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Mysterious Missile Launch Over California - 11/8/2010

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posted on Nov, 20 2010 @ 07:03 PM
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reply to post by smurfy
 


So just how far above do u go?? wow, u are very,very, good



posted on Nov, 20 2010 @ 07:34 PM
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reply to post by backinblack
 


Again, it has been covered several times in this thread and others. At the time of that news report being issued Dr Minnis and his colleagues were aware through contrail science and news agencies that there were two possible candidates. At that time the media and contrail science were favouring AWE808.

spacefellowship.com...

“Later, while viewing some blogs, I found that the contrail corresponded remarkably well with flight AWE808, which flew from Hawaii to Phoenix; it showed the change in course to the northeast at the same location, further confirming my conclusion.”

Minnis said that while he can’t prove the contrail came from an airplane, it is the “most likely” scenario based on his research'

Even Dr Minnis initially though it was a missile launch! His assessment and conclusion on the contrail and how it doesn't fit in with being a missile is the key in the report.

TJ



posted on Nov, 20 2010 @ 07:44 PM
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reply to post by tommyjo
 



I found that the contrail corresponded remarkably well with flight AWE808, which flew from Hawaii to Phoenix; it showed the change in course to the northeast at the same location, further confirming my conclusion.”

I know what you are saying but are we not talking about professionals here?
When I see phrases like "corresponded remarkably well with flight AWE808" I would assume they checked times and flightpaths etc...
Is it not possible then that this contrail they pictured was in fact flight AWA808 and the event filmed was something entirely different?? Another trail altogether??



posted on Nov, 20 2010 @ 07:55 PM
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reply to post by backinblack
 


Backinblack,

Why not e-mail or write to Dr Minnis and ask him your questions about the contrail?

www-pm.larc.nasa.gov...

What size contrail can be seen in a satellite image?

asd-www.larc.nasa.gov...

spacefellowship.com...

The article highlights the assessment of conditions on 8th Nov for persistent and non-persitent contrails.

'Minnis then consulted a contrail “forecaster” put together by SSAI’s David Duda and Rabindra Palikonda, who also work at Langley, and learned that conditions were “ripe” for persistent contrails over the Pacific west of Los Angeles, but that contrails would only survive for a short time closer to the coast.'

Contact address for Rabindra Palikonda and David Duda. It is all out there for people seeking answers to their specific questions.

www-pm.larc.nasa.gov...

experts.nasa.gov...

TJ





edit on 20-11-2010 by tommyjo because: (no reason given)



posted on Nov, 20 2010 @ 08:02 PM
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reply to post by BobAthome
 

Hi BoBa,
I am just trying to see my way through things. In fact, I only read through this post rather than make a post until yesterday/today, just having some thoughts, and so far a follower, not having started any threads of my own, despite what my profile says. In any case it's about sharing/adding info and learning from it, and there are some good people here, and more if the rest would invest.



posted on Nov, 20 2010 @ 08:08 PM
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reply to post by tommyjo
 



Why not e-mail or write to Dr Minnis and ask him your questions about the contrail?

Thanks Tommy, I may just do that..
There are a few questions I have like why does the satelitte image show a reasonably constant width contrail where as the pictures of this trail show anything but a constant width.
It is clearly many times wider at certain points, especially near the horizon..
Surely that would have shown up on the sat' images..



posted on Nov, 20 2010 @ 08:30 PM
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reply to post by smurfy
 


Welcome to a place that in times for a brief moment will astound and amaze, and the avatars are nice
Welcome



posted on Nov, 20 2010 @ 09:22 PM
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Originally posted by SarK0Y
reply to post by Arbitrageur
 


after sunlight reflecting, here: www.abovetopsecret.com...
after color fluctuations of incandescent point, here: www.abovetopsecret.com...
Where did you get your "given data" from?

And I don't think you can measure light intensity in watts, it should be lumens or something like that. I have a 100W incandescent light bulb and a 23W compact fluorescent light bulb and they both put out the same amount of light measured in lumens. So I don't think your units are right, I don't know where you got that "given data" from, and I don't see where you took the shape of an airplane into account.("area of hemispherical surface is...") what about the wings? They aren't hemispherical, and in fact have some flattish areas.

Just look at the examples people posted of airplanes with bright glare of reflections, and imagine the glare will be dimmer when the airplane is further away, it's really no stretch and plenty of examples were posted. You don't even have to do the math if you just look at those examples of plane glare.



posted on Nov, 20 2010 @ 10:18 PM
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reply to post by Arbitrageur
 



...do the engines all rotate the same direction? Or do the engines rotate in opposite directions on opposite wings?


Thanks for one of the easiest questions to answer, ever! AND, kudos for thinking "outside the box".

Each engine (in a modern airliner) rotates in the same direction. There is no "counter-rotating" installations.

***sidebar, and forgive me, a bit of education.....when it comes to what Arbitrageur appears to be referring, it gets into what is learned at the most basic levels, by pilots who fly propeller-driven airplanes.

The propeller itself, whether two, three four or six blades, is a series of airfoils. AND, there certain "torque" and "downward-moving-blades-angle-of-attack" issues that affect an airplane, and which will vary depending on thrust setting, and airspeed, etc.

A web search of terms such as "P-factor", or "propeller factor" and "adverse yaw" should bring up a lot of pages to help explain (so that I don't have to, here....).

True, in some airplane multi-engine designs (the Lockheed P-38, from WWII jumps to mind) the idea of "counter-rotating" engines were designed in....it stems from what's called, in multi-engine flight training involving airplanes powered by piston engine/propeller combinations, the "critical engine" concept....meaning, for most American-built designs, the LEFT engine.

(From your point of view, as a pilot, in most piston/propeller airplanes, the engine (and propeller) is rotating clockwise. WHEN you are flying with a high angle of attack, the propeller blade on the down stroke (RIGHT-HAND side, for clockwise rotation) has a greater "bite", or 'angle of attack' also....and it results in a "pull" to the left...a 'YAW" force. Every new pilot learns this early on, and before long it is habit, almost second nature, to know that some right rudder (rudder is applied by using one's feet) is needed, in pitch attitudes (nose high) and airspeeds (such as climb speeds, for small airplanes) with high power settings (full throttle for carburetors engines. appropriate manifold pressure/prop RPM for more advance installations.....) These "adverse yaw" situations or most noted at high angles of attack climbs) and lower airspeeds (takeoffs, acceleration to climb speeds) and so forth.

Well, basically I've just described a bit of the normal Flight Instruction anyone who wishes to learn to fly will receive, and quite early on.


NOW....to answer, again.....engines rotate the same direction (although Rolls Royce N1 Fans ARE opposite than GE or P&W!!!), on modern jets, they all rotate the same direction.

ALL of the above, about "P-factor: and such? Not the same with big jets. THRUST is what we feel, no torque.....



posted on Nov, 21 2010 @ 12:44 AM
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So I've looked a few pages back and it seems nobody posted it... so here goes.




posted on Nov, 21 2010 @ 12:51 AM
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reply to post by Vitchilo
 


I haven't seen that video but all the information has been posted before..
Good viewing anyways..I don't think we'll know the truth any time soon..



posted on Nov, 21 2010 @ 01:13 PM
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reply to post by Arbitrageur
 



Where did you get your "given data" from?

i said that was rough(read optimistic for plane supporters) data




And I don't think you can measure light intensity in watts,

more exactly, in watt/m2
however, if you want Lumens so hard, you can convert units with own efforts



and I don't see where you took the shape of an airplane into account.("area of hemispherical surface is...") what about the wings? They aren't hemispherical, and in fact have some flattish areas.

learn how electromagnetic waves travel through space(environments) & you'll get what i was talking about "area of hemispherical surface is..."



posted on Nov, 21 2010 @ 05:10 PM
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Originally posted by SarK0Y
reply to post by Arbitrageur
 


and I don't see where you took the shape of an airplane into account.("area of hemispherical surface is...") what about the wings? They aren't hemispherical, and in fact have some flattish areas.

learn how electromagnetic waves travel through space(environments) & you'll get what i was talking about "area of hemispherical surface is..."
I see. Physics was one of my majors so I don't really have to learn it. Now that I know what you're trying to do, I can explain the correct way to do the math and the physics. Here's the math you gave:


Originally posted by SarK0Y
reply to post by tommyjo
 


Reflection and glare doesn't always appear. Why do you think that every aircraft or any other object filmed in the sky will automatically generate glare or reflection?
well, let's do math of the event
i took rough data to simplify situation, but that's towards upper limit of possible numbers.
------------------------------------------------
given data.
distance to object 60, 000 m's
reflecting area Ra= 100 m2
light intensity I=1, 000 W per m2
----------------------------------------------------
so, area of hemispherical surface is S=(2*3.14)*60000^2=2.260800000*10^10; light intensity at cam from reflecting area is i=Ra*I/S=0.000004423213022 W per m2.
You assumed the light source was 60km away but this is not the distance to the sun.

What has been proposed is that the light source is the sun. The distance to the sun varies but for the sake of argument let's say it's 93 million miles or 149668992 km away from the plane. And let's say the observer is 200 miles or 322 km away from the plane. This makes the total distance from the light source to the observer 93,000,200 miles, or 149669314 km (following the path of the reflected light).

Now let's say there's a 1 meter square flat reflective surface on the plane's wing that catches sunlight and reflects it to your eye. If there's 1000 watts of sunlight radiation per square meter hitting the wing 200 miles away, then applying the inverse square law yields 1000/(93000200/93000000)^2 which says that the radiation will be 999.9956989 watts per square meter 200 miles away, if the light hitting the wing is sunlight. This only applies to a flat surface like a wing, and in reality the wing is not going to be perfectly flat nor a perfect reflector, so the reflection from the wing will be less intense than that, but if you're going to do the math and the physics, that's how it works.

If the sunlight strikes a rounded surface, then the reflected light becomes more spread out (Isotropic) and then you can start treating the light source like it's 200 miles away. But if the reflective surface is flat, you're applying the inverse square law to a very tiny difference in the distance from the light source since 200 miles is a very small fraction of 93,000,000 miles. The physical reason for this is that the light rays are already very nearly parallel at a distance of 93,000,000 miles, so they don't tend to spread out much when reflecting off a flat surface.

As this text says, the inverse square law doesn't apply to a non-isotropic light source, which would apply to an airplane wing reflecting sunlight:



The sun itself, however, is an isotropic light source so you need to make your distance calculations from that source if it's reflected sunlight.



edit on 21-11-2010 by Arbitrageur because: clarification



posted on Nov, 21 2010 @ 05:31 PM
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Originally posted by weedwhacker
NOW....to answer, again.....engines rotate the same direction (although Rolls Royce N1 Fans ARE opposite than GE or P&W!!!), on modern jets, they all rotate the same direction.

ALL of the above, about "P-factor: and such? Not the same with big jets. THRUST is what we feel, no torque.....
Thanks for the answer. I actually tried to research it before asking and I did find the propeller torque stuff, but not the answer I was looking for which you just provided. I thought they were all clockwise facing forward as the pilot sits, but you're saying the Rolls Royce N1 are counterclockwise?

In any case based on your answer, if we see 2 contrails corkscrewing in the same direction, then fan rotation could be the cause, whereas when the 2 contrails corkscrew in opposite directions, then it must be due to wingtip vortices or something like that, right?



posted on Nov, 21 2010 @ 05:57 PM
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reply to post by Arbitrageur
 


R/R fans tend to rotate the opposite of other manufacturers'.

Here is the CFM-56 (SNECMA/GE/CFM Corp) engine from a B-737:



Fan rotates clockwise, when viewed from aft.

A R/R (latest) Trent 900, from A-380:



Business end of a GE engine, on A-330 (Aer Lingus) A-330s can be ordered with GE, P/W or R/R engine options:



Still, the rotation of the contrail, from the engine N1 fan is minimal, and sometimes not even seen. Wingtip vortices even less, I think...since the vortex tends to drift downwards, after forming. Of course, situations vary....depends on many factors.



posted on Nov, 21 2010 @ 06:02 PM
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You guys see my thread www.abovetopsecret.com... ?

Retired U.S. Air Force Brig. Gen. Jim Cash, former assistant director of operations at NORAD has come out and said:


In my opinion there is absolutely no doubt that what was captured on video off the coast of California was a missile launch, was clearly observed by NORAD, assessed by a four-star General in minutes, and passed to the President immediately.


See his credentials for making this comment there.



posted on Nov, 21 2010 @ 06:14 PM
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reply to post by Arbitrageur
 



You assumed the light source was 60km away but this is not the distance to the sun.

wuhhhhhhhhhhh! why do we need distance to the Sun??
really, we need to know sunlight Power which was reached to the Earth
en.wikipedia.org...
ADD:



1000/(93000200/93000000)^2

hmmmmm.... what is that??




aw doesn't apply to a non-isotropic light source, which would apply to an airplane wing reflecting sunlight

wuhhhhh! where do you get it??????

edit on 21-11-2010 by SarK0Y because: (no reason given)



posted on Nov, 21 2010 @ 06:21 PM
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reply to post by Iamonlyhuman
 


Notice how the Brig General passes it off as a "lame excuse"? It appears to be a perfect excuse simply to slate President Obama. Not the first time that Brig Gen Jim Cash has created an article to have a political rant.

Remember that the Brig Gen is only viewing the footage as shown by CBS. Some people are questioning why CBS chose to edit the tape and are calling for the full footage to be released. Is it another 'CBS-Gate?'

www.freerepublic.com...

TJ



posted on Nov, 21 2010 @ 06:52 PM
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reply to post by tommyjo
 


I've looked but can't find a bio for Gen Cash that says what he has done since he retired from the military in 1991.. But interesting to see if he has any links to military contractors..



posted on Nov, 21 2010 @ 07:05 PM
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Originally posted by SarK0Y
wuhhhhhhhhhhh! why do we need distance to the Sun??
really, we need to know sunlight Power which was reached to the Earth
en.wikipedia.org...
You wanted to apply the inverse square law, so the sun is the isotropic light source. You compare the 1000 watt per square meter intensity at 93 million miles versus 93 million miles plus 200 miles when the sunlight reflects off a flat surface, like a mirror, or a plane wing.



1000/(93000200/93000000)^2
hmmmmm.... what is that??
That's the math of what I just said in English, applying the inverse square law.



aw doesn't apply to a non-isotropic light source, which would apply to an airplane wing reflecting sunlight
wuhhhhh! where do you get it??????
look at the page of the textbook I posted, it's the bullet that says the inverse square law only applies to isotropic light sources (which means it doesn't apply to non-isotropic light sources). The sun is an isotropic source, because the light rays go out in all directions. Sunlight reflecting from a flat mirror or a flat plane wing is not isotropic, the light rays are nearly parallel so they don't spread out hardly at all except for imperfections in the reflecting surface.

www.cs.dartmouth.edu...


Because the sun is so far away, the rays of sunlight are nearly parallel to one another.

What this means is, over a distance of 200 miles, the nearly parallel rays of light from the sun don't spread out very much. This is true whether the sunlight continues in its original direction, or if it is reflected off a flat surface like a mirror or a wing, though of course they do scatter due to atmospheric effects which is why sunsets are red, the longer red wavelengths don't scatter as much as the shorter wavelengths.



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