Posted on 05/17/2003 7:23:43 AM PDT by joesnuffy
Right--which means that a FASTER impact velocity equals (be sure you're sitting down) more FORCE on the telemetry package.
The nearest shooter was too far away for a missile to be visible on Long Island.
Som much for that theory.
It was.
They didn't.
Stop blaming the U.S. Navy!
An infrared-guided missile would hit the engines. But a laser-guided missile could be made to hit the the 747 center section.
http://www.army-technology.com/contractors/missiles/bofors_rbs70/
Why are so many so quick to blame the U.S. Navy?
You got this one right! Clintoon wanted his "four more years".
3000 feet is "Possible" but it didn't happen according to the best evidence.
Theoretically, approximately 3000 feet is "possible." If you convert 100% of the aircraft's forward vectored momentum instantly into upward vectored momentum, you can get MORE than 3000 feet of climb if you ignore drag. However if this were what happened, it would have required ALL of the forward momentum to get that much climb... and the stalled aircraft would have then fallen almost almost straight down... meaning it would not have impacted the ocean where we know it did: 2.5 miles East North East of the initiating event position. However, the plane DID impact the ocean exactly where the math for a ballistic fall from the initiating event said it should so therefore it did not exchange forward momentum for upward momentum.
From the math I did on a previous thread:
Let us play a "what if" scenario out...
What if the entire horizontal Momentum of Flight TWA-800 was converted instantly into vertical Momentum? How long would TWA-800 climb straight up and how much higher could it climb before losing ALL MOMENTUM and begin falling? For this "what if" we will ignore the drag on the climbing aircraft (which will only mean shorter time and lower total climb) and consider only the pull of gravity.
I am assuming 330 MPH and 547,000 Lbs aircraft mass. These convert to ~147 Meters per second and ~249,000 Kgs aircraft mass.
The horizontal momentum of the 747 is ~36,500,000 Kg*m/s. and the acceleration of gravity is 2,440,200Kg*m/s^2.
2,440,200Kg*m/s^2 *t = 36,500,000 Kg*m/s - 5,350,000Kg*m/s (Momentum of the missing nose)
2,440,200Kg*m/s^2 *t = 31,150,000 Kg*m/s Again, solve for time (t) when s = 1
2,440,200Kg*m * t = 31,150,000 Kg*m
t = 12.75 seconds
If ALL of the FORWARD vectored MOMENTUM was magically changed instantly to UPWARD vectored MOMENTUM, Flight TWA-800 could only continue climbing for an additional ~13 seconds or so.
How high???
The aircraft, even though climbing because of its Momentum, is being acted on only by the FORCE of gravity... essentially it is in freefall.
The formula for this is:
Vf^2 = Vi^2 + 2 * a * d
(0 m/s)^2 = (147m/s)^2 +2*(-9.8m/s^2) * d
0 m^2/m^2 = 21609 m^2/s^2 + (-19.6m/s^2) *d
(-19.6 m/s^2) * d = 0 m^2/s^2 - 21609 m^2/s^2
(-19.6 m/s^2) * d = - 21609 m^2/s^2
d = (- 21609 m^2/s^2)/(-19.6 m/s^2)
d = 1102.5m = 3617 Feet
Remember, that is ignoring the not inconsiderable force of drag. Also WHERE does the amazing force that converts the vector of the plane's momentum come from?? The CIA and NTSB and you would have us believe that the aircraft's wing maintained its proper and most efficient angle of attack and applied lift force to accomplish this... but that force would have to come from converting the Momentum to lift... and therefore there would be a lot less momentum to continue climbing.
Of course, none of this could happen... the plane will continue mostly forward, decelerating from the initial velocity of 147m/s because of air resistance... not going instantly upward at the 147m/s.
"The captain of the NOAA research ship Rude entered Flight 800's last secondary radar position, speed, heading and gross weight into his computer and it predicted the landing point by calculating a ballistic fall. He went to that spot and immediately found the main wreckage including the fuselage, wings and engines. "
Ergo, unless you want to ignore the very well understood laws of Physics, there was no "zoom" climb. NONE, NADA, ZILCH.
The average climb in those 12.75 seconds would have to be ~283 feet per second to achieve a terminal altitude of 3617 feet above the point of pitch up. However, an average velocity is deceptive... during the last second before reaching terminal altitude, the aircraft would climb only 16 feet! That means that in the first couple of seconds the climb had to be MUCH greater... and during the initial moments of conversion of forward vector to upward vector (supposedly all caused by the force of lift on the upward pitching wing) the apparent G forces on the airframe must have been astronomic. I calculate it to be around 12-13 Gs for at least one second!
In addition, IF it had climbed, the aircraft would have climbed until upward momentum was used up, reached zero upward movement and begun to fall... taking an equal amount of time to fall from its ultimate altitude back to the altitude of the start of the climb... and then fallen the rest of the way to the ocean. The CIA and NTSB cartoons allowed 8-10 seconds for the climb... which requires an additional 8-10 seconds to return to the starting altitude... a total of 16-20 seconds added.
The radar returns do not show the aircraft remaining in the sky for those additional ~16-20 seconds. The radar record shows that TWA-800 was in the sky for only ~38-40 seconds after the last transponder return which took place just before the initiating event.
If we subtract the 16 seconds for the "zoom climb" and fall back to initial pre-zoom altitude, we find that it would be necessary for TWA-800 to fall 13,800 feet AND travel approximately 2.5 miles horizontally in ~22-24 seconds, an astonishing 392-427.5 miles per hour and an equally incredible terminal velocity for the airframe of 575-627 feet per second.
Those same figures (except terminal velocity) ALSO apply to the horizontal vector... but TWA-800 was originally moving at only 330 MPH (484 feet per second) horizontally before the initiating event. WHERE did the extra velocity come from???
Of course these incredible speeds are only required IF we add in the time for the mythical CIA-NTSB "Zoom Climb" to have occured. Without that "extra" time, we find that to fall from the initiating event altitude and position to splashdown and wreckage position requires a much more explicable average velocity of 363 feet per second (247MPH) which IS consistent with an unpowered, ballistic fall of the noseless aircraft experiencing drag.
Again, ergo: NO ZOOM CLIMB OCCURED!
There is no missile range off of Long Island, Navy or otherwise.
Capability to do what?
The Navy denied there were any subs in the area. Once proven to be lying, the Navy fessed up.
Two warning areas were active and there was a P-3 from Brunswick flying at low levels in them. Was it working with the sub in an exercise? Possibly. New London is a major sub base for the Atlantic Fleet and W-105 and W-106 would be convenient. Was it testing something secret? Again, possibly. A lot of test and evaluation work goes on with the subs at New London. Is this necessarily connected with TWA 800? I don't see how.
If a missile was fired from a terrorist boat, the Navy had enough assets to know the boat was there and where it went.
If it was looking for it, perhaps.
Sonar, as I am sure you know, can detect a private class of boat many, many miles away. If 800 went down, the Navy must have known within seconds or minutes of the downing and searched for all craft.
Sonar can track surface targets, and it can classify targets by type...if it knows what it's looking for. The waters off Long Island constitute the busiest sea lane on the Atlantic seaboard with dozens of freighters and literally thousands of small pleasure craft, fishing boats, and the like. Trying to pick a single small boat out of that is the ultimate needle in a haystack. For the Navy to search with sonar it would have had to know what it was searching for.
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