# Thread: Transponders used in measuring the absolute velocity of objects moving in free space

1. A transducer is a device used daily in air traffic control systems that emit and receive EM pulseswhere the time of emission and rece ptionare recorded in the local computers and these event times are also imbedded inthe transmitted signal.
Two transducers, A and B, are separated by a constant distance. When theA transponder emits an EM pulse at time A0 the B transponder’s clocktime is unknown, hence the entry Bu. The table below is structuredsuch that the times on the same horizontal are times that are on both clocks simultaneously,even if the other time is unknown. Example: At A0 the clock time of emission from A is A0 - this time on the B transducer is unknown,hence the Bu entry.
The time of flight from A to B is B1 and isunknown, but when the pulse arrives at B the clock time is recorded as Bu + B1 where the sum is recorded. The pulse isimmediately emitted back to A arriving at A at A2 a known and recorded time.This clock time at B is Bu + A2.
 A B A and B transducers A0 Bu unknown A + B1 Bu + B1 recorded (known) A2 Bu + A2 Recorded = X1 A2 + B1 Bu + A2 + B1 recorded = x2 X2 – X1 = Bu + A2 + B1 - (Bu + A2) = B1

For c =1 a unit speed of light, the outbound leg distance is d1 = B1 - A0. The return leg of the round trip pulse is d2 = A2 – B1 .

The total distance of light travel is d1 + d2 = A2 –A0.
The difference in the length of the two legs is, d1 - d2 = B1 - A0 –(A2 – A0) = D = 2B1 – (A2 + A0)= 2B1 – (A2 + A0). For A0 =0, D = 2B1 – A2.
A difference inlengths d1 - d2 >0 is unambiguously an inference of motion during A2 – AD. Hence, V(A2 – A0) = 2B1– (A2 + A0),
V = [2B1 –(A2 + A0)]/[A2 - A0],or for A0 = 0, V = 2B1/A2 - 1.
A
0|
_______________ d11___________| B1

----D----- A2 |_______ d2_________|B1

2.

3. We have tested for this. There is no difference in the length of the two legs.

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