Re: Pseudoscience III: Each SR/GR experiment is a FRAUD!

Den 25.09.2024 23:54, skrev rhertz:

Paul. you are PROJECTING. And as a good cornered relativist, try to
divert the attention with something else, EXCEPT what I questioned.
 I put these simple calculations for you TO LEARN, but I sincerely doubt
that you may go the first lines WITHOUT CRYING FOUL, like a little girl.
  ***************************************************************
THEORETICAL TOTAL EASTWARD FLIGHT: -40  ± 23
MEASURED TOTAL EASTWARD FLIGHT: -59  ± 10
NOTE: Starting 4 October 1971, eastward flights lasted 65 hours, with 41
accumulated hours of flight.
  FOR EASTWARD FLIGHT, CALCULATIONS:

You yet again demonstrates your ability to read a text and
misinterpret what you read.
https://www.masterclock.com/cmss_files/attachmentlibrary/Archived-papers/Performance-and-Results-of-Portable-Clocks-in-Aircraft-1971.pdf
On the top of page 268:
"Although the assumption of an equatorial circumnavigation at
  constant ground speed and altitude is not essential, it does
  simplify somewhat the calculations for estimating the magnitude
  of expected relativistic effects.
  For an equatorial circumnavigation with constant ground speed v
  (m/sec) and altitude h (m), the predicted relativistic time gain
  for the flying clock over a similar reference clock kept at "rest"
  on the Earth's surface is given by:" see equation (1)
kinematic term: τₖ = (-(2RΩv + v²)/2c²)τ₀
  Eastward trip:
   τ₀ = 65.42 hours  |v| = 2πR/τ₀ = 170.16 m/s = 612.58 km/h
   v = +170.16 m/s
   τₖ = -245.32 ns
If you insert τ₀ = 2πR/|v| in (1), the result is obviously the same.
Your giant blunder:
Not recognising that this is a very simplified example with
"equatorial circumnavigation at constant ground speed and altitude".
for estimating the magnitude of expected relativistic effects."
(-245 ns is of the same order of magnitude as -184 ns)
How could you imagine that this was the equation to calculate
the kinetic terms from all the flights?

 τ - τ₀ = - τ₀ (2RΩv + v²)/2c² = - 2πR/|v| (2RΩv + v²)/2c² =  -184  ± 18

No. That formula will never give that result.
https://paulba.no/paper/Hafele_Keating.pdf
The Kinematic effect for the Eastward trip = -184±18 ns
is given in the introduction to the paper.
This value is obviously the final result when all the flights
in different direction and with different speeds are taken
into consideration.
READ THE PAPER PROPERLY!

<snip nonsense>

--
Paul
https://paulba.no/