[geocentrism] Re: acceleration calcs
- From: allendaves@xxxxxxxxxxxxxx
- To: geocentrism@xxxxxxxxxxxxx
- Date: Wed, 23 Jan 2008 09:42:09 -0800 (PST)
Regner......This is the problem with the equivalence principle argument.
1. It can only be proven to be false it has never been positivly proven only
assumed.
2. The solar system itself is in free fall around the galactic center and thus
the solar system is a inertial ref fame that we should be able to measure
within just as the earth is with the solarsytem and we can measure inertia in
earths inertail reference system.....So you cant hid behind the equivalence
principle for a lack of detection of inertia any more then you could with a car
or a airplane in free fall on earth.......inertia is still measured against all
free falling objects in every case. If you could not detect the earths inertia
then you could not detect any inertia from any object in space or on the
earth......
3. All ref frames are equivalent in relativity...you can have any number or
pick any sections for your reference frame.... individual atoms are inertial
reference frames!?
Philip...........What we are measuring is not the velocity of the earth in its
orbit but its change in velocity, in the same way that a change in velocity is
felt/ measured when you either put the brakes on a car or you give the car some
gas...you may not feel like your moving 100 mph but if you either put on the
brakes or give it some gas you will feel the change even as small as it is.....
----- Original Message ----
From: Regner Trampedach <art@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Tuesday, January 22, 2008 8:10:40 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.
As a partial reply to your (much) earlier post on pseudo 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 when your
reference frame is
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
* your colleague will see the ball follow a parabola 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
philip madsen wrote:
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?
Free 3D Marine Aquarium Screensaver
Watch dolphins, sharks & orcas on your desktop! Check it out at
www.inbox.com/marineaquarium
No virus found in this incoming message.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.19.8/1235 - Release Date: 21/01/2008
9:39 AM
Other related posts:
