Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Sue... on 22 Nov 2005 15:34 Henri Wilson wrote: > On Tue, 22 Nov 2005 09:51:18 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > > > >"Henri Wilson" <HW@..> wrote in message > >news:9vj2o15r5gf1eaenenj9ddjtggsrlmn2fe(a)4ax.com... > >> On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" > > >>>>> L' = L + v * t / sqrt(2) > >>>>> > >>>>>When the rotation is at constant acceleration, the path > >>>>>length is also constant at the length of line AE or > >>>>>roughly: > >>>>> > >>>>> L' = L + (v * t + a * t^2 / 2) / sqrt(2) > >>>>> > >>>>> _________ > >>>>> / > >>>>> / ^ > >>>>> __________/ | speed > >>>>> | > >>>>> ______________________ > >>>>> > >>>>> ------> > >>>>> time > >>>>> > >>>>>Your diagram predicts there would be an output like > >>>>>this: > >>>>> > >>>>> __ > >>>>> | | ^ > >>>>> | | | output > >>>>> | | | > >>>>> | | > >>>>> __________|__|________ > >>>>> > >>>>> ------> > >>>>> time > >>>> > >>>> No it doesn't. > >>>> > >>>> Assume that fringe movement is +ve for +ve acceleration and -ve for -ve > >>>> acceleration. > >>>> There is no -ve acceleration in the above diagram showing a +ve speed > >>>> change.. > >>> > >>>> Therefore fringe displaement moves in only one direction during the > >>>> acceleration period. > >>> > >>>The fringes ARE DISPLACED in only one direction during > >>>theacceleration period as shown by your diagram. When > >>>the acceleration is zero, the factor a t^2 / 2 is zero > >>>so there is zero displacement, not zero rate-of-change > >>>of displacement. > >>> > >>>> It doesn't suddenly revert ot zero when acceleration ceases. > >>> > >>>Your diagram says it does, if 'a' is zero then the > >>>length difference is zero. > >>> > >>>> SO: > >>>> _________ > >>>> | ^ > >>>> | | output > >>>> | | > >>>> | > >>>> __________|__________ > >>>> > >>>> ------> > >>>> time > >>>> > >>>> Is the correct result. > >>> > >>>Let's add a negative acceleration period: > >>> > >>> _________ > >>> / \ > >>> / \ ^ > >>> __________/ \______ | speed > >>> | > >>> _______________________________ > >>> > >>> ------> > >>> time > >>> > >>>Your diagram predicts there would be an output like > >>>this: > >>> > >>> __ > >>>+ve | | ^ > >>> | | | output > >>> | | | > >>> 0 ________|__|__________________ > >>> | | > >>> ------> | | > >>> time | | > >>>-ve |__| > >>> > >>>This is basically a graph of the "a t^2 / 2" factor. > >> > >> How are you defining 'output'. > >> > >> I'm using 'output' to mean 'fringe displacement' from the central > >> (non-rotating) position. Maybe you are using it as rotation angle. > > > >Well strictly it would be the voltage out of the > >equipment but that is derived from the photodiode > >and that is really the same as fringe displacement. > > > >> In either case, I think I disagree with you. > > > >Possibly, but below you say "Yes, I suppose that > >its right. It is proportional to the path length > >difference." and later "yes, something like > >that." so I don't see how you disagree. This next > >paragraph appears to be in conflict with those > >later agreements. > > > >You need to clarify this for me because if "the > >fringe displacement is a t^2 / 2 as you show in > >your diagram." then the fringes are _displaced_ > >while a is non-zero, i.e. "during acceleration > >periods" rather than moving. > > > >George > > > > > >> 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. In fact it probably is NOT. You can forget the '/2' anyway because the > effect is doubled. > > > > >> > >>> > >>>>>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? << 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 > > > 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 16:00 "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? 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? > 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
From: George Dishman on 22 Nov 2005 16:00 "Henri Wilson" <HW@..> wrote in message news:68v6o11dhue4eib2fmbm68qbmcnikv832h(a)4ax.com... > On Tue, 22 Nov 2005 09:51:18 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:9vj2o15r5gf1eaenenj9ddjtggsrlmn2fe(a)4ax.com... >>> On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" > >>>>>> L' = L + v * t / sqrt(2) >>>>>> >>>>>>When the rotation is at constant acceleration, the path >>>>>>length is also constant at the length of line AE or >>>>>>roughly: >>>>>> >>>>>> L' = L + (v * t + a * t^2 / 2) / sqrt(2) >>>>>> >>>>>> _________ >>>>>> / >>>>>> / ^ >>>>>> __________/ | speed >>>>>> | >>>>>> ______________________ >>>>>> >>>>>> ------> >>>>>> time >>>>>> >>>>>>Your diagram predicts there would be an output like >>>>>>this: >>>>>> >>>>>> __ >>>>>> | | ^ >>>>>> | | | output >>>>>> | | | >>>>>> | | >>>>>> __________|__|________ >>>>>> >>>>>> ------> >>>>>> time >>>>> >>>>> No it doesn't. >>>>> >>>>> Assume that fringe movement is +ve for +ve acceleration and -ve >>>>> for -ve >>>>> acceleration. >>>>> There is no -ve acceleration in the above diagram showing a +ve speed >>>>> change.. >>>> >>>>> Therefore fringe displaement moves in only one direction during the >>>>> acceleration period. >>>> >>>>The fringes ARE DISPLACED in only one direction during >>>>theacceleration period as shown by your diagram. When >>>>the acceleration is zero, the factor a t^2 / 2 is zero >>>>so there is zero displacement, not zero rate-of-change >>>>of displacement. >>>> >>>>> It doesn't suddenly revert ot zero when acceleration ceases. >>>> >>>>Your diagram says it does, if 'a' is zero then the >>>>length difference is zero. >>>> >>>>> SO: >>>>> _________ >>>>> | ^ >>>>> | | output >>>>> | | >>>>> | >>>>> __________|__________ >>>>> >>>>> ------> >>>>> time >>>>> >>>>> Is the correct result. >>>> >>>>Let's add a negative acceleration period: >>>> >>>> _________ >>>> / \ >>>> / \ ^ >>>> __________/ \______ | speed >>>> | >>>> _______________________________ >>>> >>>> ------> >>>> time >>>> >>>>Your diagram predicts there would be an output like >>>>this: >>>> >>>> __ >>>>+ve | | ^ >>>> | | | output >>>> | | | >>>> 0 ________|__|__________________ >>>> | | >>>> ------> | | >>>> time | | >>>>-ve |__| >>>> >>>>This is basically a graph of the "a t^2 / 2" factor. >>> >>> How are you defining 'output'. >>> >>> I'm using 'output' to mean 'fringe displacement' from the central >>> (non-rotating) position. Maybe you are using it as rotation angle. >> >>Well strictly it would be the voltage out of the >>equipment but that is derived from the photodiode >>and that is really the same as fringe displacement. >> >>> In either case, I think I disagree with you. >> >>Possibly, but below you say "Yes, I suppose that >>its right. It is proportional to the path length >>difference." and later "yes, something like >>that." so I don't see how you disagree. This next >>paragraph appears to be in conflict with those >>later agreements. >> >>You need to clarify this for me because if "the >>fringe displacement is a t^2 / 2 as you show in >>your diagram." then the fringes are _displaced_ >>while a is non-zero, i.e. "during acceleration >>periods" rather than moving. >> >>George >> >> >>> 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? > 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. >>>>>>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. George
From: George Dishman on 22 Nov 2005 16:04 "Henri Wilson" <HW@..> wrote in message news:d107o15053t20qmtjmmrs36po2q1853c5f(a)4ax.com... > On Tue, 22 Nov 2005 10:08:49 -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: >>>>> On Sun, 20 Nov 2005 10:23:12 -0000, "George Dishman" >>>>> <george(a)briar.demon.co.uk> >>>>> wrote: >>> >>>>> >That's true, you can use more turns in the fibre >>>>> >coil, but it is still quite small. >>>>> >>>>> Are you sure? >>>> >>>>Yes, but it isn't too important as you'll see later, our >>>>disagreement lies elsewhere. Consider that Sagnac >>>>used an area of nearly a square metre and got a shift >>>>of 0.07 of a fringe at 120rpm. iFOGs tend to be in a >>>>box a few cm across and can just about sense the >>>>rotation of the Earth. That implies they are measuring >>>>about 10^-6 of a fringe at the low end up to maybe a >>>>few percent of fringe at their highest rate. >>> >>> I can't find a decent description of a FoG anywhere. >> >>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. >>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. George
From: George Dishman on 22 Nov 2005 16:44
"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. >>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. >>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. > 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. 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. >>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. >>> >> The beats are counted. >>> > >>> >That would be an electronic integration which would >>> >give speed. >> >>Actually I was wrong there, the first counter gives acceleration, >>a separate integrator is needed to get speed and a second to >>get angle turned. >> >>> It gives change in speed. >> >>Both integrators need to have an initial value placed in them. > > yes Bu&*&&*r. Now you are back to agreeing with me. >>> 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? >>> >> The count IS really an integration of the path length >>> >> change (or, more correctly, the number of wavelengths in each path) >>> > >>> >Yes but counting would have to be done as part of the >>> >electronics, it is not inherent in the physical process. >>> >>> The path length change effectively integrates acceleration. No need for >>> electronics. >> >>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. >>> >> The count itself (fringe diplacement) is also electronically >>> >> integrated to >>> >> give total rotation angle. >>> > >>> >That would be a second integration. >>> >>> yes. >> >>I think we are thinking along very similar lines, the difference >>between us is really in the detail, unless you want to go back >>to sayig the acceleraton term is a transient and it is the "vt" >>part on your diagram that is responsible for the output. > > 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. George |