[geocentrism] Re: acceleration calcs

Regner said. 
"Eh, no!
We can still measure our position/velocity relative to the stars.
Then it is a matter of conviction, I guess, whether you believe the whole 
Universe
to be moving that way, or just the Earth." 

Regner my words were not phrased too well. What I meant by certainty was that 
there is no absolute way of knowing or finding a spot with absolutely NO 
MOTION. You can relate movement between the stars, and the earth. but motion 
(constant velocity no acceleration ) as such in isolated space cannot be 
detected. 

You also responded with, "Eh, they do.
You don't need more than that - you just integrate once over time to get 
velocity, and
twice to get position. [ I do not follow. What device can detect motion 
movement and velocity, except with respect to other bodies, such as the stars, 
whose own movements as a whole, can only be assumed. ] 
But they won't detect acceleration of an object free-falling in a gravitational 
field.
We both argued for that..." [ yes, and therefore I have confused myself. I had 
always assumed that a gyroscope in a satellite would show the curved motion . 
Just as I expect one to show the fall of the earth around the sun????
Could you please explain this for us. 

Philip. 





Philip. 


  ----- Original Message ----- 
  From: Regner Trampedach 
  To: geocentrism@xxxxxxxxxxxxx 
  Sent: Thursday, January 24, 2008 1:08 PM
  Subject: [geocentrism] Re: acceleration calcs


  philip madsen wrote: 
    The truck drives along a turn in the road.
    * A ball is dropped from the ceiling of the container.

    Thanks Regner.. I now glean what is meant by the "fiction".. Actually we 
had discussed the falling ball in a transparant railway carriage passing a 
station a few times here on this list. I just missed the connection as regards 
the "fiction" of observation. 

  I am glad that clarified it.

    You make a good case for it being impossible for anybody to establish 
absolute certainty for any position or movement in space. (we've been here 
before as well) 

  Eh, no!
  We can still measure our position/velocity relative to the stars.
  Then it is a matter of conviction, I guess, whether you believe the whole 
Universe
  to be moving that way, or just the Earth.
  But apart from that, you are right that there is no "absolute certainty" in 
the world
  of science.

    Its claimed inertial devices will detect acceleration. but that is only 
part of motion...
  Eh, they do.
  You don't need more than that - you just integrate once over time to get 
velocity, and
  twice to get position.
  But they won't detect acceleration of an object free-falling in a 
gravitational field.
  We both argued for that...

          Regner


    Philip. 
      ----- Original Message ----- 
      From: Regner Trampedach 
      To: geocentrism@xxxxxxxxxxxxx 
      Sent: Wednesday, January 23, 2008 1:10 PM
      Subject: [geocentrism] Re: acceleration calcs


      Exactly, Philip.
      The Earth, and we with it, are in free fall around the Sun, with the 
gravitational acceleration
      by the Sun (and towards the Sun) keeping us in our elliptic orbit.
         Without careful analysis, I actually thought that you might be able to 
detect it, but you
      are right, Philip. This is also stated in Einsteins equivalence principle 
which states that
      a free-falling reference frame is an inertial reference frame, and there 
will therefore be no
      fictitious forces (centrifugal-, Coriolis- and Euler-forces). The 
equivalence principle
      means that the orbit of Earth can just as well be seen as the Earth 
traveling along a straight
      line in a curved space - the two are equivalent - and the latter is 
described by general relativity.

         As a partial reply to your (much) earlier post on pseudo forces, I 
will note a few facts
      on them here - and there is nothing dubious about them.
         Pseudo forces, more often called fictitious forces, arise when your 
reference frame is
      being accelerated. Let's say you set up a laboratory inside a container 
on a trailer truck.
      * The truck drives along a turn in the road.
      * A ball is dropped from the ceiling of the container.
      Imagine the container turning transparent, so that your colleague can 
record the
      trajectory of the ball, as seen from the roadside
      * your colleague will see the ball follow a parabola determined by the 
speed of
         the truck when the ball was released, and the local acceleration of 
gravity.
         Only one force, gravity, acts on the ball:  F_obs = F_grav.
      * You, however, will see the ball being acted upon by another force, 
since the
         ball (and you...) will be accelerated towards the side of the 
container:
                        F_obs = F_grav + F_fict
         This force is entirely due to the truck accelerating iin the opposite 
direction,
      towards the inside of the bend in the road, and we call it a fictive 
force.
      Fictive forces are trivial (but often cumbersome) to derive as the 
opposite of
      the acceleration of your (non-interial) reference frame.

                Regards,

                    Regner


      philip madsen wrote: 
        re Alan and Regners figures.  


        On this business of "feeling" acceleration, whilst I do not pretend to 
having had enough interest in checking the figures, I still reason that its a 
matter of how forces are applied, as to whether you feel anything. 

        In a suddenly braking car you get flung forward...  because the force 
is at the wheels..  But if the breaking force was applied to every molecule of 
the vehicle including you, then I concieve no effect to be "felt" 

        If I take the orbiting space station as an example, the people inside 
and even ouside are all exposed to the same accelerating forces.. They follow 
the orbit of the vehicle..  when the man steps outside, he does not get flung 
off on a merry plunge towards the sun or the earth for that matter. He would 
not "feel" any movement. Yet he is circling the earth every few hours. Thats 
travelling a fast corner. 

        Philip. 
          ----- Original Message ----- 
          From: Regner Trampedach 
          To: geocentrism@xxxxxxxxxxxxx 
          Sent: Tuesday, January 22, 2008 4:59 PM
          Subject: [geocentrism] Re: acceleration calcs


          Alan, 
           Thanks for your calculation, but I'm afraid you made a mistake - 
it's easy to do with all those 
          crazy units Americans juggle with. You forget that your velocities 
are still per hour, while
          you have the change of velocity per second, so your result is 
actually: 
          (1)               11.43 cm/(hour*s) = 0.003175 cm/s2    (cm per 
second squared) 
           The actual change is: 
          (2)               (30.29e5 cm/s - 29.29e5 cm/s) / (year/2d0)  = 
0.006338 cm/s2 
          We agree on the velocities and the difference in velocity - I just 
use centimeter-gram-second (cgs) 
          units. One year is 365.26 days * 24 hours/day * 3600 s/hour = 
3.155693e7 s. 
          The change happens during half a year (I divide year by 2, in Eq. 
[2]) so you would actually 
          have underestimated the change (as you can see from my correction, 
Eq. [1]). 
            It is always a good idea to put your result in perspective by 
comparing with another relevant 
          quantity - the gravitational acceleration at the surface of Earth is 
about g=9.8 m/s2 on average, 
          which means that the acceleration along Earth's orbit is 
          (3)               (0.006338 cm/s2) / (980 cm/s2) = 0.000006467 
          times the average gravitational acceleration at the surface of Earth, 
g.
          or conversely, the acceleration along Earth's orbit is 154600 times 
smaller than g.  I don't 
          think you would notice that! 

            But that is obviously a tiny component of the accelerations 
actually involved.
          Remember that (in HC) the direction of the velocity has also changed 
over the 6 months 
          and the velocities in the two instances will be exactly opposite. We 
can get a rough estimate 
          of that acceleration by just adding the two velocities in Eq. (2), 
since  a-(-b) = a+b, to get: 
          (4)               (30.29e5 cm/s + 29.29e5 cm/s) / (year/2d0)  =  
0.3776 cm/s2 
          which is then 2595 times smaller than g. Absolutely measurable, but 
it wouldn't exactly 
          knock you over. 

                 Regner 


          allendaves@xxxxxxxxxxxxxx wrote: 
            Try to move 4.5 inches within one sec without feeling/ (being abel 
to detect that using current technology) it. This demonstrates the crux of the 
problem with earths inertial motion. Appealing to some imaginary reason why you 
could not detect it in the earth but you could  with anything and everything 
else is not going to work untill you can first prove that your imaginary reason 
exist in reality. NO one isarguing it could be, but if we are to arive at a 
conclusion and proclaim it logical we have to prove the variables along the way 
not make them up as we go along. that is the fundimental difference between GC 
& HC. GC accepts as proof only the effidence presented as it goes along through 
the discovery process.....HC makes it up as it goes along to save it's 
conclusions.....


             
            ----- Original Message ----
            From: Allen Daves <allendaves@xxxxxxxxxxxxxx>
            To: geocentrism@xxxxxxxxxxxxx
            Sent: Monday, January 21, 2008 8:38:51 AM
            Subject: [geocentrism] Re: acceleration calcs



            Here it is ..quick & rough....

            18.5 miles per second average speed * 60 sec for min * 60 min for 
MPH = 66600MPH or 
            The ~ avg change over the course of a year is 3.4%* 66600= 2264.4   
  / 365.4 days=6.2038 MPH per day /24 hours = .25840166 MPH change per hour   
/60min=.00430819444 MPH change per min ……There are 5,280 feet in a mile 
.0043081944 MPH = 22.747266432 feet ( or 6.933366807864
            meters) per min   /60 to convert to seconds = .3791 feet per sec/ 
per second change ( or .11554968 meters per sec per sec). This is  a change in 
velocity of  ~4.5 inches per sec/ per sec  Or 11.43 centimeters per sec per 
second

            There is now way to consider this amount to be inertial change 
negligible. The effect rate of change regardless of how fast the earth is 
supposed to be traveling because only the rate of any change from the effective 
inertail 0 is measured. 


            This means that the velocity change of the earth going around the 
sun is not just moving 4.5 inches ever second but changing by 4.5 inches per 
sec. During the earths closest approach to the sun (such as traveling in a 
moving car)if we experience 0 velocity change because we are traveling with the 
earth then whatever the current velocity is would be felt as 0. However, the 
rate of change just as in a moving vehicle would be changed if we "give it some 
gas" and in this case the rate of change would be a increase ~4.5 inches every 
sec every sec. This is to say we on second one we increase by 4.5 sec on second 
two we have increased to 9 by second three we have increased to 13.5….the rate 
makes for a exponential distance traveled curve.  In any case this is the rate 
of change. Assume for the sake of argument that your body could not  detect 
that  change rate, current instrumentation however ( acelerometers) are able to 
detect that amount of inertial change to almost infinite amounts, and they are 
not "aetheraly" depemdent).  

            Neville Jones <njones@xxxxxxxxx> wrote: 

              Not me. 


                -----Original Message-----
                From: allendaves@xxxxxxxxxxxxxx
                Sent: Fri, 18 Jan 2008 07:55:05 -0800 (PST)


                Since the earth changes its speed throught it's orbit, has 
anyone out there ever calculated the actual acceleration force changes to the 
earth as it moves back and fourth through its apogee and perigee elliptical 
orbit around the sun? 

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