# [geocentrism] Re: acceleration calcs

• To: <geocentrism@xxxxxxxxxxxxx>
• Date: Wed, 23 Jan 2008 09:03:38 +1000
```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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