Flight 77 obstacles

[jaydeehess]

Then why did R. Mackey use a constant total velocity of 781 fps?

Because it approximates the situation sufficiently. In a shallow desent the velocity along the long axis of the aircraft will not differ greatly from the ground speed.
In a 45 degree steep dive the ground speed would be 70% of the velocity along the long axis of the aircraft.

In a 4.5 degree 'dive' (which I use for illustrative purposes. 4.5 degrees being 1/10 of 45 degrees) the ground speed would be 99.7% of the velocity along the long axis of the aircraft.

(the following includes more significant digits than would be proper but I include them for illustrative purposes)

So 781 f/s X 0.997 = 778.6 f/s

As the plane pulled out of the dive the angle with the ground decreases and the difference between ground speed and longitudinal velocity becomes even closer.

For a 3 degree 'dive' it is 99.86% (779.9 f/s)
at 1 degree 'dive' it is 99.98% (780.8 f/s)


I do note that the equation that PfT used
a=v²/r
is the calculation of centripetal acceleration in circular motion.

I thought this was to be a parabolic trajectory! (r is not a constant)

I still do not see an answer as to how the radius of the circle was determined either.

 

[jaydeehess]

Further to the above;
Assume the aircraft starts with a 10 degree desent (I don't believe it was ever that steep but this is for educational puposes.

The the ground speed would be the 0.9848 of the longitudinal velocity.

Let's assume that the long.vel. is 781 f/s
ground speed is then 769.1 f/s
If the plane levels out and still has a long.vel. of 781 f/s its ground speed is now equal to that.
The plane in pulling out of a 10 degree dive (3 times that of a normal 3 degree glide slope) its ground speed (that's horizontal speed perhaps to PfT) would change only by 12 f/s


At 2500 feet (along the horizontal) out from the Pentagon a ground (horizontal) speed of 769 f/s will have the plane take 3.25 seconds
At 2500 feet out from the Pentagon, a ground (horizontal) speed of 781 f/s will have the plane take 3.20 seconds.

,,, and clearly if the plane's ground speed is changing from 769 to 781 f/s while traversing that distance then the time to travel that distance will be between 3.20 and 3.25 seconds

Is it clear now, TF, why using 781 as both the long.vel. and the ground (horizontal) velocity does not result in significant error?

 

[PhantomWolf]

What I'd like to know is why are PfT making videos to sell rather than sending all this information to KSM's Laywer.

 

[R.Mackey]

Then why did R. Mackey use a constant total velocity of 781 fps?

I justify the assumption explicitly in the derivation itself, as follows:

2. Boundary Conditions

The proposed boundary conditions of our problem are not well summarized, but can be extracted from articles and follow-up discussion on the Internet, initiated by a group known as "Pilots for 9/11 Truth." One such posting is found here. From this discussion, we derive the following boundary conditions on our aircraft:

[...]

• The aircraft's last known velocity was approximately 781 feet per second. We will assume this is the groundspeed in all cases. For shallow angles of pitch, this is approximately constant; we further have no insight into thrust or drag in the final few seconds before impact.

In other words, 781 feet per second was taken per your suggestion. Since the flight path is rather flat, this estimate is assumed to be good enough.

And, for the second time in this very thread, I specifically estimate the error due to the vertical part of the velocity vector (as you ask for) and the increase in speed from altitude recovery (which probably never occurred to you), in this post. The error is shown to be on the order of 2%, worst case. It is negligible.

Please read the answers to your questions before asking them yet again.

 

[beachnut]

When you take the FDR and try to match it to RADAR data you have 77 over 6 seconds away from impact. John Farmer plots data from mulitple radars and closes in on the same conclusion I have had for a while based on RADAR and FDR matching time and path information.

http://911files.info/blog/?p=113


With over 6 seconds to go, these are the last 6 seconds of altitude change information for 77 as the terrorist flew them. With 6 more to go, and only at 400 feet MSL, you can see the terrorist pilot has to let up some to hit the Pentagon and avoid the ground a second early prior to impact

Balsamo wants to use the FDR information but fails to show the world where 77 is. If he uses 1.5 DME and 61.2 degrees he has failed to account for resolution of storages and accuracy of the DME. The DME accuracy and resolution storage give up to a half mile uncertain. Combined that with RADAR data and you have over 6 seconds to go when 77 first hits 400 MSL, or 273 feet above the ground. Balsamo may argue the altitude being 528 to 470 feet high 6 seconds from impact based on PA, but the system could easily have a 100 foot error at 463 KIAS, and take note the maximum speed for normal operations for this 757 is 350 KCAS. The truth is Balsamo p4t crew of terrorist apologist decoded the RADAR altimeter, and 6 seconds back the altitude is 400 MSL. Look at the data, there is no problems hitting the Pentaogn with the terrorist flying even from 551 feet MSL.

The failed ideas of p4t are dashed by facts and the terrorist they try to apologize for.

Things Balsamo can't do! He can't give a position for 77.

Who saw 77 below the VDOT tower, and which side did it pass.

OMG, with normal terrorist inputs, 77 could be over the VDOT tower and still impact at the Pentagon. BTW, 77 hit in a dive, there is no pullout. So sorry non-Captain Bob.

From 307 MSL, or 407 MSL at the VDOT tower, the terrorist can hit the Pentagon within 0 to 2 Gs, very close to what the terrorist have been doing for the past 26 seconds. 77 is not more than 350 MSL at the VDOT tower, the Gs needed to hit the Pentagon are less than typical values seen on the FDR. The 10 G/34 G junk are Balsamo's stupid parameters of impossible level offs, no one can do. Balsamo has an impossible level off due to his own stupidity on flying, it can't be done with an airplane. 77 was already at 400 MSL over 6 seconds from the Pentagon. So in two seconds 77 would be below 350 MSL and could be lower than 307, there is plenty of room beside the VDOT tower! Balsamo moronic G junk is not needed, you can use the rates used by the terrorist to hit the Pentagon. Simple math, no instantaneous G loads.

Based on RADAR data and Flight 77 true track, 77 is 20 feet right of the VDOT tower. We don't have to go through Balsamo's bad math. The only pilots unable to do the 9/11 maneuvers (due to incompetence?) are the p4t pilots. They tried in simulators to hit buildings and failed. Not even as good at aiming as the terrorist they decide to apologize for, and sell implications of lies. They can't stand by their own implied conclusions.

 

[jaydeehess]

It strikes me that perhaps PfT would care to decide on the intial conditions they would like to use. In their first kick at the can they gave the conditions of the plane passing just above the VDOT tower, with a forward velocity of 781 f/s.

NOW they want to change it to a varying horizontal velocity and make a big fuss over that when in fact for shallow angles the difference between aircraft forward velocity and horizontal velocity is insignificant.

As I stated above, when I do the trig on this, a 10 degree desent and a 781 forward velocity gives a horizontal velocity of 769.1 f/s
Further to this though, such a desent would require that the aircraft have a desent rate of 135.8 f/s .

The aircraft came down 5000 feet in 2.5 minutes, an average desent rate of 33.3 f/s
So let's say that it was desending at 35 f/s as it passes over the VDOT tower and going 781 f/s forward velocity.
Doing the math has it desending at a 2.5 degree angle.

At this point if the plane begins its pull out of the dive and follows a parabolic path, we have a point on the horizontal parabola since we know the distance from the Pentagon, the height above the impact level and the slope of the parabola as it passes that point.

We also have another point on the parabola, the top of the first lamp post.

A horizontal parabola has the equation of (I'm rusty at this, so if someone can correct me please do publically)
x=ay² + b
the slope at any point is found be the equatioin that is the derivative of the above equation
dx/dy= 2ax

Lunch is over. I will try to cipher my way through this again later. It will be interesting to see how close this matches PfT's circular path of desent.

 

[jaydeehess]

Typos, trying to post too quick during lunch.

The above should read;

We also have another point on the parabola, the top of the first lamp post.

A horizontal parabola has the equation of (I'm rusty at this, so if someone can correct me please do publically)
x=ay² + b
the slope at any point is found by using the derivative of the above equation

slope = dx/dy = 2ay


Lunch is over. I will try to cipher my way through this again later. It will be interesting to see how close this matches PfT's circular path of desent.

 

[jaydeehess]

I see that the formula I used for a parabola is for the vertex in the standard position and we do not know the point of the vertex (the plane would be going straight down if it was on the parabola at the vertex)

thus the equation becomes
y=a(x-h)² + k

where k is the y coordinate of the vertex and h is the x coordinate of the vertex

This gets a lot harder then, but I am still working on it.

Place the origin (x,y)=(0,0) at the top of the VDOT tower. Slope at this point is -0.0448

The parabola opens to the horizontal. (one could redraw in which case the slope would be positive at (0,0)

The other point I would place at the bottom of the Pentagon 259 feet lower and 3416 feet away.

I am unsure if there is enough information to construct a formula for the parabola through those two points. If so then one can determine the vertical velocity at any point and then therefore the average vertical acceleration between any two points on the path.

 

[beachnut]

That would be all nice and stuff, however it would place the aircraft WELL over the tower when you correct the altitude. Don't forget about that!

BTW, if horizontal velocity is constant, how the hell is a plane going to pull
out of a dive? Where will the vertical velocity go as it decreases to zero?

PFT admitted their mistake since day one and tried to inform the 'critics' of the mistake. They chose to plug their ears and now they are trying to blame it on PFT mistakes? Are you all for real?

Check this out. Note the date:
http://pilotsfor911truth.org/forum//index.php?s=&showtopic=11360&view=findpost&p=10735436

Some truth crept in so he CLOSED THAT THREAD! The Truth-NAZI, terrorist loyalist Balsamo, the want-to-be-Capt Bob, Closed the tread! Too much truth as seeping in and dissolving the pure stupid of p4t failed ideas and faulty physics.

The p4t admitted a mistake, but it is still published! Now they produce a video that is dumber, and they think the new video makes reality base work here on Gs, wrong. But they failed. The new video is not practical, and does not show anyone at JREF wrong, it shows how stupid the new video is as they use faulty methods to produce unrealistic results.

Your VVI stupidity is classic truther trait. You have no practical knowledge on this subject, the more you struggle, you prove your lack of skills in these areas.

To prove your ideas, and p4t new video wrong, take the last seconds of actual feet lost per second by the terrorist pilot!

Use 40 MSL for impact. Back up to the tower at 4 plus seconds and 77 is at 363 feet MSL. 77 is just over the first pole by 10 feet or so. But these are actual terrorist feet lost per second, and impact is made within the terrorist flying parameters. Adjust the decent, with small changes in G, and 77 hits all the object as on 9/11. No 10 Gs, no 30 Gs. Pure stupid p4t G junk, proven wrong 7 years ago by terrorist, the terrorist Balsamo apologizes for. People actually saw 77 right nest to the tower so this example is off 40 to 60 feet too high, 77 was lower. Plus these actually feet lost were concluded 6 seconds prior to impact, the real decent from 6 seconds out from impact are not recorded on the FDR.

To hit the lamppost as done on 9/11 takes minor adjustments with no major Gs required. Balsamo is not very good at using physics, or even estimating the forces.

Thus based on past inputs from the terrorist, we have impact from above the VDOT tower.

Balsamo has no idea where 77 is. Why?

 

[Mangoose]

In the last page of this thread I posted the Steve Riskus 9/11 photo of the VHP mast and one I took in 2005:

I just realized that there is another 9/11 image of the mast, this one taken by James Ingersoll:

Since there was some question of whether the mast was damaged or not, I thought this image may be of interest.

 

[BCR]

Oh buddy Mangoose ole pal, could you provide me with a download or email me the original Ingersoll image. I seem to have missed that one. When was the Ingersoll taken?

john.farmer@spcengineer.com

 

[Mangoose]

The photo was taken just a few minutes after the collapse, so about 10:20 or so. You might have missed it because the Ingersoll photos were released on different hosting websites and neither collection was complete. The image of the mast comes from the DSC_0xxx.JPG series (several of which still have original EXIF data) while the other major collection is the DM-SD-02-03xxx.JPEG series. Ingersoll's first photo is DSC_0404.JPG (at the Navy Annex) but the first image with EXIF data is DSC_0407.JPG, taken when Ingersoll was walking down the hill from the Annex (camera model was NIKON D1X and date of the picture is 9/11/2001 at 10:46PM). The photos of the National fire units straying foam on the building were from DSC_0425.JPG to DSC_0434.JPG, and his famous photo of the piece of wreckage on the lawn was DSC_445.JPG (EXIF date was 9/11/2001 at 11:07PM). Then he took some shots of the wounded in the triage area (DSC_0447 to DSC_0451.JPG) and then turned his attention to the just-collapsed building. The image in question that shows the VHP mast (DSC_0455.JPG) intervenes between these first photos of the collapsed building. The mast photo doesn't have EXIF data, but the immediately preceding photo does, and that one gives the date as 9/11/2001 at 11:21PM. Then the evacuation follows and the next photo with EXIF data was DSC_0458.JPG taken at 11:45PM. So the time of the photo is pretty confidently set at between 10:15 and 10:20 or so.

Here is the download link: http://911digitalarchive.org/repository_object.php?object_id=88306

 

[BCR]

Thanks, I found it. The problem with it is the same as with the others (Warren did one too from the Sheraton Hotel which he sent me) is that the resolution is too low (or too distant) to verify (or dismiss) the "bent antenna" statements of the Paik brothers and others. Oh well, the mystery continues.

 

[Kent1]

Thanks, I found it. The problem with it is the same as with the others (Warren did one too from the Sheraton Hotel which he sent me) is that the resolution is too low (or too distant) to verify (or dismiss) the "bent antenna" statements of the Paik brothers and others. Oh well, the mystery continues.

I remember looking into that one a while ago. Boy that's a tough call. I lean slightly toward no hit.

 

[Mangoose]

Oh well. BTW, here is my image of the top of the mast from 2005 at full resolution, FWIW:

 

[BCR]

Oh well. BTW, here is my image of the top of the mast from 2005 at full resolution, FWIW:

[qimg]http://img516.imageshack.us/img516/739/newu.jpg[/qimg]

Don't tell CIT that you actually went to Arlington. They are the only investigooglers who ever went there to examine the scene and talk to people