Is something out there around the Sun??

[link to sohowww.nascom.nasa.gov]

look at C3 please!

lower right

look at it!

[link to sohowww.nascom.nasa.gov]

look at C3 please!

lower right

look at it!

If anyone posted this, then i am in error. But it seems some years ago I had read something about a certain planet appearing as a red cross.

Not buying into this you-know-what stuff, but thought this was quite coincidental after seeing that frame with the red cross.

[link to www.zetatalk.com]

[link to sohowww.nascom.nasa.gov]

look at C3 please!

lower right

look at it!

Are you talking about the two brightest objects? If so the one on the right is Jupiter, the one on the left is Mercury. This further proves that the Mystery Object is not Mercury. In the Stereo mpeg, the object is close enough to the Sun that the CME tentacle appears to reach out and touch it. In the C3 image Mercury is gone and out of site, no where near the Sun or the tentacle.

C3 512x512 (3.0 MB)
[link to sohowww.nascom.nasa.gov]

C3 256x256 (475K)
[link to sohowww.nascom.nasa.gov]

[link to sohowww.nascom.nasa.gov]

look at C3 please!

lower right

look at it!

Are you talking about the two brightest objects? If so the one on the right is Jupiter, the one on the left is Mercury. This further proves that the Mystery Object is not Mercury. In the Stereo mpeg, the object is close enough to the Sun that the CME tentacle appears to reach out and touch it. In the C3 image Mercury is gone and out of site, no where near the Sun or the tentacle.

C3 512x512 (3.0 MB)
[link to sohowww.nascom.nasa.gov]

C3 256x256 (475K)
[link to sohowww.nascom.nasa.gov]



So you are saying that C3 is showing Mercury and Jupiter but Stereo is not show same. To be you can't have it both ways but I don't understand how this timeline on C3 can be the same that they talked about for the 27th of Dec.

[link to sohowww.nascom.nasa.gov]

look at C3 please!

lower right

look at it!

Are you talking about the two brightest objects? If so the one on the right is Jupiter, the one on the left is Mercury. This further proves that the Mystery Object is not Mercury. In the Stereo mpeg, the object is close enough to the Sun that the CME tentacle appears to reach out and touch it. In the C3 image Mercury is gone and out of site, no where near the Sun or the tentacle.

C3 512x512 (3.0 MB)
[link to sohowww.nascom.nasa.gov]

C3 256x256 (475K)
[link to sohowww.nascom.nasa.gov]



So you are saying that C3 is showing Mercury and Jupiter but Stereo is not show same. To be you can't have it both ways but I don't understand how this timeline on C3 can be the same that they talked about for the 27th of Dec.

What I am saying is this. Here is the original link about the object. When viewing the object it is in the lower left quadrant moving right to left.

[link to stereo.gsfc.nasa.gov]

This particular view from Stereo is zoomed in closer to the Sun. Note the size of the circle blocking out the Sun. This view is very similar to the Lasco C2 view which is also zoomed in tighter on the Sun (note the circle size). The C3 image is a pulled back image that sees more of the surrounding space. This is why you can view the current C2 movie and you do not see Mercury or Mars but you can on the C3.

My point is that the object in the Stereo movie is MUCH closer to the Sun than Mercury is. The C3 movie proves that so therefore the object in the Stereo movie cannot be Mercury. Mercury is too far away.

[link to sohowww.nascom.nasa.gov]

look at C3 please!

lower right

look at it!

this has been put to bed in major thread

we got some very informed info and thats all I ever asked for

lets all move forward

god bless

ice

Ice, I don't know if you'll be able to follow, but it is possible to calculate the position of Mercury in the FOV of the STEREO-A Cor2 instrument. It isn't necessary if you think about what Asinine posted on pgs 9 & 12 of the original thread.

A quick and dirty calculation of Mercury's position in the Cor2 instrument:

The first image in the .mpg referenced in the OP of the original thread was dated 31 Dec. 2007 at 00:07:53 UT. We'll call the time 00:08 UT.

On 31 Dec. 2007 00:08 UT the J2000 heliocentric positions of STEREO_A and Mercury were:

STEREO-A
Longitude: 119.963°
Latitude: -0.125°
r: 0.967378 AU

Mercury
Longitude: 304.615°
Latitude: -6.807°
r: 0.430883 AU



Let's rotate the system around the Z axis to get STEREO-A longitude at 180°, aligned with the X axis, by adding 60.037° to both longitudes. Result:

STEREO-A
Longitude: 180.000°
Latitude: -0.125°
r: 0.967378 AU

Mercury
Longitude: 364.652° = 4.652°
Latitude: -6.807°
r: 0.430883 AU
Note: These are the exact same relative positions. Nothing has changed except the reference line.



Now let's rotate the system around the Y axis to get STEREO-A at 0.000° latitude by adding 0.125° to STEREO-A's latitude and subtracting 0.125° from Mercury's latitude. Result:

STEREO-A
Longitude: 180.000°
Latitude: 0.000°
r: 0.967378 AU

Mercury
Longitude: 4.652°
Latitude: -6.932°
r: 0.430883 AU
Note: These are the exact same relative positions. Nothing has changed except the reference line.

STEREO-A is now exactly on the X axis, as is the Sun. Mercury is now the only object with Y & Z components. We use these coordinates to determine the rectangular coordinates of STEREO-A and Mercury.

The equations are:
X = r * Cos(Lon) * Cos(Lat)
Y = r * Sin(Lon) * Cos(Lat)
Z = r * Sin(Lat)

X & Y are your familiar x,y that you plotted on a sheet of paper in secondary school. Z is how far above or below the sheet of paper the point is. Remember? Get it?

Calculate the rectangular coordinates:

STEREO-A
X = -0.967378 AU
Y = 0.000000 AU
Z = 0.000000 AU
(see why we rotated the system?)

Mercury
X = 0.426324 AU
Y = 0.034691 AU
Z = -0.052004 AU

We're just about done. We now have STEREO-A staring dead centre at the Sun from 0.967378 AU, Mercury is 0.426324 AU beyond the Sun. Therefore Mercury is 1.393702 AU distant from STEREO_A along the X axis (the STEREO-A--->Sun line). Mercury is also 0.034691 AU to the left of the Sun's centre (+Y) and 0.052004 AU below the Sun's centre (-Z).

Therefore, from STEREO-A Mercury should appear:

Y° = arcTan(0.034691/1.393702) = 1.426° to the left of the Sun's centre
and
Z° = arcTan(-0.052004/1.393702) = 2.137° below the Sun's centre.

STEREO-A's Cor2 instrument has a FOV of 15 Sun radii in all directions in the Y-Z plane from the centre of the Sun. This = 4.00° Mercury is only 2.57° from the centre of the Sun by our calculation. 2.57° = sqrt(1.426^2 + 2.137^2).

You can even plot it on the STEREO-A Cor2 image to see how close we came using non-precise values.

See this image I did of the above example:
[link to www.imagehosting.com]

The yellow lines are the Y° and Z° from above. Where they cross is where we calculated Mercury to be in the FOV. Not too bad, eh? We used 3 decimal point coordinates and didn't take into account any spacecraft attitude (roll), etc. The time to type this post has taken about ten times what it took to do the calculations on a calculator. Making the image took a few minutes in Paint, but once you see the rotated frame of reference all doubts are gone.

The point is there are a few people here who can do this stuff but why should we bother? It was obviously Mercury for the exact reasons Asinine gave in the original thread. The only reason I did it was because you were actually reasonable in your postings, although somewhat woo, and I could tell you were still confused. Plus I had some time on my hands. Asking questions and asking for help from those who are more familiar with a subject is quite rare at GLP. Making bold and absolute statements from laughable ignorance of the subject in question is de rigueur.

Have fun.


R

Ice, I don't know if you'll be able to follow, but it is possible to calculate the position of Mercury in the FOV of the STEREO-A Cor2 instrument. It isn't necessary if you think about what Asinine posted on pgs 9 & 12 of the original thread.

A quick and dirty calculation of Mercury's position in the Cor2 instrument:

The first image in the .mpg referenced in the OP of the original thread was dated 31 Dec. 2007 at 00:07:53 UT. We'll call the time 00:08 UT.

On 31 Dec. 2007 00:08 UT the J2000 heliocentric positions of STEREO_A and Mercury were:

STEREO-A
Longitude: 119.963°
Latitude: -0.125°
r: 0.967378 AU

Mercury
Longitude: 304.615°
Latitude: -6.807°
r: 0.430883 AU



Let's rotate the system around the Z axis to get STEREO-A longitude at 180°, aligned with the X axis, by adding 60.037° to both longitudes. Result:

STEREO-A
Longitude: 180.000°
Latitude: -0.125°
r: 0.967378 AU

Mercury
Longitude: 364.652° = 4.652°
Latitude: -6.807°
r: 0.430883 AU
Note: These are the exact same relative positions. Nothing has changed except the reference line.



Now let's rotate the system around the Y axis to get STEREO-A at 0.000° latitude by adding 0.125° to STEREO-A's latitude and subtracting 0.125° from Mercury's latitude. Result:

STEREO-A
Longitude: 180.000°
Latitude: 0.000°
r: 0.967378 AU

Mercury
Longitude: 4.652°
Latitude: -6.932°
r: 0.430883 AU
Note: These are the exact same relative positions. Nothing has changed except the reference line.

STEREO-A is now exactly on the X axis, as is the Sun. Mercury is now the only object with Y & Z components. We use these coordinates to determine the rectangular coordinates of STEREO-A and Mercury.

The equations are:
X = r * Cos(Lon) * Cos(Lat)
Y = r * Sin(Lon) * Cos(Lat)
Z = r * Sin(Lat)

X & Y are your familiar x,y that you plotted on a sheet of paper in secondary school. Z is how far above or below the sheet of paper the point is. Remember? Get it?

Calculate the rectangular coordinates:

STEREO-A
X = -0.967378 AU
Y = 0.000000 AU
Z = 0.000000 AU
(see why we rotated the system?)

Mercury
X = 0.426324 AU
Y = 0.034691 AU
Z = -0.052004 AU

We're just about done. We now have STEREO-A staring dead centre at the Sun from 0.967378 AU, Mercury is 0.426324 AU beyond the Sun. Therefore Mercury is 1.393702 AU distant from STEREO_A along the X axis (the STEREO-A--->Sun line). Mercury is also 0.034691 AU to the left of the Sun's centre (+Y) and 0.052004 AU below the Sun's centre (-Z).

Therefore, from STEREO-A Mercury should appear:

Y° = arcTan(0.034691/1.393702) = 1.426° to the left of the Sun's centre
and
Z° = arcTan(-0.052004/1.393702) = 2.137° below the Sun's centre.

STEREO-A's Cor2 instrument has a FOV of 15 Sun radii in all directions in the Y-Z plane from the centre of the Sun. This = 4.00° Mercury is only 2.57° from the centre of the Sun by our calculation. 2.57° = sqrt(1.426^2 + 2.137^2).

You can even plot it on the STEREO-A Cor2 image to see how close we came using non-precise values.

See this image I did of the above example:
[link to www.imagehosting.com]

The yellow lines are the Y° and Z° from above. Where they cross is where we calculated Mercury to be in the FOV. Not too bad, eh? We used 3 decimal point coordinates and didn't take into account any spacecraft attitude (roll), etc. The time to type this post has taken about ten times what it took to do the calculations on a calculator. Making the image took a few minutes in Paint, but once you see the rotated frame of reference all doubts are gone.

The point is there are a few people here who can do this stuff but why should we bother? It was obviously Mercury for the exact reasons Asinine gave in the original thread. The only reason I did it was because you were actually reasonable in your postings, although somewhat woo, and I could tell you were still confused. Plus I had some time on my hands. Asking questions and asking for help from those who are more familiar with a subject is quite rare at GLP. Making bold and absolute statements from laughable ignorance of the subject in question is de rigueur.

Have fun.


R

Here is your requested image.

[link to img232.imageshack.us]