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Phil,
The relitive motion of any two bodies can only be visualised from a third position...The relitive motion of that third position must be taken into consideration..you cannot just assume your third position is motionless because what/ where is in motion or not is the question underconsideration.
--- On Sun, 9/21/08, philip madsen <pma15027@xxxxxxxxxxxxxx> wrote:
From: philip madsen <pma15027@xxxxxxxxxxxxxx>
Subject: [geocentrism] Re: translational orbit
To: "geocentrism list" <geocentrism@xxxxxxxxxxxxx>
Date: Sunday, September 21, 2008, 4:34 PM
Those mathmatical representations mean nothing to me.. ! Well not without some study, and I will not do that.
In our context re the earth orbit of the sun ,
For the full years orbit, the angle of inclination of the pole, is fixed relative to any nominated axis of the sun. In the following diagram you see the earth position in every season as the green ball.. the arrows show the direction of tilt, which is maintained for the full year of orbit..
 That is translation without rotation.
Now if the the orbit was the edge of a fixed disc and the balls were fixed to this disc and the disc rotated.. the arrows would rotate always pointing away from the centre. That is translation with rotation
really quite simple without a single sine cos or tan.....
The former is what the HC position claims.. They claim it!
This is consistent with a stationary earth!
If the polar star makes a circle , there can be only one of two reasons.
(a) The world rotates! or
(b) The stars revolve around the world.
What the scale of distance does is this.
The view will be exactly the same , (save miniscule parallax) from any place in the solar system, or anywhere on the circumference of the alleged annual orbit.
Hence I repeat, it is indistinguishable from static earth or a orbiting earth..The Parallax does not support HC simply because it would be the same as for a rotating cosmos.. I reccommend a relook at the animations on GWW's DVD.
HC had to come up with a tilted earth ...they had no option...
Philip.
----- Original Message -----
Sent: Monday, September 22, 2008 8:42 AM
Subject: [geocentrism] Re: translational orbit
It should be clear, by now, that there is a strong analogy between rotational motion and standard translational motion. Indeed, each physical concept used to analyze rotational motion has its translational concomitant. Likewise, every law of physics governing rotational motion has a translational equivalent. The analogies between rotational and translational motion are summarized in Table 3.
| Table 3: The analogies between translational and rotational motion. |
| Translational motion |
|
Rotational motion |
|
| Displacement |
![$d{\bf r}$]() |
Angular displacement |
![$d\mbox{\boldmath$\phi$}$]() |
| Velocity |
![${\bf v} = d{\bf r}/dt$]() |
Angular velocity |
![$\mbox{\boldmath$\omega$}= d\mbox{\boldmath$\phi$}/dt$]() |
| Acceleration |
![${\bf a} = d{\bf v}/dt$]() |
Angular acceleration |
![$\mbox{\boldmath$\alpha$}= d\mbox{\boldmath$\omega$}/dt$]() |
| Mass |
![$M$]() |
Moment of inertia |
![$I =
{\scriptstyle\int \rho \vert\hat{\mbox{\boldmath$\omega$}}\times {\bf r}\vert^2 dV}$]() |
| Force |
![${\bf f} = M {\bf a}$]() |
Torque |
![$\mbox{\boldmath$\tau$}\equiv{\bf r}\times {\bf f}= I \mbox{\boldmath$\alpha$}$]() |
| Work |
![$W = \int {\bf f}\!\cdot\!d{\bf r}$]() |
Work |
![$W = \int \mbox{\boldmath$\tau$}\!\cdot\!d\mbox{\boldmath$\phi$}$]() |
| Power |
![$P = {\bf f}\!\cdot\!{\bf v}$]() |
Power |
![$P = \mbox{\boldmath$\tau$}\!\cdot\!\mbox{\boldmath$\omega$}$]() |
| Kinetic energy |
![$K = M v^2/2$]() |
Kinetic energy |
![$K = I \omega^2/2$]() | |
On Sun, Sep 21, 2008 at 3:41 PM, philip madsen <pma15027@xxxxxxxxxxxxxx> wrote:
I thought you wer out to prove that movement of the world around the sun would be detected as a rotation..
Maybe I am jumping ahead.. so bring on your first simple step and we will go from there. though I feel somehow this was al ready resolved by Regner.
Philip
----- Original Message -----
Sent: Sunday, September 21, 2008 1:01 PM
Subject: [geocentrism] Re: translational orbit
Just how does scale have anything to do with rotaional effects?..Demonstrate somthing dont just imagin..... if you take any camera and rotate it against any object at any distance , at any scale ..including real stars you get rotational effects!..That being absolutly true...please clarify & explain your objections with somthing real not just imagined.......
--- On Sat, 9/20/08, Neville Jones <njones@xxxxxxxxx> wrote:
From: Neville Jones <njones@xxxxxxxxx>
Subject: [geocentrism] Re: translational orbit
To: geocentrism@xxxxxxxxxxxxx
Date: Saturday, September 20, 2008, 5:11 PM
No go Allen.. You are not including scale.. HC propose two myths.
One that the earth moves around the sun...
Now if this were the only myth, then your camera would indeed show the rotation. and you are correct.
However the other myth makes your camera worthless..
Oh the other myth.. That is the alleged distance of your star... You see with optics, everything past a certain focal length is infinity.. It cannot be resolved.. And they have conveniently placed all the stars at a distance well beyond your power of resolution..
Hope you got all that.. I didn't .. but it sounds good..
Philip.
----- Original Message -----
Sent: Friday, September 19, 2008 1:59 AM
Subject: [geocentrism] translational orbit
|
EUREKA!!......A translational orbit still produces a rotational effect!...The punch line is ....
1. the axis of the rotation shifts from the body that is being orbited to the center of the body in the translational orbit....
2. It reverses the effects of the rotational effects. That is to say that a clockwise orbit will produce counter clockwise rotational impression on film where if the translational orbit is clockwise then the rotational effects on film will be clockwise!..
The fact that the earth's has a translational orbit around the sun cannot and will not hide a rotation around the NCP which is offset from the nightly NEP by 23 degrees.. YES, Im already fully aware of ALL the previous as well as possible objections.....i was able to isolate each and every single one.........I have now found the way to prove it as well as demonstrate how it can be accomplished in the real world ......The solution is remarkably "simple" but extremely hard to visualize due to the complexities of the kinematics........If you imagine a set of crosshairs they have a up/ down equal distant mark as well as a left and right equal distant mark....The trick is understanding that the back and fourth motion of the sun by 23 degrees annually is nothing more then up/down deviations from that up/down center mark.....The key is as long as the right/ left center mark does not deviate we can still get our
rotation around a axis that lay parallel to a axis that is perpendicular and runs through the suns (ecliptic deviation/ path) since it lay perpendicular to the up/down centerline on our cross hairs, because it lays 23 degrees offset ..this is true because any rotation around the sun or ecliptic is not dependent on the north south deviation of the sun/ecliptic ..the rotation & it's effects are around a axis that lay perpendicular to that deviation. Yes we have to have a camera that does not move with respect to that ecliptic deviation….I will show but we should already know exactly how to accomplish that… .......the proof is quite detailed i will lay out the fundamentals bit by bit so we don't get confused by all the motions........I plan to submit some diagrams and photos eventually...using real stars and demonstrating exactly how it was done...but the key is a translational motion
will still produce a rotation on the NCP .....The rub is, I kept trying to tell you guys that the clue was "hidden" in that "most powerful definition of rotation known to man" ....In fact, It would have to produce a rotational effect in order for all of the motions to be "equivalent"!.............. Oh, what fun I am going to have now....."Destruction" and "chaos" the likes of which have not been seen since the Renaissance itself..& ..It won't take me any 400 years either! | |
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