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From: "Paul.B.Andersen" <relativity@paulba.no>
Newsgroups: sci.physics.relativity
Subject: Re: Pseudoscience III: Each SR/GR experiment is a FRAUD!
Date: Thu, 26 Sep 2024 22:01:15 +0200
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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/