Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Jeff Root on 18 Nov 2005 21:01 Eric, Would you explain why you are posting in this thread? What are you trying to accomplish? Are you succeeding? -- Jeff, in Minneapolis
From: Jeff Root on 18 Nov 2005 21:02 George, Would you explain to me why you are posting in this thread? What are you trying to accomplish? Are you succeeding, or making progress? -- Jeff, in Minneapolis
From: Jeff Root on 18 Nov 2005 21:04 Paul, Would you explain why you are posting in this thread? What are you trying to accomplish? Are you succeeding? -- Jeff, in Minneapolis
From: George Dishman on 19 Nov 2005 05:18 "Henri Wilson" <HW@..> wrote in message news:7tgsn11flp2iggll8is15lrdpqck37voaj(a)4ax.com... > On Fri, 18 Nov 2005 15:19:11 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: >>"Henri Wilson" <HW@..> wrote in message >>news:g4ann1lip1klougd93bdl7o3v1gid856ec(a)4ax.com... >>> 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." <Background trimmed> >>>>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. >> >>>>> http://www.users.bigpond.com/hewn/sagnac.jpg >> >>In that case you are wrong if I understand your diagram. >>When the rotation is at constant speed, the path length >>is also constant at the length of line AD or roughly: >> >> 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. >>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. >>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. George
From: George Dishman on 19 Nov 2005 05:48
"Jeff Root" <jeff5(a)freemars.org> wrote in message news:1132365769.558671.111690(a)z14g2000cwz.googlegroups.com... > George, > > Would you explain to me why you are posting in this thread? Purely recreational. > What are you trying to accomplish? To highlight to Henri that the Sagnac effect is a major problem for Ritz's ballistic light theory and by the normal rules of science it falsifies that theory. There is no reason why he shouldn't invent an alternative based on Ritz which he could call "BaT" but the challenge is there for him to explain what the difference is that resolves the problem. > Are you succeeding, or making progress? I think Henri has realised that the problem isn't trivial and is worthy of some thought whereas he initially dismissed it so yes, progress has been made. Certainly he has a much better understanding than when we started a few months ago which I also consider a positive accomplishment. For my part I have looked at the effect of angular acceleration in Ritz for the first time and can see qualitatively how it would produce an output but I can also see that a general quantitative result would be more complex because an 'event horizon' can be formed at high rates making the result discontinuous. Henri, if 'event horizon' sounds too like relativity, consider that in your binary star program, sometimes later wavefronts can overtake earlier ones if they are emitted faster due to the c+v factor. In theory the same thing could happen round the mirrors in the Sagnac experiment but only at unrealistic accelerations. Similarly you can get gaps in the beams for rapid speed reduction and the converse would apply to the opposite beam since an acceleration for one beam is a deceleration for the other. George |