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Norway spiral - Russia accepts blame even though Norway may have been responsible ! !

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posted on Dec, 15 2009 @ 02:06 AM
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reply to post by mixmix
 

I was referring to the physicstoday article of 2009.

In February last year, HAARP unexpectedly managed to induce a strange bullseye pattern in the night sky.

blogs.physicstoday.org...
This would seem to the the 2009 experiment they were talking about (link to adsabs) but I think you're right about the date of the images.

But a little more research found this. A visible disk with an outer ring is produced.

Observations of HF-induced artificial optical emissions at the 3.6 MW HAARP facility show unexpected features not seen at the previous 960 kW level. Optical emissions often form a bright rayed ring near the 10% power contour surrounding a central disk with a sharp edge near the 50% power contour.


We conclude that the optical bullseye patterns are a refraction phenomenon and an indicator of ionization production within the transmitter beam.

www.agu.org...




posted on Dec, 15 2009 @ 02:27 AM
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reply to post by Phage
 





If the spiral were 2-3 times the size of the full moon that would make it 1.5º across, the width of the tip of your thumb held an an arms length. Obviously, in some of the images it appears much, much larger than that. In the others, it seems about right.


They are all right imo.

The size may depend on the location these pics were taken.

Those pics are the real deal.



posted on Dec, 15 2009 @ 02:32 AM
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reply to post by Phage
 





But a little more research found this. A visible disk with an outer ring is produced.


yes your link is dated from 13 July 2009

but in the freerepublic link from Oct 04 2009 05:37:18 GMT+0200
the same team cause all these links reference the same team
they state
"But in February last year, HAARP managed to induce a strange bullseye pattern in the night sky. Instead of the expected fuzzy, doughnut-shaped blob, surprising irregular luminescent bands radiated out from the centre of the bullseye"
link

BULLEYE



posted on Dec, 15 2009 @ 02:38 AM
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reply to post by mixmix
 

You found it.
A visible effect.
But the photo doesn't really look so good.



[edit on 12/15/2009 by Phage]



posted on Dec, 15 2009 @ 03:14 AM
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Originally posted by Phage
reply to post by tauristercus
 

No.
That calculator is asking for radians. Not degrees.
radians = degrees * pi/180

1º = .01745 radian
tan(1º)=.01745
Coincidence? Not really.

tan(.0175)*500=8.75



Phage, apparently you're lacking a grasp (experience ?) in the application of basic trigonometry to calculate lengths/distances.

So for your benefit (and anyone else who may be interested), I've created a very basic tutorial/primer in the use of the trig functions.

So please step inside the classroom and take a seat ... class is about to commence ...




posted on Dec, 15 2009 @ 03:16 AM
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Well well... Sorta makes everyone who said it wasnt a missile seem pretty stupid now.



posted on Dec, 15 2009 @ 03:21 AM
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Originally posted by Ghost in the Machine


Well well... Sorta makes everyone who said it wasnt a missile seem pretty stupid now.


You've lost me ... what are you referring too ?



posted on Dec, 15 2009 @ 03:40 AM
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reply to post by tauristercus
 

What's ludicrous is to believe that a circle with a 1º angular diameter would have a larger diameter than the distance from it.

Think about it. 500km away? 546km diameter?


One radian is the angle of an arc created by wrapping the radius of a circle around its circumference.

To convert radians to degrees:
radians = degrees * pi/180
(Here's just one place you can look it up. There are gazillions of others)

Solving for radians:
z = 1º x pi/180
z = pi/180
z = .017453

Solving for diameter:
diameter = z * 500
diameter = 8.727


You can also look up the tangent of 1º here
So now lets use degrees instead of radians.
tan(1º)=.0175 (from the table, it's rounded up a bit)
500 x .0175 = 8.75


You are wrong.



posted on Dec, 15 2009 @ 04:03 AM
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Originally posted by Phage
reply to post by tauristercus
 

What's ludicrous is to believe that a circle with a 1º angular diameter would have a larger diameter than the distance from it.

Think about it. 500km away? 546km diameter?


One radian is the angle of an arc created by wrapping the radius of a circle around its circumference.

To convert radians to degrees:
radians = degrees * pi/180
(Here's just one place you can look it up. There are gazillions of others)

Solving for radians:
z = 1º x pi/180
z = pi/180
z = .017453

Solving for diameter:
diameter = z * 500
diameter = 8.727


You can also look up the tangent of 1º here
So now lets use degrees instead of radians.
tan(1º)=.0175 (from the table, it's rounded up a bit)
500 x .0175 = 8.75


You are wrong.


Incredible !!!!

I've just led you by the hand through a SIMPLE trig exercise that lets you calculate an unknown length/distance based on other info that IS available ... and you have the temerity to disregard all that evidence that is mathematically correct using equations/formulas that have been around for almost "forever" ... I have no idea what your mathematical background may be (or not) but you certainly have your blinkers well and truly on.

Interesting how you didn't disprove or falsify one iota of my presentation to you ... not one equation ... not one statement ... instead your entire rebutal is obviously based on doing a very quick and cursory Google search, finding something that "looks" like what you need to "make your case" but totally ballsing it up in the process and showing that you actually DO NOT understand the basic maths behind what you've Googled.

This is one instance where you can't railroad your way over someone by continuously giving one/two liner posts in response to a well thought out and presented case ... just not going to happen here.

If you're saying my calculations and deductions are faulty or incorrect, then fine ... I would appreciate you going through my presentation examples and correcting any errors.

But at the moment, it looks to like there's only one way ... and that's the Phage way


Unimpressed to say the least !



posted on Dec, 15 2009 @ 04:22 AM
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Any updates on whether other countries in between have sightings of this phenomena?

It is a pretty weak argument to say that not a lot of folks were up at the time.

And is there an actual article that states the Russians taking responsibility for the Norway spiral and not just of the failed missile experiment?



posted on Dec, 15 2009 @ 04:23 AM
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I'm not on anyones side here but that drawing was funny,"its a missile"lol



posted on Dec, 15 2009 @ 04:29 AM
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Why does everybody keep saying that Russia has taken responsibility for the spiral (including the OP title)?
Attention to detail would show that Russia's initial denial still stands. All Russia has admitted to is a failed Bulava missile launch from a submarine in the White Sea. Nothing more. Please keep this in mind!




Originally posted by ALLis0NE...I believe stage one is higher than 40 kms.. but I will double check my source later, I am busy now.


Please double check your source and let us know how far we can expect the smoke plume to extend into the atmosphere. And, if you are feeling really tricky, how far away can we expect to see this on earth? And why the hell is it visible approximately north of where the photos were taken?




Originally posted by ALLis0NEI will answer other questions later when I get back.


I have summarised mine for you



Originally posted by GobbledokTChipeater
Shouldn't the shutter speed able to be (roughly) calculated? I don't have the inclination but perhaps you could use rotation speed of the rocket (from a video) and the arc(s) from the exhaust of the rocket in the long exposure photo.

If, as you say, the spirals are exactly overlapping each other in that photo, then the exposure should be the time it takes the rocket to rotate 180 degrees.



Originally posted by GobbledokTChipeater...obviously the shutter speed is slow, but if it was as slow as you are making it out to be then shouldn't the picture just be a single gray disc?



Originally posted by GobbledokTChipeater
Yet you don't respond to some of the more "intelligent" questions. Like how can the effects of a rocket, which was fired 800 Km's away, in a different direction to the observer be seen?


[edit on 15/12/09 by GobbledokTChipeater]



posted on Dec, 15 2009 @ 04:35 AM
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reply to post by tauristercus
 


Hi,

Very entertaining post Tauristercus, I will not side with anyone as yet but I'll do it the practical way - using Autocad as I am trying to understand using the values provided but my math is quite poor you must understand!

Let me get this right. I quickly read what values you were using: 500km from the ground level, and the angle is 0.5 degrees, right?

I simply created a line 500 units long, copied it and rotated the second line 0.5 degrees out from the bottom base point, so there is a gap at the top.

At 500km long, 0.5 degrees angle the radius at the top between the two ends of the line appear to be 4.363km (8.726km diameter) very close to Phage's result. It may not be exacat as I didn't extend the second line square off the first straight up / down line.

I have also included the diagram of my results:



i11.photobucket.com...

Hopefully this may clear things up?

Take care,

IdeaLogical.



posted on Dec, 15 2009 @ 04:56 AM
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super-interesting, thanks!
anyone read uzumaki by junjo ito?
xhail eris!



posted on Dec, 15 2009 @ 05:28 AM
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reply to post by tauristercus
 


I'm sorry, but I have to say Phage is right here.

If you use the moon as example then:

-distance to moon: 384403 km

-diameter of moon: 3474 km

That's like a factor hundred. Give and take, I don't have a calculator at hand.

So at 500 km, the spiral would've been 500:100 = 5 km.(basing it on moon size) Not entirely accurate but nowhere near your result.

[edit on 15-12-2009 by Point of No Return]



posted on Dec, 15 2009 @ 05:40 AM
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reply to post by Point of No Return
 


I think you would need an viewing angle bigger than 60 degrees, in order for the diameter to be bigger than the distance up.



posted on Dec, 15 2009 @ 06:03 AM
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I would like to restate what other posts have said already...

Where does it say that the Russians have accepted responsibility for the Norway Spiral?

This question is quite critical as a lot of the rebuttals include this unproven assumption. This proof would be appreciated as it would allow me to move on with arguments that use this info.



posted on Dec, 15 2009 @ 06:57 AM
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reply to post by tauristercus
 


Thanks for the trig lesson there....


Now how's about you address my question on page 15, top post...

You can do trig problems til the cows come home, but it doesn't do you or this thread any good to just ignore the visual proof of a rocket.... unless you have an explanation for the contrail that's clearly visible in that pic, which is clearly connected to the blue haze, which was clearly connected to the spiral effect...

or you can just do another trig problem based on assumptions



posted on Dec, 15 2009 @ 07:35 AM
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Originally posted by Phage
reply to post by His Doodness
 

I'm not an expert in photography but have dabbled in it. See if this makes any sense, the foreshortening effect caused by telephoto lenses.

When using a long lens the space between objects appears to shrink, the objects get jammed together, the more distant object appearing too close to the nearer one. Doesn't this occur because the angular size of the more distant object is increased more than the angular size of the closer object? Because the more distant object is "enlarged" more (relative to the nearer object) it appears to be closer to the nearer object than it actually is. The longer the lens, the more pronounced the effect.

In a similar way, some of the photos make the spiral appear closer to the foreground (larger) than it was, while in other photos, made with a wider angle lens it appears much smaller.

If the spiral were 2-3 times the size of the full moon that would make it 1.5º across, the width of the tip of your thumb held an an arms length. Obviously, in some of the images it appears much, much larger than that. In the others, it seems about right.


This might take a couple abstract analogies, but I think I can explain this a bit better.

Think of a scene in a movie showing heavily congested traffic on an LA freeway. There are two ways you could shoot this scene, and you want to shoot it in a way that it looks like the worst traffic in the world (which it probably is).

a.) you could stand on a low bridge and shoot towards the cars in front of you with a telephoto lens zoomed out all the way (wide). Taking a photo like this would look pretty normal. There would be a lot of cars, but it would seem like they are moving along just fine because of the obvious spaces between them. A characteristic of a large depth of field, which is the result of the wide lens.

b.)or, stand on the same bridge but point your camera a quarter mile up the road while zoomed all the way in. By zooming in, the depth of field is dramatically reduced, which makes things appear to be jammed together as you said. You can see why this shot would make the traffic seem a lot worse. The frame would be filled with cars and it would look like they are riding each others' bumper.

Now I know you have seen a shot like this in some movie or television show, so Ill ask you, did any of the cars seem disproportionately large? or small? They still look like cars, they only thing that changed was their apparent distance from each other. In reality, the distance between them does not change.

[I will try to dig through some very old tapes I have displaying this effect. The first time I saw it I was fascinated, and yet it is only basic optics. They might be lost though.]

Sure some of the spiral photos could have been taken from further away with a very long lens, and that would certainly make the spiral appear closer to the horizon, but not because of a change in angular size, only in depth of field.

With a wide lens, everything is in focus, so perceiving distance is easy, you could say natural, for us. Zooming in on something, which is not natural for our eyes, throws everything in front of a certain point and beyond a certain point out of focus. The depth of field decreases, or, the range of in-focus viewing is decreased.

I wish I could recall the actual science behind why this works. Problem is, my learning process involves discarding details once I fully understand a concept, I know what depth of field is and how it can be manipulated, so there was no need for me to retain the unnecessary details of why. Therefore I always end up having my own ways of explaining things, like the traffic example. I will look that info up when I get a chance.

Sorry, this is getting long... continued...



posted on Dec, 15 2009 @ 07:35 AM
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The point here is, trying to judge a distance on a photo that could or could not have been taken with a long lens could be near impossible without some sort of reference point. For example, if you knew exactly where the camera was, the lens length, the distance to some of the objects in the shot, then you could still only make a more educated guess. The problem with an object floating in the sky is, there is no frame of reference.

The only possible reference point is the moon, which is not helpful. We know that the moon is 1.5 degrees, or a thumb-width at an arms length. But by looking through a camera lens (as in a photo) that does not hold true. I could zoom in on the moon and print it on a 4x6 piece of photo paper and it would be larger than my thumb at an arms length.

I don't want to go out on a limb and say that there is no way to determine the distance of this spiral from the camera, but I can't think of any way to do it.


Were there any photos of the spiral with the moon in the shot as well? I have not seen any, but I would say that if there are, the spiral will appear to be the same size IN REALTION TO THE MOON in all of the photos. Unless of course they were taken during different phases of the spirals' life.

I don't know if I helped explain anything, or answer any questions. But, I unitentionally realized that it is a waste of time for anyone to make claims about the distance of the object from the camera or horizon or any thing from just photos alone. Especially since no photos seem to have surfaced from a side angle or with the moon in the shot.

As I said though, knowing the exact placement of the camera, along with all lens info, AND if the moon was also in the shot, we could accurately find the size. And by WE, I mean NOT me. I wouldn't know how to calculate that, but I know through my understanding of a camera and its lens that the calculation wouls be possible to make in the circumstances mentioned.

Hope this helps, sleep time for me.



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