## [geocentrism] Re: (no subject)

• From: Regner Trampedach <art@xxxxxxxxxxxxxx>
• To: geocentrism@xxxxxxxxxxxxx
• Date: Wed, 23 Jan 2008 13:13:34 +1100

 Allen,    I have redone my original post on this, since I discovered that the equations were completely garbled. I have now supplied the equations as in-lined gif-files. If there is any problems with that, please tell me. I have supplied answers to your questions/comments, Allen, in red.       - Regner ----------------------------------------------------------------------------------------- [Re-post from 23.12.2007] 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: 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. a is 1 astronomical unit (AU), the average radius of the Earth's orbit around 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 or 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 (1.1) therefore goes through one cycle (orbit) in one year.   Equation (1.1), 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: 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. That doesn't matter. The time variable, t, is continous - it can take on any value, including fractions of a second. The period of the first term, the orbit, is the year = 3.1557e7 s, while the period of the second term, the rotation, is the sidereal day, which is 86164 s.   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., (1.2) - (1.1) [equation (1.2) minus equation (1.1)]: 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. See my comment above, about time being continuous - no strobe-lights involved. 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. I don't know what you mean by this. However, they are not in solar time. These equations describe the translational motions of the earth in orbit in sidreal time. They describe the various parts of the Earth's motion at any time. The translational motion of the Earth's orbit is described by Eq. (1.1). but it only describes a fixed point on the earth’s orientation in sidereal time. That Eq. (1.2) describes a fixed point on the earth’s surface was my intention - not a limitation. In sidereal time it does not jump forward but in solar time it would. Time is continuous - no jumping anywhere. 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. Okay - a line connecting the Sun and the centre of the Earth does indeed rotate. The direction of the line changes with a period of one year, and that rotation is obviously in the ecliptic plane. That is all described by Eq. (1.1) But that does NOT have ANY bearing on the orientation of the Earth. The orientation of the Earth is measured with respect to the (far away) stars. The orientation of the Earth changes with a period of 1 sidereal day and the rotation is around the equatorial axis, as described by Eq. (1.3). It is a compound motion not a simple single plane single axis rotation . And what I have described with Eqs. (1.1-1.3), is exactly what you ask for. So now you actually understand composite motions - very good. The period in question is solar or 24 hour intervals, The period of the Earth's rotation is the sidereal day. We are interested in the absolute orientation  in space, i.e., with respect to the (far away) stars. these equations cannot show the orbital rotation ...because there is none - The orbit is a translation. Eq. (1.3) shows that the only thing that changes the orientation of the Earth, is the daily 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..... Eq. (1.3) shows what goes on at all times - a rotation in the equatorial plane. You can choose your "strobe" with any interval - till it looks like it is rotating backwards, if you will, but it doesn't change the underlying motion. The annual cycle is going to look the same as the nightly as long as you "strobe" the sidereal period. ( nightly). That is because the only rotation is the daily rotation. Your annual effect arise from seeing the sidereal daily rotation with a tropical daily strobe-light. 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. Sounds okay to me. 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. It would be great if you could formulate your position for all of us, so we can see exactly what the disagreement is about - as for me, I'm not quite sure anymore.   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. That's all for now.           Regner allendaves@xxxxxxxxxxxxxx wrote: Regner, Blue...... ----- Original Message ---- From: Regner Trampedach 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 earth’s 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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