# [geocentrism] Re: acceleration calcs

• From: Regner Trampedach <art@xxxxxxxxxxxxxx>
• To: geocentrism@xxxxxxxxxxxxx
• Date: Tue, 22 Jan 2008 17:59:55 +1100

```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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