Seeking a correct explanation for Stellar Abberation
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From: Paul B. Andersen on 13 Apr 2010 10:19 On 12.04.2010 23:46, Henry Wilson DSc wrote: > On Mon, 12 Apr 2010 15:21:33 +0200, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 09.04.2010 23:09, Henry Wilson DSc wrote: >>> I will simplify my question even further. >> >> OK. >> This (what is happening when we increase the length of the pole) >> is actually an interesting problem, so I will respond. >> >> (Your scenario isn't a simplification, it is much more complicated >> than you in your naivety think.) > > I told you days ago that I know the answer. Quite. You know the wrong answer. >>> You and your telescope are standing on a small nonrotating asteroid that is >>> orbiting the sun with a period of 1 year. Next to you is a short vertical pole >>> on top of which is bright object. You point your telescope vertically so the >>> object appears in the centre of your eyepiece. >> >> Let us add that the orbit is the same as the Earth, >> and let us call the asteroid 'the Earth'. >> The pole is perpendicular to the orbital plane (ecliptic), >> so it is pointing towards the ecliptic North pole. >> Let us call the bright object on top of the pole 'the star'. >> We will call the direction towards the ecliptic pole 'vertical'. >> Let v be the orbital speed of the Earth. >> >>> Q) DOES YOUR TELESCOPE HAVE TO BE TILTED AT DIFFERENT ANGLES AS YOU AND THE >>> OBJECT ORBIT THE SUN (to keep the object in view)? >>> >>> A) No of course not. >> >> Close enough, if the pole is short. >> Not exactly right, you would have to tilt the telescope a bit, > > No I don't. > >> but so little that it is many orders of magnitude less than what >> can be measured. >> >>> >>> Next, increase the length of the pole to 1 light minute. >> >> Same answer. >> The angle is unmeasurable. >> We see the 'star' on the top of the pole at the ecliptic north pole. >> >> Consider the following carefully. >> It may seem obvious, and unnecessary to view it in the Sun-frame, >> but it will be very important later: >> In the non-rotating Sun-frame, the light that hits the telescope now, >> has an angle arctan(v/c) ~= v/c in the Sun-frame. >> But this is exactly equal to the aberration between the Sun-frame >> and the Earth-frame, so the telescope will have to be vertical. >> >> Drawn in the Sun-frame: >> ----------------------- >> >> *' = position of star when light was emitted >> * = position of star now >> O = Earth now >> \ = path of the light that is received now >> >> *' * -> v >> \ >> \ >> \ >> O -> v >> >> The light is entering the telescope from 'the left', >> but since the telescope is moving to the right, >> aberration will be v/c towards the right. >> The net effect is that the telescope must be vertical. >> >>> >>> Again: >>> Q) DOES YOUR TELESCOPE HAVE TO BE TILTED AT DIFFERENT ANGLES AS YOU ORBIT THE >>> SUN (to keep the object in view)? >>> >>> (I await Paul's answer) >>> >>> Next, increase the pole's length to precisely 1Ly and answer the same question. >> >> Now the telescope will have to be tilted an angle v/c toward the right. >> The light that is observed now was emitted a year ago, when the 'star' >> was in the same position as it is now when it has made exactly one orbit >> since the light was emitted. >> But NOW IS THE PATH OF THE OBSERVED LIGHT VERTICAL IN THE SUN-FRAME: >> >> Drawn in the Sun-frame: >> ----------------------- >> >> * -> v (star is now back after having made a full orbit) >> *' -> v (star where it was a year ago when light was emitted) >> | >> | >> | >> | >> O -> v (Earth now) >> >> The light is entering the telescope 'vertically', >> but since the telescope is moving to the right, >> aberration will be v/c towards the right. >> The net effect is that the telescope must be tilted v/c to the right. > > That does not tie in with your previous diagram in which you said that the > telescope has to remain vertical. The emission point *' should still be to the > left, as before. > You are contradicting yourself. Your reading comprehension problem shows again. Why are you invariably unable to comprehend a simple text? I get a bit tired of having to repeat everything over and over. In the former case, the rod was short, so the curvature of the orbit didn't matter much. So the emission point was to the left of the star's current position (which is vertically above the Earth), the path of the received light is slanted. In the latter case, the rod is 1 LY, so when the light is received the Earth and the star will have made exactly one orbit and the star will be back at the emission point. The path of the received light is vertical. > >>> How about 100LYs? >> >> The telescope will always have to be tilted v/c in the direction >> of the orbital velocity, which half a year later is in the opposite >> direction. The star will appear to move in a circle around the ecliptic >> north pole with radius v/c. >> >> The reason for this is as I now have explained over and over. > > You have put forward two conflicting theories. > >> BUT what happens when the length of the pole is gradually increased >> is in fact quite complicated. > > That's why I phrased the question in a simple form. You couldn't even get that > right. > >> Where will we see the star when the distance (length of pole) is 1/2 LY? >> >> Drawn in the the Sun-frame in a plane perpendicular to orbital velocity: >> ------------------------------------------------------------------------ >> >> |< 2AU>| >> * *' >> / >> / >> / >> / >> / >> / >> O >> >> The star will appear to be an angle 2AU/0.5LY = 13 arcsecs >> off the vertical, towards the Sun. >> >> >> Drawn in the the Sun-frame in a plane parallel to the orbital velocity: >> -------------------------------------------------------------------- >> As for 1 LY. >> >> The net effect is that the telescope must be tilted >> v/c = 20.5 arcsecs in the direction of the orbital velocity, >> and 13 arcsecs toward the sun. >> >> If we increase the distance (length of pole) to 1.5 LY, the angle >> towards the Sun will be 2AU/1.5LY = 4.3 arcsecs. >> >> So to sum it up: >> When we increase the length of the pole, the star will appear to >> move away from vertical, one component will be in the direction >> of the orbital velocity, and one component will be towards the sun. > > ..not bad... > >> The former component will vary a bit with the distance, it will be >> exactly v/c = 20.5 arcsecs every odd number of half light years, >> but be a bit less under whole numbers of LYs (when the apparent >> position is 'behind' the true position) and a bit more above whole >> numbers of LYs. The variation will be less with the distance, >> for 100+ LY, the variation will be negligible. >> >> The latter component will have a maximum every odd number of half >> light years, and be zero for whole number of light years. It will be >> smaller as the distance increases, and for 100.5LY, it will be >> a mere 0.062 arcsecs. >> >> So we can conclude that for distances 100+ LYs, the star at the end >> of your pole will move in a circle with 20.5 arcsecs radius around >> vertical. > > Sorry Paul, you haven't analysed the problem correctly or given the right > answer. > I'll give you a clue. > > """"If the star emits a laser beam, at what precise direction does the laser > have to be pointed for the beam to strike my telescope? It depends on the distance. What you probably aren't realizing is that the laser beam (as it could be visualized in a fog chamber) isn't a straight line, but a helix. Don't confuse the path of the light with the laser beam. The path of each individual photon is a straight line, the beam is not. If the laser beam is emitted vertically and the distance is 1 LY, then the helix would have made one rotation, and would hit the Earth. The beam would hit the Earth at the angle 2pi*1AU/1LY = v/c = 20.5" from vertical. But if the laser beam is emitted vertically and the distance is 1.5 LY, the helix will have made 1.5 rotation and miss the Earth by 2 AU, The laser beam would have to be aimed differently to hit the Earth. It would have to be aimed differently whenever the distance is changed. > At what angle does my > telescoope have to be set so the beam will go straight down the centre? It will obviously have to be set parallel to the beam. In the 1 LY case, that is at an angle 20.5" from vertical. >> >> I have no doubt that you are as confused as ever, >> but it was an interesting problem, and there may be >> a lurker out there who has learned something. > > It IS and interesting question....one that also has an interesting answer...but > you haven't been able to find it. > >> But you haven't, Henry. >> Right? :-) Wasn't I right, or was I right? :-) -- Paul http://home.c2i.net/pb_andersen/
From: GogoJF on 13 Apr 2010 10:39 On Apr 12, 4:50 pm, ..@..(Henry Wilson DSc) wrote: > On Mon, 12 Apr 2010 13:25:00 -0700 (PDT), GogoJF <jfgog...(a)yahoo.com> wrote: > >On Apr 6, 1:53 pm, GogoJF <jfgog...(a)yahoo.com> wrote: > >> On Mar 6, 7:19 pm, train <gehan.ameresek...(a)gmail.com> wrote: > > >> > According to the special theory of relativity, the aberration only > >> > depends on the relative velocity v between the observer and the light > >> > from the star. > > >> >http://www.mathpages.com/rr/s2-05/2-05.htm > > >> > "relative velocity v between the observer and the light from the > >> > star." > > >> > Whic is always c , right? > > >> Looked on Wiki about "Aberration of light"- and printed it out in 12 > >> pages. On page 3, it talks of "apparent and true positions". It uses > >> the well known diagram illustrating stellar aberration. Along with > >> the diagram, it supposes that the star is sufficiently distant, so > >> that all light from the star travels in parallel paths to the Earth > >> observer, regardless of where the Earth is in its orbit. > > >> And this is where the problem starts: > > >> Wiki: On the left side of figure 1, the case of infinite light speed > >> is shown. S represents the spot where the star light enters the > >> telescope, and E the position of the eye piece. > > >> Gogo: Then wiki very subtly poses this question: > > >> Wiki: If light moves instantaneously, the telescope does not move, > >> and the true direction of the star relative to the observer can be > >> found by following the line ES. However, if light travels at finite > >> speed, the Earth, and therefore the eye piece of the telescope, moves > >> from E to E' during the time it takes light to travel from S to E. > >> Consequently, the star will no longer appear in the center of the eye > >> piece. The telescope must therefore be adjusted. > > >> Gogo: But the truth of the matter, there is no adjusting- there is > >> nothing that happens or exists which would suggest that any observable > >> or instrumental change has been performed, other than by idea, belief, > >> or faith that it was and is done this way. Aberration of light is a > >> purely man-made concept! > > >> Wiki: Consequently, the star will no longer appear in the center of > >> the eye piece. > > >> Gogo: This is complete baloney. This is pure theory- science > >> fiction. In practice, this does not accur. > > >> Wiki: The telescope must therefore be adjusted. > > >> Gogo: Again, there is no adjusting going on here. This happens only > >> "what if" light was finite. > > >> Wiki: When the telescope is at position E it must be oriented toward > >> S' so that the star light enters the telescope at spot S'. > > >> Gogo: Spot S', do you see how ridiculous this sounds, logically? > >> Again, this is 100% one way or 100% the other. Either light is > >> instant, or it is finite. This illustration or principle is all or > >> nothing. There is no step by step transformation process which starts > >> from instant and transforms to finite. > > >> Wiki: Now the star light will travel along the line S'E' (parallel to > >> SE) and reach E' exactly when the moving eye piece also reaches E'. > > >> Gogo: Again, this is describing a physical process which does not > >> exist. It begins with instantaneous light and extrapolates from this > >> reality to the reality of finite light. Do you see the major logical > >> flaw here? Either light is instant, or finite. One of these > >> realities cannot exist. > > >> Wiki: Since the telescope has been adjusted by the angle SES', the > >> star's apparent position is hence displaced by the same angle. > > >> Gogo: This is simply not the case. There has been no adjustment of > >> the telescope. Since I believe that light is instantaneous, the > >> displacement of the star's apparent position by the angle SES' must be > >> given a new interpretation and meaning. > > >I am surprised that nobody was curious enough to ask what the meaning > >of the changing of the star's position throughout the year means, if > >it is not aberration. > > >Well, if instantaneous light theory is upheld, it can only mean one > >thing- the Earth does not revolve around the sun on a ecliptic plane, > >but oscillates vertically upward and downwards, 20.5 degrees upwards, > >20.5 degrees downwards= 41 total degrees. > > Diurnal aberration is another matter. Annual changes in tilt angle is an > unfortunate complication that must be corrected out when calculating annual > aberration. > If you want to be REALLY fussy, you must also consider galatic rotation. > > Henry Wilson... > > .......A person's IQ = his snipping ability. From "general science journal forum": the oscillating ecliptical plane April 13 2010, 11:34 AM Gogo JF: Well, if instantaneous light theory is upheld, it can only mean one thing- the Earth does not revolve around the sun on a ecliptic plane, but oscillates vertically upward and downwards, 20.5 degrees upwards, 20.5 degrees downwards. ********************* Rebis: Or Sun's ecliptic possesses same amount of a nutation wink.gif Gogo: Rebis, this is a very good answer. You would think that any qualified astronomer would be able to find this out. In my paper, "A Case in Instantaneous Cosmology and the Disqualification of Jupiter's Moons Used to Measure the Speed of Light", I originally suggested that the Earth does not revolve around the sun in a plane but oscillates. Sorry, I used the word "elliptical plane" not "ecliptic plane"- my bad. In this paper, I suggested that Jupiter "pushes" the Earth downward, below the ecliptic, as it aligns with Jupiter and the sun- and on the other side it bobs back above the ecliptic. This suggests that the Earth and Jupiter "repel" each other, while, each individually are attracted to the sun. What do you think?
From: Paul B. Andersen on 13 Apr 2010 16:01 On 13.04.2010 11:54, Henry Wilson DSc wrote: > On Tue, 13 Apr 2010 09:18:11 +0200, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 12.04.2010 23:06, Henry Wilson DSc wrote: >>> On Mon, 12 Apr 2010 16:11:12 +0200, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>>>> All one has to do is move perpendicularly to the beam...and it bends.... >>>> >>>> Good grief, Henry. >>>> You never get tired of making a fool of yourself, do you? :-) >>>> >>>> You throw a ball with a horizontal velocity component v (in ground frame). >>>> Is the trajectory in the ground frame a parabola, Henry? >>> >>> It is indeed. So according to Ralph Rabbidge aka Henry Wilson: The trajectory of the ball in the ground frame is indeed a parabola. >>> >>>> (If you don't know what 'a trajectory' is, look it up.) >>>> >>>> A laser with a vertical beam is moving along the ground under the ball >>>> with the speed speed v, so that the beam always hits the ball. >>> >>> hahahahhhahaha! >>> ......a kind of tracking device....the laser has to be rotated continuously. >> >> Your reading comprehension problem shows again. :-) >> >> "You throw a ball with a horizontal velocity component v. >> A laser with a vertical beam is moving along the ground under the ball >> with the speed v, so that the beam always hits the ball." >> >> This is of course incomprehensible for a Doctor of Science, >> so we will have to draw it: > > OK, sometimes your English isn't the best. Quite. When you read: "A laser with a vertical beam is moving along the ground under the ball with the speed v, so that the beam always hits the ball." as: "A stationary laser is tracking the ball so that the beam always hits the ball." it is obviously due to my bad English, and not to your serious reading comprehension problem. >> O = ball with horizontal velocity component v >> X = laser moving along the horizontal ground with speed v >> | = vertical laser beam always hitting the ball >> .. = trajectory of ball in ground frame >> -- = ground >> >> . O->v . >> . | . >> . | . >> . X->v . >> ---------------------------- >> >> Is the laser rotated, Henry? >> >> Is the ball moving along the laser beam, Henry? >> >> Is the laser beam bent, Henry? >> >> Is the trajectory of the ball a straight line in >> the laser frame, Henry? > > Of course. So according to Ralph Rabbidge aka Henry Wilson: Of course the trajectory of the ball is a straight line in the laser frame. The trajectory of the ball in the ground frame is indeed a parabola. >> Is the trajectory of the ball a straight line in one frame >> and a parabola in another, Henry? So Ralph Rabbidge aka Henry Wilson's answer is "yes". But oooops! Ralph Rabbidge has now admitted that he was wrong! What shall he do? This is what he does: > Are you claiming that the laser beam is in the shape of a parabola? > Oh dear! I think you are... Hilarious, no? > >> The stupidity of Doctor Henry Wilson has ceased to amaze, >> but I still find it amusing that 'the fully competent physicist' >> is so ignorant about Galilean relativity and Newtonian dynamics >> that he can utter: >> << >> How is it that the object can move in a straight line in one frame but a >> parabola in another? >> >> Please explain how anything straight can bent by a moving observer. >> >> I really think you should get a job as a 'spoon bender' in a circus somewhere. >>>> >> >> This is fun, isn't it? :-) > > It certainly is. It exposes how the use of frames can be very confusing to > amateurs. It certainly does! When Paul B. Andersen wrote: | | In inertial frame A, a particle is going 'straight down', | and is reducing its speed from v1 to v2. | | o | o v1 | o | X | o | o v2 | o | The trajectory is a straight line in frame A. | | Frame B is moving 'horizontally' to the right at some speed v. | In this frame the trajectory would look something like this this: | | o | o | o | X | o | o | o | | The trajectory is bent in frame B. | | ... the "fully qualified professional physicist Doctor" Ralph Rabbidge aka Henry Wilson responded: | | Hahahhahhhahhahhaha! | | Let an object accelerate along the centre line of a long straight | tube. | Does its increase in speed wrt the tube cause the tube to bend | whenever a moving observer happens to look at it? | | ..........the mind of the relativist ccertainly operates in | very strange ways.... And "the fully qualified professional physicist Doctor" Ralph Rabbidge aka Henry Wilson wrote: | How is it that the object can move in a straight line in one | frame but a parabola in another? | Please explain how anything straight can bent by a moving observer. | I really think you should get a job as a 'spoon bender' in a circus | somewhere. Now the "fully qualified professional physicist Doctor" Ralph Rabbidge aka Henry Wilson admits: | The trajectory of the ball in the ground frame is indeed a parabola. | Of course the trajectory of the ball is a straight line in | the laser frame. and thus admitting: | An object can move in a straight line in one frame but | along a parabola in another. Indeed a very revealing exposure of the confused amateur Ralph Rabbidge aka Henry Wilson! :-) This IS fun! I simply love to rub it in! -- Paul, more amused than ever http://home.c2i.net/pb_andersen/
From: Henry Wilson DSc on 13 Apr 2010 18:00 On Tue, 13 Apr 2010 22:01:46 +0200, "Paul B. Andersen" <someone(a)somewhere.no> wrote: >On 13.04.2010 11:54, Henry Wilson DSc wrote: >> On Tue, 13 Apr 2010 09:18:11 +0200, "Paul B. Andersen" >> <paul.b.andersen(a)somewhere.no> wrote: >> >>>>> Good grief, Henry. >>>>> You never get tired of making a fool of yourself, do you? :-) >>>>> >>>>> You throw a ball with a horizontal velocity component v (in ground frame). >>>>> Is the trajectory in the ground frame a parabola, Henry? >>>> >>>> It is indeed. > >So according to Henry Wilson: >The trajectory of the ball in the ground frame is indeed a parabola. >>> "You throw a ball with a horizontal velocity component v. >>> A laser with a vertical beam is moving along the ground under the ball >>> with the speed v, so that the beam always hits the ball." >>> >>> This is of course incomprehensible for a Doctor of Science, >>> so we will have to draw it: >> >> OK, sometimes your English isn't the best. > >Quite. >When you read: > "A laser with a vertical beam is moving along the ground under the ball > with the speed v, so that the beam always hits the ball." >as: > "A stationary laser is tracking the ball so that the beam always > hits the ball." >it is obviously due to my bad English, and not to your serious >reading comprehension problem. yes >>> O = ball with horizontal velocity component v >>> X = laser moving along the horizontal ground with speed v >>> | = vertical laser beam always hitting the ball >>> .. = trajectory of ball in ground frame >>> -- = ground >>> >>> . O->v . >>> . | . >>> . | . >>> . X->v . >>> ---------------------------- >>> >>> Is the laser rotated, Henry? >>> >>> Is the ball moving along the laser beam, Henry? >>> >>> Is the laser beam bent, Henry? >>> >>> Is the trajectory of the ball a straight line in >>> the laser frame, Henry? >> >> Of course. > >So according to Ralph Rabbidge aka Henry Wilson: > Of course the trajectory of the ball is a straight line in > the laser frame. > The trajectory of the ball in the ground frame is indeed a parabola. I wouldn't use the word 'trajectory'. >>> Is the trajectory of the ball a straight line in one frame >>> and a parabola in another, Henry? > >So Henry Wilson's answer is "yes". > >But oooops! >Henry Wilson has now admitted that he was wrong! >What shall he do? >This is what he does: > >> Are you claiming that the laser beam is in the shape of a parabola? >> Oh dear! I think you are... > >Hilarious, no? Paul, if a moving object's position never deviates from a particular laser beam, how can its TRAJECTORY be anything but straight. You must know how laser beams are used to create level surfaces, for instance for farms and buildings, are you claiming that these are NOT level whenever someone climbs up a ladder? >>> The stupidity of Doctor Henry Wilson has ceased to amaze, >>> but I still find it amusing that 'the fully competent physicist' >>> is so ignorant about Galilean relativity and Newtonian dynamics >>> that he can utter: >>> << >>> How is it that the object can move in a straight line in one frame but a >>> parabola in another? >>> >>> Please explain how anything straight can bent by a moving observer. >>> >>> I really think you should get a job as a 'spoon bender' in a circus somewhere. >>>>> >>> >>> This is fun, isn't it? :-) >> >> It certainly is. It exposes how the use of frames can be very confusing to >> amateurs. > >It certainly does! > >When Paul B. Andersen wrote: >| >| In inertial frame A, a particle is going 'straight down', >| and is reducing its speed from v1 to v2. >| >| o >| o v1 >| o >| X >| o >| o v2 >| o >| The trajectory is a straight line in frame A. >| >| Frame B is moving 'horizontally' to the right at some speed v. >| In this frame the trajectory would look something like this this: >| >| o >| o >| o >| X >| o >| o >|o The trajectory is bent in frame B. .....and now you have to explain how the laser beam that shines a spot on the object is also apparently curved in frame B. >.. the "fully qualified professional physicist Doctor Henry Wilson responded: >| >| Hahahhahhhahhahhaha! >| >| Let an object accelerate along the centre line of a long straight >| tube. >| Does its increase in speed wrt the tube cause the tube to bend >| whenever a moving observer happens to look at it? >| >| ..........the mind of the relativist certainly operates in >| very strange ways.... > > >And "the fully qualified professional physicist Doctor" Henry Wilson wrote: >| How is it that the object can move in a straight line in one >| frame but a parabola in another? >| Please explain how anything straight can bent by a moving observer. >| I really think you should get a job as a 'spoon bender' in a circus >| somewhere. > >Now the "fully qualified professional physicist Doctor" Henry Wilson admits: >| The trajectory of the ball in the ground frame is indeed a parabola. >| Of course the trajectory of the ball is a straight line in >| the laser frame. I did not use the word 'trajectory'. I said if you plot the height of the ball on a time scale, a curve will result. >and thus admitting: >| An object can move in a straight line in one frame but >| along a parabola in another. > >Indeed a very revealing exposure of the confused amateur Ralph Rabbidge >aka Henry Wilson! :-) > > >This IS fun! >I simply love to rub it in! So do I. You obviously cannot fathom how the path can APPEAR to be bent yet always lies on the straight laser beam. You are stalling for time in the hope that an answer will suddenly appear from the sky. Henry Wilson... ........A person's IQ = his snipping ability.
From: Androcles on 13 Apr 2010 18:18
"Henry Wilson DSc" <..@..> wrote in message news:lro9s5puvni0vqq1avf9q5q499po6rbope(a)4ax.com... > On Tue, 13 Apr 2010 22:01:46 +0200, "Paul B. Andersen" > <someone(a)somewhere.no> > wrote: > >>On 13.04.2010 11:54, Henry Wilson DSc wrote: >>> On Tue, 13 Apr 2010 09:18:11 +0200, "Paul B. Andersen" >>> <paul.b.andersen(a)somewhere.no> wrote: >>> > >>>>>> Good grief, Henry. >>>>>> You never get tired of making a fool of yourself, do you? :-) >>>>>> >>>>>> You throw a ball with a horizontal velocity component v (in ground >>>>>> frame). >>>>>> Is the trajectory in the ground frame a parabola, Henry? >>>>> >>>>> It is indeed. >> >>So according to Henry Wilson: >>The trajectory of the ball in the ground frame is indeed a parabola. > > >>>> "You throw a ball with a horizontal velocity component v. >>>> A laser with a vertical beam is moving along the ground under the >>>> ball >>>> with the speed v, so that the beam always hits the ball." >>>> >>>> This is of course incomprehensible for a Doctor of Science, >>>> so we will have to draw it: >>> >>> OK, sometimes your English isn't the best. >> >>Quite. >>When you read: >> "A laser with a vertical beam is moving along the ground under the ball >> with the speed v, so that the beam always hits the ball." >>as: >> "A stationary laser is tracking the ball so that the beam always >> hits the ball." >>it is obviously due to my bad English, and not to your serious >>reading comprehension problem. > > yes He's got bad breath, bad algebra and bad geometry, too. http://androcles01.pwp.blueyonder.co.uk/NordicGeometry.JPG |