My analysis the Pentagon CCTV frame of the Plane before impact.
The lightpoles do indeed line up with an angle of entry consistent with the exit hole which below I first determined to be 55o and now accept that if the lightpole evidence is correct, that this angle would be 50o. There is also the issue of the spacing of the light poles and which ones were felled by the plane.
If it was indeed a B757, then the plane must have started to bank left if my calculations are correct. The result would be that the plane would still be traveling along it's trajectory, but it would start to turn towards the camera more, making it's appearance shorter if this manouvre is possible over a short distance for a plane this size.
My guess is there are 3 possible reasons why there would be a last moment change in direction.
1) Assuming it was remote controlled, it could be a correction to get back on target. The proximity to the helipad fire hydrant may have determined the point of the Pentagon to be hit. The section hit, I've read, was under renovation and therefore not fully staffed. The inference being damage limitation.
2) Assuming human suicide pilots, it could be a last second change of mind by the pilot, perhaps again because the pilot was off target and thought he had a small enough turning circle to avoid the building and come around for a second pass, or the suicide pilot turned instinctively at the last moment.
3) The plane was given rotation by hitting the light poles.
Either that or the plane was a drone made up to look like an AA plane but was something shorter as I've previously asserted. However, some of the witness evidence contradicts that hypothesis. Others appear to support it. I'm in no position to assert one way or the other. This is only an analysis of CCTV evidence in isolation.
These calculation are made by comparing the ratios between known distances and countable pixels. If you want to check this for yourself, you will need a good photo package like ADOBE PHOTOSHOP or JASC PAINTSHOP PRO and save these images to a folder.
To start off, it should be determined exactly where the point of impact was and the angles relative to the CCTV camera.
In the picture below I have determined the angles my measuring the x and y components of the red lines and using the atan function available of most modern calculators and spreadsheets. Here I've used the spreadsheet available in MSWorks 4.5a (a bit old but still does the job adequately). Here, I've rounded to the nearest 2 decimals points.
So, if we now look at the first frame of the CCTV and want to determine where the exact point of impact is, we have to add a straight line from one end of the mid-section to the other with a marker intersecting it to give the ratio of 10.34:1.31 or 100:12.63
But before we do that, it has also been pointed out to me that I haven't taken into consideration the distortion of the wide angle lens (this is a common problem with wide angled lenses) but until now have left this out to simplify matters. This is called 'barrel distortion' if you choose to look it up in your search engine.
A 'pin correction' for 'barrel distortion' correction has been forwarded to me by Ron Harvey:
I've zoomed in by x2 and added an outline of the mid-section wall.
The next step is to add the crossectional line to determine the entry hole.
I've done this below and extended the vertical line to the height of the mid-section wall.
From this I have a reference height in pixels from which to calculate determine the length of the plane. I measure this to be 112 pixels
We know the actual height to be 71 feet.
So 112 pixels : 71 ft at this distance from the camera.
From this it can be said that anything the same distance from the camera as the impact point has to be of this scale (112 pixels : 71 ft)
Below are the orthogonal dimensions of a Boeing 757.
In the above image, the length of the plane is 213 pixels. So we now have
213 pixels = 155 ft (1.374 pixel/ft) and have to
scale it so it becomes
112 pixels = 71 ft (1.577 pixel/ft)
So the new length of the plane in pixels is
155ft * 1.577 pixels/ft = 244 pixels
By rotating the plane and lining it up with the poin of impact we get:
This would be correct if the nose was at the point of impact and the direction the plane is travelling in is at right angles to the camera.
The next step is to calculate for the angle of entry.
The following angles were determined from the structural intregrity diagram overlayed on the satellite image.
Now if we look at one of the photos taken from a helicopter survey of the damage we see an exit hole.
Here I've overlayed the damage map.
Extending the line to the light poles that were damaged we get:
(I've made this using Ron Harveys plot of the light poles)
Previously I had place had overlayed on the satellite image where I calculated the angle of entry to be 55o.
To calculate an angle I used the following rule:
a = atan (w / h)
b = atan (h / w)
In Paint Shop Pro, I made a rectangular selection of a portion of line with the line passing thru the corners. I then copied and pasted the selection as a new image.
Image -> Canvas Size...
to get the height and width of the selection. Then I used a spreadsheet to return the exact angle.
(My spreadsheet returns angles in radians, so I had to extend the function to: a = atan (w / h) * 180o / π -- to convert from radians to degrees.)
So, moving on, the angle between the impact point, the camera and the trajectory is 83 + 90 - 55 = 118o.
So the initial assumption of 90o has changed by 28o, i.e. the plane is at 28o to the camera.
This means an adjustment in the apparent horizontal length adjusted by
COS (28o) = 0.883 (= 88.3%)
Our plane should now have a new pixel length.
244 pixels * cos(28o) = 244* 0.883 = 215 pixels
and should have a perspective adjustment to the height of the tail by 88.3%.
Now if we place this horizontally with the nose lined up at the point of impact we get:
There is another step that should be added to be exact. That is the additional reduction in size to account for extra feet that the plane is away from the camera. Again, we can use triangles to calculate this.
First we need the distance from the camera to the point of impact. For this we need to go back to:
We know that the length of one side of the pentagon is 920 ft, which in this image is equivalent to 280 pixels. If we now rotate this image 7o clockwize, we can measure the distance between the camera and the point of impact in pixels. I make this 191 pixels. So the distance from the camera to the impact point is
920 ft * (191 / 280) = 627.57 ft
(Also, knowing that the plane is 155 ft we can calculate
the length of it in pixels if it were in this satellite picture -
280 pixels * (155 / 920) = 47 pixels)
At the moment, if the nose of the plane is at the point of impact, the image is correct, but if we reverse the plane back so that it is in line with tail of the plane in the original image we get the following image with the angle between the plane's nose, the camera and the point of impact at 12o.
So if we rotate the image again, this time 5o anti-clockwise we can measure it to be 215 pixels.
920 ft * (215 / 280) = 706.43ft
So we can now calculate the addition scaling factor to be
627.57 / 706.43 = 0.88 or 88 %
(also = COS (28o))
So what do I have left when I apply this to the original image.
So it looks like we have a working model, if the plane is flying straight. Now assuming that the plane hit at the far corner of the mid-section we get a height of 97 pixels as apposed to 112 pixel measured at where I place the point of impact. That means that the size of the plane is reduced by 88.6%. However, looking at the satellite picture, we can see that from the collapse section corner to the mid-section corner is halfway to the opposite corner of the collapse section where I measured the angle to the camera from. So if it hit the other corner of the midsection at this angle, the size reduction would be halfway between 88.6% and 100%, i.e. 93.3%
(Below I have moved up the previous results and placed the next underneath)
If we take the shallowest angle possible (using what's 'missing' red dots as a guide), we get 50o, from which we can then say that instead of cos(28) we should have cos(33) as our persective correction, i.e. 84% (rather than 88%)
(Again I have moved up the previous results and placed the next underneath.)
So clearly, based on the assumption that it's flying straight, we cannot conclude that this is a match against what we are seeing.
Now if we assume that it is a B757 or of equal length, then the plane, I estimate, has to have spun round about 30o towards the camera at the moment in time that this frame was taken. Otherwise, if the plane is still flying straight along the line of trajectory, then it has to be about 20% shorter than a B757.