Motor Daddy
Valued Senior Member
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.
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.
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