[geocentrism] Re: acceleration calcs
- From: "philip madsen" <pma15027@xxxxxxxxxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Thu, 24 Jan 2008 07:29:21 +1000
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.
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)
Its claimed inertial devices will detect acceleration. but that is only part of
motion... Philip.
----- Original Message -----
From: Regner Trampedach
To: 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.
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?
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