[geocentrism] Re: (no subject)
- From: "allendaves@xxxxxxxxxxxxxx" <allendaves@xxxxxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Sun, 23 Dec 2007 10:21:49 -0800 (PST)
Regner,
Blue......
----- Original Message ----
From: Regner Trampedach <art@xxxxxxxxxx>
To: geocentrism@xxxxxxxxxxxxx
Sent: Sunday, December 23, 2007 7:50:22 AM
Subject: (no subject)
Allen,
You have asked for vectors - I'll let the Christmas spirit prevail
and give you some vectors:
I simplify a bit by describing The Earth's orbit around the Sun as a circle
- not an ellipse. That is an approximation, but a good one for this purpose
- it will have no effect on the argument at hand.
The position vector of the centre of the Earth is then given by:
/ xc \ / COS(t/year*2*pi) \
rc = | yc | = | SIN(t/year*2*pi) | (1)
\ zc / \ 0 /
where we have put then Sun at (0,0,0), and the ecliptic plane (plane of the
orbit) is the x-/y-plane and the z-axis is perpendicular to that plane and goes
through the centre of the Sun.
The first equality can also be written rc=(xc,yc,zc) , but the column
vector notation is more compact.
t is time, measured in the same unit as year (e.g., seconds, hours, days)
and the factor of 2*pi is there because the sine and the cosine go through one
full oscillation when the argument changes by 2*pi - Equation (1) therefore
goes through one cycle (orbit) in one year.
Equation (1), above, describes the translational orbit of the Earth around
the Sun. It describes how the position of the centre of the Earth changes
with time. Only position! - it says nothing about orientation.
The position vector of a particular point on the surface of Earth, at
the equator, is given by:
/ xe \ / COS(t/year*2*pi) \ / COS(t/day*2*pi) \
re = | ye | = | SIN(t/year*2*pi) | + | COS(23.5°)*SIN(t/day*2*pi) | (2)
\ ze / \ 0 / \ SIN(23.5°)*SIN(t/day*2*pi) /
where the 23.5° is the tilt of the equatorial plane compared to the ecliptic.
The consequence of that tilt is that the point on the equator will go above
and then below the ecliptic during one day, as seen in the z-component of
the second term. The day is here the sidereal day. The orbital rotaion does not
take place in sidereal time.
The orientation of the Earth is described by the vector going from the
centre of the Earth, to our chosen point on the equator, i.e., (2) - (1)
[equation (2) minus equation (1)]:
/ xo \ / COS(t/day*2*pi) \
ro = | yo | = | COS(23.5°)*SIN(t/day*2*pi) | (3)
\ zo / \ SIN(23.5°)*SIN(t/day*2*pi) /
This describes the orientation of Earth with respect to the stars. In sidereal
time
It changes in a cyclic manner, repeating itself every sidereal day.
There is no reference to anything happening on a yearly cycle.
Yes, in sidereal day/ time....which is not under consideration.
If you take a strobe light at a a rotating blade the period of the strobe will
determine whether or not you see the blade to rotate clockwise or
counterclockwise.
It should be obvious, that both the rotation described by Eq. (3) and
the translational orbital motion described by (1) are continuous motions
that happen all the time - e.g., the Earth does not jump forward every
time a sidereal day has been completed, instead the Earth moves ahead in
its orbit all the while it also spins on it's celestial axis. As long as we are
looking at sidereal time the two sets of vectors are balenced/ equivocated wrt
nightly and annualy motions. However, they are not in solar time. These
equations describe the translational motions of the earth in orbit in sidreal
time but it only describes a fixed point on the earthʼs orientation in sidereal
time. In sidereal time it does not jump forward but in solar time it would.
Using a point at the center of the earth only makes my case. A fixed point on
the earth progressively and radial moves around that point at the center while
that point progressively and radial moves around the sun. It is a compound
motion not a simple single plane single axis rotation . The period in question
is solar or 24 hour intervals, these equations cannot show the orbital rotation
because there is none in the sidereal "strobe"/ interval. This would be true
regaudless of distance a few
feet or "infinity". The rotation is and can only be found in the solar
"strobe" / interval/ 24 hours.....The annual cycle is going to look the same as
the nightly as long as you "strobe" the sidereal period. ( nightly). When you
"strobe" a different interval the other and different effect will be seen if
the cause of the effect realy exist. If the effect exist then the cause exist.
However, no effect then no cause.
I am going to have to settle this with somthing more then just logic and
mathematics. Formal photo graphic experimental proof of what would/ should be
seen is needed . This is what I will attempt to get accomplished. When I
complete that task, then I will give you the correct mathematical equations to
describe the then "proven" effect. Feel free and I suggest you pursue other
discussion interest on the forum in the meantime. I will defer any other time
spent on this particular discusion "how to observe the/ any orbital rotation of
the earth" to the other issuse untill I can get this done.
Have a mery Christmas and happy new year to everyone! :-)
Regner
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