[geocentrism] Re: acceleration calcs

  • From: allendaves@xxxxxxxxxxxxxx
  • To: geocentrism@xxxxxxxxxxxxx
  • Date: Wed, 23 Jan 2008 15:01:33 -0800 (PST)

Philip It makes no differece!...You did not hear my post on gravitaion and 
inertia....the same principle applies in a moving care or a ball on a teather 
or a orbiting spacstation.or a ride up or down/ toward or away from center of a 
gravitational field......ie...in a elevator ..you are trying to make a 
distiction that does not exist even in MS....inertia and gravity are one and 
the same/ dependent.......Inertia  felt in a car due to breaks is just another 
affect/ change to the normal "non felt" gravity  we experiance every 
day......Again it is the change in gravity/ inertia not the constant velocity 
of the vehicle or the constant acceleration of Gravity.......the cause or how 
the cause is applied is irrelevant!....if the gravitation feild was qual on all 
parts(molecules) of the earth at one time then there would be no "centrifical 
forces" ficticious or real. The feild is not and thus a drop of water on a 
teather ball will fly off if it orbits fast
 enouph..................again it is the change from a zero or normal condition 
gravity or inertia every particle all at once or not.......... :-)

----- Original Message ----
From: philip madsen <pma15027@xxxxxxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Wednesday, January 23, 2008 3:22:59 PM
Subject: [geocentrism] Re: acceleration calcs

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..... 
You did not hear me Allen.. read again. The brakes on the car is not an example 
of your case. In this case in space on earth the forces act on every molecule 
in the object. What is accelerating the world is also accelerating you. Except 
on earth gravity is a major influence. 
Again use the space station with an elliptical orbit.. As it orbit changes up 
or down the spaceman outside being under the same forces also makes the same 
changes.  He does not have to hang on to the steering wheel...  Unless the 
pilot was making course corrections using his rockets..  That would be 
----- Original Message ----- 
From: allendaves@xxxxxxxxxxxxxx 
To: geocentrism@xxxxxxxxxxxxx 
Sent: Thursday, January 24, 2008 3:42 AM
Subject: [geocentrism] Re: acceleration calcs

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 
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 
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 
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 
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.



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. 
----- Original Message ----- 
From: Regner Trampedach 
To: geocentrism@xxxxxxxxxxxxx 
Sent: Tuesday, January 22, 2008 4:59 PM
Subject: [geocentrism] Re: acceleration calcs

 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 
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 
knock you over. 


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 

----- 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 
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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