On Einstein's explanation of the invariance of c

Discussion in 'Pseudoscience Archive' started by RJBeery, Dec 8, 2010.

  1. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Here it is Ladies and Gentlemen, the moment you've been waiting for. 3D absolute velocity!!!

    Using the equation v=(ct-l)/t you can find the absolute velocity and x,y,z position in space from INSIDE THE BOX!!!

    Simply mount a light source at the center of the cube and a receiver on the center of each wall and measure the one-way light travel times from the center of the cube to the center of the walls.


    The box in this example is a perfect cube with all sides measuring 299,792,458 meters in length and height.

    v=(ct-l)/t

    The length in the equation is 149,896,229 meters in this example (center of cube to center of walls).

    Times as measured from the source at the center of the cube to the receiver at the center of indicated walls:

    The center of the cube starts at coordinates (0,0,0)

    x-component = .6667 seconds (.25c)
    y-component = 1.0 seconds (.5c)
    z-component = 2.0 seconds (.75c)

    Positions of center of cube at times indicated:

    2 seconds: (.5,1.0,1.5)

    Travel distances:
    x = 2.0(.25c) = .5
    y = 2.0(.5c) = 1.0
    z = 2.0(.75c) = 1.5


    4 seconds: (1.0,2.0,3.0)

    Travel distances:
    x = 4.0(0.25c) = 1.0
    y = 4.0(0.5c) = 2.0
    z = 4.0(0.75c) = 3.0


    6 seconds: (1.5,3.0,4.5)

    Travel distances:
    x = 6.0(.25c) = 1.5
    y = 6.0(.5c) = 3.0
    z = 6.0(.75c) = 4.5

    Since you know where the center of the cube is, and the size of the cube, you also know where each corner is at all times simply by adding or subtracting the distances from the known center of the cube.

    So for example, the center of the cube traveled from (0,0,0) to (1.5,3.0,4.5) in 6 seconds, which is 1,682,580,997.46 meters, an absolute velocity of 280,430,166.24 m/s. The x corner is located at (2.0,2.5,4.0) at 6 seconds.
     
    Last edited: Jan 10, 2011
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. Neddy Bate Valued Senior Member

    Messages:
    1,403

    (0.0, 0.0, 0.0) center of cube
    (0.5, 0.0, 0.0) center of "x" face of cube
    (0.0, 0.5, 0.0) center of "y" face of cube
    (0.0, 0.0, 0.5) center of "z" face of cube

    (0.600) seconds for light to reach "x" face center
    (0.668) seconds for light to reach "y" face center
    (0.741) seconds for light to reach "z" face center

    What is the absolute velocity of the cube?
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. arfa brane call me arf Valued Senior Member

    Messages:
    5,305
    Why Motor Daddy's model is naive.

    What it assumes is that a perfect cube can be constructed with equal length sides or edges. If he wants them to be the distance light travels in 1 second, that is determined by units of ct or if t is 1, units become "the velocity of light".

    Velocity has a direction and each vertex of the cube will have to 'send' some light to three others, in 'beams' which are not only straight (like a kind of rigid material), but perpendicular. He can't see any problem, but there is one.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    The center of the cube starts at coordinates (0,0,0)

    x-component = .600 seconds (.1666c)
    y-component = .668 seconds (.2514c)
    z-component = .741 seconds (.3252c)

    Positions of center of cube at times indicated:

    2 seconds: (.3332,.5028,.6504)

    Travel distances:
    x = 2.0(.1666c) = .3332
    y = 2.0(.2514c) = .5028
    z = 2.0(.3252c) = .6504


    4 seconds: (.6664,1.0056,1.3008)

    Travel distances:
    x = 4.0(.1666c) = 0.6664
    y = 4.0(.2514c) = 1.0056
    z = 4.0(.3252c) = 1.3008


    6 seconds: (.9996,1.5084,1.9512)

    Travel distances:
    x = 6.0(.1666c) = 0.9996
    y = 6.0(.2514c) = 1.5084
    z = 6.0(.3252c) = 1.9512


    The distance the center of the cube traveled from (0,0,0) to (.9996,1.5084,1.9512) is 797,788,916.49 meters. The absolute velocity from (0,0,0) to (.9996,1.5084,1.9512) is 132,964,819.42 m/s
     
  8. AlexG Like nailing Jello to a tree Valued Senior Member

    Messages:
    4,304
    This is still going on. I'm impressed.


    Please Register or Log in to view the hidden image!

     
  9. Neddy Bate Valued Senior Member

    Messages:
    1,403
    Is this supposed to be an answer to my question? The times in my question are not 2, 4, or 6 seconds.
     
  10. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    I gave you everything you need to know. The absolute velocities of each component are noted for the times you gave. I also gave a travel direction, total distance the center traveled in 6 seconds, and the absolute velocity of the cubes center. What more do you want?
     
  11. Neddy Bate Valued Senior Member

    Messages:
    1,403
    Is this your answer for the velocity components?

    If so, where do you think the center of the x-face located at time 0.600 seconds?

    (??, ??, ??)

    Can you show that the light does indeed travel from (0,0,0) to (??, ??, ??) in a time of 0.600 seconds?
     
  12. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Oh, maybe you wanted me to do the math for the distance each component traveled in the times you gave?

    x-component = .600 seconds (.1666c) 29,967,254.10 meters
    y-component = .668 seconds (.2514c) 50,345,706.39 meters
    z-component = .741 seconds (.3252c) 72,241,947.94 meters
     
  13. Neddy Bate Valued Senior Member

    Messages:
    1,403
    Tell me how you got .1666c
     
  14. Neddy Bate Valued Senior Member

    Messages:
    1,403
    Nevermind, I see it:

    (ct - l) / t
    v(x) = (0.600 - 0.500) / 0.600 = 0.1666c

    Yeah, that's the same equation we already agreed doesn't work for 3D.
     
  15. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    x-component = .600 seconds (.1666c) 29,967,254.10 meters
    y-component = .668 seconds (.2514c) 50,345,706.39 meters
    z-component = .741 seconds (.3252c) 72,241,947.94 meters
     
    Last edited: Jan 10, 2011
  16. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Your way is wrong. I've showed you the correct way using my equation. I can tell you everything you need to know. It is dead accurate! Prove me wrong!
     
  17. Neddy Bate Valued Senior Member

    Messages:
    1,403
    OK.


    If you say so. Now tell me where the center of the "x-side" is at that time. I'm talking about the point which was originally located at (0.50, 0.00, 0.00) in units of light seconds.
     
  18. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105


    At .6 seconds the center of the cube is located at (0.0999,.1508,.1951) and the point that was at (0.50, 0.00, 0.00) is located at (.5999,.1508,.1951)
     
  19. Neddy Bate Valued Senior Member

    Messages:
    1,403

    You've changed your numbers. Where is the center again?
     
  20. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    That was a mistake, I used the wrong numbers to calculate it.

    At .6 seconds the center of the cube is located at (0.0999,.1508,.1951) and the point that was at (0.50, 0.00, 0.00) is located at (.5999,.1508,.1951)
     
  21. Neddy Bate Valued Senior Member

    Messages:
    1,403

    And how far did the light travel in 0.60 seconds?
     
  22. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Light traveled 179,875,474.8 meters in .6 seconds to reach the wall and the center of the cube and the wall traveled 29,967,254.10 meters in the x direction.
     
  23. Neddy Bate Valued Senior Member

    Messages:
    1,403
    You say the cube traveled 29,967,254.10 meters in the x direction. How far did it travel in the y direction in .6 seconds?
     

Share This Page