Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: "Androcles" <Androcles@ on 9 Oct 2005 19:21 "Timo Nieminen" <timo(a)physics.uq.edu.au> wrote in message news:Pine.LNX.4.50.0510100807360.7418-100000(a)localhost... | Sure. Especially the physics of complex systems. Yeah, a CRT is a really complex system. ROFLMAO! Androcles.
From: bz on 9 Oct 2005 19:06 H@..(Henri Wilson) wrote in news:ke1jk11hal7bl7v36010jpemc8oik68ngr(a)4ax.com: > On Sat, 8 Oct 2005 23:51:52 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> > wrote: > >>H@..(Henri Wilson) wrote in >>news:4gjgk11be4aadfp4vu9a0hehnc4hmie7sm(a)4ax.com: >> >>> On Sat, 8 Oct 2005 17:24:03 +0000 (UTC), bz >>> <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >>> > >>>>Ad hominem attacks are the last refuge of desperate people. >>>>When you make such attacks, you put yourself into that catagory. >>>> >>>>Download starlight pro from >>>>http://www.physics.sfasu.edu/astro/binstar.html and play with it a >>>>bit. >>> >>> Can't you see the flat sections between the eclipses? >> >>Of course. But the 'flat sections between the eclipses' disappear when >>the stars are large and close together, although the region may look >>rather flat. > > Between the eclipses it is dead flat (apart maybe from the 'day/night' > effect). > > You are refering to the partial eclipse condition. I am refering to stars that are not spherical. Download that program and hit the 'slide show' button'. There are many non spherical binaries. >>> I don't know what you are trying to say here Bob. If the star is not >>> being eclipsed, its brightness should remain almost dead constant >>> whatever its shape and spin. >> >>Just as the earth receives different amount of illumination from the sun >>when the sun is close to the horizon vs when it is overhead, our sun and >>other stars emit different amounts of light at different angles of >>radiation. > > Rubbish. Stars are all homogeneous balls of gas. Strange, I had the impression that even our star had starspots and that many stars were not homogeneous. However even a homogeneous star radiates less energy in our direction from the limbs of the star than from the central region of the surface that is perpendicular to us. >>This produces an effect called 'limb darkening'. > > How the hell does the ecplise by a second distant object affect the > properties of the star itself? It doesn't. Just the light we see. > If you are claiming that the star becomes ellipsoidal or something, then > that would only happen if the orbiting object was extremely close. Yep. Yep. > Generally, the star's brightness remains constant outside the eclipse > zone. Generally. >>This makes a body eclipsing the central region of the star 'intercept >>more of the star light destended to head in our direction' than the same >>body does when it eclipses an off-central region of the star. > > You appear confused. > > We were taking about the non-eclipse region. Even then there is often variations in brightness. >>>>If you want the troughs flat bottomed, >>>>turn off limb darkening and make sure the plane of the orbit allows >>>>total or annular occultation to take place. >>> >>> I don't want the troughs flat bottomed. It is obvious how that can be >>> achieved. >> >>Is it? > > yes, quite obvious. How? >>>>>>> ..unlike Algol's , which has a distinct >>>>>>> concavity....just as the BaT predicts. >>>>>>> -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: bz on 9 Oct 2005 19:14 H@..(Henri Wilson) wrote in news:782jk19hi71c9p9iec724febljiqsfnoq0(a)4ax.com: > The fact that electric fields operate at c is the reason why charges > cannot be accelerated beyond c in cyclotrons etc. Charges can not be accleerated to c in cyclotrons, etc. The phase velocity of the RF field is what accelerates the charges. It is quite possible to have the phase velocity in the accelerator move faster than light. Non the less, the charges won't move even as fast as light. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: bz on 9 Oct 2005 19:25 H@..(Henri Wilson) wrote in news:rm4jk1trn1pjobaipmlluur1jiq49hfju5(a)4ax.com: > On Sun, 9 Oct 2005 15:09:55 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> > wrote: > >>"George Dishman" <george(a)briar.demon.co.uk> wrote in news:dib6vp$472$1 >>@news.freedom2surf.net: >> >>> Consider a non-rotating frame centred on the mirror. >>> It is not inertial because the origin is accelerated >> >>agreed. >> >>> but the source orbits the mirror at the same rate that >>> the mirror rotates. >> >>In the mirror's FoR, the universe is rotating in a rather eccentric way, >>but the source is fixed in position at a contant distance and angle. >>This must be true as all mirrors and source are fastened securely to a >>base- plate of some sort. >> >>You (Henri or George) seem choosing a POINT on the mirror, perhaps a >>point at the exact center of the mirror and are using that as your >>origin, but looking at things around that point from a FoR that is >>parallel to some reference in the lab FoR. >> >>Thus you have NOT chosen the mirror's FoR, but a hybred FoR that is >>related to the turntable AND to the lab in a rather complex way. >> >>> >>> It's hard to imagine a more awkward frame to choose >> >>Agreed, except, perhaps, for the photons FoR. > > The moon is rotating around the earth, which is also rotating daily. > The moon's radial speed wrt Earth is zero. It wouldn't make any > difference if the earth rotated in 1 day, 28 days or 1000 days. I think you have it backwards. The Earths radial speed wrt the moon is zero. If you replace the earth with a satelite in a lunar syncronous orbit, there would be no doppler shift on signals between that satelite and the moon. On the other hand, the moon swings around the earth every 28 days. The earth rotates every day. Those motions will give doppler shifts on signals from moon to earth. > Sagnac DOES NOT refute the BaTh. I think you are wrong. > Sagnac clearly supports either my transverse doppler theory' my 'photon > axis' theory or good old fashioned aether, of which SR is a subset. > . I think you are right, one of the above. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: George Dishman on 9 Oct 2005 19:53
"Henri Wilson" <H@..> wrote in message news:db4jk1hda18n1gno0lhsmto6rr7hajprip(a)4ax.com... > On Sun, 9 Oct 2005 14:53:39 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message >>news:Xns96EA50DFCC736WQAHBGMXSZHVspammote(a)130.39.198.139... >>> H@..(Henri Wilson) wrote in >>> news:f4mgk1ls96rgfjoplja5l2munbhkctkaii(a)4ax.com: >>> >>>> In the mirror frame, the source is moving in a circle around that >>>> mirror. >>> >>> >>> False. >>> >>> In the mirror's frame, the location of the source is constant and the >>> angle >>> of incidence of light from the source is constant. >>> >>> Proof? If what you stated were true, the source would orbit around the >>> mirror and the angle of incidence of the light would change through 360 >>> degrees with every rotation of the platform. >> >>Consider a non-rotating frame centred on the mirror. >>It is not inertial because the origin is accelerated >>but the source orbits the mirror at the same rate that >>the mirror rotates. >> >>It's hard to imagine a more awkward frame to choose > > Now that you have gotten the picture right , please explain it to the > other > morons here. Lucky guess, you didn't say whether you meant a rotating or non-rotating frame. > We have an approximate example of this in the case of the moon orbiting > the > earth...with the earth circling the sun and also rotating about its own > axis. > > All the time, there is no radial motion between earth and moon....but > light > from the moon is 'transversely doppler shifted'. Transverse Doppler is another name for time dilation so doesn't occur in Ritzian theory (nor I presume in your BaT). > Similarly, there is no radial motion between any mirror of the Sagnac and > the > previous one. > > If you want to write a joint paper on this I am quite willing. I think the > world should know why Sagnac DOES NOT refute the BaTh. But it does Henri. Doppler would produce a second order output while the Sagnac Effect is first order, and transverse Doppler doesn't exist in Ritzian theory. Try again. George |