Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 22 Nov 2005 05:08 "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 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
From: Henri Wilson on 22 Nov 2005 15:18 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. George, out of pure desperation, tried to make he fringe movement proportional to da/dt. > >Jim G >c'=c+v >(G'day George---------just watching :-) ) 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 15:24 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? 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 15:32 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. >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. 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 15:41
<jgreen(a)seol.net.au> wrote in message news:1132652040.894970.163580(a)g47g2000cwa.googlegroups.com... > > 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. Jim, it's clear you didn't try to work out what would happen from the structure of the equipment. Interferometric devices have no means of storing the location in this way. In fact your description is quite a good match for a ring laser gyro where the standing wave essentially stays at a fixed orientation and fringes could be counted to indicate the change of direction, but there is no similar wave in the iFOG design. In your example, from North to North-east is a change of 45 degrees. Suppose that happens smoothly over a period of 10 seconds. What actually happens is that while the plane is flying constantly north, there is no signal on the primary output which is the rate of turn. When the plane starts changing direction, the output would go to a voltage that indicates 4.5 degrees per second. Once the plane is moving North-east and is heading in a straight line, the output goes back to zero. That is, the output is proportional to the rate of change of heading. To get the bearing, you put that signal through something called an integrator. > 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) Your prediction is not only wrong, it is obviously impossible to even derive that prediction from the equipment design, you are obviously just guessing. The prediction from SR for constant rate of turn based on the known structure of the device matches what actually happens. The discussion between Henri and I is as follows: I said that the prediction based on Ritz's model (what you call "c'=c+v") for a constant rate of turn is no output. Henri was arguing for that it predicts an output proportional to speed, the same as SR. Lately Henri has prompted discussion of the effect of _varying_ rate of turn. Having looked at that, I agree with the diagram he drew which showed no output for constant rate of turn but an output proportional to the rate at which it varied, the angular acceleration. Henri is interpreting the same diagram differently, saying that it again suggests an output proportional to speed (rate of turn). The reason we differ is because the diagram doesn't have a way of showing the speed of the light, it only shows the path lengths. This is his diagram: http://www.users.bigpond.com/hewn/sagnac.jpg George |