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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