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Common Arguments Of "Topography" Article Addressed

rob balsamo
post Mar 14 2008, 11:04 AM
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Since those with "Common Arguments" refuse to sign up here to discuss our latest article, i thought i would address them here for reference. Will update this as required.


Claim -
QUOTE
The reason your flightpath makes no sense (I assume this is the work of Mr. Balsamo) is because you've assumed the aircraft was flat and level until the DOT mast, then dove to hit the light pole, then pulled up to avoid contact with the ground. This scenario has your maximum rate of descent at the light pole. If you propose a strawman flightpath with such ridiculous assumptions, it's no surprise you get ridiculous answers.

In order to disprove it was Flight 77 -- in order for your statement quoted above to be correct -- you have to show that the evidence is inconsistent with any flightpath, not just the stupid one you came up with. A more reasonable scenario has the aircraft diving before coming upon the DOT mast, then pulling up the rest of the way. In this scenario, the maximum rate of descent is at the DOT mast.

For sake of argument, let's assume your measurements are correct, although frankly we have no reason to trust you or "Pilots for 9/11 Truth." It is also possible that the aircraft missed the DOT mast entirely, or contacted it after all. In fact, there is circumstantial evidence that Flight 77 did hit the pole; if done with a wingtip, depending on roll angle, this means the height restriction at that distance is relaxed substantially.

But anyway, taking your data means we know the following:

* The aircraft contacted the Pentagon wall at 45 feet AGL. This is the minimum height.
* The aircraft was at 60 feet AGL 1016 feet away from the Pentagon.
* The aircraft was at 304 feet AGL at 2400+1016 = 3416 feet from the Pentagon.

We'll assume that the aircraft followed a parabola going through the latter two points, flattening out at the vertex and proceeding flat and level from there until hitting the wall. This gives us two points and an intercept, which uniquely defines our flight path.

The flight path takes the following form:

y = s (x - z)2 + h

where s is the steepness of the parabola, z is the horizontal location of the vertex (the lowest point of flight), and h is the height at the vertex, defined here as 45 feet.

We know this equation is satisfied for our two points, namely {1016, 60} and {3416, 304}. Substituting these in gives us a solution for z, and then we can find s using either point. The solution is left as an exercise -- if you can't do this, you aren't qualified to question the NTSB hypothesis. The solution has the vertex at z = 537 feet from the Pentagon, and the steepness s = 1 / 15296 feet.

Now we can calculate the vertical acceleration, by taking a time derivative. However, there is no time in this equation, since we've referenced the height y to the horizontal position x. If we accept the speed of 781 feet per second, this means:

x = x0 + 781 t feet / second

Using this substitution, we find that the vertical acceleration is given by the second time derivative of the height as a function of time, y(t):

d2/dt2 (y(t)) = (2) (781 feet/second)2 s

"y double dot" = 79.8 ft/s2 = 2.49 g

2.49 g is not a gentle ride, but it's well within the performance of a 757.

----

That kills your hypothesis right there, but it's important to note there's yet another problem with your model.

You stated that the impact height corresponded to the center of the hole. But there's no reason to assume the light pole was hit by the center of the aircraft. We've got apples and oranges here. Since the poles are referenced to the bottom of the aircraft, the impact point should be so referenced too, and that is at 33 feet, not 45.

If you repeat the above but substitute in 33 feet, you find that the vertex is actually at -90 feet, i.e. the aircraft is still descending at impact. And the acceleration now is only 26.9 ft/s2, or a mere 0.84 g.

That's peanuts. Even the Wright Flyer could handle that.

I remind you, I'm using your figures. Your figures are consistent with a totally viable flightpath. Therefore, Mr. Balsamo is either spectacularly incompetent, or a total fraud. Whichever it is, you have been duped.


Reply -
QUOTE
First problem is that he is not very observant. We reference MSL throughout our entire article, not AGL as he claims.

Next problem is that he assumes we are saying the path is level prior to the VDOT Antenna. This is false and a complete strawman. Usual tactic from the "critical thinkers". We are basing our vertical acceleration "assumption" between the VDOT Antenna and Pole 1 using FDR Data which averages just over 1 G for this segment. Its also interesting to note that the segment (Pole 1 to Pentagon) where the above claim requires 2.49 G's, and our math requires 11.2 G's, the FDR shows less than 1.

Further, he calculates 2.49 G's for the duration of the distance/time. He is saying 2.49 G's was sustained for 4.3 seconds till impact. Only one problem, this is not shown anywhere in the FDR data. Also, he forgot to add earths gravity. So he needs 3.49 G's sustained for 4.3 seconds during his "parabola", through 5 light poles. Further distancing his calculations from the FDR data provided by the NTSB. Also note, 3.49 G's exceed Transport Category G Load limits.

I dont blame him for not wanting to come to P4T to discuss this. It appears he too cannot escape discussion without ad hom attack.



Claim -
QUOTE
Arresting a 75 fps descent within 1.3 seconds requires arresting 75/1.3 = 57.7 feet per second of velocity, per second. He's multiplied where he should have divided.


Reply -
QUOTE
We are calculating vertical height for a 1.3 second duiration. a.k.a distance, here. Not vertical acceleration/velocity. The vertical accelration is 75 fps. The G's needed to arrest a vertical height of 97.5 over a 1.3 second time inteval based on 75 fps descent rate is 4 G's. If we need to arrest 75 feet in one second, how do you figure that height gets lower at 1.3 seconds? You just lowered vertical velocity to 57.7 when 75 is required.
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rob balsamo
post Mar 14 2008, 12:51 PM
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Keep in mind the height we used is the lowest possible based on the antenna. If we really want to get technical, we could use the FDR altitude at this position which would cause the G loads required for the "pull level" to be astronomical. Right now we are using 304 MSL - 80 MSL. Feel free to punch in the numbers using FDR Altitude of 699 MSL - 80 MSL. Wow! Or even try 273 AGL (radar altitude) + 135 ground elevation. = 408 MSL. Both will require much more than 11.2 G's.

This article was written on the hypothetical that the aircraft was as low as the VDOT antenna. Basically giving the govt story the benefit of the doubt on altitude. According to the FDR, it was much higher.
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Domenick DiMaggi...
post Mar 14 2008, 03:56 PM
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QUOTE (rob balsamo @ Mar 12 2008, 04:51 PM) *
Keep in mind the height we used is the lowest possible based on the antenna. If we really want to get technical, we could use the FDR altitude at this position which would cause the G loads required for the "pull level" to be astronomical. Right now we are using 304 MSL - 80 MSL. Feel free to punch in the numbers using FDR Altitude of 699 MSL - 80 MSL. Wow! Or even try 273 AGL (radar altitude) + 135 ground elevation. = 408 MSL. Both will require much more than 11.2 G's.

This article was written on the hypothetical that the aircraft was as low as the VDOT antenna. Basically giving the govt story the benefit of the doubt on altitude. According to the FDR, it was much higher.



The analysis he linked to at the PfT site bizarrely assumes that after 77 passed over the tall antenna, it continued at a constant, straight-line descent rate until it hit the first light pole a half-mile later, and only at that moment did it start to level off - the slope from the light pole to the Pentagon's wall is a much shallower descent rate. Balsamo's reasoning is that a plane couldn't pull out of a 4500 ft/min descent when it's only 40 feet off the ground.
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rob balsamo
post Mar 14 2008, 04:06 PM
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QUOTE (Domenick DiMaggio CIT @ Mar 14 2008, 04:56 PM) *
Balsamo's reasoning is that a plane couldn't pull out of a 4500 ft/min descent when it's only 40 feet off the ground.


That is exactly correct. Well, i dont feel a 757 could do it. Im not alone. http://pilotsfor911truth.org/core.html. The list grows regularly. Whoever feels they can pull a 757 out of a 4500 fpm dive at less than 40 feet above the ground, i would like to know if they would be willing to put their name on such a claim and be willing to attempt such a maneuver.

Also, the FDR shows for that segment a relatively uniform/linear descent rate and uniform average G load just above 1 G.. Unfortunately, it is no where near the forces needed as claimed by even the so called "critical thinkers" (eg. 3.49). I'd like to see them input 699MSL into their calculations as shown by the FDR for this position. Then compare their results to the vertical accelerations shown in the data provided by the NTSB. Their claims/requirements are already too great at 304 MSL as compared to the FDR data and Transport Category G limits. They'll shoot themselves in the foot even further using data supplied by the govt claimed to be from AA77.
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rob balsamo
post Mar 14 2008, 04:29 PM
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Claim -
QUOTE
You have 75 hamburgers to fry.
If you fry 57.7 hamburgers per hour,
you'll fry 75 hamburgers in 1.3 hours.


Reply -
QUOTE
You are frying 75 hamburgers per hour. How many hamburgers do you fry in 1.3 hours?


75 fps is the descent rate. How many feet will you descend in 1.3 seconds? How many G's are required to stop this descent rate within this distance (height) and time based on a 75 fps descent rate?


However, i will say he did correct my terminology. Im pretty beat. Was out till 4am last night and didnt get much sleep.


Claim-
QUOTE
You have 75 fps of vertical velocity to arrest (reduce to 0 fps).
If you arrest 57.7 fps of vertical velocity per second,
you'll arrest 75 fps of vertical velocity in 1.3 seconds.

So, the acceleration needed to arrest 75 fps of vertical velocity in 1.3 seconds, is 57.7 feet per second per second.


Reply -
QUOTE
75 feet per second is happening each second. Hence the "per second". Its not 75 feet per every 1.3 seconds.


By the way, im tired of this forum cross posting. If they want to address the issues. Tell them to stop being a coward and sign up.
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amazed!
post Mar 14 2008, 09:20 PM
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One must be a pilot to appreciate what a 5000fpm descent is. Or 4000 or 3000. The number itself barely matters.

Doing the required profile might be done in a Pitts or an F-15, but not in a Boeing. No matter who is flying.
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mrodway
post Mar 14 2008, 11:28 PM
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Having never been a pilot myself…

Can I ask if the skill of maintaining a parabolic flight path in order to minimize g-forces while attempting insane pullups using passenger aircraft is included in basic flight training?
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riv
post Mar 15 2008, 12:07 AM
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The first comment is totally hilarious.
He even provided some FORMULAS!!
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rob balsamo
post Mar 15 2008, 07:28 AM
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We will be looking into this further of course. We have gained quite a bit of (expected) criticism regarding this article and many calculations have been attempted by others. The parabola calculation (when corrected as shown above) makes the most sense out of all the common arguments. However, the FDR data does not reflect these G Load requirements. We will revise our article if required, but keep in mind, if revised, we will include calculations based on data provided by the NTSB further increasing the G Load.
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Aldo Marquis CIT
post Mar 17 2008, 01:01 PM
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Isn\'t the top flight path supposed to show a plane low and level across the lawn?

This post has been edited by rob balsamo: Mar 17 2008, 01:50 PM
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rob balsamo
post Mar 17 2008, 01:47 PM
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Claim -
QUOTE
"FDR figures supplied by them along with the most challenging altitude at both light pole and impact, requires only 4.0 g of load in the airframe for a mere 4.4 seconds"

"there is absolutely no case to be made that (1) the obstacles are inconsistent with the impact of Flight 77, (2) the FDR data is inconsistent with the impact of Flight 77, or (3) the FDR data is inconsistent with impacts to the obstacles themselves."


Reply -
QUOTE
If his first paragraph is accurate (which is probably not the case considering his initial calculations), his second paragraph statements are entirely false. We will write up a response to his calcluations as time permits.

We currently have people working on this issue however there are many scheduling conflicts. If we find we have to revise the article, it will be done.
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riv
post Mar 17 2008, 04:01 PM
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QUOTE (Aldo Marquis CIT @ Mar 15 2008, 04:01 PM) *


Isn\'t the top flight path supposed to show a plane low and level across the lawn?


Where are these graphs from?
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rob balsamo
post Mar 17 2008, 04:15 PM
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the govt loyalist site known for spin, obfuscation, ad homs and personal attacks, with some analysis mixed in based on their bias... smile.gif

We prefer not to promote their BS, but will post critical analysis. Im sure you know which site it is...
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magic_rat
post Mar 19 2008, 09:06 PM
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Unfortunately I think he is correct regarding the 57.7 f/s^2 acceleration needed to arrest a 75f/s descent within 1.3 seconds. By multiplying the 75fps figure by 1.3s rather than divide it, you came up with 97.5 in feet-per-1.3-seconds rather than feet-per-second. Continuing with that in mind, dividing it by 32fps^2 will not give a correct answer since the acceleration of gravity is in terms of seconds instead of 1.3-seconds. If you were to change that acceleration of gravity into terms of 1.3-seconds you would have to multiply it by 1.3^2 making it 54.08 feet-per-1.3seconds-squared. Then it would be proper to divide the velocity by the acceleration to get it in G's, and doing so yields the same 1.8G figure (without the addition of 1G) as 57.7/32fps^2 does. (97.5/54.08=57.7/32)

I just hope this helps clarify the issue. By no means do I feel that an abstract arithmetic error negates all the other potent research you've done, but by repeatedly endorsing a flawed figure you may inadvertently make this impression on others who are less convinced of the issue and upon whom the slightest indication of error may be enough to solidify their opinion either way.
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rob balsamo
post Mar 19 2008, 09:43 PM
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Hi Magic_Rat,

We are working extensively on this issue and will be publishing a revision to the original article with the proper math and calculations within the context of the original article. It is taking some time because some of the people we have working on it have an abundance of scheduling conflicts and others are not available at the moment.

It appears the math used by all (including myself and "the opposition") is not the proper way to tackle this problem. The math you cite above is especially not the proper way to do such a problem. The formula used in your example was distance for an accelerating object. In laymans terms, the formula used for the example you cite is based on a car accelerating or stopping uniformly over a distance. Not a plane pulling up from a dive at altitude with changing velocity vectors within a distance horizontally.

For instance, the formula used in your example can be demonstrated with the following calculator:

http://tutor4physics.com/calculators.htm

As an example, plug in 3000 feet as altitude at pole 1. Plug in 75 f/s for initial velocity. Plug in 0 for final. Note the G's. The G load decreases with distance for same velocity, which makes sense for a car stopping over a distance. Increase distance with same speed, G's get less. But is completely opposite of what you would expect if you increase altitude (distance) at the same point for an aircraft.

Again, we're working on it. This problem is extremely complex and not as simple as adjusting altitude to local pressure.
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UnderTow
post Mar 19 2008, 09:52 PM
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What you don't see in the graph.

The speed of the maneuver, the path before it, the inputs, the real world view of the approach.

What you do see in the graph.

A near perfect parabolic computer generated curve of perfect precision showing an impossible final approach.


For giggles, I would love to hear when they think Hani started to pull up?
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rob balsamo
post Mar 19 2008, 10:04 PM
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QUOTE (UnderTow @ Mar 19 2008, 10:52 PM) *
For giggles, I would love to hear when they think Hani started to pull up?



They kind of did. However their statements are false. We will be covering that as well. wink.gif
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magic_rat
post Mar 19 2008, 10:25 PM
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Well, I supposed the horizontal component be of some consequence.... whistle.gif

However, the fact that my example was simply a rewording of the same method each of you used doesn't makes it "especially" wrong as far as I'm concerned, and was hardly deserving of the condescending link tongue.gif

But here I'm only defending myself rather than anything of consequence, and I don't mean to be at all divisive so I'll let you get on with your calculations.
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rob balsamo
post Mar 19 2008, 10:33 PM
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It was especially wrong because it doesnt take into account the fact increased height (distance) will also increase G load (and from what i understand, is exponential). Its not inversely related as suggested by your example. Are you the original author of such calculations? I see you registered here a long time ago, but this is the first time you posted.
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magic_rat
post Mar 19 2008, 10:45 PM
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My registration date is the result of my lurking since a time that you had to be logged in to view the forums, and the return to that format prompted me to go ahead and post something for once.

I don't follow what you mean by increased height, and certainly don't see where that was taken into account by the other two calculations. The fact that I was able to get the same result as the dissenter by manipulating my equations the way you would if they were, in fact, the correct ones to use speaks for itself. However, considering my approach is quite likely way off base, that is not necessarily a bad thing for our purposes...
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