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From: Paul B. Andersen on 26 Apr 2010 17:03 On 25.04.2010 00:42, Henry Wilson DSc wrote: > On Sat, 24 Apr 2010 23:58:07 +0200, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: > >> On 20.04.2010 23:57, Henry Wilson DSc wrote: >>> >>> I maintain that 'true trajectory' is y versus t for constant x....and that can >>> only be meaningful in the object's own frame. >> >> A true gem. :-) >> >> The only 'true trajectory' is a point in the object's own rest frame. >> >> A true keeper! :-) > > Is you definition of 'trajectory' a plot of y-t or y-x? > > Come on! Answer! How come "a fully qualified physicist" and Doctor of science havs to be taught what a trajectory is? Here follows an elementary lesson for the Doctor. A trajectory is the path in space followed by an object. So it is a line, or curve, in _space_. (You could of course also talk about a trajectory in space-time, but the above is the traditional definition of a trajectory.) It can be expressed by one of the coordinates as a function of the two others: z = f(x,y) (or more generally f(x,y,z) = 0) (this form has its limitations, it isn't always possible to express it this way) or it can be expressed on parametric form: z = f_z(p) y = f_y(p) x = f_z(p) where p is the parameter. The most common parameter is the time, of course, but there is always an infinite number of possible parameters which can be used. An example: Let z be the vertical axis, and x and y the horizontal axes. A projectile is launched with the horizontal velocity component v_x along the x-axis, v_y = 0 along the y-axis, and the vertical velocity component v_z along the z-axis. Let the projectile be launched from the origin at time t=0. Using the differential operator D = d/dt, we can write the differential equations: D^2.z = -g giving: z = -gt^2/2 + Dz(0).t + z(0) Dz(0) = v_z and z(0) = 0 z = -gt^2/2 + v_z.t and: Dx = v_x giving: x = v_x.t (x(0) = 0) So the trajectory on parametric form is: z = v_z.t - gt^2/2 x = v_x.t y = 0 ...or with a different parameter p: z = 2v_z.p - 2gp^2 x = 2v_x.p y = 0 Eliminating the parameter, this trajectory can be written: z = (v_z/v_x)x + (g/2.v_x^2)x^2 which is a parabola. In _space_. Homework: What is the trajectory of the above projectile in a frame of reference (x',y',z') defined by the Galilean transform: x' = x - v_x.t y' = y z' = z t' = t Use math, please. -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 26 Apr 2010 18:35 On Mon, 26 Apr 2010 23:03:32 +0200, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 25.04.2010 00:42, Henry Wilson DSc wrote: >> On Sat, 24 Apr 2010 23:58:07 +0200, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >> >>> On 20.04.2010 23:57, Henry Wilson DSc wrote: >>>> >>>> I maintain that 'true trajectory' is y versus t for constant x....and that can >>>> only be meaningful in the object's own frame. >>> >>> A true gem. :-) >>> >>> The only 'true trajectory' is a point in the object's own rest frame. >>> >>> A true keeper! :-) >> >> Is you definition of 'trajectory' a plot of y-t or y-x? >> >> Come on! Answer! > >How come "a fully qualified physicist" and Doctor of science >havs to be taught what a trajectory is? > >Here follows an elementary lesson for the Doctor. > >A trajectory is the path in space followed by an object. >So it is a line, or curve, in _space_. > >(You could of course also talk about a trajectory > in space-time, but the above is the traditional > definition of a trajectory.) > >It can be expressed by one of the coordinates as >a function of the two others: > z = f(x,y) (or more generally f(x,y,z) = 0) >(this form has its limitations, it isn't always possible > to express it this way) > >or it can be expressed on parametric form: > z = f_z(p) > y = f_y(p) > x = f_z(p) >where p is the parameter. >The most common parameter is the time, of course, >but there is always an infinite number of possible >parameters which can be used. What you are calling 'trajectory' I would call a 'plot'. You still cannot explain how the path of an object, whose movement is restricted to the axis of a straight tube, can ever be curved. >An example: >Let z be the vertical axis, and x and y the horizontal axes. >A projectile is launched with the horizontal velocity >component v_x along the x-axis, v_y = 0 along the y-axis, >and the vertical velocity component v_z along the z-axis. >Let the projectile be launched from the origin at time t=0. > >Using the differential operator D = d/dt, we can write >the differential equations: > D^2.z = -g > giving: > z = -gt^2/2 + Dz(0).t + z(0) > Dz(0) = v_z and z(0) = 0 > z = -gt^2/2 + v_z.t > >and: > Dx = v_x > giving: > x = v_x.t (x(0) = 0) > > >So the trajectory on parametric form is: > z = v_z.t - gt^2/2 > x = v_x.t > y = 0 > >..or with a different parameter p: > z = 2v_z.p - 2gp^2 > x = 2v_x.p > y = 0 > > >Eliminating the parameter, this trajectory can be written: > z = (v_z/v_x)x + (g/2.v_x^2)x^2 > >which is a parabola. In _space_. > >Homework: >What is the trajectory of the above projectile >in a frame of reference (x',y',z') defined by the Galilean transform: > x' = x - v_x.t > y' = y > z' = z > t' = t > >Use math, please. That transform just changes the x scale. Henry Wilson... ........A person's IQ = his snipping ability.
From: Henry Wilson DSc on 26 Apr 2010 18:43 On Mon, 26 Apr 2010 14:54:10 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 26.04.2010 00:37, Henry Wilson DSc wrote: >> On Sun, 25 Apr 2010 23:33:21 +0200, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >> >>> On 25.04.2010 01:33, Henry Wilson DSc wrote: >>>> On Sat, 24 Apr 2010 23:35:12 +0200, "Paul B. Andersen"<someone(a)somewhere.no> >>>> wrote: >>>>> On Fri, 07 Apr 2006 "Paul B. Andersen" wrote: >>>>> | There is but one type of what you call "time compression", >>>>> | and it is basically the same phenomenon as the Doppler shift. >>>>> | And it isn't hard to calculate at all. >>>>> | If an event happens at the source at the time t, >>>>> | and this event is observed by the observer at the time to, >>>>> | then the "time compression" simply is : dto/dt. >>>>> | >>>>> | Let me first demonstrate this "time compression" >>>>> | when we assume that the speed of light is c in >>>>> | the observer's rest frame. (In a Galilean world.) >>>>> | The observed object is moving with the radial speed v. >>>>> | We define positive v as approaching. >>>>> | At t = 0, the distance to the object is D. >>>>> | At time t, the distance to the object is >>>>> | D - (integral from 0 to t of) v(t)*dt >>>>> | I will write the latter term as I(vdt). >>>>> | So an event happening at the time t will be observed >>>>> | by the observer at the time: >>>>> | to = t + (D - I(vdt))/c >>>>> | dto/dt = 1 - v/c = (c - v)/c >>>>> | This is observed "time compression". >>>>> | Since a period of the emitted light is "compressed" >>>>> | by the factor dto/dt, the Doppler shift is the inverse >>>>> | of this: DS = c/(c - v) >>>>> | >>>>> | Now, let us do the same using the ballistic theory. >>>>> | The difference is only that now the speed of light >>>>> | in the observer's rest frame is c+v. >>>>> | Thus: >>>>> | to = t + (D - I(vdt))/(c+v) >>>>> | If we assume that v/c<< 1, we can write the approximation: >>>>> | to = t + (D - I(vdt))*(1 - v/c)/c >>>>> | dto/dt = 1 - v/c + (v/c)^2 - ((D + I(vdt))/c^2)*dv/dt >>>>> | We can ignore the (v/c)^2 term. >>>>> | And if we assume that that the object is in orbit, >>>>> | the displacement I(vdt) can be ignored compared to D. >>>>> | (The radius of the orbit is small compared to the distance >>>>> | to the star.) >>>>> | So we get: >>>>> | dto/dt = 1 - v/c - (D/c^2)*dv/dt >>>>> | >>>>> | Note one very important point, though. >>>>> | This equation says that the light that is >>>>> | emitted at the time t will be observed >>>>> | at the rate dto/dt, it says nothing about >>>>> | _when_ the observation is done. >>>>> | (Meaning that dto/dt doesn't have to vary >>>>> | sinusoidally with to even if v(t)is sinusoidal.) >>>>> | >>>>> | The Doppler shift is the inverse of this: >>>>> | DS = 1/(1 - v/c - (D/c2)*dv/dt) >>>>> | >>>>> | If we can assume that (D/c^2)*dv/dt<< 1, >>>>> | then this can be written: >>>>> | DS = 1 + v/c + (D/c^2)*dv/dt >>>>> | >>>>> | Note however that according to the ballistic >>>>> | theory, this assumption can NOT generally be done. >>>>> | (Se remark below.) >>>>> | >>>>> | If you analyse under which conditions the ballistic >>>>> | theory predicts that the intensity will be infinite, >>>>> | you will find that it is when it predicts that >>>>> | the Doppler shift also is infinite. >>>>> | It is a good reason for that. When light emitted >>>>> | at the time t is received at the same time as >>>>> | the light emitted a time dt later, the "time compression" >>>>> | dto/dt = 0, which means that a period is compressed to nothing, >>>>> | the observed frequency is infinite, and the intensity >>>>> | is infinite. >>>>> | If the light emitted at the time t is received _after_ >>>>> | the light emitted dt later, the "time compression" >>>>> | dto/dt is negative (time is reversed), and the Doppler >>>>> | shift is negative. >>>>> | If this happens, we will at the same time observe >>>>> | light emitted at another time (multiple stars), >>>>> | with a different Doppler shift. >>>>> | >>>>> | But the bottom line is that it is a one to one >>>>> | relationship between the predicted brightening of >>>>> | the star and the Doppler shift. >>>> >>>> No it isn't. ADoppler is never as large as brightness variation. >>>> >>>> Compression WITHIN a photon is different from compression BETWEEN photons. >>>> >>>> Think of a photon as a damped inelastic coil spring. >>>> If photon is emitted by an accelerating source, then its rear end moves up on >>>> its front end....but not for long. >>>> Individual photons, on the other hand, continue to move up on each other ad >>>> infinitum...or at least till external factors tend to unify their speeds. >>> >>> Nonsensical babble. >>> >>> _All_ frequencies are Doppler shifted exactly the same. >>> It doesn't matter what mechanism causes the Doppler shift. >>> f = 1/T. So if D is the observed Doppler shift f' = Df, then >>> the observed duration T' = T/D, and the energy emitted during >>> the time T is received during the time T/D, so the brightening is D. >>> >>> It doesn't matter which theory you use, this is a fundamental >>> relationship true for _all_ theories. So even SR predicts that >>> the brightness of binaries should vary a little due to the Doppler >>> shift, but this variation will hardly ever be measurable. >> >> Oh dear, Norwegian science is indeed in a bad way. >> You obviously don't have the faintest idea of what I'm talking about. >> Do you know what radial acceleration is...for an elliptical orbit? >> >> ADoppler is a wavelenght shift due to acceleration. You are talking about >> conventional VDoppler. >> >>> It is bloody obvious that _no_ variable star has a brightness >>> variation exactly equal to the Doppler shift. >>> That means that even according to the emission theory, the variation >>> of _all_ variable stars must be caused by either an intrinsic variation, >>> or by being eclipsed by another body. >> >> Oh dear...I thought you might have learnt something by now. Your old colleague >> George Dishman knew exactly what ADopppler implied. It is a pity George isn't >> with us now. He probably would have been the first relativist to admit he and >> Einstein had been completely wrong. >> >>> And it is equally obvious that if such a thing as a binary star >>> exists, then we do not observe the Doppler shift predicted >>> by the emission theory. The Doppler shift due to acceleration >>> should be giant. We never observe it. >> >> It would be of the same order of magnitude as the linear brightness variation >> if for instance light was just a 'wave in the aether'. But my theory explains >> it with the 'coiled spring' model. >> >> The bunching effect WITHIN a photon is limited whilst that BETWEEN photons is >> continuous. >> >> It also states that the equation E= h.c/L' does not hold for ADoppler shifted >> photons since their individual energy cannot change due to 'internal bunching'. >> It should be E = h.c/Lo >> >>>>>> Tell me Paul, how is it that astronomers cannot explain why cepheid velocity >>>>>> curves are a virtual mirror image of their brightness curves? >>>>> >>>>> But they can. >>>> >>>> No they cannot. >>>> >>>>>> I know the answer, Paul. ......ADoppler...or 'WaSh' (the Wilson >>>>>> acceleration Shift) >>>>> >>>>> You are dead wrong. Yet another demonstration of your failure >>>>> to understand what the emission theory predicts. >>>>> Remember that 'the velocity curve' is calculated from >>>>> the assumption that the Doppler shift is f = fo(1 + v/c). >>>>> It the Doppler shift that is observed. >>>>> -------------------------------------- >>>>> For delta Cep the observed Doppler shift is about f = fo(1 +/- 0.7E-4). >>>>> >>>>> If we assume that the Cepheid really is an orbiting >>>>> star with Doppler shift as observed, then the emission theory >>>>> predicts that the brightness variation should be 1.4E-4, and >>>>> the brightness should be maximum when the Doppler shift is maximum. >>>> >>>> We can assume that many cepheids are indeed pulsating stars. Certainly those >>>> with harmonics present must be. >>>> The radial velocities of pulsating stars is by nature very similar to those of >>>> a star in elliptical orbit with a small yaw angle. So it is probable that many >>>> supposed cepheids are ordinary orbiting stars. (Androcles claims 100%...but he >>>> cannot explain the presence of harmonics) >>>> >>>>> The observed brightness variation is 2 (0.8 magnitude variation), >>>>> and the brightness is minimum when the Doppler shift is maximum. >>>>> >>>>> So the emission theory gets the brightness variation wrong by >>>>> more than 4 orders of magnitude, and the phase wrong by 180 degrees. >>>> >>>> It does not. You have your velocity signs back to front. >>>> BaTh predicts maximum blue shift at maximum brightness or slightly after. >>> >>> OK. >> >> So you admit you were wrong with a simple 'OK'? >> When are you going to admit to all your other mistakes? >> >>> But it is still wrong by more than 4 orders of magnitude. >> >> George and I looked into that problem years ago. That's why I came up with the >> 'sawtooth' or 'damped spring' models. They are only models...but they describe >> the principle. The 'wavelength' of light is determined by a spatial pattern on >> each photon. If a source is accelerating, the back end moves up on the front >> end BUT ONLY FOR A VERY LIMITED TIME. Individual photons continue their >> relative movement virtually forever...or until speed-unified by some process. >> >> My latest variable star program already produces predicted spectral shifts by >> adding ADoppler to VDoppler in different proportions. It is not on my website >> yet but produces curves that are a better match of RT Aur's velocity curve than >> the published one. >> >>>> Maximum brightness occurs at maximum acceleration TOWARDS Earth. >>>> That coincides with MINIMUM radius....or in the case of an orbiting star, when >>>> the periastron is furthest from Earth. >>>> >>>> Of course the observed spectral line shifts are a sum of ADoppler and VDoppler >>>> which are 90 degress apart.. ..Usually, but not necesarily, the former will >>>> dominate but the presence of the latter can affect the phasing. >>>> >>>> >>>> That's what is observed. >>>> Here is an example: >>>> http://mb-soft.com/public2/cepheid.html >>>> >>>> Quote: >>>> "If a horizontal line is drawn across the velocity graph about halfway up (at >>>> around +21.6 km/sec) then the portion of the velocity graph that is below that >>>> line represents movement relatively toward us, and the portion above that line >>>> represents movement relatively away from us. " >>> >>> Right. >>> So the observed Doppler shift is f = fo(1 +/- 0.7E-4), >>> which is equal to the brightness variation predicted by the emission >>> theory. So the emission theory predicts that the brightness variation >>> should be unmeasurable. Wrong by more than four . orders of magnitude. > >This pretty much sums it up: > >> No. Put simply, my theory predicts that, if a star's brightness varies by a >> factor of, say 10 (linear) due to cyclical light bunching, then its ADoppler >> line shift should be 10/W, where W is Wilson's 'photon compression factor'. W >> is normally quite large. I have been investigating to see if its value is >> constant for a range of stars...but data availability is a problem. > >"Wilson's 'photon compression factor' " indeed. :-) >Doppler shift without frequency shift. How did you reach that conclusion? Frequency is 'wavecrest arrival rate'. If a photon is emitted by an accelerating source, it will shrink. Its final velocity (relative to an observer) will be the average of its source's speed during emission. So the wavelength shortens and the frequency increases. >Hilarious, no? :-) No. >>>> So from that, the bottom half is blue shifted (and assumed to be moving towards >>>> us when in fact it isn't). >>>> Maximum blue shift is almost in phase with maximum brightness. >>>> It lags slightly because VDoppler lags ADoppler by 90. >>> >>> But it shouldn't lag! >> >> Yes it should. >> >>> According to the emission theory, you know nothing about the velocity. >>> The 'velocity curve' isn't a velocity curve at all, it shows >>> the observed Doppler shift. >> >> My BaTh program SIMULATES observed brightness curves by adjusting orbit >> parameters until a good match is achieved. The true source orbit >> characteristics and velocities are therefore assumed known. It follows that the >> VDoppler factor is also known. It is usually small compared with the linear >> brightness variation. >> >> The value of 'W' can be roughly calculated for a particular star...and I have >> been looking into this. >> >>> It doesn't matter if it is caused by the >>> acceleration or the velocity, when the 'velocity curve' shows a maximum, >>> the Doppler shift is maximum, and the emission theory predicts maximum >>> brightness. Four orders of magnitude too small. >> >> Paul, consider a star in elliptical orbit with its periastron furthest from us. >> Its maximum radial ACCELERATION towards Earth occurs at the periastron. >> It maximum radial VELOCITY occurs somewhere in the first quadrant AFTER >> periastron...typically about 70 degrees for cepheids. ADoppler might be 100 >> times a large as VDoppler, which lags by 70 degrees. > >But we observe the Doppler shift, and in the real world >the Doppler shift and the brightening are exactly equal. >No predicted lag. They are not. There can be a small lag. RT Aur has one. >Predicted magnitude four orders of magnitudes wrong for delta Cep. >For binaries, we do not observe the huge Doppler shift >predicted by the emission theory. W >But in Wonderland, where frequencies can be Doppler shifted >without shift of frequency, "the BaTH" works just fine. No doubt you would think W stands for 'wonderland' because that's where your theory belongs. Henry Wilson... ........A person's IQ = his snipping ability.
From: Paul B. Andersen on 3 May 2010 07:38 On 27.04.2010 00:35, Henry Wilson DSc wrote: > On Mon, 26 Apr 2010 23:03:32 +0200, "Paul B. Andersen"<someone(a)somewhere.no> > wrote: > >> On 25.04.2010 00:42, Henry Wilson DSc wrote: >>> On Sat, 24 Apr 2010 23:58:07 +0200, "Paul B. Andersen"<someone(a)somewhere.no> >>> wrote: >>> >>>> On 20.04.2010 23:57, Henry Wilson DSc wrote: >>>>> >>>>> I maintain that 'true trajectory' is y versus t for constant x....and that can >>>>> only be meaningful in the object's own frame. >>>> >>>> A true gem. :-) >>>> >>>> The only 'true trajectory' is a point in the object's own rest frame. >>>> >>>> A true keeper! :-) >>> >>> Is you definition of 'trajectory' a plot of y-t or y-x? >>> >>> Come on! Answer! >> >> How come "a fully qualified physicist" and Doctor of science >> havs to be taught what a trajectory is? >> >> Here follows an elementary lesson for the Doctor. >> >> A trajectory is the path in space followed by an object. >> So it is a line, or curve, in _space_. >> >> (You could of course also talk about a trajectory >> in space-time, but the above is the traditional >> definition of a trajectory.) >> >> It can be expressed by one of the coordinates as >> a function of the two others: >> z = f(x,y) (or more generally f(x,y,z) = 0) >> (this form has its limitations, it isn't always possible >> to express it this way) >> >> or it can be expressed on parametric form: >> z = f_z(p) >> y = f_y(p) >> x = f_z(p) >> where p is the parameter. >> The most common parameter is the time, of course, >> but there is always an infinite number of possible >> parameters which can be used. > > What you are calling 'trajectory' I would call a 'plot'. Quite. But now you have learned what a 'trajectory' is in English. Don't you speak English down under? > > You still cannot explain how the path of an object, whose movement is > restricted to the axis of a straight tube, can ever be curved. That's what your homework is about. Your straight tube is stationary in the frame of reference described in your homework. > >> An example: >> Let z be the vertical axis, and x and y the horizontal axes. >> A projectile is launched with the horizontal velocity >> component v_x along the x-axis, v_y = 0 along the y-axis, >> and the vertical velocity component v_z along the z-axis. >> Let the projectile be launched from the origin at time t=0. >> >> Using the differential operator D = d/dt, we can write >> the differential equations: >> D^2.z = -g >> giving: >> z = -gt^2/2 + Dz(0).t + z(0) >> Dz(0) = v_z and z(0) = 0 >> z = -gt^2/2 + v_z.t >> >> and: >> Dx = v_x >> giving: >> x = v_x.t (x(0) = 0) >> >> >> So the trajectory on parametric form is: >> z = v_z.t - gt^2/2 >> x = v_x.t >> y = 0 >> >> ..or with a different parameter p: >> z = 2v_z.p - 2gp^2 >> x = 2v_x.p >> y = 0 >> >> >> Eliminating the parameter, this trajectory can be written: >> z = (v_z/v_x)x + (g/2.v_x^2)x^2 >> >> which is a parabola. In _space_. >> >> Homework: >> What is the trajectory of the above projectile >> in a frame of reference (x',y',z') defined by the Galilean transform: >> x' = x - v_x.t >> y' = y >> z' = z >> t' = t >> >> Use math, please. > > That transform just changes the x scale. And the trajectory is? Math too difficult for you? That's why you still don't see how the object with a parabolic trajectory in the ground frame still can move within a straight tube without touching the walls. No physicist would fail to understand that. (Hardly anybody would fail to understand that.) So much for the 'fully qualified physicist'. :-) -- Paul http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 3 May 2010 17:51
On Mon, 03 May 2010 13:38:34 +0200, "Paul B. Andersen" <paul.b.andersen(a)somewhere.no> wrote: >On 27.04.2010 00:35, Henry Wilson DSc wrote: >> On Mon, 26 Apr 2010 23:03:32 +0200, "Paul B. Andersen"<someone(a)somewhere.no> >> wrote: >>> It can be expressed by one of the coordinates as >>> a function of the two others: >>> z = f(x,y) (or more generally f(x,y,z) = 0) >>> (this form has its limitations, it isn't always possible >>> to express it this way) >>> >>> or it can be expressed on parametric form: >>> z = f_z(p) >>> y = f_y(p) >>> x = f_z(p) >>> where p is the parameter. >>> The most common parameter is the time, of course, >>> but there is always an infinite number of possible >>> parameters which can be used. >> >> What you are calling 'trajectory' I would call a 'plot'. > >Quite. >But now you have learned what a 'trajectory' is in English. >Don't you speak English down under? > >> >> You still cannot explain how the path of an object, whose movement is >> restricted to the axis of a straight tube, can ever be curved. > >That's what your homework is about. >Your straight tube is stationary in the frame of reference >described in your homework. It is straight in ALL frames of reference. Tell me this. If an object moves randomly inside a straight tube, what is the direction of the force vectors acting on it? A) all force vectors are aligned with the tube in ALL frames. Therefore the TRUE trajectory of the object is confined to a straight line in ALL frames. The straight line just happens to move sideways in other frames, resulting in curved plots of the object's path. >>> An example: >>> Let z be the vertical axis, and x and y the horizontal axes. >>> A projectile is launched with the horizontal velocity >>> component v_x along the x-axis, v_y = 0 along the y-axis, >>> and the vertical velocity component v_z along the z-axis. >>> Let the projectile be launched from the origin at time t=0. >>> >>> Using the differential operator D = d/dt, we can write >>> the differential equations: >>> D^2.z = -g >>> giving: >>> z = -gt^2/2 + Dz(0).t + z(0) >>> Dz(0) = v_z and z(0) = 0 >>> z = -gt^2/2 + v_z.t >>> >>> and: >>> Dx = v_x >>> giving: >>> x = v_x.t (x(0) = 0) >>> >>> >>> So the trajectory on parametric form is: >>> z = v_z.t - gt^2/2 >>> x = v_x.t >>> y = 0 >>> >>> ..or with a different parameter p: >>> z = 2v_z.p - 2gp^2 >>> x = 2v_x.p >>> y = 0 >>> >>> >>> Eliminating the parameter, this trajectory can be written: >>> z = (v_z/v_x)x + (g/2.v_x^2)x^2 >>> >>> which is a parabola. In _space_. >>> >>> Homework: >>> What is the trajectory of the above projectile >>> in a frame of reference (x',y',z') defined by the Galilean transform: >>> x' = x - v_x.t >>> y' = y >>> z' = z >>> t' = t >>> >>> Use math, please. >> >> That transform just changes the x scale. > >And the trajectory is? >Math too difficult for you? The true trajectory is dtermined by the direction of the force vectors. >That's why you still don't see how the object with >a parabolic trajectory in the ground frame >still can move within a straight tube without touching >the walls. > >No physicist would fail to understand that. >(Hardly anybody would fail to understand that.) > >So much for the 'fully qualified physicist'. :-) It is your use of the word 'trajectory' that is the issue here. Henry Wilson... ........A person's IQ = his snipping ability. |