Over the years of these forums we have always asserted or assumed that there can be no absolute stationary reference point. Some said it didn't matter, others that it couldn't be known or observed. Yet since the MM debate , I think the means was right there before our eyes all of the time. In the list below, I posited point No 2. , 2. Once a beam is transmitted, it travels in a straight line, independent of any forces whatsoever. This means the beam as such during its travel time is stationary in space. Its path is a fixed and stationary true straight line from the earth, provided the transmitter is stationary. It is essential this property be recognised for my experiment below. Once the "photon" exits the transmitter it is free of all external forces, having only its forward velocity in a straight line. This line represents a fixed position in space, observable for the duration of its existence, and will, given enough time, show the relative movements of physical bodies proximate to it. I think Einstein really worried about this hence his desire to prove photons were like mass and affected by gravity.. we cannot be forced to accept photons (a name given to EMR) as particles. I am sure that any experiments used to confirm his relativity theory as regards light bending around Jupiter etc are nebulous. "Einstein's Equivalence Principle, upon which General Relativity rests, claims that all forms of mass-energy experience the same acceleration in response to an external gravitational force. This is to say that the inertial mass and gravitational mass are equivalent for any form of mass and/or energy. This is very difficult to verify for gravitational energy itself, because laboratory masses have no appreciable gravitational binding energy. One needs bodies as large as the earth to have any measurable self-energy content. Even then, the self-energy contribution to Earth's total mass-energy is less than one part-per-billion." enough said, they still have this goal, and its not proven... My experiment. Figures I used. orbit: 384,400 km from Earth diameter: 3476 km 2416228 kilometers travelled in 24 hours or 709 hours = 100,676kph or 3,408kmh respectfully. Earth moves forward 30km/sec or does not move at all. Between Marc, and Paul, their discourse opens up some points of experiment that need simplification, clarification, and confirmation.. I would like to ennumerate these, hoping to enable all to get an actual mental image of what is happening when we shoot a beam at the moon. I want to ignore the special reflectors Paul speaks about for the moment, and Marcs ball and trains analogy, which can cause confusion as indicated by Paul. Now these are my presumptions, and I hope any errors will be corrected and weaknesses resolved. 1. We will call the transit time earth to moon for simplicity 1 second each way. This has no bearing on the relationships we are exploring. It means our "photon" has just one second of existence, or more for those reflected back off the surface of the moon. . 2. Once a beam is transmitted, it travels in a straight line, independent of any forces whatsoever. This means the beam as such during its travel time is stationary in space. Its path is a fixed and stationary true straight line radial from the earth, provided the transmitter is stationary. It is essential this property be recognised. Note: If the transmitter is moving for 10ms then we essentially still have a sequence of parallel lines of "photons". 3. To make the movement of the transmitter irrelevant and relatively stationary, or independent of any alleged rotation of the earth or its orbital speed around the sun we must make the beam have no more than a few, nominated 10 milliseconds of duration. 4. We need to know how wide the laser beam will be by the time it reaches the moon. What size spot will it make. NASA's beam was 2km across. Only one in 30 million photons were reflected from their reflector array. se pic below . Another said:When pointing a well focussed laser at the moon, its beam will widen to well over 100 meters in radius by the time it reaches the surface. 5. For our purpose we need to shoot during a new moon to make the flash/spot more visible. We can use a killer laser to make sure it is visible. Telescope and camera. 6. As we have effectively stopped any rotation of the earth by using only a few milliseconds, we can assume the moon to be travelling radially, as observed, for aiming purposes. (We are not measuring reflection times, but visible patterns on the surface) 7. The reality of the moons speed will be proven because we are making a comparison of the light pattern drawn on the moon by a truly stationary 10ms beam in space, with either a moon travelling daily at 168,000mph or travelling monthly at 5,680mph in space. 8. It should be a simple calculation to compare the distance the beam moves across the moonscape in 10milliseconds at either speed. 9. If the firing was done on a trajectory that was at right angles to the forward motion of the earth around the sun, then again considering the absolute stationary property of our light beam, during the time of our 1 second journey , both the moon and earth travelling at 30km/sec. will have moved forward 30km. from the firing line. (no pun) by the time it marks the moon. 10. From (9) then the spot will hit the moon 30 km behind where it would hit during an inline with the earths orbit shot, 90 degrees of moon orbit earlier. Also the 10ms beam will have a more elongated oval shape. I look forward to any comments. ----- Original Message ----- From: Paul Deema To: Geocentrism@xxxxxxxxxxxxx Sent: Tuesday, July 31, 2007 4:08 AM Subject: [geocentrism] Re: Moving Earth Deception Marc V There are very significant differences between a ball and a laser beam so I think using one as an analogy for the other is likely to to be counter productive. Regarding the ball. First instance. The ball has no E-W motion and the angle that it is deflected W will depend heavily on coefficients of friction. Second instance. The ball has motion W-E but relative to the stationary train it seems to me the situation is similar to the first instance and it will deflect W as you state, relative to your moving train. Friction will again be a significant parameter. If you repeat this with a laser however, the physical position is quite different. There is no friction to start with and the travel time is enormously reduced. You wouldn't detect any deflection is my guess, but this is just a guess -- I don't have the tools to grapple successfully with this problem. All that aside, there is a major difference between the LRRR-Luna experiment and your train analogy. The LRRR experiment relies for its success on the reflectors used. They are 'cube-corner' reflectors -- accurate cubes (of quartz if I remember correctly). A beam of light shining into the cubes at within +/- 45 deg of a line through opposite corners will reflect directly to the source regardless of the angle from which it arrives. To achieve the same effect with the ball and train experiment, you would need to cover the reflecting train with vertical surfaces at 90 deg to each other (like throwing an 'o' at a lot of BIG 'W"s. (This is a two dimentional approximation but I think that is sufficient in this context). In the case of the real object -- the Moon, reflectors, Earth origin and Earth destinatiion, we are concerned largely with angles. I've tried to illustrate what I'm thinking in the attached drawing. In the geostatic case, the laser/camera remains on a fixed bearing, but must lead the target by some small angle so as to hit it when the Moon moves into the laser pulse just as it arrives. Then the beam is returned along the same path from the point where the reflector was hit and the camera is waiting directly in line. In the heliocentric case, the laser is rotating but as the Moon is not, it does not need to lead. (Strictly it needs to lead but only by less than one fiftieth part.) However, as the laser/camera are offset from the Earth's centre of rotation, during the out and back transit time -- some 2 1/2 seconds -- they have also moved away from the line of fire. If the signal strength and the detector sensitivity were equal to the task, theoretically this should be detectable but sadly they seem not to be. I suspect that an array of detectors strung out along the path of the reflected signal and analyzed statistically would tell the story. Paul D ------------------------------------------------------------------------------ Yahoo!7 Mail has just got even bigger and better with unlimited storage on all webmail accounts. Find out more. ------------------------------------------------------------------------------ No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.5.476 / Virus Database: 269.11.0/927 - Release Date: 30/07/2007 5:02 PM