Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 9 Nov 2005 04:01 On Wed, 9 Nov 2005 04:28:15 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >HW@..(Henri Wilson) wrote in >news:2l62n1l33o967babc5vralbrk0lrfgllmd(a)4ax.com: > >> On Tue, 8 Nov 2005 02:38:24 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> >> wrote: >> >>>HW@..(Henri Wilson) wrote in >>>news:nlhvm11gdhmpjgvo1qv8f81rg6t6nch7i6(a)4ax.com: >>> >> >>> >>>What questions of your have I refused to answer? >>> >>>Were they answerable questions or 'why' questions of the type that >>>science can not answer? >>> >>>Is my ability to answer them critical to my being able to do develope my >>>theory? The answer must be no, because I do not have a theory that I am >>>trying to develop and sell to the world. >>> >>>Your ability to answer our questions is critical to the development of >>>your theory. >>> >>>Without answers to those questions, you have no theory, only a >>>conjecture. >>> >>>If your conjecture explained many unexplained pheonomina and were backed >>>by sound math, it would stand a chance. >>> >>>Unfortunately, you must keep adding new ad hoc 'fixes' to your >>>conjecture to explain data that is easily explained by other theories. >>> >>>Unfortunately, you do NOT have sound math behind your conjecture. >>> >>>Unfortunately, you grow angry, defensive and abusive when presented with >>>questions that your theory MUST be able to answer if it is ever to stand >>>on its own. >>> >>>You have written some 'interesting' computer programs that draw pretty >>>pictures. Computer program writing is very well adapted to adding ad hoc >>>fixes to 'make things work'. >>> >>>Unfortunately physics doesn't work the same way. >> >> But aeroplanes still fly even though they are designed with computer >> simulations. > >IF aeroplanes were designed by programs as firmly based upon theory and >data as your are, the pilots would sit in the tail of the plane and steer >by shouting orders to the helmsmen riding on each wing. The rubber band >that turns the propellors would be wound up by teams of horses before the >planes were launched by giant catapults. I thought they still were. > >By the way, modern fighter planes are designed to be UNSTABLE in the air. >The pilot can NOT possibly fly them without the computer to continually >compensate for the instabilities. Modern aeroplanes are FIRMLY based on >theory and data. Don't think so Bob. 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: bz on 9 Nov 2005 09:00 HW@..(Henri Wilson) wrote in news:1ne3n11h53shhf6rrg39cfbikba5oik65p(a)4ax.com: f> On Wed, 9 Nov 2005 04:28:15 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> > wrote: > >>HW@..(Henri Wilson) wrote in >>news:2l62n1l33o967babc5vralbrk0lrfgllmd(a)4ax.com: >> >>> On Tue, 8 Nov 2005 02:38:24 +0000 (UTC), bz >>> <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >>> >>>>HW@..(Henri Wilson) wrote in >>>>news:nlhvm11gdhmpjgvo1qv8f81rg6t6nch7i6(a)4ax.com: ..... >>> But aeroplanes still fly even though they are designed with computer >>> simulations. >> >>IF aeroplanes were designed by programs as firmly based upon theory and >>data as your are, the pilots would sit in the tail of the plane and >>steer by shouting orders to the helmsmen riding on each wing. The rubber >>band that turns the propellors would be wound up by teams of horses >>before the planes were launched by giant catapults. > > I thought they still were. > >> >>By the way, modern fighter planes are designed to be UNSTABLE in the >>air. The pilot can NOT possibly fly them without the computer to >>continually compensate for the instabilities. Modern aeroplanes are >>FIRMLY based on theory and data. > > Don't think so Bob. > > so. [quote http://www.centennialofflight.gov/essay/Evolution_of_Technology/Computers/Tec h37.htm] The General Dynamics (now Lockheed-Martin) F-16, which entered service in the late 1970s and has been built in large numbers, was the first operational jet fighter to use an analog flight control system. The pilot steers the rudder pedals and joystick, but these are not directly connected to the control surfaces such as the rudder and ailerons. Instead, they are connected to a "fly-by-wire" flight control system. Three computers on the aircraft constantly adjust the flight controls to maintain the aircraft in flight and reply to the commands from the pilot. The F-16 is inherently unstable by design, meaning that it would fly out of control if the computers failed (which is why there are three of them). The designers made it unstable in order to improve its maneuverability. The computers constantly readjust the flight surfaces to keep the plane flying. Initially, pilots often referred to the F-16 as "the electric jet." But computer control systems have become so common that they are no longer unusual. [unquote] -- 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 Nov 2005 16:25 Henri Wilson wrote: > On 7 Nov 2005 05:59:38 -0800, "george(a)briar.demon.co.uk" > <george(a)briar.demon.co.uk> wrote: > > > > >Henri Wilson wrote: > >> On Sun, 6 Nov 2005 10:31:41 -0000, "George Dishman" <george(a)briar.demon.co.uk> > >> wrote: > >> >"Henri Wilson" <HW@..> wrote in message > >> >news:h3rfm11mk9sladl6qt8le2093f8l7ik3qe(a)4ax.com... > >> >> On Tue, 1 Nov 2005 19:18:26 -0000, "George Dishman" > >> >> <george(a)briar.demon.co.uk> > >... > >> Let's go back to the non-rotating table frame. > >> > >> Here, the beam moves at very nearly c+v/root2v towards the mirror''s new > >> position when the beam arrives. After each reflection, it continues at > >> c+v/root2 wrt the table. The second beam moves at c-v/root2 towards the third > >> mirror's new position. > > > >You have an extra "v" in the first formula, just a typo I assume. > >Barring that, you have this part correct. > > > >> This approach gives a similar answer to your classical analysis except that the > >> fringe shift should be about 35% of the 'aether prediction', (I think). > > > >No, you are forgetting that, in the non-rotating table frame, the > >detector moves during the time the light is moving. The distance > >in c+v/sqrt(2) direction will be increased while that in the > >c-v/sqrt(2) > >direction is decreased. The fractional change in distance is the > >same as the change of speed hence times are unaffected. > > There is another consideration that I (and you) have omitted. > The ray that moves diagonally from the source to the point where the mirror is > located on its arrival DOES NOT start out heading in that direction. You may have missed it, I didn't. That was one point of my answer to your question that I asked you to consider: > > >>>I thought we had agreed that several times before. Perhaps > > >>>I wasn't clear so I'll go through it again in more detail. > > >>>In Ritzian theory, the light is emitted at some speed greater > > >>>than c from the source. The speed can be found by taking the > > >>>magnitude of the vector sum of the mirror velocity and a > > >>>vector of magnitude c whose direction is such that the light > > >>>eventually reaches the detector. > > > > > >I agree what you say next, do you agree what I say above? > > Henri, can you confirm the above please, I didn't think it > was contentious but you seem to be avoiding it. Can you say if you agree with my statement of the behaviour. > In the table frame, the whole beam appears to move at 45 degrees, like the > green beam in my demo. We aren't interested in "the beam" but individual wavefronts. Think of the links on the chainsaw in your other analogy. > I'll have to think about this a bit more before I enlarge on it. > Can't draw it here. You need to think about the path of an individual representative wavefront. Each subsequent one just follows the same relative path but from a starting point farther round the circumference. > >> >> No your car example was wrong. Try again. > >> > > >> >The motion of the mirror in the source frame is at a fixed > >> >radius so the distance doesn't change. > >> > >> your car example was still wrong. > > > >No, my car analogy explained what "closing speed" means. > >If you think the conclusion was wrong, it was your attempt to > >use closing speed which was an inappropriate method. I can > >agree with that in the multiple mirror setup since the light, > >source/detector and mirrors don't follow the same path but > >it is perfectly correct for the fibre gyro setup where the light > >follows a circumferential path. > > You used 'closing component in the direction of the line joining the two cars'. There is no "component", it isn't a vector or even a speed in the conventional sense. Have another read of what I said before and see if you can get the idea, I've edited it slightly to try to make it clearer: Closing speed is meaningful when one object is following the same path as another but is some distance behind, for example cars on a race track. You measure the distance not along the straight line between the cars but along the path they have to travel along the track. Or think of the two cars each holding constant speed but on a winding mountain road. Closing speed is then the rate at which that distance between them is decreasing, the numerical difference between their individual ground speeds, not the vector difference. If you take the initial separation and divide by the closing speed, you get the time for the second car to catch the first. > You have to move the leading car diagonally at 45 deg. > Then your figures will be different. No, if the first car is moving at 30m/s and the second is following the same path at 40m/s then their separation measured _along_the_track_ decreases by 10m each second. This applies OK to the circular path in an iFOG but isn't really applicable to the discrete mirror case. George
From: Henri Wilson on 9 Nov 2005 16:34 On Wed, 9 Nov 2005 14:00:07 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >HW@..(Henri Wilson) wrote in >news:1ne3n11h53shhf6rrg39cfbikba5oik65p(a)4ax.com: > >>> >>>By the way, modern fighter planes are designed to be UNSTABLE in the >>>air. The pilot can NOT possibly fly them without the computer to >>>continually compensate for the instabilities. Modern aeroplanes are >>>FIRMLY based on theory and data. >> >> Don't think so Bob. >> >> > >so. > >[quote >http://www.centennialofflight.gov/essay/Evolution_of_Technology/Computers/Tec >h37.htm] >The General Dynamics (now Lockheed-Martin) F-16, which entered service in the >late 1970s and has been built in large numbers, was the first operational jet >fighter to use an analog flight control system. The pilot steers the rudder >pedals and joystick, but these are not directly connected to the control >surfaces such as the rudder and ailerons. Instead, they are connected to a >"fly-by-wire" flight control system. Three computers on the aircraft >constantly adjust the flight controls to maintain the aircraft in flight and >reply to the commands from the pilot. The F-16 is inherently unstable by >design, meaning that it would fly out of control if the computers failed >(which is why there are three of them). The designers made it unstable in >order to improve its maneuverability. The computers constantly readjust the >flight surfaces to keep the plane flying. Initially, pilots often referred to >the F-16 as "the electric jet." But computer control systems have become so >common that they are no longer unusual. >[unquote] I think that is probably a little out of context. I'm sure the planes would be stable in normal flight. I think such a servo control system would only apply during fast maveuvering. ....but any remote electronic system must work on the principle of sensing and reducing an error. One of the big problems is to find a decent compromise between reaction time, senstivity and damping. One must prevent the thing from building up an oscillation. 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: Henri Wilson on 9 Nov 2005 17:22
On 9 Nov 2005 13:25:38 -0800, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >Henri Wilson wrote: >> On 7 Nov 2005 05:59:38 -0800, "george(a)briar.demon.co.uk" >> <george(a)briar.demon.co.uk> wrote: >> >> > > >You may have missed it, I didn't. That was one point of my answer >to your question that I asked you to consider: > >> > >>>I thought we had agreed that several times before. Perhaps >> > >>>I wasn't clear so I'll go through it again in more detail. >> > >>>In Ritzian theory, the light is emitted at some speed greater >> > >>>than c from the source. The speed can be found by taking the >> > >>>magnitude of the vector sum of the mirror velocity and a >> > >>>vector of magnitude c whose direction is such that the light >> > >>>eventually reaches the detector. >> > > >> > >I agree what you say next, do you agree what I say above? >> >> Henri, can you confirm the above please, I didn't think it >> was contentious but you seem to be avoiding it. > >Can you say if you agree with my statement of the behaviour. > >> In the table frame, the whole beam appears to move at 45 degrees, like the >> green beam in my demo. > >We aren't interested in "the beam" but individual wavefronts. >Think of the links on the chainsaw in your other analogy. > >> I'll have to think about this a bit more before I enlarge on it. >> Can't draw it here. > >You need to think about the path of an individual >representative wavefront. Each subsequent one >just follows the same relative path but from a starting >point farther round the circumference. Nah! >> >No, my car analogy explained what "closing speed" means. >> >If you think the conclusion was wrong, it was your attempt to >> >use closing speed which was an inappropriate method. I can >> >agree with that in the multiple mirror setup since the light, >> >source/detector and mirrors don't follow the same path but >> >it is perfectly correct for the fibre gyro setup where the light >> >follows a circumferential path. >> >> You used 'closing component in the direction of the line joining the two cars'. > >There is no "component", it isn't a vector or even a speed >in the conventional sense. Have another read of what I >said before and see if you can get the idea, I've edited it >slightly to try to make it clearer: > > Closing speed is meaningful when one object is following > the same path as another but is some distance behind, for > example cars on a race track. You measure the distance > not along the straight line between the cars but along the > path they have to travel along the track. Or think of the two > cars each holding constant speed but on a winding mountain > road. Closing speed is then the rate at which that distance > between them is decreasing, the numerical difference between > their individual ground speeds, not the vector difference. If you > take the initial separation and divide by the closing speed, you > get the time for the second car to catch the first. >> You have to move the leading car diagonally at 45 deg. >> Then your figures will be different. > >No, if the first car is moving at 30m/s and the second is >following the same path at 40m/s then their separation >measured _along_the_track_ decreases by 10m each >second. This applies OK to the circular path in an iFOG >but isn't really applicable to the discrete mirror case. I wouldn't agree it applies to an iFOG because that really involves an infinite number of 'mirrors'. You can still use it in the four mirror system but you have to accomodate the sideways movement. > >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". |