GETTING RID OF EINSTEIN RELATIVITY
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From: Dr. Henri Wilson on 18 Oct 2007 16:26 On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... >>>It gives you the answer you WANT. It isn't the CORRECT >>>answer, you need to learn how an interferometer works. >>>Start with the applet and instructions in my other post >>>and I'll try to educate you when you immediately leap >>>to the wrong conclusion. >> >> George, the path lengths are different..and change with rotation speed. > >And the waves are not in phase at the start point of >that path, they are in phase at the end. Thiis might help. www.users.bigpond.com/hewn/toothwheel.exe (It is not complete and doesn't seem to want to run properly on windows Vista...but not many things do). The toothed wheel represents a snapshot of BOTH rays when they are in phase at the emission point. The teeth represent absolute wavelengths. It is not shown moving because the positions of the teeth at that instant is independent of light speed. The red and blue lines show the rays moving at c+v and c-v. They travel for the same time and meet at the detector together. If you vary the ring rotation speed via the combo box, you will see that the phases of the rays when they reunite are not the same. If you want to argue against this approach, I remind you that it gives the experimentally verified answer. >> At constant speed, no matter what that speed is, the fringe pattern is the >> same >> and static......and the number of waves between the source and detector is >> the >> same in both paths and constant. > >Exactly. > >> The two path lengths change and fringes move during an acceleration. This >> is >> due to the fact that photons and their 'wavelengths' shrink or expand >> during an >> acceleration. > >You forgot your "K" factor which says the photons _don't_ >get compressed. I didnt forget. >> The movement is monitored. In iFoGs there is a constant >> integration over very short time intervals to turn current fringe >> displacement >> into total angle turned. > >Integrating something proportional to angular acceleration >doesn't give you the angle turned. I said the current DISPLACEMENT is integrated George. Can't you read? The time periods for integration are also very small to account for any acceleration. Displacement is proportional to ring speed. If you integrate, for instance, 0.001 radian per second wrt TIME over a particular time interval, you get total ANGLE moved during that interval. >> I know you wont be able to understand any of this until I demonstrate it. >> An >> animation is on the way. > >I know you won't be able to understand any of this until >you learn basic calculus. ....you obviously don't have a clue about integration. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Androcles on 18 Oct 2007 18:29 "Dr. Henri Wilson" <HW@....> wrote in message news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... : On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" <george(a)briar.demon.co.uk> : wrote: : : > : >"Dr. Henri Wilson" <HW@....> wrote in message : >news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... : : >>>It gives you the answer you WANT. It isn't the CORRECT : >>>answer, you need to learn how an interferometer works. : >>>Start with the applet and instructions in my other post : >>>and I'll try to educate you when you immediately leap : >>>to the wrong conclusion. : >> : >> George, the path lengths are different..and change with rotation speed. : > : >And the waves are not in phase at the start point of : >that path, they are in phase at the end. : : Thiis might help. : www.users.bigpond.com/hewn/toothwheel.exe : (It is not complete and doesn't seem to want to run properly on windows : Vista...but not many things do). It's wrong.
From: Dr. Henri Wilson on 18 Oct 2007 19:01 On Thu, 18 Oct 2007 22:29:36 GMT, "Androcles" <Engineer(a)hogwarts.physics> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... >: On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" ><george(a)briar.demon.co.uk> >: wrote: >: >: > >: >"Dr. Henri Wilson" <HW@....> wrote in message >: >news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... >: >: >>>It gives you the answer you WANT. It isn't the CORRECT >: >>>answer, you need to learn how an interferometer works. >: >>>Start with the applet and instructions in my other post >: >>>and I'll try to educate you when you immediately leap >: >>>to the wrong conclusion. >: >> >: >> George, the path lengths are different..and change with rotation speed. >: > >: >And the waves are not in phase at the start point of >: >that path, they are in phase at the end. >: >: Thiis might help. >: www.users.bigpond.com/hewn/toothwheel.exe >: (It is not complete and doesn't seem to want to run properly on windows >: Vista...but not many things do). > >It's wrong. that's funny. It produces the right answer. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 18 Oct 2007 19:38 "Clueless Henri Wilson" <HW@....> wrote in message news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... > On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... > >>>>It gives you the answer you WANT. It isn't the CORRECT >>>>answer, you need to learn how an interferometer works. >>>>Start with the applet and instructions in my other post >>>>and I'll try to educate you when you immediately leap >>>>to the wrong conclusion. >>> >>> George, the path lengths are different..and change with rotation speed. >> >>And the waves are not in phase at the start point of >>that path, they are in phase at the end. > > Thiis might help. > www.users.bigpond.com/hewn/toothwheel.exe > (It is not complete and doesn't seem to want to run properly on windows > Vista...but not many things do). > > The toothed wheel represents a snapshot of BOTH rays when they are in > phase at > the emission point. Useless, the whole point is that they move at different speeds. Show that and see what happens. > The teeth represent absolute wavelengths. That part is correct but just make each part move at the right speed and it falls out automatically. > It is not shown > moving because the positions of the teeth at that instant is independent > of > light speed. That's your error, the waves have to get from source to detector and that takes time. > The red and blue lines show the rays moving at c+v and c-v. They travel > for the > same time and meet at the detector together. You need to show them as moving waves. > If you vary the ring rotation speed via the combo box, you will see that > the > phases of the rays when they reunite are not the same. > > If you want to argue against this approach, I remind you that it gives the > experimentally verified answer. Irrelevant, it is wrong. As I said before, it gives you the answer you WANT. It isn't the CORRECT answer. I would also remind you that SR gives the right answer if you want to use that argument. >>> At constant speed, no matter what that speed is, the fringe pattern is >>> the >>> same >>> and static......and the number of waves between the source and detector >>> is >>> the >>> same in both paths and constant. >> >>Exactly. >> >>> The two path lengths change and fringes move during an acceleration. >>> This >>> is >>> due to the fact that photons and their 'wavelengths' shrink or expand >>> during an >>> acceleration. >> >>You forgot your "K" factor which says the photons _don't_ >>get compressed. > > I didnt forget. Yes, you just conveniently ignored it. >>> The movement is monitored. In iFoGs there is a constant >>> integration over very short time intervals to turn current fringe >>> displacement >>> into total angle turned. >> >>Integrating something proportional to angular acceleration >>doesn't give you the angle turned. > > I said the current DISPLACEMENT is integrated George. Can't you read? Yes, and the displacement is proportional to acceleration according to ballistic theory, not speed. > The time > periods for integration are also very small to account for any > acceleration. > > Displacement is proportional to ring speed. Wrong. George
From: Androcles on 18 Oct 2007 19:48
"George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1oCdnQCQyo-RcIranZ2dnUVZ8q2dnZ2d(a)pipex.net... : : "Clueless Henri Wilson" <HW@....> wrote in message : news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... : > On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" : > <george(a)briar.demon.co.uk> wrote: : >>"Clueless Henri Wilson" <HW@....> wrote in message : >>news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... : > : >>>>It gives you the answer you WANT. It isn't the CORRECT : >>>>answer, you need to learn how an interferometer works. : >>>>Start with the applet and instructions in my other post : >>>>and I'll try to educate you when you immediately leap : >>>>to the wrong conclusion. : >>> : >>> George, the path lengths are different..and change with rotation speed. : >> : >>And the waves are not in phase at the start point of : >>that path, they are in phase at the end. : > : > Thiis might help. : > www.users.bigpond.com/hewn/toothwheel.exe : > (It is not complete and doesn't seem to want to run properly on windows : > Vista...but not many things do). : > : > The toothed wheel represents a snapshot of BOTH rays when they are in : > phase at : > the emission point. : : Useless, Knee jerking fuckhead, cant stand being beaten. |