# [geocentrism] Re: acceleration calcs

• From: Regner Trampedach <art@xxxxxxxxxxxxxx>
• To: geocentrism@xxxxxxxxxxxxx
• Date: Mon, 28 Jan 2008 17:25:43 +1100

```Regner's reply to Philip's in blue...

```
```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.
```To summarize:
* You cannot measure absolute velocity
- or more importantly it doesn't matter to physics.
* You _can_ measure absolute acceleration.
```
```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. ]
If you get your velocities from integration over acceleration, there is still an integration constant which is undetermined - which means you still don't have an absolute velocity.
```
```
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.
```
A gyroscope keeps it's orientation in space - therefore we can measure rotations.
```We cannot measure translations with a gyroscope, whether accelerated or not.
...Unless, of course, you use the gyroscope in a manner akin to this:
```
How to measure the height of a building using a barometer <http://www.snopes.com/college/exam/barometer.asp>
```
Regner
```
Philip. Philip.
```
----- Original Message -----
*From:* Regner Trampedach <mailto:art@xxxxxxxxxxxxxx>
*To:* geocentrism@xxxxxxxxxxxxx <mailto:geocentrism@xxxxxxxxxxxxx>
*Sent:* Thursday, January 24, 2008 1:08 PM
*Subject:* [geocentrism] Re: acceleration calcs

```
```    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 <mailto:art@xxxxxxxxxxxxxx>
*To:* geocentrism@xxxxxxxxxxxxx
<mailto: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/.

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
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
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

```
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 <mailto:art@xxxxxxxxxxxxxx>
*To:* geocentrism@xxxxxxxxxxxxx
<mailto:geocentrism@xxxxxxxxxxxxx>
*Sent:* Tuesday, January 22, 2008 4:59 PM
*Subject:* [geocentrism] Re: acceleration calcs

Alan,
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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