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stirling
What we are discussing is a dead aircraft however way you look at it......
I am a believer in the rumoured wedgie hanging off a KC 135 over the North Sea, a few years back....
Then theres the sausage contrail............and strange doings in California skies....
The Unca Sams got hardware even some top brass doesn't know about......
Originally posted by boomer135
stirling
What we are discussing is a dead aircraft however way you look at it......
I am a believer in the rumoured wedgie hanging off a KC 135 over the North Sea, a few years back....
Then theres the sausage contrail............and strange doings in California skies....
The Unca Sams got hardware even some top brass doesn't know about......
That photo was a hoax and even the creator of it said it was. It was made for a newspaper or something like that.
"Multi-frequency" stealth when discussing radar is probably referring to the fact that low-frequency radiowaves act a bit differently than the higher frequencies of search, track and targeting phase of most radars.
One of the interesting things is that larger aircraft actually defeat low-frequency returns more effectively which at first glance seems counter-intuitive.
Of course, at lower frequencies, say targeting, the smaller surfaces are obvious an advantage. Which results in some interesting trade-offs when discussing LO and different wave-lengths.
Sorry off topic
But we are agreed the guy who saw it (Royal Observer Core) actually did see something ?
Just found this ATS post on the subject
www.abovetopsecret.com...edit on 16/8/13 by macpdm because: (no reason given)
Originally posted by crazyewok
reply to post by macpdm
Might be another option. It wasnt Aurora, and it doesnt have a methan fuel system.
Maybe it was another unknown aircraft.
Originally posted by neformore
reply to post by boomer135
I disagree that he saw nothing.
I think there is fairly compelling circumstantial evidence for an exotic to have been flying at that time in and around the North Sea and the Scottish and US coastlines, including radar operator reports, "skyquakes" and Gibson's sightings.
You may have seen aircraft flying at altitudes of 35,000' to 45,000' above you. What you notice first is the vapor trail, typically only visible at altitudes above 35,000'. Even with a plane at 35,000' or almost 7 miles, it's difficult to detect the plane -- even from the ground, where you're stationary. There are many attributes that reduce the visibility of another aircraft, including the structure of the eye itself. The Airman's Information Manual (AIM) notes, in its recommendations on proper scanning techniques, that "foveal" vision -- the smaller center in the rear of the eye -- provides only about a 10 degree angle for sharpest vision. In warning pilots to set up a series of narrow scan in successive 10-degree arcs, the AIM "An aircraft at a distance of 7 miles which appears in sharp focus with the foveal center of vision would have to be as close as 7/10 of a mile in order to be recognized if it were outside of foveal vision."
If the plane is directly overhead then you aren't going to see the fin to ID the airline. So lets say it's at 45 degrees (because that's easy and you would get some view of the fin.
That aircraft then at 30000ish feet (10,000m) that's about 6 miles high and 6 miles (10km) away along the ground.
Or a slant range of around 8.5 miles (13.6 km) away.
The fin of a 747 is about 30 feet or 10m high. So the fin at 13.6km subtends an angle of arctan(10/13600) = 7.3e-04 degrees = 2.6 seconds of arc.
So to make that appear a decent size (say 10 degrees in AFOV ... about the width of your fist at arms length) you need to magnify it about 13600 times.
Even to make the fin appear 1 degree in size ( a thumb thickness at arms length) you need 1360x. So IDing the airline of a plane is not going to happen.
You can do the same for the wingspan viewed from underneath to get a feel for how much magnification you need to ID the aircraft.
The fin of a 747 is about 30 feet or 10m high. So the fin at 13.6km subtends an angle of arctan(10/13600) = 7.3e-04 degrees = 2.6 seconds of arc.
Originally posted by gariac
I have to agree with you here. I have print out from an ATC screen and it looks like they are covering at least a 200nm radius. So we're talking 230 statute miles. Take the top speed at 2200mph. Call it 2300mph to make the math easier. So you have 230/2300 which means 0.1hr to cross the screen. Actually twice that since 200mn is the radius. So it would be on the screen for up to 12 minutes. So not only would there not be a sound, but they would have plenty of time to track it.
gariac
reply to post by boomer135
The fin of a 747 is about 30 feet or 10m high. So the fin at 13.6km subtends an angle of arctan(10/13600) = 7.3e-04 degrees = 2.6 seconds of arc.
I don't know why I try to think after midnight, but I think you did the math in rads versus degrees.
But say we did the small angle approximation using s=r*theta rather than the more accurate arctangent calculation. For s=r*theta, the answer would be in rads. So we have
10=13600*theta, or theta is 10/13600 in rads. Around 7.353e-4 rads.
If I'm right, I hope you weren't thinking of knocking out an incoming meteor. ;-)