Why doesnt Nasa have any detailed pictures of the Moon anomally Shard?

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posted on Feb, 14 2013 @ 12:40 PM
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Originally posted by AthlonSavage
Anyone ever see that 70s space show space 1999, all the reruns episodes are on Youtube great show. Heres what the moon base could look like.






Have the entire series on DVD... along with U.F.O..

Love those shows.




posted on Feb, 14 2013 @ 02:26 PM
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The following is my opinion as a member participating in this discussion.


Originally posted by AthlonSavage
You did wonder why i debate with members and not moderators. Well now you know!

I do not.


As an ATS Staff Member, I will not moderate in threads such as this where I have participated as a member.



posted on Feb, 14 2013 @ 03:19 PM
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reply to post by ArMaP
 





I do not.


You know what they say...

Knowledge is Power..



posted on Feb, 14 2013 @ 03:23 PM
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reply to post by ArMaP
 

I'll explain.
By "debate" he means talking to people who disagree with him and present logical arguments to support their position.



posted on Feb, 14 2013 @ 03:35 PM
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reply to post by Phage
 


Hi Phage,
Could you check out this feature and estimate the height of the peak with the long shadow...
The thing to look for is this...


And you can find it here...
Image Data
Time (DOY:181) 2009-06-30T16:04:33 Orbit 73 Center Longitude -8.84825° Center Latitude 72.18645° Resolution 1.85 m/pixel

wms.lroc.asu.edu...

I've no idea of how to scale it up... it looks big but it could be small.
I think it was you that suggested, in regards to finding tall towers on the moon, looking for the shadows with the sun at a low angle, and this fits the bill. I think...



posted on Feb, 14 2013 @ 03:47 PM
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reply to post by manmental
 

Can't do it.
Without knowing the slopes on uneven terrain it's not possible. That "tower" is on the edge of a crater. Not a place one would expect to find level ground.
www.vgl.org...

But I don't know why every tall rock has to be a tower to so many "anomaly searchers". The shadows cast by objects do not have to be any indication of the true shape of that object. It all depends on how the light strikes the object and the area where the shadow falls.

edit on 2/14/2013 by Phage because: (no reason given)



posted on Feb, 14 2013 @ 03:53 PM
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reply to post by Phage
 


Thank you sir. I guess we want to find evidence so we see evidence in maybe the most mundane of things?
It's folks like you that keep me sane.
Cheers again.



posted on Feb, 14 2013 @ 05:40 PM
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reply to post by draknoir2
 





Have the entire series on DVD... along with U.F.O..

Love those shows.



Did they ever have an episode where they explore the moon, that could of made for some interesting epsiode Moon base alpha versus the Shard Ufo.

Moon base alpha is what i imagine a moon base would look like, Nasa should use its blue print for building one on the moon. The eagles as well while their at it.
edit on 14-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 14 2013 @ 08:46 PM
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reply to post by manmental
 





wms.lroc.asu.edu...
edit on 14-2-2013 by wolveriine because: (no reason given)



posted on Feb, 14 2013 @ 09:01 PM
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reply to post by wolveriine
 


That is certainly an interesting picture for a few reasons at least which i have noticed looking at the picture.

1. The direction of the shadow is consistent with the source of light where we can see looking into picture the light source is greatest at the left of picture.
2. There orange arrow i added in picture points to the tip of the shadow, so this thing is tall compared to a scale of the crater size its in.
3. The red circle and red arrow point to another tower structure shadow beneath the taller one.

There is definitely tall natural or artifical structures in this picture. In consideration of points above, arguments the Nasa hawks might make that light is in wrong direction for shadow or the tower is a photo defect wont wash the anomally out in this picture. Lets wait see what the Nasa hawks come up with next.





It would be nice to have a distance scale on picture. I looked on the Nasa LRO for details of this crater and the information on the Anaxagoras crater in the quote below comes from the LRO link

www.nasa.gov...





Anaxagoras is a 50 km diameter Copernican impact crater at 73.4°N, 10.1°W, with an extensive ray system (reaching over 900 km from the crater rim in some directions), and a central peak of pure anorthosite. The central peak and the material ejected and deposited onto the floor of Goldschmidt crater to the east, indicates that the Anaxagoras crater-forming impact excavated pure anorthosite. Consequently it is one of the NASA Program regions of interest targeted by LROC.


Therefore the picture of the tower is taken at the rim of a very large moon crater.

edit on 14-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 14 2013 @ 10:11 PM
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reply to post by AthlonSavage
 


Therefore the picture of the tower is taken at the rim of a very large moon crater.

What? You believe NASA? Of course you could have just looked at the image link posted above and seen that for yourself. But a 50km lunar crater is not really "very large" relatively speaking.


The direction of the shadow is consistent with the source of light where we can see looking into picture the light source is greatest at the left of picture.
Yes, there is no doubt the brightly lit object is creating the shadow. It is completely consistent with what is seen (unlike the "shard"). It is also clear that the elevation of the Sun is low, creating very long shadows, as described.

Last Tuesday, LRO's orbit was just above the lunar terminator (the day/night boundary), so huge shadows highlight topography and render many relatively normal areas of the Moon nearly unrecognizable.
Anaxagoras


It would be nice to have a distance scale on picture. I looked on the Nasa LRO for details of this crater and the information on the Anaxagoras crater in the quote below comes from the LRO link
Well the page you linked has a scale on it but you can get more accurate information here. The resolution is 1.85 meters/pixel so that makes the shadow about 650 meters long.
wms.lroc.asu.edu...

Now, if we knew the elevation of the Sun and if we knew how the surface was sloped, we could calculate the height of the "tower" (rock). But we don't have that information so we can't. But as an example. On a flat surface, with the Sun 5º above the horizon (at sunrise as described above), an object 56 meters tall would cast such a shadow.

A tower? Or a 180 foot high pile of rubble? catching the light of the rising sun.



Therefore the picture of the tower is taken at the rim of a very large moon crater.

Yes, once again as described here in case you ignored it the first time.
Anaxagoras

But of course, you ignore those who you disagree with so none of this matters. For you it is and always will be a tower.
www.abovetopsecret.com...
edit on 2/14/2013 by Phage because: (no reason given)



posted on Feb, 14 2013 @ 10:35 PM
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reply to post by AthlonSavage
 


Can you please tell me if the towers pointed out in the videos below is the same as that we were talking from the beginning in this thread ?





Edit :


edit on 14-2-2013 by wolveriine because: (no reason given)



posted on Feb, 14 2013 @ 10:49 PM
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reply to post by wolveriine
 

There are no towers in those videos and neither one is near the "shard"
www.abovetopsecret.com...

The second one? Are you kidding? That is the solar wind collector
www.lpi.usra.edu...
edit on 2/14/2013 by Phage because: (no reason given)



posted on Feb, 14 2013 @ 10:56 PM
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reply to post by Phage
 





Well the page you linked has a scale on it but you can get more accurate information here. The resolution is 1.85 meters/pixel so that makes the shadow about 650 meters long.




I understand the calculation its simple trigonometry. With an unknown variable of the sun angle we cant relay on trigonmetry to calculate length of shadow.

Do you have the LRO raw data that gives scale number of pixels on the X and Y axis of the zoomed out picture (that shows rim of crater). This is the accurate way to measure length as we have a known dimension pixels/per meter.

i admit i am being difficult but i do like working with accurate data, particulary where its being used for expolation and calculations.



edit on 14-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 14 2013 @ 11:01 PM
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reply to post by AthlonSavage
 




Do you have the LRO raw data that gives scale number of pixels on the X and Y axis of the zoomed out picture (that shows rim of crater). This is the accurate way to measure length as we have a known dimension pixels/per meter.

Did you read my post? At all?



posted on Feb, 14 2013 @ 11:13 PM
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reply to post by Phage
 


Yes i read your posts thats what gave me the idea for asking this next question.

In the picture linked i have marked X and Y lengths. How many pixels are along the X and how many pixels along the Y. This may be straight forward to answer to some, but its a relevant question I consider we should clear on when discussing scale of size of objects in the picture.

The scroll bar works in picture (need to move it to see the Y), first time i have seen this feature for pictured linked

Once we have accurate dimension scale in place for measuring the X and Y lengths of objects in the picture we are then in good position to debate the sun angle. I will admit the sun is definitely low and will be happy to work within a (5-15) degree margin.


edit on 14-2-2013 by AthlonSavage because: (no reason given)
edit on 14-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 14 2013 @ 11:34 PM
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reply to post by AthlonSavage
 




The resolution is 1.85 meters/pixel so that makes the shadow about 650 meters long.



posted on Feb, 14 2013 @ 11:43 PM
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The raw data for this picture resolution identifies a scale of resolution is 1.85 meters/pixel. Does anyone on Ats here know how from the picture to work out how many pixels run across the horizontal "X" length of the picture, and how many pixels along the "Y". Theres probably an easy way to do this but im no guru with picture files. The answer will also come in handy knowing the method for any other debates where its important t work out scale of length of objects in picture.




edit on 14-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 14 2013 @ 11:45 PM
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reply to post by AthlonSavage
 

1) Download the image.
2) Open it with your image viewing software
3) Move your mouse around and note the pixel locations

It's about 350 pixels long.
edit on 2/14/2013 by Phage because: (no reason given)



posted on Feb, 15 2013 @ 12:45 AM
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reply to post by Phage
 



My calculations working with the raw picture data of Width 5064 pixels
Height 5120 pixels and knowing 1.85 metres/pixel place the the lengh of the tower shadow 375 meters.

This is now where the interesting part comes in. Here i list the length of the tower feature in picture and calculate using the trigometric relationship of Tan angle with the opposite (height of tower object ) and adjacent (length of shadow), to work out the height of tower. The true height of tower is course based on the true elevation angle of the sun.

at 5 degrees angle of sun - 32 metres height
at 10 degrees angle of sun - 66 meters
at 15 degrees angle of sun - 100 meters
at 20 degrees angle of sun - 136 meters
at 22.5 degrees angle of sun - 155 meters


I stopped at 22.5 degrees elevation of sun relative to the region area of interest around anomally. Why because there is a very noticable crater below the object which has 1/4 of its inner illuminated. Therefore 90 degrees / 4 = 22.5 and i dont consider the sun is at higher elevation that this.

So heres where the debate resides now. What is the angle of the sun because if its 22.5 degrees thats quite a tall object. For example its height would be the same to that of the Twintowers of Deutsche Bank Headquarter in Frankfurt that rise to 155metres. I included a picture of towers for a visual reference.







 
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