After a rapid reading of your post (sorry, not much time), it seems that 
it might be some variation of known 'paradox' of special relativity:

See for example:
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/barn_pole.html
http://www.einsteins-theory-of-relativity-4engineers.com/ladder-paradox.html

Thibaut




>   Let's see if I can describe this in words only:
> 
>   As we all know, the speed of light in vacuum is the same for all inertial
> frames of reference, and because of that, movement in inertial frames of
> reference is subject to Lorentz transformations.
> 
>   For this reason if there are two spaceships travelling to opposite
> directions at speed s (from an external observer's point of view), if
> ship A measures the speed of ship B as it perceives it, it's not 2*s,
> but something less (and always less than c). It happens because ship B
> appears contracted in the direction of the movement.
> 
>   Let's assume that ship A measures the speed of ship B by starting a clock
> when the bows of both ships coincide, and stops the clock when the bow
> of ship A coincides with the stern of ship B. For the sake of simplicity,
> let's assume that, while both ships have the same length, ship A measures
> the apparent length of ship B to be exactly half of its own length (due to
> Lorentz contraction), and that the time measurement was exactly one second.
> This gives some speed s' which is less than 2*s, as mentioned earlier.
> 
>   Now, let's assume that ship A wants to estimate what ship B would measure
> (by using the same method) for the speed of A in this situation.
> 
>   Since it took 1 second for the bow of ship A to go from the bow of ship B
> to its stern, and since A measured B to be exactly half the length of A,
> it would mean by reciprocity that it would take 2 seconds for the bow of
> ship B to go from bow to stern of ship A.
> 
>   This would seem contradictory (as the ships have the same length and
> are travelling at the same speed). However, this would be from the time
> reference of ship A, not the time reference of ship B. If ship A wants to
> estimate what ship B is measuring, it has to use the time reference of ship
> B, not its own.
> 
>   If there was a big clock visible on the hull of ship B, which ship A
> could observe, it would see that it would seem to go at half speed
> compared to the clock in ship A. Hence by using the clock of ship B,
> it would take one second for the bow of ship B to go from the bow to the
> stern of ship A, hence reconciling the speed measurement.
> 
>   However, by using the clock of ship B, it takes only a half second for
> the bow of ship A to go from the bow to the stern of the ship B. Since
> this is from the time reference of ship B, it would seem that ship B is
> measuring a half second for this, rather than two seconds (which is what
> ship A measured for the bow of ship B to go from the bow to the stern of
> ship A, when using its own clock).
> 
>   How is this apparent contradiction between two seconds vs. a half second
> reconciled?
>

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