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

Den 26.09.2024 18:04, skrev rhertz:

Paul B. Andersen wrote:
 

Den 25.09.2024 23:54, skrev rhertz:

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

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)
 

 Paul, I quote this SHAMEFUL part of your post. You are becoming a disgraceful LIAR and DECEIVER, as it correspond to a relativist.
 

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?
>

 THE CORRECT FORMULA, FROM THE HAFELE PAPER, IS:
 kinematic term: (τ - τ₀) = (-(2RΩv + v²)/2c²)τ₀

But it isn't (τ - τ₀) because it is only the kinematic term.
That's why I wrote τₖ. But Δτₖ would possibly be better.
So: Δτₖ = (-(2RΩv + v²)/2c²)τ₀
But thanks for pointing out that my formula was THE CORRECT FORMULA
for an equatorial circumnavigation with constant ground speed v.

 where τ₀ is the USNO ELAPSED TIME after the eastward round trip. This is a GROSS estimation, which gives
τ₀ = 65.42 hours = 235,512,000,000,000 theoretical nanoseconds
elapsed at USNO clocks!

Thanks again for pointing out the bleeding obvious.
The USNO clocks have advanced τ₀ = 235,512,000,000,000 ns
during the 65.42 hours trip.
You are sharp today, Richard! :-D

 You CAN´T (unless you are a fraudster) to calculate a theoretical USNO
elapsed time of 235.51E+12 nsec (out of thin air), and ESTIMATE a
difference of 245 nsec between USNO and "flying clocks". This represents a fraction of about 1,000,000,000,000 parts between
both clocks, and calculated for trips around the Equator.

The USNO clocks advances τ₀ = 65.42 hours during the 65.42 hours trip,
but I CAN'T (unless I am a fraudster) calculate a theoretical USNO
elapsed time of 235.51E+12 nsec (out of thin air).
And I can't put the values:
τ₀ = 65.42 h = 235512 s = 235.512E+12 ns
R = 6378137 m
Ω = 7.2921159e-5 rad/s
c = 299792458 m/s
v = 2πR/τ₀ = +170.16 m/s
into THE CORRECT FORMULA  Δτₖ = (-(2RΩv + v²)/2c²)τ₀
and get Δτₖ = -245.32 ns.
This is the kinematic term, so it is only part of the difference
between the USNO clock and the "flying" clock.
But can you please explain why doing what I did made me a fraudster?

 Of course that, if you are a CROOK used to hack and cook, are used to
LIE and DECEIVE, and have a bunch of people that support your SCAM, then you can produce an HOAX like this one.

Are you frustrated about something, Richard?
Do you have a bad toothache?

 Your emphasis in supporting this entire FARCE shows your true colors.
You are not different from these people or others who committed fraud in widely published "experiments", like Gravity Probe A, Pound-Rebka,
Cassini and so many others.
 Shame on you, as you don't have a bit of it.

Of course we know that all physicists born after 1900 are
members of a MAFFIA, and profit from it. This is because
the different results are COOKED with the help of statistical
manipulations, fraud, cooking and peer complicity.
But I am not a physicist, so why am I a fraud?
Is it because I several times have pointed out your errors?
BTW, her is yet another error of yours:

>
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!

 From whence did you get the stupid idea that the equation
Δτₖ = - τ₀ (2RΩv + v²)/2c²  for an equatorial circumnavigation
at the constant ground speed v = +170.16 m/s would give
Δτₖ = -184±18 ns ?
--
Paul, having fun
https://paulba.no/