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

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posted on Feb, 15 2013 @ 08:21 PM
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reply to post by ArMaP
 

Fantastic.
Thanks. I wonder why it's not in the IMG header. Or is it?
edit on 2/15/2013 by Phage because: (no reason given)




posted on Feb, 15 2013 @ 08:27 PM
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reply to post by Phage
 




No. The lighting is not speculative. The parameters are provided with the image. Perhaps you can tell us which of those parameters indicates the lighting was from the left.



Again let readers to decide who arguments yours or mine are correct on the direction of lighting for the shard shadow. The pages of this thread which concentrate on discussion of the shadow are in 29, 25, 20, 21, 22.

I would like to hear readers opinions.



The location is somewhat speculative but the general area has been known by using identifiable landmarks. Saint Exupery did an excellent job of refining the location. Have you reviewed his work? He seems to have a deep familiarity with the terrain.



Yes Saint Exupery has provided some interesting input which i have looked at once, and want to look back through again as i agree the location is very speculative. Im still no sold on location but havent ruled out hes correct, may be he is.


edit on 15-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 15 2013 @ 09:56 PM
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reply to post by ArMaP
 






Red circle slope is rising out the crater. Therefore shadows are short as like your picture immediately below.



Green circle is at rim or crater so i will make a reasonable assumption its flat. So thats as per next picture.



Blue circle the anomally is in the middle of a crater. The crater starts at rim to its left and does an obvious downward slope because its filled with shadow. The downward slope of crater as per this picture.



The left of the Pink circle shows obviously the slope rising again as it goes from dark to light again. Therefore this indicates the object is in the central spot of crater, where its neither on a downward or upward slope. Therefore it resides in pic below.



73 metres high by 53 metres wide is a big pile of rocks, but my calculations above tell me that's a 4 or 5 metres high by 53 wide pile of rocks, and that's not unusual.


Therefore at an angle of 11 degrees ( and this is still a subjective value) the object which from the photo evidence appears to be not on sloping part of the ground is at least 73 metres tall. And i say at least because the bottom on crater it is sitting in is lower than the rim outer crater and therefore some of its real height may be masked by depth of crater.

You see i have answered you three times at least in this thread so no need no more complaints about i dont read your posts.
edit on 15-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 15 2013 @ 11:06 PM
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reply to post by AthlonSavage
 


Therefore at an angle of 11 degrees ( and this is still a subjective value) the object which from the photo evidence appears to be not on sloping part of the ground is at least 73 metres tall.

No need to be subjective about the Sun's elevation. We can be quite objective since Armap found the actual figure, 0.48º. Sounds like the Sun is just coming above the horizon. Sunrise. That 11º figure has nothing to do with the elevation of the Sun at the crater.

But you can be as subjective as you wish about the slopes and make as many assumptions as you like. It doesn't make them correct.



edit on 2/15/2013 by Phage because: (no reason given)



posted on Feb, 15 2013 @ 11:19 PM
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I thought you guys might find this guide highly relevant and useful. It shows how to accurately calculate the actual height of objects found in LROC Narrow Angle Camera images using a practical example.

The guide goes into how the surface slope can affects the object's apparent shadow length, which is a well known fact. Perhaps the most useful part of the guide for the interested reader may be the demonstration of quick and accurate method of finding the surface slope value for nearly any area of the Moon.

Using this guide, I don't see why any motivated reader wouldn't be able to accurately calculate the actual height or depth of any shadow casting object found in LROC NAC images.


How to Calculate Height of Moon Objects Like A Pro

edit on 15-2-2013 by PINGi14 because: (no reason given)



posted on Feb, 16 2013 @ 06:30 AM
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Originally posted by Phage
reply to post by ArMaP
 

Fantastic.
Thanks. I wonder why it's not in the IMG header. Or is it?

It's not, the header has just a small subset of the 83 parameters included in the index file.



posted on Feb, 16 2013 @ 06:58 AM
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The following is my opinion as a member participating in this discussion.


Originally posted by AthlonSavage
Therefore at an angle of 11 degrees ( and this is still a subjective value) the object which from the photo evidence appears to be not on sloping part of the ground is at least 73 metres tall. And i say at least because the bottom on crater it is sitting in is lower than the rim outer crater and therefore some of its real height may be masked by depth of crater.

But the angle is not 11 degrees, it's 0.48 (where did you get that 11º value?). Now that I know the "product name" (M101017141LE) I could find the photo on the LROC image search site, it's this one.

You can see that the incidence angle is 89.52º, which means that the Sun was 0.48º above the horizon.


You see i have answered you three times at least in this thread so no need no more complaints about i dont read your posts.

I haven't complained for some time and I agree with what you said about the shadows.


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



posted on Feb, 16 2013 @ 07:47 AM
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The following is my opinion as a member participating in this discussion.

reply to post by PINGi14
 


Thanks for that.


Armed with that new knowledge I went looking for the "tower", but it's hard to get it on the elevation chart. I tried it several times but the results were not consistent, it's probably too small for the resolution of the height data (I think it's usually worse than the photos' resolution).



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



posted on Feb, 16 2013 @ 09:09 AM
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Originally posted by ArMaP
The following is my opinion as a member participating in this discussion.

reply to post by PINGi14
 


Thanks for that.


Armed with that new knowledge I went looking for the "tower", but it's hard to get it on the elevation chart. I tried it several times but the results were not consistent, it's probably too small for the resolution of the height data (I think it's usually worse than the photos' resolution).



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



What you want to do when calculating the relevant slope data for the shadow in question is to use the Path tool to place the line starting from the base of the object, along the direction of the original shadow, ending at some point beyond the original shadow. This way, the elevation plot will most accurately reflect the slope of the surface where the original shadow was being cast upon.


So for the object in the guide, start by locating the object on Quickmap. The coordinate of the object is Latitude: 57.534 degrees and Longitude: -112.752 degrees. Use the real time coordinate of your mouse pointer located at bottom right of the Quickmap screen to locate the exact position of the object on Quickmap.

Since we know the original shadow was pointing towards the relatively large crater nearby (to the north-west), Place your line starting from base of the object and ending up somewhere near rim of the crater. To do this, click the Path tool at the top of the screen, then move your mouse pointer over the object and single-click. Now your line has started. Move the mouse pointer north-west towards the crater and you will see a line extending from the base of the object. With your mouse pointer somewhere near the rim of the crater, double click. Now you will get a pop up asking if you want to get the slope along the line you placed. Click Submit button to get the elevation plot for your line.


I wish I could explain it clearer and I hope I didn't confuse the matter further.
edit on 16-2-2013 by PINGi14 because: (no reason given)



posted on Feb, 16 2013 @ 12:46 PM
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The following is my opinion as a member participating in this discussion.

reply to post by PINGi14
 


I understand how to do it, but I wasn't able to get the measurement of the slope for that small area near the "tower" on photo M101017141LE, not the one from your example.

I will try again.


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



posted on Feb, 16 2013 @ 11:30 PM
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reply to post by Phage
 


I have doing been some celestial thinking of how the moon is illuminated by the sun and what angle the sun will be seen in the sky if standing on the moon.

This is how i see it. The moon is sphere shape that completes one orbit of the earth every 28 days. Its not spinning like the earth so therefore for a person seen standing on the centre Latitiude 0.00°, and Longitude 0.00° on the lunar map; will observe the rise and fall of the sun along the longitude line.




At this 0.00 coordinate an observer will should see the sun in the horzion is tilted on the horizion in Latitiude about 20 degrees°. This is due to the Moon orbiting closer with the eliptical equator and not the celestial equator.

The celestial equator differs by about 23 degrees from its eliptic plane as shown in the picture. The moon orbits facing directly the earth at an angle close to the eliptic plane and therefore I would expect the moons Longitude 0.00 coordinare on the LRO lunar map is aligned closely with the eliptic plane (or very close to within 5 degrees).





Another variable to consider is the Moon takes 28 days to go once around the earth. There are 360 degrees angle spread in a circle circumference. Therefore in 1 day (24 hour period) there is 360/28 = 12.85 degrees. A person standing at the lunar coordinate of longtitude and lattitude 00, 00 will therefore see the sun move in the horizion about 0.5 dgrees per hour. That can explain the variations in sun angle position as observed by Astronauts while walking on lunar surface. The link below takes you to a web bage which documents sun elevation angles observed on the lunar landing sites for each Apoolo mission.

www.hq.nasa.gov...

I will work with Latitiude 20.00° , and Longitude 0.00 as a reference position on the lunar surface or considering rise and fall of the sun along the longitude line as observed by an observed at this coordinate. The highest position/height of the sun in the lunar sky as observed from this cooridnate position will allow determination of the maximum sun elevation angle possible at Anaxagoras crater 73.4°N 10.1°W.

Since the moon is sphere that goes around the earth every 28 days, in a 360 degrees cycle. Then at one particular quarter of this cycle (i will arbitrarily set a reference for this as 0 degrees) a person standing at Latitiude 20.00° , and Longitude 0.00 on the lunar suface will see the sun directly over head. At +/-90 degree movements of the moon moving around earth from this reference; the lunar observer will see the sun falling behind the east or west horizions of moon.

If it had been a +/-45 degree moon/earth position around the reference then the observer standing at the lunar surface at Latitiude 20.00° and Longitude 0.00 will see the moon at 45 degrees over the west or eastern horizions.

For example on Apollo 15 26.13222° N latitude, 3.63386° E longitude (which are close to the coordinates i using as a reference) the sun elevation was recorded being 44.3 deg at completion of the astronauts lunar walk around.

The working out the "maximum possible" sun elevation possible at Anaxagoras crater 73.4°N 10.1°W is actually very simple. 90 - 73.4 = 14 degrees.

Well actually for the more accurate calculation >>

The sun in its highest altitude for year at Latitiude 20.00° , and Longitude 0.00 on the lunar suface, requires rescaling the lattitude coordinate line to compensate for the Latitiude 20.00°. Therefoe 90-20 = 70 degrees arc between Latitiude 20.00° and Longitude 0.00 position and where it goes into darkness at North pole Lat 90, Long 0 degrees.

Therefore 70/90 x 14 = 10.888 degrees or rounding off 11 degrees. My calculation lookings surprising familiar.




Since the photo of Anaxagoras crater rim was taken directly overhead of the crater rim I have ignored as negligible the 10 degrees West coordinate the sun illumination has on objects located around the rim of the crater.

Therefore depending on where the moon was in its orbit with relation to earth means that a sun elevation angle of 11 degrees at sunrise is possible for the 73.4°N 10.1°W location.

To put into a practical context what the above all means, is that when Nasa planned to send their satellite to take photos of Anaxagoras crater, they picked the right time of year where the sun has its maximum elevation around the Central latitude 0.00 and longitude 0.00 coordinates. And it makes sense for Nasa to do that for obtaining best sunlight angle on the crater which allows for the clearest photo picture of the crater.


edit on 17-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 16 2013 @ 11:58 PM
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reply to post by AthlonSavage
 

In the early 90's I was a modelmaker at a famous model makers in London.
We worked on tons of famous stuff...
I have had on set experience with pretty much every sort of camera and photographic system from Hasselblads for stills to motion control monsters on films.
I said all that because I just had a thought about a way, albeit a difficult way, to work out the height of the tower that we all seem to be focussing on now.

That is to build a scale model based on all the dimensions, visual references and depth info of the area and recreate the photograph' shadows based on the light source (sun).
I'm thinking of building in plaster/clay... but I guess a 3d studio / Maya digital model would suffice....
I know that is hard/ impossible on no budget... or rather, its not going to happen in the real world.

To me it, the new tower, looks like a tall and thin tower and not, as Phage compares to... a hill... whose base is as wide as the hill is tall. The base of this 'shard' looks like a much smaller % in width/length than the height of the anomaly. Is it not on Google Moon? I'm presuming not as that is the obvious look... I can't do Google moon where i am....

In the original video I posted about this the YT poster says he can see a cubic base to this 'structure' and I am inclined to see that cubic shape too... perhaps I am reading too much into it but it looks odd. And I like odd.
edit on 16-2-2013 by manmental because: hmmm



posted on Feb, 17 2013 @ 01:10 AM
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That is to build a scale model based on all the dimensions, visual references and depth info of the area and recreate the photograph' shadows based on the light source (sun).
I'm thinking of building in plaster/clay... but I guess a 3d studio / Maya digital model would suffice....
I know that is hard/ impossible on no budget... or rather, its not going to happen in the real world.


A model could be made. The most appropriate model to use software or a physical recreation would have to be decided on based on the physical parameters and variables the model needs to simulate. I agree with you it looks more like a tower than hill. It reminds me of microwave tranmission tower.



posted on Feb, 17 2013 @ 06:07 AM
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The following is my opinion as a member participating in this discussion.


Originally posted by AthlonSavage
The working out the "maximum possible" sun elevation possible at Anaxagoras crater 73.4°N 10.1°W is actually very simple. 90 - 73.4 = 14 degrees.

Well actually for the more accurate calculation >>

The sun in its highest altitude for year at Latitiude 20.00° , and Longitude 0.00 on the lunar suface, requires rescaling the lattitude coordinate line to compensate for the Latitiude 20.00°. Therefoe 90-20 = 70 degrees arc between Latitiude 20.00° and Longitude 0.00 position and where it goes into darkness at North pole Lat 90, Long 0 degrees.

Therefore 70/90 x 14 = 10.888 degrees or rounding off 11 degrees. My calculation lookings surprising familiar.

For "maximum possible sun elevation".


Therefore depending on where the moon was in its orbit with relation to earth means that a sun elevation angle of 11 degrees at sunrise is possible for the 73.4°N 10.1°W location.

Now it sounds like you are talking about a different thing.


What are you calling "sun elevation"?

If it's the position of the Sun over the horizon, then, by definition, at sunrise it's zero, the Sun is just appearing over the horizon.

But if you are talking about the apparent position of the Sun compared to the Moon's equator, then it's a different thing, not needed to make a calculation of the size of the "tower", as for that we just need the Sun's position above the horizon, regardless of being closer or farther away from the equator, as that only changes the direction of the shadow.

That would be needed for getting an idea of the direction of the shadow in the case of the photo of the "shard", but not to get and idea of it's size.

PS: I didn't completely understood your explanation (my brain is not fully working yet
), but I think that you got the angle of between the Moon's orbit and the ecliptic (not eliptic), according to Wikipedia that angle is 5.145°, so the Moon has a smaller variation of the Sun's apparent position above or below the equator than the Earth.



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



posted on Feb, 17 2013 @ 06:19 AM
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The following is my opinion as a member participating in this discussion.


Originally posted by manmental
That is to build a scale model based on all the dimensions, visual references and depth info of the area and recreate the photograph' shadows based on the light source (sun).

That would be a good idea, but I don't think we have all the data needed, as we don't have depth information with enough resolution to make a perfect reproduction of the area.

From my own calculations you can see that I got a height of a little over 4 metres, so I wouldn't even call it a hill.


I'm thinking of building in plaster/clay... but I guess a 3d studio / Maya digital model would suffice....
I know that is hard/ impossible on no budget... or rather, its not going to happen in the real world.

For those that have access to a 3D program (like the free Blender), it only takes time.




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



posted on Feb, 17 2013 @ 06:21 AM
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reply to post by ArMaP
 





Now it sounds like you are talking about a different thing.

What are you calling "sun elevation"?

If it's the position of the Sun over the horizon, then, by definition, at sunrise it's zero, the Sun is just appearing over the horizon.


The height of sun at as it apears sitting over the east or east horizon, will vary according to where moon is round is 360 Degree cycle 28 cycle around earth. At Longititude 00 and latitude 20, depending where the moon is in its cycle around earth the height of the sun will be different. The maximum height of sun possible is overhead a at Longititude 00 and latitude 20 .

This height forms the opposite of the triange, the latitude north is the adjacent length. At the coordinate position of the Anax crater 73 degrees Lat north and 10 Long i calculated using trignometry the angle of the sun seen by observer standing at Anax crater 73 degrees Lat north and 10 west will be 11 degrees. The observer will see the sun sitting over the south horizon by 11 degrees.

That is the maxium sunrise angle possible at the crater coordinate position 73 degrees Lat north and 10 Long, when the sun is directly overhhead as seen by an obsevrer standing in the Longititude 00 and latitude 20 spot.

I dont care about your quoted definition of sunrise. I used mathematics and the maths dont lie, the sunrise angle is dependant on the height of sun.

Nasa would of sent the satelite over the crater at exactly the window of time when the sun as at its maximum height over the longtitunal horizion. They did this because its simple photograpy more sun light on crater better the picture.

This stuff is starting to make much sense now. and im happy to communicate in mathematics compared to jargon.
.... definitions at sunrise its zero....Please Stop throwing that jargon at me in replies its meaningless. You can throw it to other posters, but if you reply to me with arguments on sun angle at a coordinate position on moon use mathematics to back up your argument.

edit on 17-2-2013 by AthlonSavage because: (no reason given)



posted on Feb, 17 2013 @ 07:03 AM
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The following is my opinion as a member participating in this discussion.


Originally posted by AthlonSavage
That is the maxium sunrise angle possible at the crater coordinate position 73 degrees Lat north and 10 Long, when the sun is directly overhhead as seen by an obsevrer standing in the Longititude 00 and latitude 20 spot.

OK, I understand what you mean.


I dont care about your quoted definition of sunrise. I used mathematics and the maths dont lie, the sunrise angle is dependant on the height of sun.

Mathematics is also based on definitions, as are all things, ignoring a common definition is the best way of getting wrong results.

This stuff is starting to make much sense now. and im happy to communicate in mathematics compared to jargon.
.... definitions at sunrise its zero....Please Stop throwing that jargon at me in replies its meaningless. You can throw it to other posters, but if you reply to me with arguments on sun angle at a coordinate position on moon use mathematics to back up your argument.
So, instead of trying to agree in what we are talking about you just say that I am "throwing jargon"...

Let's start again:
To see if we are talking about the same thing or not, don't we need to, first of all, agree on some definitions? For example, if we don't agree in a common language to use in our conversations, we could spend the rest of our life's discussing with each other without understanding that we are saying the same thing.

In this case, if both of us are using the same words, the only way of knowing if we are using them in the same way is to agree on their definitions.

So, what's your definition of "sunrise"? To me, it's the time when the Sun starts to appear over a horizon, whatever it may be, natural or artificial.

From Wikipedia:

Angle

Sunrise occurs before the Sun actually reaches the horizon because the Sun's image is refracted by the Earth's atmosphere. The average amount of refraction is 34 arcminutes, though this amount varies based on atmospheric conditions.[1]

Also, unlike most other solar measurements, sunrise occurs when the Sun's upper limb, rather than its center, appears to cross the horizon. The apparent radius of the Sun at the horizon is 16 arcminutes.[1]

These two angles combine to define sunrise to occur when the Sun's center is 50 arcminutes below the horizon, or 90.83° from the zenith.[1]




And, about the Location on the horizon

Neglecting the effects of refraction and the Sun's non-zero size, whenever and wherever sunrise occurs, it is always in the northeast quadrant from the March equinox to the September equinox and in the southeast quadrant from the September equinox to the March equinox.[4] Sunrises occur due east on the March and September equinoxes for all viewers on Earth.[5] Exact calculations of the azimuths of sunrise on other dates are complex, but they can be estimated with reasonable accuracy by using the analemma.


It's about sunrise on Earth but most of it can be applied to sunrise on any planet.



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



posted on Feb, 18 2013 @ 03:31 PM
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Nasa may not, but JAXA might. Here is Anaxagoras and Anaxagoras A.

shelf3d.com... taking "Anaxagoras" by HDTV (wide camera)

A does have the shadow, but don't see the high structure.



posted on Feb, 18 2013 @ 04:02 PM
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reply to post by ArMaP
 





I got a height of a little over 4 metres


No way... as in... surely not... that would make the edge of the craters mere centimetres..... that's way too small.
One can tell from the topography a vague sense of scale (unless i'm tripping) and that doesn't make sense... to me... i could be wrong but your estimation seems very small indeed.



posted on Feb, 18 2013 @ 04:13 PM
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reply to post by manmental
 

Trigonometry trumps "looks like".



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