Ballistic Theory, Progress report...Suitable for 5yo Kids
in [Relativity]
|
Prev: OWLS is not equal to c
Next: Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
From: Henri Wilson on 22 Nov 2005 19:11 On Tue, 22 Nov 2005 23:21:43 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:un77o1d04j3mucd353ud969ohinls6gc5c(a)4ax.com... >> On Tue, 22 Nov 2005 21:04:50 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >>>>> >>>>>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. > > >http://www.users.bigpond.com/hewn/sagnac.jpg > >If you think I am confused by your drawing, answer this >question and it will clear it up. > >Your diagram shows the length from C to E for constant >acceleration as "vt + 1/2 a t^2", right? Apply that to >the example speed profile and what do you get? Your speed profile has identical + and - accelerations. You end up at the starting speed. The fringe displacement returns to the original, determined by v. > >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: donstockbauer on 22 Nov 2005 19:15 "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong". Let's hear the details.
From: jgreen on 23 Nov 2005 00:27 George Dishman wrote: > "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 Verb or noun?Displacement indicating motion, or position (stationary after motion)? There is "displacement" occurring while the airframe rotates, and the (new position) "displacement" (eg 2 o'clock). It seems that your sagnac has additional electronic gadgetry which defaults the 2 o'clock to 12 o'clock, once the new heading stabilises. Jim G c'=c+v
From: donstockbauer on 23 Nov 2005 07:28 Just upping this one to 1120 posts. Yall have a very nice day.
From: Paul B. Andersen on 23 Nov 2005 10:38
Henri Wilson wrote: > On Mon, 21 Nov 2005 23:48:23 +0100, "Paul B. Andersen" > <paul.b.andersen(a)hiadeletethis.no> wrote: > > >>Henri Wilson wrote: >> >>>On Thu, 17 Nov 2005 16:54:34 +0100, "Paul B. Andersen" >>><paul.b.andersen(a)hiadeletethis.no> wrote: >>> > > >>>>>>Look Henri. >>>>>>Pick a distant star, say 500 LY away. >>>>>>Point your telescope at it, so that the image is at the centre. >>>>>>Measure the absolute angle of your telescope. >>>>>>Repeat 6 month later. >>>>>>The telescope will now point in a direction 22 arcsecs >>>>>>different from the first time. >>>>>> >>>>>>The parallax is negligible. The light path is the same, >>>>>>nameley a straight line from the star to the Earth. >>>>>>So why are the angle of the light path different? >>>>>>It is caused by the different velocity of the Earth >>>>>>at the two occations. We are observing the star from >>>>>>two different frames of reference. >>>>>>Both frames are moving relative to the star. >>>>> >>>>> >>>>>So what? Ligth leaves the star spherically. >>>> >>>>Still drunk? >>>>Didn't you get it? >>>>There is but one light path - the path from the Star to the Earth. >>> >>> >>>Are you under the impression that the star is emitting all its light in one >>>particular direction, as with a narrow laser beam? >> >>Still drunk? >>Or are you too stupid to get the simple point? >> >>The light that does not hit the CCD in our telescope is of no >>interest whatsoever. It might as well have been a laser beam, >>it would make no difference. >>(The width of the beam would have to be at as wide >>as the parallax angle, though. But as this is much smaller >>than the aberration angle, we can forget it.) >> >>Ignore the parallax. >>We can in our thought example imagine that the star >>is in the equatorial plane, and that we are observing >>it when the Sun, Earth and star are all in line. >>(That means that the star will be behind the Sun >>at one occasion, but it doesn't matter in principle) >> >>Then we have the picture drawn in solar frame: >> >> * star >> | >> | the one and only >> | light path >> | >> | >> O-> v observer >> | >> S >> | >> v<-O >> >>At one occasion, the observer is moving to the right, >>an will have to point his telescope like this: >> >> / >> / light path down the tube >> / >> ------ >>Six month later, the observer is moving to the left, >>and will have to point his telescope like this: >> >> >> \ >> \ light path down the tube >> \ >> ------ >> >>At both occasions, the angle is v/c radians from vertical. >>v/c = 10^-4 radians = 11 arcsecs >> >>The difference is 22 arcsecs. > > > I'm not disputing the fact tat you have to tilt the telescope...but you are > very confused concerning the path of the beam. It remains vertical. > > >>>>(We can neglect the small parallax angle which is only 0.0006 of >>>>the aberration angle) >>>>It is obviously utterly irrelevant that the star emits light >>>>in all other directions that don't hit the Earth, >>>>so why the hell are you stating this stupidity? >>>> >>>>The only light path of interest is the one that hits the Earth! >>>>The _direction_ of that light path is down the middle of >>>>our telescope. >>> >>> >>>You really are funny today. >>> >>> >>> >>>>The direction of that single light path changes throughout the year >>>>because the velocity of the frame of reference (Earth) changes >>>>throughout the year. >>> >>> >>>Very good Paul. You are improving. >>> >>> >>> >>>>>This is in no way related to our discussion. You are diverting attention from >>>>>the fact that SR is proved to be nonsense. >>>> >>>>It is related to your incredible stupid statements: >>>>"Wavefronts really exist only in the source frame." >>>>and: >>>>"Whatever is moving diagonally isn't light. It is >>>>infinitesimal points." >>> >>> >>>That is right. Of course I was refering to the plotting, in my frame, of the >>>paths of individual 'points' inside a vertical laser beam as I move >>>horizontally past it. >> >>You are evading the point, Henri. >>There is no important difference from your vertical beam. >>The observer is moving perpendicular to the light beam from >>the star that hits the Earth. That light path from the star is >>"vertical" in our "stationary" solar reference system, >>which is the same as the "source frame". >>But the light path that is going down the observer's >>telescope is at an angle v/c radians from vertical. > > > That's what a raw amateur would say. > The tilted telescope has to move sideways to keep the incoming vertical beam at > its centre. > The beam, as a whole, is v/c radians from the telescope tilt... which means it > is actually vertical...but moving sideways in the observer frame. > > Do you know how to move a vertical object sideways Paul? > > >>So what is it that is moving at an angle down the tube >>and hitting the CCD? Is it light, or is it just infinitesimal >>points? How can the CCD detect infinitesimal points? > > > It is vertical light moving down the sideways moving tilted tube. > That should be obvious to all but raw beginners. Quite. It is light moving vertically in the frame where the tube is moving sideways. And it is light moving diagonally in the frame where the tube is stationary. That should indeed be obvious to all but raw beginners. >>>>"WHATEVER IS COMING FROM THAT STAR, MOVING WITH CHANGING >>>>DIRECTION, IS NOT LIGHT, IT IS ONLY INFINITESIMAL POINTS." >>> >>> >>>You are quoting me completely out of context and you know it. >>>tat has nothing whatsoever to do with the topic or anything I have said. >> >>No, I am not. >>You claimed that the light is moving vertically in the "source frame", >>and "whatever is moving diagonally in the observer frame is not light." > > > that's correct. > And your telescope example shows that I am right. > > >>The light from the star IS moving vertically in the source frame. >>But our observer have to point is telescope "diagonally" to make >>the "whatever that is moving" go down his telescope tube. >> >>So what is the "whatever" that is moving diagonally down the tube? > > > You are forgetting that the tube is tilted...and you are now making a complete > fool of yourself. > > Try this: > > ^ v L > t======.............................................................u<-b.......---O > > You are at O with a gun. You want to fire a bullet radially at velocity u, at a > target (t) that lies at the centre of a pipe (L), which is spinning around you > at v. > > Q1) Is it possible to hit the target if the pipe remains aligned radially? What > if you tilt your gun wrt the pipe axis? > Q2) If not, at what angle from the radius vector should the pipe be tilted? > Q3) How is the concept of 'vertical' defined for you or the pipe? > >>You claim it is not light. So what is it then? > > > An infinitesimal element of the vertical beam. > > >>Whatever it is, CCDs detects it as if it were light. > > > Because the CCD is moving sideways so that it will continuously intercept an > infinitesimal element of a DIFFERENT section of the vertical beam in successive > infinitesimal time intervals. Come again? :-) Whatever it is that is moving in a continuous stream along the diagonal path down the telescope tube and hits the CCD is not light because? >>>>Now Henri, what is it that hits the CCD in our telescope? >>>>Is it light? >>>>It cannot be, can it? >>>>Because whatever moves along paths with different directions >>>>in different frames cannot be light, can it? >>> >>> >>>The star emits a sphere of light Paul. The wavefronts are spherical. Didn't you >>>know that. >>>When my telescope moves sideways, a different radius vector of the sphere goes >>>down the middle of my telescope. What could be more simple? I cannot see why >>>you should have any trouble understanding that. >> >>Please, Henri. >>You are not really THIS stupid, are you? :-) > > > A raw beginner might think I am stupid...just as a raw beginner like Einstein > would think that raindrops fall diagonally when he looks at them through a > moving train window. A very revealing statement. :-) The raw beginner consider the ground frame to be absolute. So he will claim that the raindrops are falling vertically even in the rest frame of a train moving in the ground frame. Our raw beginner is rather naive, isn't he? :-) >>(I am beginning to believe you are. >> I didn't think it possible!) >> >>Don't you understand that if you had two telescopes going in >>opposite direction, looking at the same star, they would have to >>point in opposite directions as they pass each other? >> >> * >> | >> 100 LY >> | >> / \ >> / \ >> O->v<-O > > > Yes. That is because the beam is vertical wrt earth's surface and THEY are > moving sideways. > > >>It is the very same one and only light path that has different >>angles in the two observers' frames. > > > You fail to see that the successive elements that strike the centre of the > telescope do not come from the same infinitesimally narrow section of the > vertical beam. No, I don't fail to see that. I never mentioned a "light beam". I said "light path". The point is that whatever hits the CCD has moved along a diagonal path down the telescope tube. Is it light? > > I accept that this might be a bit hard for you. > > >>A little movie for you. The "*" is a photon, or a bit of LIGHT. >>Two telescope tubes going through each other in opposite directions. >> >> >> enter X * X >> tubes / \ / \ >> / X \ >> / / \ \ >> / / \ \ >> -ccd- -ccd- >> >> \ / \ / >> X * X >> going / \ / \ >> down / X \ >> / / \ \ >> -ccd- -ccd- >> >> \ X / >> and \ / \ / >> down X * X >> / \ / \ >> / X \ >> -ccd-ccd- >> >> \ \ / / >> \ X / >> \ / \ / >> X * X >> / \ / \ >> -ccccd- >> >> \ \ / / >> \ \ / / >> \ X / >> \ / \ / >> X * X >> -ccd- >> >> \ \ / / >> \ \ / / >> hit \ \ / / >> ccds \ X / >> \ / \ / >> Xc*dX >> >>Our single photon - or piece of LIGHT - goes >>down both tubes. This single photon has obviously >>but one path. This single light path is vertical >>in the screen frame, AND the very same light path >>is "diagonal" in the two telescope frames. > > > ...which proves the photon as moved vertically wrt Earth. > Throw away the Earth and put the telescopes on oppositely moving vehicles. The Earth IS the moving vehicle. It is six month between each time the telescope is moving in opposite direction. Both times the photon has moved diagonally wrt Earth. >>The photon - or piece of light- is moving vertically >>in the screen frame, AND the very same photon - or piece >>of light- is moving diagonally in the telescope frames. > > > ...and the telescope is tilted by the same diagonal in the Earth frame. > >>It is obviously mindless babble to claim that the very >>same physical object both is and isn't light. > > > The CCD continuously detects an infinitesimal element of differnet light beams > from the star in each infinitesimal time interval. There is an infinite number > of them. So you have confirmed my words: Paul B. Andersen wrote: | If you from the cosmic ship Tellus ever look through | a telescope, be sure to notice the stellar aberration. | Having done so, you can kick and shout: | WHATEVER IS COMING FROM THAT STAR, MOVING WITH CHANGING | DIRECTIONS, IS NOT LIGHT, IT IS ONLY INFINITESIMAL POINTS. | | That will make you look exactly as intelligent as you are. Thanks. Paul |