Shanksville Deconstructed - Part Two... , page 7
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reply posted on 25-10-2009 @ 11:33 PM by Rewey
reply to post by weedwhacker



Hey weedwhacker - thanks for the detail you've gone to with those replies.

A couple of things, just to humour me...

1. Based on your experience (although not flying in the same manner, I assume!), ignoring the 40º pitch from the FDR, what would YOU estimate the actual impact angle to be (relative to the horizon). Or is it simply not possible to even estimate because there are too many other variables involved?

2. Although it's technically irrelevant to the argument being raised, what's your view on the plane crash pic in the OP? Moving bottom to top? Left to right? Flat spin?

3. With respect to your comments on the time factor meaning there would be no horizontal momentum transferred to the ground, I have a little trouble accepting this.

Like I've said in other posts, I can't imagine Mr. Laws of Physics saying "You know, normally I'm pretty reliable, but golly, everything just happened so fast!" I remember many debates with my Physics lecturers years ago about this exact sort of thing. His answer was always the same - the laws of physics are constants - it doesn't matter how long they have to act.

What I'm getting at is that the Shanksville crater makes it appear as though the nose of the plane impacted along the same line as the wings, even though roughly one-third of the plane sits forward of the wings.

Regardless of how little time there is involved in the crash, this one-third of the plane would still have to leave SOME evidence of horizontal momentum (regardless of how little) on the surface prior to the wings reaching it. Even you pointed out:

Originally posted by weedwhacker
They are, firstly, attached as a unit, so won't instantly break free...


I don't believe, as hooper suggested, that the entire fuselage shattered the point the nose contacted the surface. Therefore the fuselage would compress and break apart in sections as it continues to travel towards the surface (albeit in fractions of a second).

With regards to everything happening 'too fast', the simple formula for momentum is P=mv (P=momentum, m=mass, v=velocity). Therefore the higher the velocity, the higher the momentum. This means that even though the forces will not be acting over a longer period of time, their IMPACT will compensate as the resulting momentum is greater. Although the forces have less time to act, they react with a greater magnitude... Therefore the result is basically the same...

Rew


reply posted on 26-10-2009 @ 11:06 AM by weedwhacker
reply to post by Rewey



Thanks, I'll see if I can be more clear:

...ignoring the 40º pitch from the FDR, what would YOU estimate the actual impact angle to be (relative to the horizon). Or is it simply not possible to even estimate because there are too many other variables involved?


Well, we only have the FDR data to go on....however, (and this occurs to me now) it is such an unusual attitude I am not entirely certain the EXACT pitch attitude, relative to the horizon, was recorded. Again, I assume it was similar to the 40 degrees NOSE DN...(that would be below the horizon) but being inverted, and with the speed (yes, the speed of the event --- the rapidity of occurence --- not the airplane's velocity) it is conceivable that the laser-ring gyros and accelerometers that constitute the guts of the Inertial Reference System could have some lag induced by the quick attitude changes....IF the ground had not been there, then there would have been time for the system to sort itself out with greater accuracy. I know, I know, I was going to try to be clear...but this is me, thinking out loud.

In other words, the ~40-degrees recorded at impact could have been more, as I would assume the airplane's attitude would have been continuously increasing throughout the dive.



3. With respect to your comments on the time factor meaning there would be no horizontal momentum transferred to the ground, I have a little trouble accepting this.


I understand that you do. I will repeat a simplified version of the arrow analogy...or, just even something as simple as a rock. Try it...throw a rock into sand at an approximate 40-degree angle, and see the pattern of ejecta. maybve not very scientific, but sometimes experimental observation is better than just doing the math....

Back to my arrow, for a minute...my point there was that the arrow, impacting at ~40-degrees, doesn't have the rear end break off and/or fold over as any part of it's horizontal vector component, at impact.


Like I've said in other posts, I can't imagine Mr. Laws of Physics saying "You know, normally I'm pretty reliable, but golly, everything just happened so fast!"


Yes...that's why they are referred to as "Laws" and not 'theories'.

But, as I tried to point out, there are other 'Laws' at work, as well, besides just the Laws of motion. There is inertia, and with it, momentum.

AND, yes, the timing involved.

Think --- The airplane at ~500 MPH (let's convert to Knots, I use those) so for argument, let's use 435 Kts. (conversion factor is about 15%)

That works out to 733 fps. Correct? The B757 is 155 feet long...so tell me, from the moment the nose HIT the ground, until the tail HIT...how many seconds of time elapsed?

Remember, the 733 fps is the velocity along the vector component of the airplane's longitudinal axis. Same as in the analogy, along the shaft of the arrow. Capice?

You can do a vector diagram, and determine the 90-degree (horiz. and vert. velocity components...) and calculate an approximate velocity horizontally over the ground given the angle relative to the ground.

Also, it will give the vertical velocity component...which should be far, far higher. (Acceleration due to gravity is a fctor, here --- in addition to the thrust provided by the engines).

*side note* I noticed that the N1 was 70%, from your OP. That is a fairly normal cruise power setting, while they were at 5,000 MSL in level flight, before rolling over and crashing. It did not seem to increase, which indicates that no action was taken to add power in the descent.

Taking a look at the FDR tracings, I see a few things:

ALL tracings stop at time index 10:03:15 EDT (approx). All at same time.

The last recorded HDG is 235 degrees.

The chart tops out at 500 KIAS....so speed was AT LEAST 500 knots.

The vertical velocity was AT LEAST -35,000 fpm. (That is so outside normal range, I'm surprised it calculated it...)

The autopilot was DISCONNECTED at about 10:00:20. (There was some maneuvering, shown by some erratic tracings just PRIOR to disconnect...tells me that the struggle had begun, and the terrorists began the maneuvers to try to disoriente the passengers, who were not seatbelted in). Just prior to 10 o'clock, the altitude was 5,000 MSL, and KIAS was about 290 Kts. In all of the maneuvering, the airplane climbed another 5,000 feet UP, reaching 10,000 MSL at about 10:02:10, then began the final dive.


So, the reast....the "horizontal motion" given the short span of time for the entire impact sequence...perhaps you can calculate it for us.



2. Although it's technically irrelevant to the argument being raised, what's your view on the plane crash pic in the OP? Moving bottom to top? Left to right? Flat spin?


Can't tell, but I would certainly rule out a flat spin, because of the impact velocity from a flat spin isn't much greater than terminal velocity.


reply posted on 27-10-2009 @ 04:06 AM by Rewey
Originally posted by weedwhacker
…it is conceivable that the laser-ring gyros and accelerometers that constitute the guts of the Inertial Reference System could have some lag induced by the quick attitude changes....


I get what you’re saying, and have put together a quick sketch to illustrate your point. For a plane moving at an arc (as I imagine the case would be with Flight 93), the ‘pitch’ is really a measurement of the plane’s angle with reference to the horizon at a specific point only, much like a tangent on a curve. As you say, the attitude would be continuously increasing through the curve.



If the plane travelling along this curve is moving fast enough, it might only take fractions of a second to move between the red position and the blue position. If there is any ‘lag’ in the processing of the pitch angle for the FDR, a plane might be at the blue position, but be reporting the red position as its pitch for the FDR.

Therefore, as with Flight 93, the ‘official’ pitch of 40º could be more, could be less. In reality, there’s no way of knowing how accurate it is.

But I believe the processing lag for these systems was debated on another thread, with reference to the FDR recording the altitude for the plane heading into the Pentagon. Apparently those who fly the planes (which I guess would include you) say the lag time for these processes is negligible. I guess 40º is as accurate as we’re going to get for now…


Originally posted by weedwhacker
I will repeat a simplified version of the arrow analogy...or, just even something as simple as a rock. Try it...throw a rock into sand at an approximate 40-degree angle, and see the pattern of ejecta. maybve not very scientific, but sometimes experimental observation is better than just doing the math....


Firstly with the arrow - that's a little different. The arrow uses the principle of the 'wedge' to drive the surface apart and lodge itself. According to many sources, the cockpit either shattered on impact, or broke off and bounced into the woods. Either way, an arrow would be expected to remain intact in that case. A plane, not so.

Secondly... Indeed, but that’s what this thread was really about. The photo in the Op shows the ejecta pattern – it’s quite obvious. That’s what is missing in the Shanksville crater, particularly along the wing lines. The photo I posted above shows a fair bit of green around the left wing imprint (left relative to the pic, of course), which means that the grass growing around where the wing allegedly hit was not covered by this dirt and ejecta.

But here’s the kicker – according to the ‘OS’ and many of its supporters, the fragments of the plane were spread down towards the barn. The fuel was thrown forward to the trees (quite some distance) leaving the grass around the actual crater unscathed. So why would the sand also not be sprayed upon impact, as with the photo in the OP? After all, according to the ‘OS’ this sand was very soft and loose, was it not? Enough for much of the plane to bury itself, and soft enough for the sand to fill in the hole over the top of the plane.

BOTH the debris and the fuel being flung FORWARD of the crater (particularly to the point that the fireball left the grass immediately around the wing fuel tank unburnt) indicates a large amount of HORIZONTAL momentum.

Yet the plane ITSELF left virtually no evidence of this on the ‘soft sand’? That’s simply contradictory.

Originally posted by weedwhacker
Can't tell, but I would certainly rule out a flat spin, because of the impact velocity from a flat spin isn't much greater than terminal velocity.


Forget about the flat spin – it was tongue in cheek. A reference to all the Top Gun fans out there…

Rewey



reply posted on 29-10-2009 @ 02:10 AM by Rewey
Originally posted by Joey Canoli
And what's with this "sand" claim?
Where'd you get that from?


You know, all this 'OS' nonsense...

Originally posted by thedman

www.theage.com.au...


To the casual eye, it looked like solid, consolidated ground but in reality the reclaimed expanse was loose and uncompacted...

...The rest of the 757 continued its downward passage, the sandy loam closing behind it like the door of a tomb...

Eventually these pieces...came to rest against solid rock, 23 metres below the surface...


Sooo... sand that was so loose and soft that the pieces of the plane travelled 23 metres through it (around 76 feet), and loose enough to spill back over the top of it burying it neatly, yet none of this loose sand sprayed over the green grass directly in front of the wing area here?



All this sand that was displaced for tonnes of plane pieces? Where did that go? Where was the sand sitting whilst the plane pieces slipped under the surface, for it to fall neatly back over the top? Just sitting off to the side?

Rew


reply posted on 29-10-2009 @ 02:46 PM by hooper
reply to post by Rewey



You do know there is a difference between "sandy loam" and sand, correct?

Also, you do know that the Australian newspaper called "The Age" is not an official spokesperson for the government of the United States of America, correct?

And that most reporters are not licensed geologists, correct?

I just think it is kind of funny when people refer to the OS or "official story" and start quoting foreign press.
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