[geocentrism] Re: World/Moon system

  • From: "philip madsen" <pma15027@xxxxxxxxxxxxxx>
  • To: "geocentrism list" <geocentrism@xxxxxxxxxxxxx>
  • Date: Mon, 26 May 2008 17:24:05 +1000

First up Paul I did not perceive any inadequacies.. I just did not percieve 
anything.. You gave me too much to visualise, and without any objective up 
front.  I will put some questions into the text..   
  ----- Original Message ----- 
  From: Paul Deema 
  To: Geocentrism@xxxxxxxxxxxxx 
  Sent: Sunday, May 25, 2008 5:32 AM
  Subject: [geocentrism] Re: World/Moon system


  Greetings all.



  Due to perceived inadequacies, I here submit for your consideration a revised 
version of my earlier post -- From Paul Deema Fri May 23 15:16:51 2008.

  World/Moon system From Neville Jones Mon May 19 20:02:46 2008

  "Diagrams, comments, thoughts, one-way tickets to the Gulag, ... toss them 
all into the pot and let's see what comes out."

  Version 2.



  First a few simplifications - 



  The Earth's orbit is a circle; the Moon's orbit is a circle; The Moon's orbit 
lies in the plane of the ecliptic; the Earth/Moon barycentre is at the centre 
of gravity of the Earth. 

  That picture I get..  You are making simple assumptions..  but why put the 
barycentre where you did..  ?  Is it going to figure in this? 

  Next, some approximate dimensions - 

  One day = 86400 s; one year = 365.25days; the Earth's orbit has a radius of 
150 * 10^6 km; the Moon's orbit has a radius of 384 * 10^3 km; the Earth 
revolves at 360/365.25 = 0.986 deg/day; the Moon revolves at 360/27.322 = 13.2 
deg/day; the Earth has an orbital velocity of 30 km/s; the Moon has an orbital 
velocity of 1.0 km/s. 

  A body in orbit about the Sun at a distance 150 * 10^6 km, which does not 
rotate on its axis (inclined at 0 deg wrt the plane of its orbit) experiences a 
'day' of one year duration with the Sun rising over the western horizon and 
setting over the eastern horizon. The shadow of a sundial will record a 
traverse of 0.986 deg/day from East to West.  Not having experienced a sundial 
, I'll take your word for that..  You mean that the sun will traverse the sky 
.A simpler transposition..  

  Should the body of the previous paragraph rotate in a prograde direction on 
its axis once per year, then a sundial shadow will neither advance nor retreat 
from a fixed position.

  You mean the sun will not traverse the sky, but remain fixed in position..  
By Joshua thats good! 

  Again the same body, but rotating in the same direction at a rate of twice 
per year -- the shadow will now advance from West to East at a rate of 0.986 
deg/day.

  Now you baffled me. but then prograde might mean something I don't know..  We 
are still prograde, so if the rpm is doubled, There are two days in the time 
there was once only one day. the new day is half a year long. the angular 
motion of the sun traversing the sky will be faster..  I think

  Change the body's rotation rate to once per 27.322 days ie 13.2 deg/day and 
the shadow will move from East to West at a rate of 13.2 - 0.986 = 12.214 
deg/day. This second term -- 0.986 -- is the 360 deg of the orbit divided by 
the number of days to achieve one revolution. This body will have an orbital 
velocity of 30 km/s.

  Now I need a diagram.. If you make the body rotate once, thats one day.. How 
do you get 27 days out of it.. sorry but I could find no cohesion here..  my 
fault of course, but you are talking to my abilities I hope. That paragraph is 
a total loss to me, and the rest. What was your objective in presenting this 
exercise.. 27 . 322 days???

  A simple statement such as at the end of this experiment you will understand  
 bla bla..  at the beginning would perhaps help..  But 2 paragraphs down I get 
your plot...  so down to the meat..  

  If we increase the orbital velocity to 31 km/s, then the time for one orbit 
is reduced to 365.25 * 30/31 = 353.468 days and the orbital radial velocity 
increases to 1.018 deg/day. If the body is still rotating at once per 27.322 
days, then the shadow will advance at a rate of 13.2 - 1.018 = 12.182 deg/day. 
At 29 km/s orbital velocity, the corresponding figures are 365.25 * 30/29 = 
377.845 days orbital period and 0.953 deg/day revolution. Shadow advance then, 
will be 13.2 - 0.953 = 12.247 deg/day. The difference is therefore 12.247 - 
12.182 = 0.065 deg/day -- an eminently measurable difference.

  Totally confusing but I never did like maths. Algebra ok  May be algebra 
would have helped. 

  Now the purists may cry foul because we all know that if an orbiting body is 
travelling faster, then its orbit radius will be less, and conversely, if 
slower then the radius will be greater. However, if the body we have been 
discussing is the Moon and it is orbiting the Earth at a distance of 384 * 10^3 
km, and the Earth/Moon barycentre is orbiting the Sun, then at full moon it 
will be travelling at Earth orbital velocity plus Moon orbital velocity, while 
at new moon it will be travelling at Earth orbital velocity minus Moon orbital 
velocity. Its average velocity would therefore be 30 km/s. Under these 
conditions we should see the rate of advance of the shadow at new moon to be 
0.065 deg/day faster than at full moon.

  If I knew where the full moon and the New moon were at , and I never did,  
except it was always never in the sky when I looked for it.. 

  Hey you bible believin geocentrists,  God got that wrong didn't he , if the 
Bible is literal.?.  This lesser light  He tells us is for our illumination, 
"to illumine the night," is never there most nights.  Doesn't that language 
sound more poetic than literally speaking .fact? . Sorry Paul, a moment needed 
to help poor Jack..  Now Jack will say I'm being off side, and nasty to God 
with this sort of talk..  

  Not so!  I'm saying that because the Bible says Jesus set up a church to 
guide us in our understanding of Scripture..  

  Now back to Paul. 

  Should the Universe however be centred on the Earth with the Moon and the Sun 
orbiting it, the Moon's radial velocity would be constant and so the rate at 
which the shadow advances would be constant.

  I think you ignored or missed something. You insisted above for the exercise 
a moon orbit of a circle. The relative velocities to each other in either 
scenario are the same. We already tried this a year ago to prove our case and 
flunked..  I think that was what caused Allen to go insane looking for a 
vibrating gravity.  The moons radial velocity relative to the earth is constant 
in the helio system..  This is where Allen added to his insanity by bringing in 
frames of reference. Perhaps you and Allen are soul mates!  

  A real test would of course need to take into account the fact that both 
orbits in question are in fact ellipses, that they are not in the same plane 
and that the orientation of the Luna orbit major axis remains fixed while its 
angle to the Sun is constantly varying. I'm sure there will be other things 
which have not occurred to me but I am also sure that -- if I have the 
mechanics properly sorted -- this effect is eminently measurable. That we 
cannot immediately rush out and do the experiment matters not -- science is in 
the habit of making predictions which routinely take decades or centuries to 
demonstrate or destroy. 

  I'm lucky I will not need to wait that long.. I'm assuming of course that 
even the HELL bound will have access to all truth, in that universal day of 
truth when all sins are revealed to all. on Judgement Day.  Surely He will not 
be so cruel as to not let us know where or why we erred ( in science) and if 
there is life on Mars or not. 

  Finally, the Earth/Moon barycentre is not at the Earth's centre of gravity -- 
there is an offset of about 4700 km. While the effect is greatly reduced, it 
seems to me that the rate of shadow advance could, under these conditions, also 
be shown to differ slightly between new moon and full moon on the Earth.

  I know .. Thats why I asked the question first up at the top!  In any 
scenario you presented, I know, not assume, I know, that the shadow timing is 
the exact same time as observed and measured. It will be correct in either 
calculating system, geocentric or heliocentric. Why? because it is observed and 
timed. Bernies funny figures not withstanding..  You cannot leave out the 
unknown quantities of inertia and aether....., and thus till they are 
understood and figured in, no real explanation is possible..  except on 
Judgement day, if any soul, good or bad would be interested any more..  I hope 
I am. I do not want to be something I am not..  said the grub to the 
butterfly....  I mean it!!!!

  Philip and   Paul D



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