Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 22 Nov 2005 17:26 On Tue, 22 Nov 2005 21:00:01 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:i5v6o1tb7oo9lh9juv3htlon684djo5ans(a)4ax.com... >> On 22 Nov 2005 01:34:00 -0800, jgreen(a)seol.net.au wrote: >> >>> >>>Henri Wilson wrote: >>>> On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" >>>> <george(a)briar.demon.co.uk> >>>> wrote: >>>> >>>> > >>>> >"Henri Wilson" <HW@..> wrote in message >>>> >news:7tgsn11flp2iggll8is15lrdpqck37voaj(a)4ax.com... >>> >>>The sagnac is "zeroed" to read the fringe at 12 o'clock, with the >>>airframe travelling north, OK? >>>When the plane alters course (to say NE), the fringe is logged at (say) >>>2 o'clock, as an accelleration (in rotation of the craft) took place. >>>Now the observed fringe shift will REMAIN at 2 o'clock, UNLESS a >>>reverse rotation takes place. It was the change of direction (read >>>rotational accelleration) which caused the alteration to the shift >>>position. I predict that, in ACCORDANCE with c'=c+v, while the plane >>>maintains the NE heading, there will be no FURTHER shift (2 o'clock >>>maintained) >>>The end >> >> 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. >> 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.. In other words, you believe the fringe moves back to zero every time the RATE OF acceleration decreases. > >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 17:27 On 22 Nov 2005 12:34:43 -0800, "Sue..." <suzysewnshow(a)yahoo.com.au> wrote: > >Henri Wilson wrote: >> On Tue, 22 Nov 2005 09:51:18 -0000, "George Dishman" <george(a)briar.demon.co.uk> >> wrote: >> >> >> >> yes, something like that. >> >> 'a' IS the acceleration during flight time. Where is the problem? ><< Well, that was where the mistake was, there was no field. >It was just that when you shook one charge, another would >shake later. There was a direct interaction between charges, >albeit with a delay. The law of force connecting the motion of one >charge with another would just involve a delay. Shake this one, >that one shakes later. The sun atom shakes; my eye electron >shakes eight minutes later, because of a direct interaction across. > >Now, this has the attractive feature that it solves both >problems at once. >> >--Richard P. Feynman - Nobel Lecture >Nobel Lecture, December 11, 1965 >http://nobelprize.org/physics/laureates/1965/feynman-lecture.html We're talking about sagnac. get off the hard stuff..... 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 17:42 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. Anyway it IS wrong. Look at the diagram again. The path length change is AE-AD which, for small angles is at^2/(2root2) >> 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. > >>>>>>>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? Read what Jim Greenfielfd said. The fringe displacement only changes during an acceleration. If acceleration varies with time, the fringe movement automatically integrates that. It fringes stay where they are when the acceleration ceases. 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. > >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 17:45 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. > >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: George Dishman on 22 Nov 2005 18:00
"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: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:i5v6o1tb7oo9lh9juv3htlon684djo5ans(a)4ax.com... >>> On 22 Nov 2005 01:34:00 -0800, jgreen(a)seol.net.au wrote: >>> >>>> >>>>Henri Wilson wrote: >>>>> On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" >>>>> <george(a)briar.demon.co.uk> >>>>> wrote: >>>>> >>>>> > >>>>> >"Henri Wilson" <HW@..> wrote in message >>>>> >news:7tgsn11flp2iggll8is15lrdpqck37voaj(a)4ax.com... >>>> >>>>The sagnac is "zeroed" to read the fringe at 12 o'clock, with the >>>>airframe travelling north, OK? >>>>When the plane alters course (to say NE), the fringe is logged at (say) >>>>2 o'clock, as an accelleration (in rotation of the craft) took place. >>>>Now the observed fringe shift will REMAIN at 2 o'clock, UNLESS a >>>>reverse rotation takes place. It was the change of direction (read >>>>rotational accelleration) which caused the alteration to the shift >>>>position. I predict that, in ACCORDANCE with c'=c+v, while the plane >>>>maintains the NE heading, there will be no FURTHER shift (2 o'clock >>>>maintained) >>>>The end >>> >>> 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. >>> 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'. If Jim used 'fringe displacement' it would clear up what he meant. George |