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,
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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