woodfoe
New member
There’s been a lot of information and disinformation about the the celestial body know as Nibiru, the destroyer, wormwood, planetX, or a number of other names that it goes by. Of course ancient carvings on rock walls and the like only goes to add more evidence that this celestial body did visit us in the past. I’m not going to talk about it’s history you can Google that if you want. I’m going to present what I think is proof positive that Nibiru is out there and is right on schedule to swing around the sun around the 21st of December 2012. I first want to present a picture that was taken in 2007 by someone using the South Pole telescope, who posted it on Youtube which was soon taken down.
This is the picture that allegedly was taken. Nibiru with the coordinates : 5h 53m 27s-6 10′ 58. Now a very interesting thing happens if you plug these coordinates into Google sky. You will find when you hit search that you will come to a dark area. When you get there and slowly zoom out you will see that the area has been blocked out in the shape of a rectangle. This is the only place in Google sky where it was done. This begs the question why would they black out this picture? As with most cases there is always a work around and someone later found Nibiru on the world wide telescope using these coordinates: 06 08 55 22 27′ 24. The picture was exactly the same, a red object with the moons in the same place. After finding it on the world wide telescope people then used the same coordinate to find it on Google sky and although it had moved a bit they found it there at these coordinates: 06 08 55 22 47 14. You can check out these coordinates for yourself if you have Google Earth or The World Wide Telescope. Free downloads. From where it was first found in the above picture in Orion’s belt it has moved all the way over to Gemini.
After finding Nibiru I did some calculations to try and determine it’s distance and size and here is how I went about it.
Fig.1 above is the geometric schematic showing the necessary parameters to solve for d (distance ). The simple equation is :
d=R/tan Ø
In order to solve d, R ( the diameter of the object) must be known. There are ways to figure this out using triangulation where you get a measurement from one point on the earth and another from as far a distance from the first as possible. Then by using trigonometry you can calculate the size of object by knowing it’s distance. What I did in order to calculate the size of Nibiru was to take a measurement using the ruler on Google sky and measure the distance in km of Nibiru at a certain Ra and Dec and then at approximately the same Ra and Dec measure nearby Venus. Since the size of Venus is already known I simply took an approximation of the size of Nibiru. The size of Venus at about 8h and 25m was .25 km. While Nibiru was 1.01km, thus making Nibiru four times as large as Venus which is about the same size of the earth, making Nibiru about four times the size of earth or the size of Neptune. This diameter measures approx. 48,416 km or 4x the diameter of Venus which is 12,104 km. In order to solve for Ø I used the finder in world wide telescope to get the distance in arcsecs by traversing the diameter of the object then taking the difference in the readings. I converted this to degrees.
Now that I have R to work with I can proceed to solve for d(distance).
d=48,416 km/tanØ
Ø=.02 degrees
d=48,416/0.000349066
d=138,701,564.0 km=.927 au
As far as calculating the speed of Nibiru I’ll save that for later if it’s possible. I heard that it was traveling at 35,000 mph if true that would be about right for a Dec 21 encounter.
138,701,564 km = 86185156.2 miles/35,000 mph= 102.6 days
That means it would be here by Nov. 15 2012 again not sure of the speed but that is close enough.
I haven’t been able to actually come up with an accurate idea of where Nibiru is in relation to the solar system and other planet only that is indeed close. If anyone cares to add to this post I’d appreciate some more information concerning the where about of Nibiru.
This is the picture that allegedly was taken. Nibiru with the coordinates : 5h 53m 27s-6 10′ 58. Now a very interesting thing happens if you plug these coordinates into Google sky. You will find when you hit search that you will come to a dark area. When you get there and slowly zoom out you will see that the area has been blocked out in the shape of a rectangle. This is the only place in Google sky where it was done. This begs the question why would they black out this picture? As with most cases there is always a work around and someone later found Nibiru on the world wide telescope using these coordinates: 06 08 55 22 27′ 24. The picture was exactly the same, a red object with the moons in the same place. After finding it on the world wide telescope people then used the same coordinate to find it on Google sky and although it had moved a bit they found it there at these coordinates: 06 08 55 22 47 14. You can check out these coordinates for yourself if you have Google Earth or The World Wide Telescope. Free downloads. From where it was first found in the above picture in Orion’s belt it has moved all the way over to Gemini.
After finding Nibiru I did some calculations to try and determine it’s distance and size and here is how I went about it.
Fig.1 above is the geometric schematic showing the necessary parameters to solve for d (distance ). The simple equation is :
d=R/tan Ø
In order to solve d, R ( the diameter of the object) must be known. There are ways to figure this out using triangulation where you get a measurement from one point on the earth and another from as far a distance from the first as possible. Then by using trigonometry you can calculate the size of object by knowing it’s distance. What I did in order to calculate the size of Nibiru was to take a measurement using the ruler on Google sky and measure the distance in km of Nibiru at a certain Ra and Dec and then at approximately the same Ra and Dec measure nearby Venus. Since the size of Venus is already known I simply took an approximation of the size of Nibiru. The size of Venus at about 8h and 25m was .25 km. While Nibiru was 1.01km, thus making Nibiru four times as large as Venus which is about the same size of the earth, making Nibiru about four times the size of earth or the size of Neptune. This diameter measures approx. 48,416 km or 4x the diameter of Venus which is 12,104 km. In order to solve for Ø I used the finder in world wide telescope to get the distance in arcsecs by traversing the diameter of the object then taking the difference in the readings. I converted this to degrees.
Now that I have R to work with I can proceed to solve for d(distance).
d=48,416 km/tanØ
Ø=.02 degrees
d=48,416/0.000349066
d=138,701,564.0 km=.927 au
As far as calculating the speed of Nibiru I’ll save that for later if it’s possible. I heard that it was traveling at 35,000 mph if true that would be about right for a Dec 21 encounter.
138,701,564 km = 86185156.2 miles/35,000 mph= 102.6 days
That means it would be here by Nov. 15 2012 again not sure of the speed but that is close enough.
I haven’t been able to actually come up with an accurate idea of where Nibiru is in relation to the solar system and other planet only that is indeed close. If anyone cares to add to this post I’d appreciate some more information concerning the where about of Nibiru.
