[geocentrism] Re: acceleration calcs
- From: "philip madsen" <pma15027@xxxxxxxxxxxxxx>
- To: <geocentrism@xxxxxxxxxxxxx>
- Date: Tue, 29 Jan 2008 06:30:09 +1000
"We cannot measure translations with a gyroscope, whether accelerated or not."
Regner.
Of course! I should be ashamed for not applying what I knew from experience,
and my recent grasp of the term translation.
In a two body system excluding all other reference points, there seems to be
know way of feeling "detecting" which body is moving around which except by the
logic of known physics as regards the influence of gravity and centrepetal
force. Experimental proof would need to be obtained of the relative masses of
the bodies, which of course has been done betwen the sun and the earth. There
is no way anyone can claim the sun is some light airy gas ball.
Thanks Regner.
Philip.
----- Original Message -----
From: Regner Trampedach
To: geocentrism@xxxxxxxxxxxxx
Sent: Monday, January 28, 2008 4:25 PM
Subject: [geocentrism] Re: acceleration calcs
Regner's reply to Philip's in blue...
philip madsen wrote:
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
Regner
Philip.
Philip.
----- Original Message -----
From: Regner Trampedach
To: geocentrism@xxxxxxxxxxxxx
Sent: Thursday, January 24, 2008 1:08 PM
Subject: [geocentrism] Re: acceleration calcs
philip madsen wrote:
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
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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