Nov 2005 23:48:50 +0100, "Paul B. Andersen" > <paul.b.andersen(a)hiadeletethis.no> wrote: > > >>Jerry wrote: >> >>>Henri Wilson wrote: >>> >>> >>>>On Sat, 12 Nov 2005 10:35:04 -0000, "George Dishman" <george(a)briar.demon.co.uk> >>>>wrote: >>>> >>>> >>>> >>>>>"Henri Wilson" <HW@..> wrote in message >>>>>news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... >>>> >>>>George, George, George. >>>> >>>>I have finally woken up to your (and MY) complete misinterpretation of the >>>>problem. >>>> >>>>We have both been arguing about whether or not the fringes will move during >>>>constant angular rotation...and of course they don't. >>> >>> >>>Huh? >>> >>>http://www.physics.berkeley.edu/research/packard/Competition/Gyros/LaserRingGyro/Steadman/StedmanReview1997.pdf >>> >> >>Hm. >>Be aware that a ring laser is quite different from a Sagnac ring. >>In a Sagnac ring, the source (laser) and the detector is fixed to the ring. >>The phase difference between the two waves depend on the angular >>velocity, but the phase difference is constant at a constant rotation. >>That is, the fringe does not move, it is the position of the fringe >>that give the information about the rotation. >>A fibre-optic gyro is a Sagnac ring. >> >>But a ring laser is just that - a laser arranged as a ring. >>(Actually a square or a triangle). In a normal laser, >>a wave is bouncing back and forth between two parallel mirrors. >>So there are two waves going in opposite directions within >>the laser. We get a standing wave pattern, like a pearl necklace, >>within the laser. >>In a ring laser, there are also two waves going in opposite >>directions within the laser, and we get the same standing >>wave pattern. But this time, the "necklace" have no ends, >>it is a ring. Note that there is no localized source; >>the whole ring is lasing. >>The important point is that if the apparatus is rotated, >>the "necklace" will NOT rotate along. So the detector, >>which is fixed to the apparatus and rotating along with it, >>can actually "see" the "pearls" in the "necklace" passing by. >>So in a ring laser, the phase difference between the opposing >>waves are continuously changing. Or the fringes are moving >>with a speed proportional to the rotation, if you like. > > > Are you sure of that? Yes. > >>The problem with the fibre-optic gyro is that to measure >>the rotation, you have to compare the position of the fringes >>to their position when the gyro was not rotating. >>So the gyro must be calibrated when the gyro is known not >>to rotate. (Not very easy on a rotating Earth). And >>the calibration will drift with temperature, etc. > > > It is easy to compensate for those. > >>The ring laser do not have this problem. >>(Of course it has to be collimated to lase at all. >> But any laser must.) >>Because of its very principle, it is inherently more >>precise that the fibre-optic gyro. >> >>That's why ring lasers are used in the inertial navigation >>system in planes. These gyros are so sensitive that when >>the INS is started, it takes only ten minutes before it has >>have figured out how the plane is oriented. It does that >>by comparing the rotation around the three axes due >>to the rotation of the Earth. > > > That isn't very clear. Maybe not. What I tried to say is that the INS is aligned when the plane is stationary on the ground. A "strap down" system have three laser gyros, one for each of the three axes fixed to the plane. Imagine it is standing on equator, heading north. The gyros will then detect rotation around the roll axis only, and none around the other axes. If it is heading east, the rotation will be around the pitch axis only. So no rotation around the yaw axis means latitude zero. The ratio of rotation around the pitch and roll axes will give the heading. If the plane is on the north pole, the rotation will be around the yaw axis only. So in short, the rotation around the yaw axis give the latitude. The rotation around the pitch and roll axes give the heading. And this takes only ten minutes. The earth rotates only 2.5 degrees during that time. Impressive, eh? :-) > > I don't htink we will discuss ring lasers becasue they might operate on an > entirely different principle to the four mirror type. I don't think you will discuss ring lasers because they so obviously falsifies the BaT. I have explained why before, and you have fled the discussion before because you were unable to refute my arguments. In fact any gas laser falsifies the BaT. Coherent light means all the light is going at the same speed. In a gas laser, the gas atoms which are the sources of the light, are moving fast relative to each other. BaT falsified. > FoGs are similar but effectively have an infinitie number of mirrors which > reflect at infinitesimal angle. > We aren't going to get anywhere multiplying zero by infinity. Not unless you know some math, of course. Hint: limits. What is sin(x)/x for x = 0? And light in a mono-mode fibre is never reflected. It's a wave guide. BaT falsified. > So let's just stick with the four mirror sagnac eh? > > I think by now you will have realised that it fully supports the BaTh and > probably refutes SR. Any Sagnac ring falsifies the BaT. No question about it. > Poor old George has spent years proving that according to the BaTh, fringes > will not move during constant rotation. That is of course what happens. Of course it is. But I have in a much shorter time proved that according to the BaT, the phase relationship between the two waves will be the same regardless of the rate of a constant rotation. That is NOT what happens. The BaT falsified. And you never refuted my proof. You fled it by claiming that interferometers works in an entirely different manner than physicists think. You are unable to explain _how_ you think it works, thought. All you know is that it works in some mysterious way which make the fringes appear at different positions even when the phase relationship between the two waves are the same. A typical Wilsonian escape by inventing new laws of nature for every new phenomenon that must be explained away. Paul |