[geocentrism] Re: acceleration calcs

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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